Algebraic Geometry I and II - Universiteit Utrechtlooij101/AG2013v0.pdfAlgebraic Geometry I and II...

Algebraic Geometry I and II(2013 edition)

Eduard Looijenga

Some conventions. In these notes rings are always supposed to be commutativeand to possess a unit and a ring homomorphism is required to take unit to unit.We allow that 1 = 0, but in that case we get of course the zero ring 0 and therecannot be any ring homomorphism going from this ring to a nonzero ring, as itmust take unit to unit. Since a prime ideal of a ring is by definition not the wholering, the zero ring has no prime ideals and hence also no maximal ideals. When Rand R′ are two rings, then the R × R′ is also one for componentwise addition andmultiplication, the unit being (1, 1). The projections onto its factors are admitted asring homomorphims, but not an inclusion obtained by putting one coordinate zero,as this is not unital, unless in that coordinate we have the zero ring (“×” defines acategorical product but not a categorical sum).

We say that a ring is a domain if its zero ideal is a prime ideal, in other words,if the ring is not the zero ring (1 6= 0) and has no zero divisors.

Given a ring R, then an R-algebra is a ring A endowed with a ring homomor-phism φ : R → A. When is φ is understood, then for every r ∈ R and a ∈ A, theproduct φ(r)a is often denoted by ra. In case R is a field, φ will be injective so thatR may be regarded as a subring of A, but this need not be so in general. We say thatA is finitely generated as an R-algebra if we can find a1, . . . , an in A such that everyelement of A can be written as a polynomial in these elements with coefficients inR; in other words, if the R-algebra homomorphism R[x1, . . . , xn]→ A which sendsthe variable xi to ai is onto. This is not to be confused with the notion of finitegeneration of an R-module M which merely means the existence of a surjectivehomomorphism of R-modules Rn →M for some n ≥ 0.

Similarly, a field L is said to be finitely generated as a field over a subfield K ifthere exist b1, . . . , bn in L such that every element of L can be written as a fractionof two polynomials in these elements with coefficients in K.

We denote the multiplicative group of the invertible elements (units) of a ringR by R×.

Contents

Chapter 1. Affine varieties 51. The Zariski topology 52. Irreducibility and decomposition 83. Finiteness properties and the Hilbert theorems 144. The affine category 185. The sheaf of regular functions 246. The product 277. Function fields and rational maps 298. Finite morphisms 349. Dimension 4010. Nonsingular points 4311. The notion of a variety 5212. Constructible sets 55

Chapter 2. Projective varieties 571. Projective spaces 572. The Zariski topology on a projective space 593. The Segre embeddings 624. Blowing up and projections 635. Elimination theory and projections 676. The Veronese embeddings 697. Grassmannians 718. Fano varieties and the Gauß map 759. Multiplicities of modules 7710. Hilbert functions and Hilbert polynomials 81

Chapter 3. Schemes 871. Presheaves and sheaves 872. The spectrum of a ring as a locally ringed space 923. The notion of a scheme 974. Formation of products 1025. Elementary properties of schemes 1036. Separated and proper morphisms 1077. Quasi-coherent sheaves 1178. When is a scheme affine? 1229. Divisors and invertible sheaves 12410. The projective setting 13211. Cohomology of coherent modules on a projective scheme 13512. Ample and very ample sheaves 13913. Blowing up 142

3

4 CONTENTS

14. Flatness 14615. Serre duality 149

Chapter 4. Appendix 1531. The basics of category theory 1532. Abelian categories and derived functors 1593. Sheaf cohomology 169

Bibliography 181

CHAPTER 1

Affine varieties

Throughout these notes k stands for an algebraically closed field. Recall thatthis means that every polynomial f ∈ k[x] of positive degree has a root x1 ∈ k:f(x1) = 0. This implies that f is divisible by x − x1 with quotient a polynomial ofdegree one less than f . Continuing in this manner we then find that f decomposessimply as f(x) = c(x − x1) · · · (x − xd) with c ∈ k× = k r 0, d = deg(f) andx1, . . . , xd ∈ k. Since an algebraic extension of k is obtained by the adjunction ofcertain roots of polynomials in k[x], this also shows that the property in question isequivalent to: every algebraic extension of k is equal to k.

A first example you may think of is the field of complex numbers C, but aswe proceed you should become increasingly aware of the fact that there are manyothers: it is shown in a standard algebra course that for any field F an algebraicclosure F is obtained by adjoining to F the roots of every polynomial f ∈ F [x].1

So we could take for k an algebraic closure of the field of rational numbers Q, ofthe finite field Fq, where q is a prime power2 or even of the field of fractions of anydomain such as C[x1, . . . , xr].

1. The Zariski topology

Any f ∈ k[x1, . . . , xn] determines in an evident manner a function kn → k.In such cases we prefer to think of kn not as vector space—its origin and vectoraddition will be irrelevant to us—but as a set with a weaker structure. We shallmake this precise later, but it basically amounts to only remembering that elementsof k[x1, . . . , xn] can be understood as k-valued functions on it. For that reasonit is convenient to denote this set differently, namely as An (or as Ank , if we feelthat we should not forget about the field k). We refer to An as the affine n-spaceover k. A k-valued function on An is then said to be regular if it is defined bysome f ∈ k[x1, . . . , xn]. We denote the zero set of such a function by Z(f) and itscomplement (the nonzero set) by Anf ⊂ An.

A principal subset of An is any subset of the form Anf and a hypersurface of Anis any subset of the form Z(f), with f nonconstant (that is, f /∈ k).

EXERCISE 1. Prove that f ∈ k[x1, . . . , xn] is completely determined by the reg-ular function it defines. (Hint: do first the case n = 1.) So the ring k[x1, . . . , xn]

1This can not be done in one step: it is an infinite process which involves in general many choices.This is reflected by the fact that the final result is not canonical, although it is unique up to a (in generalnonunique) isomorphism; whence the use of the indefinite article in ‘an algebraic closure’.

2Since the elements of any algebraic extension of Fq of degree n ≥ 2 are roots of x(qn) − x, we

only need to adjoin roots of such polynomials.

5

6 1. AFFINE VARIETIES

can be regarded as a ring of functions on An under pointwise addition and mul-tiplication. Show that this fails be so had we not assumed that k is algebraicallyclosed (e.g., for the finite field Fq).

EXERCISE 2. Prove that a hypersurface is nonempty.

It is perhaps somewhat surprising that in this rather algebraic context, the lan-guage of topology proves to be quite effective: algebraic subsets of An shall appearas the closed sets of a topology, albeit a rather peculiar one.

LEMMA-DEFINITION 1.1. The collection of principal subsets of An is a basis of atopology on An, called the Zariski topology. A subset of An is closed for this topologyif and only if it is an intersection of zero sets of regular functions.

PROOF. Recall that a collection U of subsets of a set X may serve as a basis fora topology on X (and thus determines this topology) if and only if the intersectionof any two its members is a union of members of U. As the collection of principalsubsets is even closed under finite intersection: Anf1∩A

nf2

= Anf1f2 , the first assertionfollows. Since an open subset of An is by definition a union of subsets of the formAnf , a closed subset must be an intersection of subsets of the form Z(f).

EXAMPLE 1.2. The Zariski topology on A1 is the cofinite topology: its closedsubsets 6= A1 are the finite subsets.

EXERCISE 3. Show that the diagonal in A2 is closed for the Zariski topology,but not for the product topology (where each factor A1 is equipped with the Zariskitopology). So A2 does not have the product topology.

We will explore the mutual relationship between the following two basic maps:

subsets of An I−−−−→ ideals of k[x1, . . . , xn]

∪ ∩

closed subsets of An Z←−−−− subsets of k[x1, . . . , xn].where for a subset X ⊂ An, I(X) is the ideal of f ∈ k[x1, . . . , xn] with f |X = 0 andfor a subset J ⊂ k[x1, . . . , xn], Z(J) is the closed subset of An defined by ∩f∈JZ(f).Observe that

I(X1 ∪X2) = I(X1) ∩ I(X2) and Z(J1 ∪ J2) = Z(J1) ∩ Z(J2).

In particular, both I and Z are inclusion reversing. Furthermore, the restriction of Ito closed subsets defines a section of Z: if Y ⊂ An is closed, then Z(I(Y )) = Y . Wealso note that by Exercise 1 I(An) = (0), and that any singleton p ⊂ An is closed,as it is the common zero set of the degree one polynomials x1 − p1, . . . , xn − pn.

EXERCISE 4. Prove that I(p) is equal to the ideal generated by these degreeone polynomials and that this ideal is maximal.

EXERCISE 5. Prove that the (Zariski) closure of a subset Y of An is equal toZ(I(Y )).

Given Y ⊂ An, then f, g ∈ k[x1, . . . , xn] have the same restriction to Y if andonly if f − g ∈ I(Y ). So the quotient ring k[x1, . . . , xn]/I(Y ) (a k-algebra) can beregarded as a ring of k-valued functions on Y . Notice that this k-algebra does notchange if we replace Y by its Zariski closure.

1. THE ZARISKI TOPOLOGY 7

DEFINITION 1.3. Let Y ⊂ An be closed. The k-algebra k[x1, . . . , xn]/I(Y ) iscalled the coordinate ring of Y and we denote it by k[Y ]. A k-valued function on Yis said to be regular if it lies in this ring.

So with notation, k[An] = k[x1, . . . , xn]. Given a closed subset Y ⊂ An, thenfor every subset X ⊂ An we have X ⊂ Y if and only if I(X) ⊃ I(Y ), and in thatcase IY (X) := I(X)/I(Y ) is an ideal of k[Y ]: it is the ideal of regular functions onY that vanish on X. Conversely, an ideal of k[Y ] is of the form J/I(Y ), with J anideal of k[x1, . . . , xn] that contains I(Y ), and such an ideal defines a closed subsetZ(J) contained in Y . So the two basic maps above give rise to such a pair on Y :

subsets of Y IY−−−−→ ideals of k[Y ]

∪ ∩

closed subsets of Y ZY←−−−− subsets of k[Y ].

We ask: which ideals of k[x1, . . . , xn] are of the form I(Y ) for some Y ? Clearly,if f ∈ k[x1, . . . , xn] is such that some positive power vanishes on Y , then f vanisheson Y . In other words: if fm ∈ I(Y ) for some m > 0, then f ∈ I(Y ). This suggests:

PROPOSITION-DEFINITION 1.4. Let R be a ring (as always commutative and with1) and let J ⊂ R be an ideal. Then the set of a ∈ R with the property that am ∈ J forsome m > 0 is an ideal of R, called the radical of J and denoted

√J .

We say that J is a radical ideal if√J = J .

We say that the ring R is reduced if the zero ideal (0) is a radical ideal (in otherwords, R has no nonzero nilpotents: if a ∈ R is such that am = 0, then a = 0).

PROOF. We show that√J is an ideal. Let a, b ∈

√J so that am, bn ∈ J for

certain positive integers m,n. Then for every r ∈ R, ra ∈√J , since (ra)m =

rmam ∈ J . Similarly a − b ∈√J , for (a − b)m+n is an R-linear combination of

monomials that are multiples of am or bn and hence lie in J .

EXERCISE 6. Show that a prime ideal is a radical ideal.

Notice that J is a radical ideal if and only if R/J is reduced. The precedingshows that for every Y ⊂ An, I(Y ) is a radical ideal, so that k[Y ] is reduced. Thedictionary between algebra and geometry begins in a more substantial manner with

THEOREM 1.5 (Hilbert’s Nullstellensatz). For every ideal J ⊂ k[x1, . . . , xn] wehave I(Z(J)) =

√J .

We inclusion ⊃ is clear; the hard part is the opposite inclusion (which says thatif f ∈ k[x1, . . . , xn] vanishes on Z(J), then fm ∈ J for some positive integer m).We postpone its proof and first discuss some of the consequences.

COROLLARY 1.6. Let Y ⊂ An be closed. Then the maps IY and ZY define inclusionreversing bijections

closed subsets of Y ↔ radical ideals of k[Y ]

that each others inverse and restrict to bijections

points of Y ↔ maximal ideals of k[Y ].


PROOF. We first prove this for Y = An. We already observed that for everyclosed subset X of An we have Z(I(X)) = X. The Nullstellensatz says that for aradical ideal J ⊂ k[x1, . . . , xn], we have I(Z(J)) = J .

If X = p with p = (p1, . . . , pn) ∈ An, then mp := (x1 − p1, . . . , xn − pn) is amaximal ideal of k[x1, . . . , xn] by Exercise 4. Since I(X) ⊃ mp we must then haveI(X) = mp (for I(X) = k[x1, . . . , xn] is clearly excluded). Conversely let m be amaximal ideal of k[x1, . . . , xn]. Such an ideal is certainly radical as it is a primeideal. Hence it is of the form I(X) for a closed subset X. Since the empty subset ofAn is defined by the radical ideal k[x1, . . . , xn], the preceding implies that X will benonempty and (as m is a maximal ideal) minimal for this property. In other words,X is a singleton p. So this yields a bijection between the points of An and themaximal ideals of k[x1, . . . , xn].

The general case now also follows, because an ideal of k[Y ] is of the formJ/I(Y ) and this is a radical ideal if and only if J is one; a maximal ideal of k[Y ]corresponds to a maximal ideal of An which contains I(Y ).

Via this (or a very similar) correspondence, algebraic geometry seeks to expressgeometric properties of Y in terms of algebraic properties of k[Y ] and vice versa.In the end we want to forget about the ambient An.

2. Irreducibility and decomposition

We introduce a property which for most topological spaces is of little interest,but as we will see, is useful and natural for the Zariski topology.

DEFINITION 2.1. Let Y be a topological space. We say that Y is irreducible ifit is nonempty and cannot be written as the union of two closed subsets 6= Y (thislast property is equivalent to: any nonempty open subset of Y is dense in Y ).

An irreducible component of Y is a maximal irreducible subset of Y .

EXERCISE 7. Prove that an irreducible Hausdorff space must consist of a singlepoint. Prove also that an infinite set with the cofinite topology is irreducible.

EXERCISE 8. Let Y1, . . . , Ys be closed subsets of a topological space Y whoseunion is Y . Prove that every irreducible subset of Y is contained in some Yi. Deducethat Yisi=1 is the collection of irreducible components of Y if each Yi is irreducibleand Yi ⊂ Yj implies Yi = Yj .

LEMMA 2.2. Let Y be a topological space. If Y is irreducible, then every nonemptyopen subset of Y irreducible. Conversely, if C ⊂ Y is an irreducible subspace, then Cis also irreducible. In particular, an irreducible component of Y is always closed in Y .

PROOF. Suppose Y is irreducible and let U ⊂ Y be open and nonempty. Anonempty open subset of U is dense in Y and hence also dense in U . So U isirreducible.

Let now C ⊂ Y be irreducible (and hence nonempty). Let V ⊂ C be nonemptyand open in C. Then V ∩C is nonempty. It is also open in C and hence dense in C.But then V ∩ C is also dense in C and so V is dense in C. So C is irreducible.

Here is what irreducibility means in the Zariski topology.

PROPOSITION 2.3. A closed subset Y ⊂ An is irreducible if and only if I(Y ) is aprime ideal (which we recall is equivalent to: k[Y ] = k[x1, . . . , xn]/I(Y ) is a domain).

2. IRREDUCIBILITY AND DECOMPOSITION 9

PROOF. Suppose Y is irreducible and f, g ∈ k[x1, . . . , xn] are such that fg ∈I(Y ). Then Y ⊂ Z(fg) = Z(f) ∪ Z(g). Since Y is irreducible, Y is contained inZ(f) or in Z(g). So f ∈ I(Y ) or g ∈ I(Y ), proving that I(Y ) is a prime ideal.

Suppose that Y is the union of two closed subsets Y1 and Y2 that are both 6= Y .Then I(Y ) is not a prime ideal: since Yi 6= Y implies that there exist fi ∈ I(Yi) −I(Y ) (i = 1, 2) and then f1f2 vanishes on Y1 ∪ Y2 = Y , so that f1f2 ∈ I(Y ).

One of our first aims is to prove that the irreducible components of any closedsubset Y ⊂ An are finite in number and have Y as their union. This may notsound very surprising, but we will see that this reflects some nonobvious algebraicproperties. Let us first consider the case of a hypersurface. Since we are goingto use the fact that k[x1, . . . , xn] is a unique factorization domain, we begin withrecalling that notion.

2.4. UNIQUE FACTORIZATION DOMAINS. Let us first observe that in a ring R without zerodivisors two nonzero elements a, b generate the same ideal if and only if b is a unit times a.

DEFINITION 2.5. A ringR is called a unique factorization domain if it has no zero divisorsand every principal ideal (a) := Ra in R which is neither the zero ideal nor all of R is inunique manner an (unordered) product of principal prime ideals: (a) = (p1)(p2) · · · (ps) (sothe ideals (p1), . . . , (ps) are unique up to order).

Note that last property amounts to the statement that a can be written as a producta = p1p2 · · · ps such that each pi generates a prime ideal and this is unique up to order andmultiplication by units: if a = q1q2 · · · qt is another such way of writing a, then t = s andqi = uipσ(i), where σ ∈ Sn is a permutation and u1u2 · · ·us = 1.

For a field (which has no proper principal ideals distinct from (0)) the imposed conditionis empty and hence a field is automatically a unique factorization domain. A more substantialexample (that motivated this notion in the first place) is Z: a principal prime ideal of Z is ofthe form (p), with p a prime number. Every integer n ≥ 2 has a unique prime decompositionand so Z is a unique factorization domain.

A basic theorem in the theory of rings asserts that if R is a unique factorization domain,then so is its polynomial ring R[x]. This implies (with induction on n) that R[x1, . . . , xn] isone. This applies to the case when R is a field (such as our k): a nonzero principal idealof this ring is prime precisely when it is generated by an irreducible polynomial of positivedegree and every f ∈ R[x1, . . . , xn] of positive degree then can be written as a productof irreducible polynomials: f = f1f2 · · · fs, a factorization that is unique up to order andmultiplication of each fi by a nonzero element of R.

The following proposition connects two notions of irreducibility.

PROPOSITION 2.6. Let f ∈ k[x1, . . . , xn] have positive degree. If f = f1f2 · · · fsis a factoring into irreducible polynomials, then Z(f1), . . . , Z(fs) are the irreduciblecomponents of Z(f) and their union equals Z(f) (but we are not claiming that theZ(fi)’s are pairwise distinct). In particular, a hypersurface is the union of its ir-reducible components; these irreducible components are hypersurfaces and finite innumber (so that f is irreducible if and only if Z(f) is irreducible).

PROOF. We first note that when g ∈ k[x1, . . . , xn] is irreducible, then g gener-ates a prime ideal and so Z(g) is an irreducible hypersurface by Proposition 2.3.

It follows that if f = f1f2 · · · fs is as in the proposition, then Z(f) = Z(f1) ∪· · · ∪ Z(fs) with each Z(fi) irreducible. To see that Z(fi)si=1 is the collectionof irreducible components of Z(f), it suffices, in view of Exercise 8, to observethat any inclusion relation Z(fi) ⊂ Z(fj) is necessarily an identity. Since fi is


irreducible it generates a prime ideal. A prime ideal is a radical ideal and so by theNullstellensatz, fj ∈ (fi). But fj is irreducible also and so fj is a unit times fi. Thisproves that Z(fj) = Z(fi).

The discussion of irreducibility in general begins with the somewhat formal

LEMMA 2.7. For a partially ordered set (A,≤) the following are equivalent:(i) (A,≤) satisfies the ascending chain condition: every ascending chain a1 ≤

a2 ≤ a3 ≤ · · · becomes stationary: an = an+1 = · · · for n sufficiently large.(ii) Every nonempty subset B ⊂ A has a maximal element, that is, an element

b0 ∈ B such that there is no b ∈ B with b > b0.

PROOF. (i)⇒(ii). Suppose (A,≤) satisfies the ascending chain condition andlet B ⊂ A be nonempty. Choose b1 ∈ B. If b1 is maximal, we are done. If not, thenthere exists a b2 ∈ B with b2 > b1. We repeat the same argument for b2. We cannotindefinitely continue in this manner because of the ascending chain condition.

(ii)⇒(i). If (A,≤) satisfies (ii), then the set of members of any ascending chainhas a maximal element, in other words, the chain becomes stationary.

If we replace ≤ by ≥, then we obtain the notion of the descending chain condi-tion and we find that this property is equivalent to: every nonempty subset B ⊂ Ahas a minimal element. These properties appear in the following pair of definitions.

DEFINITIONS 2.8. We say that a ring R is noetherian if its collection of idealssatisfies the ascending chain condition.

We say that a topological space Y is noetherian if its collection of closed subsetssatisfies the descending chain condition.

EXERCISE 9. Prove that a subspace of a noetherian space is noetherian. Provealso that a ring quotient of a noetherian ring is noetherian.

EXERCISE 10. Prove that a noetherian space is quasi-compact: every coveringof such a space by open subsets contains a finite subcovering.

The interest of the noetherian property is that it is one which is possessed byalmost all the rings we encounter and that it implies many finiteness propertieswithout which we are often unable to go very far.

We give a nonexample first. The ring H(D) of holomorphic functions on theunit disk D ⊂ C is not noetherian: choose fo ∈ H(D) such that fo has simple zeroesin a sequence (zi ∈ D)i≥1 whose terms are pairwise distinct (e.g., sin(π/(1 − z))).Let In denote the ideal of f ∈ H(D) having a zero in zi for all i ≥ n. Thenfo(z)(z − z1)−1 · · · (z − zn)−1 defines an element of In+1 − In and so I1 ⊂ I2 ⊂ · · ·is a strictly ascending chain of ideals in H(D).

On the other hand, the ring of convergent power series Cz is noetherian(we leave this as a little exercise). Obviously a field is noetherian. The ring Z isnoetherian: if I1 ⊂ I2 ⊂ · · · is an ascending chain of ideals in Z, then ∪∞s=1Is isan ideal of Z, hence of the form (n) for some n ∈ Z. But if s is such that n ∈ Is,then clearly the chain is stationary as of index s. (This argument only used thefact that any ideal in Z is generated by a single element, i.e., that Z is a principalideal domain.) That most rings we encounter are noetherian is a consequence ofthe following theorem.

THEOREM 2.9 (Hilbert’s basis theorem). If R is a noetherian ring, then so isR[x].


As with the Nullstellensatz, we postpone the proof and discuss some of itsconsequences first.

COROLLARY 2.10. If R is a noetherian ring (for example, a field) then so is everyfinitely generated R-algebra. Also, the space An (and hence any closed subset of An)is noetherian.

PROOF. The Hilbert basis theorem implies (with induction on n) that the ringR[x1, . . . , xn] is noetherian. By Exercise 9, every quotient ring R[x1, . . . , xn]/I isthen also noetherian. But a finitely generated R-algebra is (by definition) isomor-phic to some such quotient and so the first statement follows.

Suppose An ⊃ Y1 ⊃ Y2 ⊃ · · · is a descending chain of closed subsets. Then(0) ⊂ I(Y1) ⊂ I(Y2) ⊂ · · · is an ascending chain of ideals. As the latter becomesstationary, so will become the former.

PROPOSITION 2.11. If Y is noetherian space, then its irreducible components arefinite in number and their union equals Y .

PROOF. Suppose Y is a noetherian space. We first show that every closed sub-set can be written as a finite union of closed irreducible subsets. First note that theempty set has this property (despite the fact that an irreducible set is nonempty bydefinition), for a union with empty index set is empty. Let B be the collection ofclosed subspaces of Y for which this is not possible, i.e., that can not be written asa finite union of closed irreducible subsets. Suppose that B is nonempty. Accordingto 2.7 this collection has a minimal element, Z, say. This Z must be nonempty andcannot be irreducible. So Z is the union of two proper closed subsets Z ′ and Z ′′.The minimality of Z implies that neither Z ′ nor Z ′′ is in B and so both Z ′ and Z ′′

can be written as a finite union of closed irreducible subsets. But then so can Z andwe get a contradiction.

In particular, there exist closed irreducible subsets Y1, . . . , Ys of Y whose unionis Y (if Y = ∅, take s = 0). We may of course assume that no Yi is containedin some Yj with j 6= i. An application Exercise 8 then shows that the Yi’s are theirreducible components of Y .

If we apply this to An (endowed as always with its Zariski topology), thenwe find that every subset Y ⊂ An has a finite number of irreducible components,the union of which is all of Y . If Y is closed in An, then so is every irreduciblecomponent of Y and according to Proposition 2.3 any such irreducible componentis defined by a prime ideal. This allows us to recover the irreducible components ofa closed subset Y ⊂ An from its coordinate ring:

COROLLARY 2.12. Let Y ⊂ An be a closed subset. If C is an irreducible componentof Y , then the image IY (C) of I(C) in k[Y ] is a minimal prime ideal of k[Y ] and anyminimal prime ideal of k[Y ] is so obtained: we thus get a bijective correspondencebetween the irreducible components of Y and the minimal prime ideals of k[Y ].

PROOF. Let C be a closed subset of Y and let IY (C) be the corresponding idealof k[Y ]. Now C is irreducible if and only if I(C) is a prime ideal of k[x1, . . . , xn],or what amounts to the same, if and only if IY (C) is a prime ideal of k[Y ]. It isan irreducible component if C is maximal for this property, or what amounts to thesame, if IY (C) is minimal for the property of being a prime ideal of k[Y ].


EXAMPLE 2.13. First consider the set C := (t, t2, t3) ∈ A3 | t ∈ k. This is aclosed subset of A3: if we use (x, y, z) instead of (x1, x2, x3), then C is the commonzero set of y−x2 and z−x3. Now the inclusion k[x] ⊂ k[x, y, z] composed with thering quotient k[x, y, z] → k[x, y, z]/(y − x2, z − x3) is easily seen to be an isomor-phism. Since k[x] has no zero divisors, (y− x2, z− x3) must be a prime ideal. So Cis irreducible and I(C) = (y − x2, z − x3).

We now turn to the closed subset Y ⊂ A3 defined by xy−z = 0 and y3−z2 = 0.Let p = (x, y, z) ∈ Y . If y 6= 0, then we put t := z/y; from y3 = x2, it follows thaty = t2 and z = t3 and xy = z implies that x = t. In other words, p ∈ C in thatcase. If y = 0, then z = 0, in other words p lies on the x-axis. Conversely, any pointon the x-axis lies in Y . So Y is the union of C and the x-axis and these are theirreducible components of Y .

We begin with recalling the notion of localization and we do this in the gener-ality that is needed later.

2.14. LOCALIZATION. Let R be a ring and let S be a multiplicative subset of R: 1 ∈ Sand S closed under multiplication. Then a ring S−1R, together with a ring homomorphismR → S−1R is defined as follows. An element of S−1R is by definition written as a formalfraction r/s, with r ∈ R and s ∈ S, with the understanding that r/s = r′/s′ if and only ifs′′(s′r − sr′) = 0 for some s′′ ∈ S. This is a ring indeed: multiplication and subtraction isdefined as for ordinary fractions: r/s.r′/s′ = (rr′)/(ss′) and r/s− r′/s′ = (s′r− sr′)/(ss′);it has 0/1 as zero and 1/1 as unit element and the ring homomorphism R→ S−1R is simplyr 7→ r/1. Observe that the definition shows that 0/1 = 1/1 if and only if 0 ∈ S, in whichcase S−1R is reduced to the zero ring. We also note that any s ∈ S maps to an invertibleelement of S−1R, the inverse of s/1 being 1/s (this is also true when 0 ∈ S, for 0 is its owninverse in the zero ring). In a sense (made precise in part (b) of Exercise 11 below) the ringhomomorphism R → S−1R is universal for that property. This construction is called thelocalization away from S.

It is clear that if S does not contain zero divisors, then r/s = r′/s′ if and only ifs′r − sr′ = 0; in particular, r/1 = r′/1 if and only if r = r′, so that R → S−1R is theninjective. If we take S maximal for this property, namely take it to be the set of nonzerodivisors of R (which is indeed multiplicative), then S−1R is called the fraction ring Frac(R)of R. When R is a domain, S = Rr0 and so Frac(R) is a field, the fraction field of R. Thisgives the following corollary, which hints to the importance of prime ideals in the subject.

COROLLARY 2.15. An ideal p of a ring R is a prime ideal if and only if it is the kernel of aring homomorphism from R to a field.

PROOF. It is clear that the kernel of a ring homomorphism from R to a field is alwaysa prime ideal. Conversely, if p is a prime ideal, then it is the kernel of the composite R →R/p → Frac(R/p).

Of special interest is when S = sn |n ≥ 0 for some s ∈ R. We then usually writeR[1/s] for S−1R. Notice that the image of s in R[1/s] is invertible and that R[1/s] is thezero ring if and only if s is nilpotent.

EXERCISE 11. Let R be a ring and let S be a multiplicative subset of R.(a) What is the the kernel of R→ S−1R?(b) Prove that a ring homomorphism φ : R → R′ with the property that φ(s)

is invertible for every s ∈ S factors in a unique manner through S−1R.(c) Consider the polynomial ring R[xs : s ∈ S] and the homomorphism of

R-algebras R[xs : s ∈ S] → S−1R that sends xs to 1/s. Prove that thishomomorphism is surjective and that its kernel consists of the f ∈ R[xs :


s ∈ S] which after multiplication by an element of S lie in the idealgenerated the degree one polynomials sxs − 1, s ∈ S.

EXERCISE 12. Let R be a ring and let p be a prime ideal of R.(a) Prove that the complement R − p is a multiplicative system. The result-

ing localization (R − p)−1R is called the localization at p and is usuallydenoted Rp.

(b) Prove that pRp is a maximal ideal of Rp and that it is the only maximalideal of Rp. (A ring with a unique maximal ideal is called a local ring.)

(c) Prove that the localization mapR→ Rp drops to an isomorphism of fieldsFrac(R/p)→ Rp/pRp.

(d) Work this out for R = Z and p = (p), where p is a prime number.(e) Same for R = k[x, y] and p = (x).

LEMMA 2.16. Let R be a ring. Then the intersection of all the prime ideals of Ris the ideal of nilpotents

√(0) of R. Equivalently, for every nonnilpotent a ∈ R, there

exists a ring homomorphism from R to a field that is nonzero on a.

PROOF. It is easy to see that a nilpotent element lies in every prime ideal. Nowfor nonnilpotent a ∈ R consider the homomorphism R → R[1/a]. The ring R[1/a]

is nonzero, hence has a maximal ideal3 m so that F := R[1/a]/m is a field. Thenthe kernel of the composite φ : R → R[1/a] → F is a prime ideal and a is not inthis kernel (for φ(a) ∈ F is invertible with inverse the image of 1/a).

EXERCISE 13. Let R be a ring. Prove that the intersection of all the maximalideals of a ring R consists of the a ∈ R for which 1 + aR ⊂ R× (i.e., 1 + ax isinvertible for every x ∈ R). You may use the fact that every proper ideal of R iscontained in a maximal ideal.

We can do better if R is noetherian. The following proposition is the algebraiccounterpart of Proposition 2.11. Note the similarity between the proofs.

PROPOSITION 2.17. Let R be a noetherian ring. Then any radical ideal in R isan intersection of finitely many prime ideals. Also, the minimal prime ideals of R arefinite in number and their intersection is equal to the ideal of nilpotents

√(0).

PROOF. We first make the rather formal observation that R is a radical idealand indeed appears as a finite (namely empty) intersection of prime ideals. So thecollection B of the radical ideals I ⊂ R that can not be written as an intersection offinitely many prime ideals does not contain R. We prove that B is empty. Supposeotherwise. Since R is noetherian, it will have a maximal member I0 6= R. We thenderive a contradiction as follows.

Since I0 cannot be a prime ideal, there exist a1, a2 ∈ R − I0 with a1a2 ∈ I0.Consider the radical ideal Ji :=

√I0 +Rai. Since Ji strictly contains I0, it does not

belong to B. In other words, Ji is an intersection of finitely many prime ideals. Wenext show that J1 ∩ J2 = I0, so that I0 is an intersection of finitely many primeideals also, thus arriving contradiction. The inclusion ⊃ is obvious and ⊂ is seenas follows: if a ∈ J1 ∩ J2, then for i = 1, 2, there exists an ni > 0 such thatani ∈ I0 +Rai. Hence an1+n2 ∈ (I0 +Ra1)(I0 +Ra2) ⊂ I0, so that a ∈ I0.

3Every nonzero ring has a maximal ideal. For noetherian rings, which are our main concern, thisis obvious, but in general this follows with transfinite induction, the adoption of which is equivalent tothe adoption of the axiom of choice.


We thus find that√

(0) = p1 ∩ · · · ∩ ps for certain prime ideals pi. We mayof course assume that no pi contains some pj with j 6= i (otherwise, omit pi). Itnow remains to prove that every prime ideal p of R contains some pi. If that isnot the case, then for i = 1, . . . , s there exists a ai ∈ pi − p. But then a1a2 · · · as ∈p1 ∩ · · · ∩ ps =

√(0) ⊂ p and since p is a prime ideal, some factor ai lies in p. This

is clearly a contradiction.

EXERCISE 14. Let J be an ideal of the ring R. Show that√J is the intersection

of all the prime ideals that contain J . Prove that when R is noetherian, its minimalprime ideals are finite in number and that their common intersection is still

√J .

What do we get for R = Z and J = Zn?

EXERCISE 15. Let R be a ring, S ⊂ R be a multiplicative system and denote byφ : R → S−1R the natural homomorphism. Prove that the map which assigns toevery prime ideal of S−1R its preimage in R under φ defines a bijection betweenthe prime ideals of S−1R and the prime ideals of R disjoint with S. Prove also thatif S has no zero divisors, then the preimage of the ideal of nilpotents of S−1R isthe ideal of nilpotents of R.

3. Finiteness properties and the Hilbert theorems

The noetherian property in commutative algebra is best discussed in the con-text of modules, even if one’s interest is only in rings. We fix a ring R and first recallthe notion of an R-module.

The notion of an R-module is the natural generalization of a K-vector space (where Kis some field). Let us observe that if M is an (additively written) abelian group, then theset End(M) of group homomorphisms M → M is a ring for which subtraction is pointwisedefined and multiplication is composition (so if f, g ∈ End(M), then f − g : m ∈ M 7→f(m) − g(m) and fg : m 7→ f(g(m))); clearly the zero element is the zero homomorphismand the unit element is the identity. It only fails to obey our convention in the sense that thisring is usually noncommutative. We only introduced it in order to be able state succinctly:

DEFINITION 3.1. An R-module is an abelian group M , equipped with a ring homomor-phism R→ End(M).

So any r ∈ R defines a homomorphism M → M ; we usually denote the image ofm ∈ M under this homomorphism simply by rm. If we write out the properties of an R-module structure in these terms, we get: r(m1−m2) = rm1−rm2, (r1−r2)m = r1m−r2m,1.m = m, r1(r2m) = (r1r2)m. If R happens to be field, then we see that an R-module is thesame thing as an R-vector space.

The notion of an R-module is quite ubiquitous, once you are aware of it. A simpleexample is an ideal I ⊂ R. Any abelian group M is in a natural manner a Z-module. And aR[x]-module can be understood as an real linear space V (an R-module) endowed with anendomorphism (the image of x in End(V )). A more involved example is the following: ifX is a manifold, f is a C∞-function on X and ω a C∞-differential p-form on X, then fω isalso a C∞ differential p-form on X. Thus the linear space of C∞-differential forms on X ofa fixed degree p is naturally a module over the ring of C∞-functions on X.

Here are a few companion notions, followed by a brief discussion.

3.2. In what follows M is an R-module. A map f : M → N from M to an R-module Nis called a R-homomorphism if it is a group homomorphism with the property that f(rm) =rf(m) for all r ∈ R and m ∈ M . If f is also bijective, then we call it an R-isomorphism; inthat case its inverse is also a homomorphism of R-modules.

3. FINITENESS PROPERTIES AND THE HILBERT THEOREMS 15

For instance, given a ring homomorphism f : R → R′, then R′ becomes an R-moduleby rr′ := f(r)r′ and this makes f a homomorphism of R-modules.

A subset N ⊂ M is called an R-submodule of M if it is a subgroup and rn ∈ N for allr ∈ R and n ∈ N . Then the group quotient M/N is in a unique manner a R-module in sucha way that the quotient map M →M/N is a R-homomorphism: we let r(m+N) := rm+Nfor r ∈ R and m ∈ M . Notice that a R-submodule of R (here we regard R as a R-module)is the same thing as an ideal of R.

Given a subset S ⊂ M , then the set of elements m ∈ M that can be written as r1s1 +· · · + rksk with ri ∈ R and si ∈ S is a R-submodule of M . We call it the R-submodule ofM generated by S and we shall denote it by RS. If there exists a finite set S ⊂ M such thatM = RS, then we say that M is finitely generated as an R-module.

DEFINITION 3.3. We say that an R-module M is noetherian if the collection ofR-submodules of M satisfies the ascending chain condition: any ascending chainof R-submodules N1 ⊂ N2 ⊂ · · · becomes stationary.

It is clear that then every quotient module of a noetherian module is also noe-therian. The noetherian property of R as a ring (as previously defined) coincideswith this property of R as an R-module.

The following two propositions provide the passage from the noetherian prop-erty to finite generation:

PROPOSITION 3.4. AnR-moduleM is noetherian if and only if everyR-submoduleof M is finitely generated as an R-module.

PROOF. Suppose that M is a noetherian R-module and let N ⊂ M be a R-submodule. The collection of finitely generated R-submodules of M contained inN is nonempty. Hence it has a maximal element N0. If N0 = N , then N is finitelygenerated. If not, we run into a contradiction: just choose x ∈ N−N0 and considerN0 + Rx. This is a R-submodule of N . It is finitely generated (for N0 is), whichcontradicts the maximal character of N0.

Suppose now that every R-submodule of M is finitely generated. If N1 ⊂N2 ⊂ · · · is an ascending chain of R-modules, then the union N := ∪∞i=1Ni is aR-submodule. Let s1, . . . , sk be a finite set of generators of N . If sκ ∈ Niκ , andj := maxi1, . . . , ik, then it is clear that Nj = N . So the chain becomes stationaryas of index j.

PROPOSITION 3.5. Suppose that R is a noetherian ring. Then every finitely gen-erated R-module M is noetherian.

PROOF. By assumption M = RS for a finite set S ⊂ M . We prove the propo-sition by induction on the number of elements of S. If S = ∅, then M = 0 andthere is nothing to prove. Suppose now S 6= ∅ and choose s ∈ S, so that our induc-tion hypothesis applies to M ′ := RS′ with S′ = S r s: M ′ is noetherian. But sois M/M ′, for it is a quotient of the noetherian ring R via the surjective R-modulehomomorphism R→M/M ′, r 7→ rs+M ′.

Let now N1 ⊂ N2 ⊂ · · · be an ascending chain of R-submodules of M . ThenN1 ∩ M ′ ⊂ N2 ∩ M ′ ⊂ · · · becomes stationary, say as of index j1. Hence weonly need to be concerned for k ≥ j1 with the stabilization of the submodulesNk/(Nj1 ∩M ′) = Nk/(Nk ∩M ′) ∼= (Nk +M ′)/M ′ of M/M ′. These stabilize indeed(say as of index j2), sinceM/M ′ is noetherian. So the original chainN1 ⊂ N2 ⊂ · · ·stabilizes as of index j2.


We are now sufficiently prepared for the proofs of the Hilbert theorems. Theyare gems of elegance and efficiency.

We will use the notion of initial coefficient of a polynomial, which we re-call. Given a ring R, then every nonzero f ∈ R[x] is uniquely written as rdxd +rd−1x

d−1 + · · · + r0 with rd 6= 0. We call rd ∈ R the initial coefficient of f anddenote it by in(f). For the zero polynomial, we simply define this to be 0 ∈ R.Notice that when in(f) in(g) nonzero, then it is equal to in(fg).

PROOF OF THEOREM 2.9. The assumption is here that R is a noetherian ring.In view of Proposition 3.4 we must show that every ideal I of R[x] is finitely gener-ated. Consider the subset in(I) := in(f) : f ∈ I of R. We first show that this isan ideal of R. If r ∈ R, f ∈ I, then r in(f) equals in(rf) or is zero and since rf ∈ I,it follows that r in(f) ∈ I. If f, g ∈ I, then in(f)− in(g) equals in(xdeg gf − xdeg fg)or is zero. So in(I) is an ideal as asserted.

Since R is noetherian, in(I) is finitely generated: there exist f1, . . . , fk ∈ Isuch that in(I) = R in(f1) + · · · + R in(fk). Let di be the degree of fi, d0 :=maxdeg(f1), . . . ,deg(fk) and R[x]<d0 the set of polynomials of degree < d0. SoR[x]<d0 is the R-submodule of R[x] generated by 1, x, . . . , xd0−1. We claim that

I = R[x]f1 + · · ·+R[x]fk + (I ∩R[x]<d0),

in other words, that every f ∈ I is modulo R[x]f1 + · · · + R[x]fk a polynomial ofdegree < d0. We prove this with induction on the degree d of f . Since for d < d0

there is nothing to prove, assume that d ≥ d0. We have in(f) = r1 in(f1) + · · · +rk in(fk) for certain r1, . . . rk ∈ R, where we may of course assume that every termri in(fi) is nonzero and hence equal to in(rifi). Since in(f) is nonzero, it thenequals

∑i in(rifi) = in(

∑i rifix

d−deg(fi)). So f −∑i rifix

d−deg(fi) is an elementof I of degree < d and hence lies in R[x]f1 + · · · + R[x]fk + (I ∩ R[x]<d0) by ourinduction hypothesis. Hence so does f .

Our claim implies the theorem: R[x]<d0 is a finitely generated R-module andso a noetherian R-module by Proposition 3.5. Hence the R-submodule I ∩R[x]<d0is a finitely generated R-module by Proposition 3.4. If fk+1, . . . , fk+l is a set ofR-generators of I ∩R[x]<d0 , then f1, . . . , fk+l is a set of R[x]-generators of I.

For the Nullstellensatz we need another finiteness result.

PROPOSITION 3.6 (Artin-Tate). Let R be a noetherian ring, B an R-algebra andA ⊂ B an R-subalgebra. Assume that B is finitely generated as an A-module. ThenA is finitely generated as an R-algebra if and only if B is so.

PROOF. By assumption there exist b1, . . . , bm ∈ B such that B =∑mi=1Abi.

If there exist a1, . . . , an ∈ A which generate A as an R-algebra (which meansthat A = R[a1, . . . , an]), then a1, . . . , an, b1, . . . , bm generate B as an R-algebra.

Suppose, conversely, that there exists a finite subset of B which generates Bas a R-algebra. By adding this subset to b1, . . . , bm, we may assume that b1, . . . , bmalso generate B as an R-algebra. Then every product bibj can be written as anA-linear combination of b1, . . . , bm:

bibj =

m∑k=1

akijbk, akij ∈ A.

Let A0 ⊂ A be the R-subalgebra of A generated by all the (finitely many) coeffi-cients akij . This is a noetherian ring by Corollary 2.10. It is clear that bibj ∈

∑k A0bk

3. FINITENESS PROPERTIES AND THE HILBERT THEOREMS 17

and so∑k A0bk is an R-subalgebra of B. Since the b1, . . . , bm generate B as an R-

algebra, it then follows this is all of B: B =∑k A0bk. So B is finitely generated

as an A0-module. Since A is an A0-submodule of B, A is also finitely generatedas an A0-module by Proposition 3.4. It follows that A is a finitely generated R-algebra.

This has a consequence for field extensions:

COROLLARY 3.7. A field extension L/K is finite if and only if L is finitely gener-ated as a K-algebra.

PROOF. It is clear that if L is a finite dimensional K-vector space, then L isfinitely generated as a K-algebra.

Suppose now b1, . . . , bm ∈ L generate L as a K-algebra. It suffices to show thatevery bi is algebraic over K. Suppose that this is not the case. After renumberingwe can and will assume that (for some 1 ≤ r ≤ m) b1, . . . , br are algebraically inde-pendent over K and br+1, . . . , bm are algebraic over the quotient field K(b1, . . . , br)of K[b1, . . . , br]. So L is a finite extension of K(b1, . . . , br). We apply Proposition3.6 to R := K, A := K(b1, . . . , br) and B := L and find that K(b1, . . . , br) is asa K-algebra generated by a finite subset S ⊂ K(b1, . . . , br). If g is a common de-nominator for the elements of S, then clearly K(b1, . . . , br) = K[b1, . . . , br][1/g].Since K(b1, . . . , br) strictly contains K[b1, . . . , br], g must have positive degree. Inparticular, g 6= 1, so that 1/(1 − g) ∈ K(b1, . . . , br) can be written as f/gN , withf ∈ K[b1, . . . , br]. Here we may of course assume that f is not divisible by g inK[b1, . . . , br]. From the identity f(1− g) = gN we see that N ≥ 1 (for the left handside has positive degree). But then f = g(f + gN−1) shows that f is divisible by g.We thus get a contradiction.

COROLLARY 3.8. Let A be a finitely generated k-algebra. Then for every maximalideal m ⊂ A, the natural map k → A→ A/m is an isomorphism of fields.

PROOF. Since m is maximal, A/m is a field that is also finitely generated as ak-algebra. By corollary 3.7, k → A/m is then a finite extension of k. Since k isalgebraically closed, this extension will be the identity.

EXERCISE 16. Prove that a field which is finite generated as a ring (i.e., isisomorphic to a quotient of Z[x1, . . . , xn] for some n) is finite.

We deduce from the preceding corollary the Nullstellensatz.

PROOF OF THE NULLSTELLENSATZ 1.5. Let J ⊂ k[x1, . . . , xn] be an ideal. Wemust show that I(Z(J)) ⊂

√J . This amounts to: for every f ∈ k[x1, . . . , xn]−

√J

there exists a p ∈ Z(J) for which f(p) 6= 0. Consider k[x1, . . . , xn]/J and denote byf ∈ k[x1, . . . , xn]/J the image of f . Since f is not nilpotent,

A := (k[x1, . . . , xn]/J)[1/f ].

is not the zero ring and so has a maximal ideal m ⊂ A. Observe that A is a finitelygenerated k-algebra (we can take the images of x1, . . . , xn and 1/f as generators)and so the map k → A/m is by Corollary 3.8 an isomorphism. Denote by φ :k[x1, . . . , xn] → A → A/m = k the corresponding surjection and put pi := φ(xi)and p := (p1, . . . , pn) ∈ An. So if we view xi as a function on An, then φ(xi) is thevalue of xi at p. The fact that φ is a homomorphism of k-algebras implies that it isthen given as ‘evaluation in p’: for any g ∈ k[x1, . . . , xn] we have φ(g) = g(p). Since


the kernel of φ contains J , every g ∈ J will be zero in p, in other words, p ∈ Z(J).On the other hand, f(p) = φ(f) is invertible, for it has the image of 1/f in A/m = kas its inverse. So f(p) 6= 0.

4. The affine category

We begin with specifying the maps between closed subsets of affine spaces thatwe wish to consider.

DEFINITION 4.1. Let X ⊂ Am and Y ⊂ An be closed subsets. We say that amap f : X → Y is regular if the components f1, . . . , fn of f are regular functionson X (i.e., are given by the restrictions of polynomial functions to X).

Composition of a regular function on Y with f yields a regular function on X(for if we substitute in a polynomial of n variables g(y1, . . . , yn) for every variableyi a polynomial fi(x1, . . . , xm) of m variables, we get a polynomial of m variables).So f then induces a k-algebra homomorphism f∗ : k[Y ] → k[X]. This property isclearly equivalent to f being regular. The same argument shows that if f : X → Yand g : Y → Z are regular maps, then so is their composite gf : X → Z. Sowe have a category (with objects the closed subsets of some affine space An andregular maps as defined above). In particular, we have a notion of isomorphism: aregular map f : X → Y is an isomorphism if is has a two-sided inverse g : Y → Xwhich is also a regular map. This implies that f∗ : k[Y ] → k[X] has a two-sidedinverse g∗ : k[X]→ k[Y ] which is also an homomorphism of k-algebras, and henceis an isomorphism of k-algebras.

There is also a converse:

PROPOSITION 4.2. Let be given closed subsets X ⊂ Am and Y ⊂ An and ak-algebra homomorphism φ : k[Y ] → k[X]. Then there is a unique regular mapf : X → Y such that f∗ = φ.

PROOF. The inclusion j : Y ⊂ An defines a k-algebra homomorphism j∗ :k[y1, . . . , yn] → k[Y ] with kernel I(Y ). Put fi := φj∗(yi) ∈ k[X] (i = 1, . . . , n)and define f = (f1, . . . , fn) : X → An, so that f∗yi = fi = φj∗yi. Since the k-algebra homomorphisms f∗, φj∗ : k[y1, . . . , yn] → k[X] coincide on the generatorsyi, they must be equal: f∗ = φj∗. It follows that f∗ is zero on the kernel I(Y ) ofj∗, which means that f takes its values in Z(I(Y )) = Y , and that the resulting mapk[Y ]→ k[X] equals φ. The proof of uniqueness is left to you.

In particular, an isomorphism of k-algebras k[Y ] → k[X] comes from a uniqueisomorphism X → Y . In the special case of an inclusion of a closed subset Z ⊂ Y ,the induced map k[Y ] → k[Z] is of course the formation of the quotient algebrak[Z] = k[Y ]/IY (X). So f : X → Y is an isomorphism of X onto a closed subsetof Y (we then say that f is a closed immersion) if and only if f∗ : k[Y ] → k[X] is asurjection of k-algebras (with ker(f∗) being the ideal defining the image of f).

We complete the picture by showing that any finitely generated reduced k-algebra A is isomorphic to some k[Y ]; the preceding then shows that Y is uniqueup to isomorphism. Since A is finitely generated as a k-algebra, there exists a sur-jective k-algebra homomorphism φ : k[x1, . . . , xn] → A. If we put I := Ker(φ),then φ induces an isomorphism k[x1, . . . , xn]/I ∼= A. Put Y := Z(I) ⊂ An. SinceA is reduced, I is a radical ideal and hence equal to I(Y ) by the Nullstellensatz. Itfollows that φ factors through a k-algebra isomorphism k[Y ] ∼= A.

4. THE AFFINE CATEGORY 19

We may sum up this discussion in categorical language as follows.

PROPOSITION 4.3. The map which assigns to a closed subset of some An its coor-dinate ring defines an anti-equivalence between the category of closed subsets of affinespaces (whose morphisms are the regular maps) and the category of reduced finitelygenerated k-algebras (whose morphisms are k-algebra homomorphisms). It makesclosed immersions correspond to epimorphisms of such k-algebras.

EXAMPLE 4.4. Consider the regular map f : A1 → A2, f(t) = (t2, t3). The mapsA1 bijectively onto the hypersurface (curve) C defined by x3 − y2 = 0: the imageis clearly contained in C and the inverse sends (0, 0) to 0 and is on C r (0, 0)given by (x, y) 7→ y/x. The Zariski topology on A1 and C is the cofinite topologyand so this is even a homeomorphism. In order to determine whether the inverseis regular, we consider f∗. We have k[C] = k[x, y]/(x3 − y2), k[A1] = k[t] andf∗ : k[C] → k[t] is given by x 7→ t2, y 7→ t3. This algebra homomorphism is notsurjective for its image misses t ∈ k[t]. In fact, f identifies k[C] with the subalgebrak + t2k[t] of k[t]. So f is not an isomorphism.

EXAMPLE 4.5. An affine-linear transformation of kn is of the form x ∈ kn 7→g(x) + a, where a ∈ kn and g ∈ GL(n, k) is a linear transformation. Its inverse isy 7→ g−1(y−a) = g−1(y)−g−1(a) and so of the same type. When we regard such anaffine linear transformation as a map from An to itself, then it is regular: its coordi-nates (g1, . . . , gn) are polynomials of degree one. So an affine-linear transformationis also an isomorphism of An onto itself. When n ≥ 2, there exist automorphisms ofAn not of this form. For instance σ : (x, y) 7→ (x, y + x2) defines an automorphismof A2 with inverse (x, y) 7→ (x, y − x2) (see also Exercise 18). This also shows thatthe group of affine-linear translations in An is not a normal subgroup, for conjuga-tion by σ takes the transformation (x, y) 7→ (x + y, y) to an automorphism that isnot affine-linear (check this). Hence the group of affine-linear transformations ofAn is not a “natural” subgroup of the automorphism group of An (this makes thatthe name affine n-space for An is a bit unfortunate).

EXERCISE 17. Let C ⊂ A2 be the ‘circle’, defined by x2 + y2 = 1 and let p0 :=(−1, 0) ∈ C. For every p = (x, y) ∈ C r p0, the line through p0 and p has slopef(p) = y/(x+ 1). Denote by

√−1 ∈ k a root of the equation t2 + 1 = 0.

(a) Prove that when char(k) 6= 2, f defines an isomorphism4 onto A1 r±√−1.

(b) Consider the map g : C → A1, g(x, y) := x +√−1y. Prove that when

char(k) 6= 2, g defines an isomorphism of C onto A1 r 0.(c) Prove that when char(k) = 2, the defining polynomial x2 + y2 − 1 for C

is the square of a degree one polynomial so that C is a line.

EXERCISE 18. Let f1, . . . , fn ∈ k[x1, . . . , xn] be such that f1 = x1 and fi − xi ∈k[x1, . . . , xi−1] for i = 2, . . . , xn. Prove that f defines an isomorphism An → An.

4We have not really defined yet what is an isomorphism between two nonclosed subsets of an affinespace. Interpret this here as: f∗ maps k[x, y][1/(x+ 1)]/(x2 + y2− 1) (the algebra of regular functionson C r p0) isomorphically onto k[t][1/(t2 + 1)] (the algebra of regular functions on A1 r ±

√−1).

This will be justified by Proposition 4.8.


EXAMPLE 4.6. QUADRATIC HYPERSURFACES IN CASE char(k) 6= 2. Let H ⊂ Anbe a hypersurface defined by a polynomial of degree two:

f(x1, . . . , xn) =∑

1≤i≤j=n

aijxixj +

n∑i=1

aixi + a0.

By means of a linear transformation the quadratic form∑

1≤i≤j=n aijxixj canbe brought in diagonal form (this involves splitting off squares, hence requiresthe existence of 1/2 ∈ k). This means that we can make all the coefficientsaij with i 6= j vanish. Another diagonal transformation (which replaces xi by√aiixi when aii 6= 0) takes every nonzero coefficient aii to 1 and then renumber-

ing the coordinates (which is also a linear transformation) brings f into the formf(x1, . . . , xn) =

∑ri=1 x

2i +

∑ni=1 aixi + a0 for some r ≥ 1. Splitting off squares

once more enables us to get rid of∑ri=1 aixi so that we get

f(x1, . . . , xn) =

r∑i=1

x2i +

n∑i=r+1

aixi + a0.

We now have the following cases.If the nonsquare part is identically zero, then we end up with the equation∑r

i=1 x2i = 0 for H.

If the linear part∑ni=r+1 aixi is nonzero (so that we must have r < n), then an

affine-linear transformation which does not affect x1, . . . , xr and takes∑ni=r+1 aixi+

a0 to −xn yields the equation xn =∑ri=1 x

2i . This is the graph of the function∑r

i=1 x2i on An−1 and so H is then isomorphic to An−1.

If the linear part∑ni=r+1 aixi is zero, but the constant term a0 is nonzero, then

we can make another diagonal transformation which replaces xi by√−a0xi) and

divide f by a0: then H gets the equation∑ri=1 x

2i = 1.

In particular, there are only a finite number of quadratic hypersurfaces up toisomorphism. (This is also true in characteristic two, but the discussion is a bitmore delicate.)

The previous discussion (and in particular Proposition 4.3) leads us to associateto any finitely generated k-algebra A in a direct manner a space (which we shalldenote by Spm(A)) which forA = k[X] yields a space homeomorphic toX. Since inthat case the points of X correspond to maximal ideals of k[X], we simply choosethe underlying set of Spm(A) to be the collection of maximal ideals of A. Forx ∈ Spm(A), we shall denote the corresponding maximal ideal of A by mx. Since Ais finitely generated as a k-algebra, A/mx can be identified with k by Corollary 3.8.We denote the resulting k-algebra homomorphism A→ k by ρx. It is clear that anyk-algebra homomorphism A → k has a maximal ideal of A as its kernel and so wemay also think of Spm(A) as the set of k-algebra homomorphisms A→ k.

Any f ∈ A defines defines a ‘regular function’ f : Spm(A) → k which takesin x ∈ Spm(A) the value ρx(f) ∈ k. So its zero set Z(f) ⊂ Spm(A) is the setof x ∈ Spm(A) with f ∈ mx. We denote the complement of Spm(A) − Z(f) bySpm(A)f . We have Z(ff ′) = Z(f) ∪ Z(f ′) (for ρx(ff ′) = ρx(f)ρx(f ′)) and henceSpm(A)ff ′ = Spm(A)f ∩Spm(A)f ′ . So the collection of Spm(A)ff∈A is the basisof a topology on Spm(A). Note that a subset Spm(A) is closed precisely if it is anintersection of subsets of the form Z(f); this is equal to the common zero set ofthe set of functions defined by an ideal of A. For A = k[x1, . . . , xn]/I, the above


discussion shows that Spm(A) can be identified with Z(I) ⊂ An as a topologicalspace and that under this identification, A/(0) becomes the ring of regular functionson Z(I). More generally, for any finitely generated k-algebra A, the map f 7→ fmaps A onto a subalgebra of the algebra of k-valued functions on Spm(A) withkernel the ideal of nilpotents (Exercise 20).

The space Spm(A) is called the maximal ideal spectrum5 of R (but our notationfor it is less standard). In case A is a reduced finitely generated k-algebra, we referto Spm(A) as an affine variety (we will give a more complete definition later). Wethen recover A as its algebra of regular functions.

We observe for later reference:

LEMMA 4.7. The maximal ideal spectrum Spm(A) is quasi-compact: every opencovering of Spm(A) admits a finite subcovering.

PROOF. It suffices to verify this for an open covering by principal open subsets.So let S ⊂ A be such that Spm(A) = ∪g∈S Spm(A)g. This means that ∩s∈SZ(g) =∅. So the ideal generated by S is not contained in any maximal ideal and hencemust be all of A. In particular, 1 =

∑ni=1 gifi for certain fi ∈ A and gi ∈ S. It

follows that gini=1 generates A so that Spm(A) = ∪ni=1 Spm(A)gi .

A homomorphism φ : A→ B of finitely generated k-algebras gives rise to a mapSpm(φ) : Spm(B) → Spm(A): if y ∈ Spm(B), then the composite homomorphismρyφ : A → k is the identity map when restricted to k so that φ−1my is a maximalideal of A with residue field k. We thus get a map

Spm(φ) : Spm(B)→ Spm(A).

characterized by mSpm(φ)(y) = φ−1my. For g ∈ A, the preimage of Z(g) underSpm(φ) is Z(φ(g)) and hence the preimage of Spm(A)g is Spm(B)φ(g). This showsthat Spm(φ) is continuous. We call the resulting pair (Spm(φ), φ) a morphism.

PROPOSITION 4.8. Let A be a finitely generated k-algebra. Then for every g ∈ A,A[1/g] is a finitely generated k-algebra (which is reduced when A is) and the naturalk-algebra homomorphism A → A[1/g] induces a homeomorphism of Spm(A[1/g])onto Spm(A)g = X − Z(g). Moreover, for g, g′ ∈ A the following are equivalent:

(i) Spm(A)g ⊂ Spm(A)g′ ,(ii) g′ divides some positive power of g,

(iii) there exists a A-homomorphism A[1/g′] → A[1/g] (which must then beunique).

PROOF. It is clear that A[1/g] is a k-algebra and is as such finitely generated(just add to a generating set for A the generator 1/g). We show that if A is reduced,then so is A[1/g]. For this we may suppose that g is not nilpotent (otherwise A[1/g]is the zero ring). Suppose that f/gr ∈ A[1/g] is nilpotent: (f/gr)m = 0 for somem ≥ 1. This means that there exists an n ≥ 0 such that fmgn = 0. Then (fgn)m = 0and since A is reduced it follows that fgn = 0. So f/gr = fgn/gr+n = 0 in A[1/g].

A point of A[1/g] is given by a k-algebra homomorphism A[1/g] → k. Thisis the same thing as to give a k-algebra homomorphism A → k that is nonzeroon g, in other words a point of Spm(A)g. So the map A → A[1/g] induces an

5I. Gelfand was presumably the first to consider this, albeit in the context of functional analysis:he characterized the Banach algebras that appear as the algebras of continuous C-valued functions oncompact Hausdorff spaces. So it might be appropriate to call this the Gelfand spectrum.


injection of Spm(A[1/g]) in Spm(A) with image Spm(A)g. The map Spm(A[1/g])→Spm(A) is a morphism and hence continuous. To see that it is also open, notethat a principal open subset of Spm(A[1/g]) is of the form Spm(A[1/g])f/gn , withf ∈ A. By the preceding discussion we may identify this with Spm(A[1/g][gn/f ]) =Spm(A[1/(fg)]) and so its image in Spm(A) is the open subset Spm(A)fg.

We check the equivalence of the three conditions.(i) ⇒ (ii) If Spm(A)g ⊂ Spm(A)g′ , then Z(g) ⊃ Z(g′) and so by the Nullstel-

lensatz, g ∈√

(g′). This implies that we can write gn = fg′ for some f ∈ A andsome n ≥ 1 and (ii) follows.

(ii) ⇒ (iii) If (ii) holds, then we have defined a A-homomorphism A[1/g′] →A[1/(fg′)] = A[1/gn] = A[1/g] that is easily checked to be independent of thechoices made for n and f ′ and so (iii) follows.

(iii) ⇒ (i) If we have an A-homomorphism A[1/g′] → A[1/g], then we get amorphism Spm(A[1/g])→ Spm(A[1/g′]) whose composition with the identificationof Spm(A[1/g′]) with the open subset Spm(A)g′ ⊂ Spm(A) yields the identifica-tion of Spm(A[1/g]) with the open subset Spm(A)g ⊂ Spm(A). This means thatSpm(A)g ⊂ Spm(A)g′ and shows at the same time that such an A-homomorphismis unique.

From now on we identify a principal open subset Xg = Spm(A)g with themaximal ideal spectrum Spm(A[1/g]).

EXERCISE 19. Give an example of ring homomorphism φ : S → R and a maxi-mal ideal m ⊂ R, such that φ−1m is not a maximal ideal of S. (Hint: take a look atExercise 12.)

EXERCISE 20. Prove that if A is a finitely generated k-algebra, then the mapf ∈ A 7→ f is a k-algebra homomorphism from A onto the algebra of k-valuedfunctions on Spm(A) with kernel

√(0). Show that for every subset X ⊂ Spm(A),

the set I(X) of f ∈ A with f |X = 0 is a radical ideal of A.

Let f : X → Y be a morphism between affine varieties. Since f is continuous,a fiber f−1(y), or more generally, the preimage f−1Z of a closed subset Z ⊂ Y , willbe closed in X. It is the zero set of the ideal in k[X] generated by f∗I(Z). In fact,any ideal I ⊂ k[X] can arise this way, for if (f1, . . . , fr) ∈ k[X] generate I, thentake f = (f1, . . . , fr) : X → Ar = Y and y = 0. We will later see that the failure off∗I(Z) to be a radical ideal is sometimes a welcome property, as it can be exploitedto define multiplicity. Here is a very simple example.

EXAMPLE 4.9. Let f : X = A1 → A1 = Y be defined by f(x) = x2. Thenf∗ : k[y] → k[x] is given by f∗y = x2. If we assume k not to be of characteristic 2,and we take a ∈ Y r 0, then the fiber f−1(a) is defined by the ideal generatedby f∗(y − a) = x2 − a. It consists of two distinct points that are the two roots ofx2 = a, denoted ±

√a and the pair of evaluation maps (ρ√a, ρ−

√a) identifies the

coordinate ring k[x]/(x2 − a) with k⊕ k. However, the fiber over 0 ∈ Y = A1 is thesingleton 0 ⊂ X = A1 and the ideal generated by f∗y = x2 is not a radical ideal.This example indicates that there might good reason to accept nilpotent elementsin the coordinate ring of f−1(0) by endowing f−1(0) with the ring of functionsk[f−1(0)] := k[x]/(x2). This is a k-vector space of dimension 2 (a k-basis is definedby the pair 1, x) and we thus retain the information that two points have cometogether. The fiber should indeed be thought of as a point with multiplicity 2.


Example 4.4 shows that a continuous bijection (and even a homeomorphism)of affine varieties need not be an isomorphism. We next discuss a class of examplesof an entirely different nature. It involves a notion that plays a central role inalgebraic geometry when the base field k has positive characteristic.

EXAMPLE 4.10 (THE FROBENIUS MORPHISM). Assume that k has positive char-acteristic p and consider the morphism Φp : A1 → A1, a 7→ ap. If we rememberthat A1 can be identified with k, then we observe that under this identification, Φpis a field automorphism: Φp(a − b) = (a − b)p = ap − bp = Φp(a) − Φp(b) (and ofcourse Φp(ab) = (ab)p = Φp(a)Φp(b)). This shows that Φp is injective. Since k isalgebraically closed, every element of k has a pth root and so Φp is also surjective.But the endomorphism Φ∗p of k[x] induced by Φp sends x to xp and has thereforeimage k[xp]. Clearly, Φ∗p is not surjective.

The fixed point set of Φp (the set of a ∈ A1 with ap = a) is via the identificationof A1 with k just the prime subfield Fp ⊂ k and we therefore denote it by A1(Fp) ⊂A1. Likewise, the fixed point set A1(Fpr ) of Φrp is the subfield of k with pr elements.Since the algebraic closure Fp of Fp in k is the union of the finite subfields of k,the affine line over Fp equals ∪r≥1A1(Fpr ). This generalizes in a straightforwardmanner to higher dimensions: by letting Φp act coordinatewise on An, we get amorphism An → An (which we still denote by Φp) which is also a bijection. Thefixed point of Φrp is An(Fpr ) and An(Fp) = ∪r≥1An(Fpr ).

EXERCISE 21. Assume that k has positive characteristic p. Let q = pr be a powerof p with r > 0 and denote by Fq ⊂ k the subfield of a ∈ k satisfying aq = a. Wewrite Φq for Φrp : a ∈ An 7→ aq ∈ An.

(a) Prove that f ∈ k[x1, . . . , xn] has its coefficients in Fq if and only if Φqf =fq.

(b) Prove that an affine-linear transformation of An with coefficients in Fqcommutes with Φq.

(c) Let Y ⊂ An be the common zero set of a subset of Fq[x1, . . . , xn] ⊂k[x1, . . . , xn]. Prove that Φq restricts to a bijection ΦY,q : Y → Y and thatthe fixed point set of ΦmY,q is Y (Fqm) := Y ∩ An(Fqm).

(d) Suppose that k is an algebraic closure of Fp. Prove that every closedsubset Y ⊂ An is defined over a finite subfield of k and hence is invariantunder some positive power of Φp.

REMARK 4.11. After this exercise we cannot resist to mention the Weil zetafunction. This function and its relatives—among them the Riemann zeta function—codify arithmetic properties of algebro-geometric objects in a very intricate manner.In the situation of Exercise 21, we can use the numbers |Y (Fqm)| (= the number offixed points of Φm in Y ) to define a generating series

∑m≥1 |Y (Fqm)|tm. It appears

to be more convenient to work with the Weil zeta function:

ZY (t) := exp( ∞∑m=1

|Y (Fqm)| tm

m

),

which has the property that t ddt logZY yields the generating series above. This se-ries has remarkable properties. For instance, a deep theorem due to Bernard Dwork(1960) asserts that it represents a rational function of t. Another deep theorem, dueto Pierre Deligne (1974), states that the roots of the numerator and denominator


have for absolute value a nonpositive half-integral power of q and that moreover,these powers have an interpretation in terms of an ‘algebraic topology for algebraicgeometry’, as was predicted by Andre Weil in 1949. (This can be put in a broadercontext by making the change of variable t = q−s. Indeed, now numerator anddenominator have their zeroes when the real part of s is a nonnegative half-integerand this makes Deligne’s result reminiscent of the famous conjectured property ofthe Riemann zeta function.)

EXERCISE 22. Compute the Weil zeta function of affine n-space relative to thefield of q elements.

REMARK 4.12. The Frobenius morphism as defined above should not be confused withpth power map FA : a ∈ A 7→ ap ∈ A that we have on any k-algebra A (with k of char-acteristic p > 0) and that is sometimes referred to as the absolute Frobenius. This is a ringendomorphism but not a k-algebra endomorphism, for it is on k also the pth power map(the usual Frobenius Fk) and so not the identity. We can in a sense remedy this by replacingthe ring homomorphism i : k → A that makes A a k-algebra by its precomposite with theFk, i.e., by replacing i : k → A by iFk : k → A (so this sends λ to i(λp) = i(λ)p). Theresulting k-algebra is called the Frobenius twist of A, and is denoted A(p). We now have afactorization

FA : AFk⊗1−−−→ A(p)

FA(p)/k−−−−→ A,

where the first map is essentially the identity (but is not k-linear), whereas the second mapis homomorphism of k-algebras. The induced map Spm(A(p))→ Spm(A) is the identity, butthe map induced by FA(p)/k, Spm(A)→ Spm(A(p)) is in general not. Observe that the othercomposite (Fk ⊗ 1)FA(p)/k : A(p) → A(p) is just the pth power map of A(p), FA(p) (if we putB := A(p) and write B(1/p) for A, this reads as FB = (F−1

k ⊗ 1B)FB/k : B → B(1/p) → B).Iterating this r times yields a k-algebra A(q) with q = pr. If it so happens that A is given

to us as obtained from a Fq-algebra Ao by extension of scalars: A = k ⊗Fq Ao (where wehave identified Fq with a subfield of k), then A(q) = k(q) ⊗Fq Ao, and since the qth powermap k → k(q) is a field isomorphism, we can use that isomorphism to identify A(q) with A asa k-algebra. So the qth power map then does determine a k-algebra homomorphism A→ A(given by c ⊗Fq ao 7→ c ⊗Fq a

qo). It called the geometric Frobenius. For Ao = Fq[x1, . . . , xn]

(and more generally for the coordinate ring of an Y as above), this yields our Φ∗q .

5. The sheaf of regular functions

In any topology or analysis course you learn that the notion of continuity islocal: there exists a notion of continuity at a point so that a function is continuousif it is so at every point of its domain. We shall see that in algebraic geometry theproperty for a function to be regular is also local in nature.

Let A be a reduced finitely generated k-algebra. We abbreviate X := Spm(A).A principal neighborhood of x ∈ X is of the form Xg with g ∈ A−mx and a regularfunction f on that neighborhood is an element of A[1/g]. Let us say that two suchpairs (Xg, f) and (Xg′ , f

′) have the same germ at x if there exists a neighborhoodU of x in Xg ∩ Xg′ such that f |U = f ′|U . This is an equivalence relation; anequivalence class is called a germ of a regular function on X at x. The germs ofregular functions on X at x form an k-algebra, which we shall denote by OX,x. Infact OX,x is nothing but the localization Amx (so that OX,x is a local ring): any φ ∈Amx is represented by a fraction f/g with f ∈ A and g ∈ A−mx, hence comes froma regular function on the principal neighborhood Xg of x. And if φ is also given as

5. THE SHEAF OF REGULAR FUNCTIONS 25

f ′/g′, then we have (fg′− f ′g)g′′ for some g′′ ∈ A−mx, which just means that f/gand f ′/g′ define the same element of A[1/(gg′g′′)], where we note that gg′g′′ /∈ mxso that Spm(A[1/(gg′g′′)]) = Xgg′g′′ is a principal neighborhood of x in X. Observethat ρx defines a surjective ‘evaluation homomorphism’ ρX,x : OX,x → k: it takesany φ ∈ OX,x as above to ρx(f)/ρx(g). So φ is invertible in OX,x precisely whenρx(f) 6= 0, or equivalently, when ρX,x(φ) 6= 0 (with its inverse represented by g/f)and hence the kernel of ρX,x is the maximal ideal mX,x of the local ring OX,x.

DEFINITION 5.1. A k-valued function φ defined on an open subset U of X issaid to be regular at x ∈ U if it defines an element of OX,x, in other words, if itdefines a regular function on some principal neighborhood of x in X. We denoteby O(U) the set of k-valued functions U → k that are regular at every point of U .

Note that is O(U) is in fact a k-algebra. We would like to call an element ofO(U) a regular function on U , but we have that notion already defined in caseU = X, or more generally, when U is a principal open subset. Fortunately, there isno conflict here:

PROPOSITION 5.2. LetX be an affine variety andXg ⊂ X a principal open subset.Then the natural k-algebra homomorphism k[X][1/g]→ O(Xg) is an isomorphism.

A map φ : X → Y between affine varieties is a morphism if and only if φ iscontinuous and for any f ∈ O(V ) (with V ⊂ Y open) we have f∗φ = φf ∈ O(f−1V ).

PROOF. The map k[X][1/g] → O(Xg) is injective: if f/gr ∈ k[X][1/g] is in thekernel (with f ∈ A and r ≥ 0), then f must be identically zero as a function on Xg.Hence fg is zero everywhere and so f/gr is zero in k[X][1/g].

For surjectivity, let φ : Xg → k be regular in every point of Xg. We must showthat φ is representable by f/gr ∈ k[X] for some f ∈ k[X] and r ≥ 0. By assumptionthere exist for every x ∈ X, gx ∈ k[X] r mx and fx ∈ k[X] such that φ|Xgx ∩ Xg

is representable as fx/gx. Upon replacing gx by gxg and fx by fxg, we may assumethat Xgx ⊂ Xg so that fx/gx represents φ|Xgx .

Since Xg is quasicompact (Lemma 4.7), the covering Xgxx∈X of Xg has afinite subcovering Xgxi

Ni=1. Let us write fi for fxi and gi for gxi . Then fi/giand fj/gj define the same regular function on Xgi ∩ Xgj = Xgigj and so gifj −gjfi is annihilated by (gigj)

mij for some mij ≥ 0. Put m := maxmij, so thatgm+1i gmj fj = gm+1

j gmi fi for all i, j. Upon replacing fi by figmi and gi by gm+1i , we

may then assume that in fact gifj = gjfi for all i, j.Since the ∪iXgi = Xg, we have ∩iZ(gi) = Z(g), and so by the Nullstellensatz

there must exist an r > 0 such that gr =∑Ni=1 higi for certain hi ∈ k[X]. Now

consider f :=∑Ni=1 hifi ∈ k[X]. We have for every j,

fgj =

N∑i=1

hifigj =

N∑i=1

higifj = grfj

and so the restriction of f/gr to Xgj is equal to fj/gj . As this is also the restrictionof φ to Xgj and ∪jXgj = Xg, it follows that φ is represented by f/gr.

The last statement is left as an exercise.

Let us denote by OX the collection of the k-algebras O(U), where U runs overall open subsets ofX. The preceding proposition says thatOX is a sheaf of k-valuedfunctions on X, by which we mean the following:


DEFINITION 5.3. Let X be a topological space and R a ring. A sheaf O of R-valued functions6 on X assigns to every open subset U of X an R-subalgebra O(U)of the R-algebra of R-valued functions on U with the property that

(i) for every inclusion U ⊂ U ′ of open subsets of X, ‘restriction to U ’ mapsO(U ′) in O(U) and

(ii) given a collection Uii∈I of open subsets ofX and a function f : ∪i∈IUi →R, then f ∈ O(∪iUi) if and only if f |Ui ∈ O(Ui) for all i.

If (X,OX) and (Y,OY ) are topological spaces endowed with a sheaf of R-valued functions, then a continuous map f : X → Y is called a morphism if forevery open V ⊂ Y , composition with f takes OY (V ) to OX(f−1V ).

This definition simply expresses the fact that the functions we are consider-ing are characterized by a local property—just as we have a sheaf of continuousR-valued functions on a topological space, a sheaf of differentiable R-valued func-tions on a manifold and a sheaf of holomorphic C-valued functions on a complexmanifold. In fact for the definition below of an affine variety (and the one of avariety that we shall give later), we take our cue from the definition of a manifold.

With the notion of a morphism, we have a category of topological spaces en-dowed with a sheaf of R-valued functions. In particular, we have the notion ofisomorphism: this is a homeomorphism f : X → Y which for every open V ⊂ Ymaps OY (V ) onto OX(f−1V ).

Note that a sheaf O of R-valued functions on X restricts to a sheaf O|U forevery open U ⊂ X. We are now ready to introduce the notion of an affine varietyin a more proper fashion.

DEFINITION 5.4. A topological space X endowed with a sheaf OX of k-valuedfunctions is called an affine variety when it is isomorphic to a pair (Spm(A),OSpm(A))as above. We refer to OX(X) as its coordinate ring and usually denote it by k[X].

We call (X,OX) a quasi-affine variety if it is isomorphic to an open subset ofsome pair (Spm(A),OSpm(A)).

Thus a reduced finitely generated k-algebra defines an affine variety and con-versely, an affine variety determines a reduced finitely generated k-algebra. Thesetwo assignments are inverses of each other. The present definition lends itself betterthan the previous one to immediate generalization (e.g., when we will introducethe notion of a variety) and has other technical advantages as well. Here is anexample.

EXAMPLE 5.5. Here is an example an affine open subset of an affine variety that isnot principal. Take the cuspidal plane cubic curve C ⊂ A2 of Example 4.4 defined byy2 = x3 and assume that k is of characteristic zero. As we have seen, the parametrizationf : t ∈ A1 7→ (t2, t3) ∈ C identifies k[C] with the subalgebra k + t2k[t] of k[t]. Now leta ∈ A1 r 0. So C r f(a) is quasi-affine. But C r f(a) is not a principal open subset:it is not of the form Cg for some g ∈ k[x, y]. For then f∗(g) would have a as its only zero, sothat f∗g is a nonzero constant times (t−a)n. But the coefficient of t in (t−a)n is n(−a)n−1,and hence nonzero. This contradicts the fact that f∗g ∈ k + t2k[t].

We claim however that C r f(a) is even affine with coordinate ring via f∗ identifiedwith Spm k[t2, t3, t2/(t−a)]. Let us first observe that k[t2, t3, t2/(t−a)] is a finitely generatedk-algebra contained in the reduced k-algebra k[t][1/(t − a)]. So it defines an affine variety

6We give the general definition of a sheaf later. This will do for now. A defect of this definition isthat a sheaf of R-valued functions on a space X need not restrict to one on a subspace of X.

6. THE PRODUCT 27

C and the inclusion k[t2, t3] ⊂ k[t2, t3, t2/(t− a)] defines a morphism j : C → C. We provethat j is an isomorphism of C onto Cr f(a) (it will then follow that Cr f(a) is affine).

We do this by localization: the ideal generated by f∗x = t2 in k[C] defines f(0) andso k[C r f(0)] = k[C][1/t2] = k[t2, t3, t−2] = k[t, t−1]. On the other hand,

k[C r j−1f(0)] = k[t2, t3, t2/(t− a)][t−2]) =

= k[t, t−1, (t− a)−1] = k[C r f(0)][1/(t− a)] = k[C r f(0), f(a)].

From this it is clear that C r j−1f(0) maps isomorphically onto C r f(0), f(a). Itremains to prove that there is neighborhood U of (0, 0) ∈ A2 r f(a) such that j mapsj−1(C ∩ U) isomorphically onto C ∩ U . We take for U the principal open subset A2

x−a2 .Then k[C ∩ U ] = k[t2, t3][1/(t2 − a2)] and k[j−1(C ∩ U)] = k[t2, t3, t2/(t− a)][1/(t2 − a2)].But these k-algebras are the same as t2/(t− a) ∈ k[t2, t3][1/(t2 − a2)].

Other such examples (among them nonsingular plane cubic curves) that are also validin positive characteristic are best understood after we have discussed the Picard group.

6. The product

Let m and n be nonnegative integers. If f ∈ k[x1, . . . , xm] and g ∈ k[y1, . . . , yn],then we have a regular function f ∗ g on Am+n defined by

f ∗ g(x1, . . . , xm, y1, . . . , yn) := f(x1, . . . , xm)g(y1, . . . , yn).

It is clear that Am+nf∗g = Amf × Ang , which shows that the Zariski topology on Am+n

refines the product topology on Am×An. Equivalently, if X ⊂ Am and Y ⊂ An areclosed, then X × Y is closed in Am+n. We give X × Y the topology it inherits fromAm+n (which is finer than the product topology when m > 0 and n > 0). For thecoordinate rings we have defined a map:

k[X]× k[Y ]→ k[X × Y ], (f, g) 7→ f ∗ g

which is evidently k-bilinear (i.e., k-linear in either variable). We want to provethat the ideal I(X × Y ) defining X × Y in Am+n is generated by I(X) and I(Y )(viewed as subsets of k[x1, . . . , xm, y1, . . . yn]) and that X × Y is irreducible whenX and Y are. This requires that we translate the formation of the product into al-gebra. This centers around the notion of the tensor product, the definition of whichwe recall. (Although we here only need tensor products over k, we shall define thisnotion for modules over a ring, as this is its natural habitat. This is the setting thatis needed later anyhow.)

If R is a ring and M and N are R-modules, then we can form their tensor product overR, M ⊗R N : as an abelian group M ⊗R N is generated by the expressions a ⊗R b, a ∈ M ,b ∈ N and subject to the conditions (ra)⊗R b = a⊗R (rb), (a+ a′)⊗R b = a⊗R b+ a′ ⊗R band a ⊗R (b + b′) = a ⊗R b + a ⊗R b′. So a general element of M ⊗R N can be writtenlike this:

∑Ni=1 ai ⊗R bi, with ai ∈ M and bi ∈ N . We make M ⊗R N an R-module if we

stipulate that r(a⊗R b) := (ra)⊗R b (which is then also equal to a⊗R (rb)). Notice that themap

⊗R : M ×N →M ⊗R N, (a, b) 7→ a⊗R b,

is R-bilinear (if we fix one of the variables, then it becomes an R-linear map in the othervariable).

In case R = k we shall often omit the suffix k in ⊗k.


EXERCISE 23. Prove that ⊗R is universal for this property in the sense that every R-bilinear mapM×N → P ofR-modules is the composite of⊗R and a uniqueR-homomorphismM ⊗R N → P . In other words, the map

HomR(M ⊗R N,P )→ BilR(M,N ;P )), f 7→ f ⊗R

is an isomorphism of R-modules.

EXERCISE 24. Let m and n be nonnegative integers. Prove that Z/(n)⊗Z Z/(m) can beidentified with Z/(m,n).

If A is an R-algebra and N is an R-module, then A ⊗R N acquires the structure of anA-module which is characterized by

a.(a′ ⊗R b) := (aa′)⊗R b.

For instance, if N is an R-vector space, then C⊗RN is a complex vector space, the complexi-fication of N . If A and B are R-algebras, then A⊗RB acquires the structure of an R-algebracharacterized by

(a⊗R b).(a′ ⊗R b′) := (aa′)⊗R (bb′).

Notice that A→ A⊗R B, a 7→ a⊗R 1 and B → A⊗R B, b 7→ 1⊗R b are R-algebra homo-morphisms. For example, A⊗RR[x] = A[x] as A-algebras (and hence A⊗RR[x1, . . . , xn] =A[x1, . . . , xn] with induction).

EXERCISE 25. Prove that C⊗R C is as a C-algebra isomorphic to C⊕C with componen-twise multiplication.

PROPOSITION 6.1. For closed subsets X ⊂ Am and Y ⊂ An the bilinear mapk[X]× k[Y ]→ k[X × Y ], (f, g) 7→ f ∗ g induces an isomorphism µ : k[X]⊗ k[Y ]→k[X × Y ] of k-algebras (so that in particular k[X]⊗ k[Y ] is reduced).

If X and Y are irreducible, then so is X×Y , or equivalently, if k[X] and k[Y ] aredomains, then so is k[X]⊗ k[Y ].

PROOF. Since the obvious map

k[x1, . . . , xm]⊗ k[y1, . . . , yn]→ k[x1, . . . , xm, y1, . . . , yn]

is an isomorphism, it follows that µ is onto. In order to prove that µ is injective, letus first observe that every φ ∈ k[X] ⊗ k[Y ] can be written φ =

∑Ni=1 fi ⊗ gi such

that g1, . . . , gN are k-linearly independent. Given p ∈ X, then the restriction ofµ(φ) =

∑Ni=1 fi∗gi to p×Y ∼= Y is the regular function φp :=

∑Ni=1 fi(p)gi ∈ k[Y ].

Since the gi’s are linearly independent, we have φp = 0 if and only if fi(p) = 0 forall i. In particular, the subset X(φ) ⊂ X of p ∈ X for which φp = 0, is equal to∩Ni=1Z(fi) and hence closed.

If µ(φ) = 0, then φp = 0 for all p ∈ X and hence fi = 0 for all i. So φ = 0. Thisproves that µ is injective.

Suppose now X and Y irreducible. We prove that k[X] ⊗ k[Y ] is a domain sothat X × Y is irreducible. Let φ, ψ ∈ k[X] ⊗ k[Y ] be such that φψ = 0. Since therestriction of φψ = 0 to p × Y ∼= Y is φpψp and k[Y ] is a domain, it follows thatφp = 0 or ψp = 0. So X = X(φ) ∪X(ψ). Since X is irreducible we have X = X(φ)or X(ψ). This means that φ = 0 or ψ = 0.

EXERCISE 26. Let A and B be finitely generated k-algebras. Prove that A⊗k Bis a finitely generated algebra and define a natural map Spm(A⊗kB)→ Spm(A)×Spm(B) and show that this is a bijection (hint: do not use Proposition 6.1).

7. FUNCTION FIELDS AND RATIONAL MAPS 29

EXERCISE 27. Let X and Y be closed subsets of affine spaces. Prove that eachirreducible component of X × Y is the product of an irreducible component of Xand one of Y .

It is clear that the projections πX : X×Y → X and πY : X×Y → Y are regular.We have observed that the space underlying X × Y is usually not the topologicalproduct of its factors. Still it is the ‘right’ product in the sense of category theory: ithas the following universal property, which almost seems too obvious to mention: ifZ is a closed subset of some affine space, then any pair of regular maps f : Z → X,g : Z → Y defines a regular map Z → X × Y characterized by the property that itscomposite with πX resp. πY yields f resp. g (this is of course (f, g)).

7. Function fields and rational maps

In this section we interpret the fraction ring of an algebra of regular functions.LetX be an affine variety. Recall that an element φ ∈ Frac(k[X]) is by definition

represented by a fraction f/g with f, g ∈ k[X] and g not a zero divisor in k[X]. Weneed the following lemma.

LEMMA 7.1. LetX be an affine variety and let C1, . . . , Cr be its distinct irreduciblecomponents. Then for any g ∈ k[X] the following are equivalent: (i) g is a zero divisorof k[X], (ii) Ci ⊂ Z(gi) for some i and (iii) Xg is not dense in X. Moreover, any open-dense subset of X contains a principal open-dense subset defined by a nonzero divisor.

The restriction maps (well-defined in view of the above) defines an isomorphismof k-algebras

R = (Ri)ri=1 : Frac(k[X])→ ⊕ri=1 Frac(k[Ci]), f/g 7→ (f/g

∣∣Ci

)ri=1

(note that the right hand side is a direct sum of fields).

PROOF. Choose for i = 1, . . . , r a nonzero hi ∈ I(∪j 6=iCj).(i)⇒ (ii) Let g ∈ k[X] be a zero divisor. Then there exists a nonzero g′ ∈ k[X]

with gg′ = 0. Let Ci be such that g′|Ci is nonzero. Then we must have g|Ci = 0.(ii) ⇒ (iii) If Ci ⊂ Z(g), then clearly, Xg (and hence its closure) is contained

in the proper closed subset ∪j 6=iCj 6= X.(iii) ⇒ (i) If Xg is not dense in X, then its closure is a proper closed subset

of X and so there exists a nonzero g′ ∈ k[X] with Xg ⊂ Z(g′). This implies thatgg′ = 0 and so g is a zero divisor.

To prove the second assertion, let U ⊂ X be open-dense. Then for every i,(Ci r ∪j 6=iCj) ∩ U is nonempty and so contains a nonempty principal open subsetXfi . Then fi|Cj is nonzero if and only j = i. This is also so for hi|Cj: it is nonzeroif and only if i = j. We put g :=

∏i(fi +

∑j 6=i hj). Its restriction to Ci is fihr−1

i |Ciand so Ci ∩Xg = Ci ∩Xfi ∩Xhi . The latter is nonempty and open (hence dense)in Ci and contained in Ci ∩ U . So Xg is an open-dense in X and contained in U .

As to the last assertion, if f/g ∈ Frac(k[X]) is such that Ri(f/g) = 0 for all i,then f |Ci = 0 for all i and hence f = 0. So R is injective. To see that R is onto, letfor i = 1, . . . , r, fi/gi ∈ Frac(k[Ci]). Let fi, gi ∈ k[X] map to fi, gi ∈ k[Ci] and putf =

∑i hifi and g :=

∑i higi. Then f |Ci = hi|Ci.fi and g|Ci = hi|Ci.gi and the

latter is nonzero. So g is not a zero divisor of k[X] so that f/g ∈ Frac(k[X]) andwe have Ri(f/g) = fi/gi.


COROLLARY 7.2. If φ is a regular function defined on an open-dense subset U ofX, then φ determines an element of Frac(k[X]). All elements of Frac(k[X]) are thusobtained and two pairs (U, φ) and (U ′, φ′) determine the same element of Frac(k[X])if and only if φ|U ∩ U ′ = φ′|U ∩ U ′.

PROOF. Let φ ∈ Frac(k[X]) be represented as a fraction f/g. Since g is anonzero divisor, Lemma 7.1 tells us that f/g defines a regular function on an prin-cipal open-dense subset of X (namely Xg). If φ is also represented by f ′/g′, thenfg′ − f ′g = 0 and so the two associated regular functions on Xg and Xg′ have thesame restriction to the principal open-dense subset Xgg′ = Xg ∩Xg′ of X.

Conversely, suppose φ is a regular function on an open-dense subset U of X.According to Lemma 7.1, U ⊃ Xg for some nonzero divisor g and hence φ|Xg is byProposition 5.2 given by f/gr ∈ Frac(k[X]) for some f ∈ k[X].

There is in general no best way to represent a given element of Frac(k[X]) as afraction (as there is in a UFD), and so we must be content with the corollary above.Note that it is essentially equivalent to the assertion that

Frac(k[X]) = lim−→g nonzero divisor in k[X]

k[X][1/g] = lim−→U open-dense in X

OX(U).

An element of Frac(k[X]) is often called a rational function on X. This is thealgebro-geometric analogue of a meromorphic function in complex function the-ory. When X is irreducible, then Frac(k[X]) is a field, called the function field of X,and will be denoted k(X).

We shall now give a geometric interpretation of finitely generated field exten-sions of k and the k-linear field homomorphisms between them.

DEFINITION 7.3. Let X and Y be affine varieties. A rational map from X to Yis given by a pair (U,F ), where U is an open-dense subset of X and F : U → Y isa morphism, with the understanding that a pair (U ′, F ′) defines the same rationalmap if F and F ′ coincide on an open-dense subset of U ∩U ′ (then they coincide onall of U ∩ U ′ by continuity, but formulated in this way we see that we are dealingwith an equivalence relation). We denote such a rational map as f : X 99K Y .

We say that the rational map is dominant if for a representative pair (U,F ),F (U) is dense in Y . (This is then also so for any other representative pair. Why?)

So a rational map f : X 99K A1 is same thing as a rational function on X.Observe that for a morphism of affine varieties f : X → Y , g ∈ ker(f∗) is

equivalent to f(X) ⊂ Zg and hence equivalent to the closure f(X) being containedin Zg. It follows that f is dominant if and only if f∗ is injective.

PROPOSITION 7.4. Any finitely generated field extension of k is k-isomorphic tothe function field of an irreducible affine variety. If X and Y are irreducible affinevarieties, then a dominant rational map f : X 99K Y determines a k-linear fieldembedding f∗ : k(Y ) → k(X) and conversely, every k-linear field embedding k(Y )→k(X) is of this form for a unique dominant rational map f .

PROOF. Let K/k be a finitely generated field extension of k. This means thereexist elements a1, . . . , an ∈ K such that every element of K can be written as afraction of polynomials in a1, . . . , an with coefficients in k. So the k-subalgebra ofKgenerated by a1, . . . , an is a domain A ⊂ K (sinceK is a field) that hasK as its fieldof fractions. Since A is the coordinate ring of a closed irreducible subset X ⊂ An


(defined by the kernel of the obvious ring homomorphism k[x1, . . . , xn] → A), itfollows that K can be identified with k(X).

Suppose we are given a principal open-dense subset U ⊂ X and a morphismF : U → Y with F (U) dense in Y . Now F ∗ : k[Y ] → k[U ] will be injective, for ifg ∈ k[Y ] is in the kernel: F ∗(g) = 0, then F (U) ⊂ Z(g). Since F (U) is dense in Y ,this implies that Z(g) = Y , in other words, that g = 0. Hence the composite mapk[Y ] → k[U ] → k(U) = k(X) is an injective homomorphism from a domain to afield and therefore extends to a field embedding k(Y ) → k(X).

It remains to show that every k-linear field homomorphism Φ : k(Y )→ k(X) isso obtained. For this, choose generators b1, . . . , bm of k[Y ]. Then Φ(b1), . . . ,Φ(bm)are rational functions on X and so are regular on a principal nonempty subsetXh ⊂ X. Since b1, . . . , bm generate k[Y ] as a k-algebra, it follows that Φ maps k[Y ]to k[Xh] = k[X][1/h] ⊂ k(X). This k-algebra homomorphism defines a morphismF : Xh → Y such that F ∗ = Φ|k[Y ]. The image of F will be dense by the argumentabove: if g ∈ k[Y ] vanishes on F (Xh), then Φ(g) = F ∗(g) = 0, which implies g = 0,since Φ is injective. It is clear that Φ is the extension of F ∗ to the function fields.

As to the uniqueness: if (U ′, F ′) is another solution, then choose a nonemptyprincipal subset U ′′ ⊂ U ∩ U ′ such that F and F ′ both define morphisms U ′′ → Y .These must be equal since the associated k-algebra homomorphisms k[Y ]→ k[U ′′]are the same (namely the restriction of Φ).

The following exercise explains the focus on irreducible varieties when consid-ering rational maps.

EXERCISE 28. Let X and Y be an affine varieties with distinct irreducible com-ponents X1, . . . , Xr resp. Y1, . . . , Ys. Prove that to give a rational map f : X 99K Yis equivalent to giving a rational map fi : Xi 99K Yji for i = 1, . . . , r. Show that f isdominant if and only if for each j ∈ 1, . . . , s, there exists an ij ∈ 1, . . . , r suchthat f maps Xij to Yj as a dominant map.

EXERCISE 29. Let f ∈ k[x1, . . . , xn+1] be irreducible of positive degree. Its zeroset X ⊂ An+1 is then closed and irreducible. Assume that the degree d of f in xn+1

is positive.(a) Prove that the projection π : X → An induces an injective k-algebra

homomorphism π∗ : k[x1, . . . , xn]→ k[X] = k[x1, . . . , xn]/(f).(b) Prove that π is dominant and that the resulting field homomorphism

k(x1, . . . , xn)→ k(X) is a finite extension of degree d.

COROLLARY 7.5. Two dominant maps f : X 99K Y and g : Y 99K Z betweenirreducible affine varieties can be composed to yield a dominant map gf : X 99K Zso that we have a category with the irreducible affine varieties as objects and therational dominant maps as morphisms. Assigning to an irreducible affine variety itsfunction field makes this category anti-equivalent to the category of finitely generatedfield extensions of the base field k.

PROOF. The dominant maps yield k-linear field extensions f∗ : k(Y ) → k(X)and g∗ : k(Z) → k(Y ) and these can be composed to give a k-linear field extensionf∗g∗ : k(Z) → k(X). Proposition 7.4 says that this is induced by a unique rationalmap X 99K Z. This we define to be gf . The rest of the corollary now follows.

PROPOSITION-DEFINITION 7.6. A rational map f : X 99K Y is an isomorphismin the above category (that is, induces a k-linear isomorphism of function fields) if and


only if there exists a representative pair (U,F ) of f such that F maps U isomorphicallyonto an open subset of Y . If these two equivalent conditions are satisfied, then f iscalled a birational map. If a birational map X 99K Y merely exists (in other words,if there exists a k-linear field isomorphism between k(X) and k(Y )), then we say thatX and Y are birationally equivalent.

PROOF. If f identifies a nonempty open subset of X with one of Y , then f∗ :k(Y )→ k(X) is clearly a k-algebra isomorphism.

Suppose now we have a k-linear isomorphism k(Y ) ∼= k(X). Represent thisisomorphism and its inverse by (U,F ) and (V,G) respectively. Since F−1V is anonempty open subset of U , it contains a nonempty principal open subset Xg.Since GF induces the identity on k(X), its restriction to Xg must be the inclusionXg ⊂ X. Since Xg is dense in F−1V , and GF is continuous, the same is then trueof GF : it is the inclusion F−1V ⊂ X. This implies that F maps F−1V injectivelyto G−1U . For the same reason, G maps G−1U injectively to F−1V . So F defines anisomorphism between the open subsets F−1V ⊂ X and G−1U ⊂ Y .

EXERCISE 30. Assume k not of characteristic 2. Let X ⊂ A2 be defined byx2

1 + x22 = 1. Prove that X is birationally equivalent to the affine line A1. (Hint:

take a look at Exercise 17.) More generally, prove that the quadric in An+1 definedby x2

1 + x22 + · · ·x2

n+1 = 1 is birationally equivalent to An. What about the zero setin An+1 of a quadric function?

In case k(X)/k(Y ) is a finite extension, one may wonder what the geometricmeaning is of the degree d of that extension, perhaps hoping that this is just thenumber of elements of a general fiber of the associated rational map X 99K Y . Wewill see that this is often true (namely when the characteristic of k is zero, or moregenerally, when this characteristic does not divide d), but not always, witness thefollowing example.

EXAMPLE 7.7. Suppose k has characteristic p > 0. We take X = A1 = Y and letf = Φp be the Frobenius map: f : a ∈ A1 7→ ap ∈ A1. Then f is homeomorphism,but f∗ : k[Y ] = k[y] → k[x] = k[X] is given by y 7→ xp and so induces the fieldextension k(y) = k(xp) ⊂ k(x), which is of degree p. From the perspective of Y , wehave enlarged its algebra of regular functions by introducing a formal pth root of itscoordinate y (which yields another copy of A1, namely X). From the perspectiveof X, k[Y ] is just the subalgebra f∗k[X].

This is in fact the basic example of a purely inseparable field extension, i.e.,an algebraic field extension L/K with the property that every element of L has aminimal polynomial in K[T ] that has precisely one root in L Such a polynomialmust be of the form T p

r − c, with c ∈ K not a p-th power of an element of K whenr > 0, where p is the characteristic of K (for p = 0 we necessarily have L = K).So if L 6= K, then the absolute Frobenius map a ∈ K 7→ ap ∈ K is not surjective.Purely inseparable extensions are not detected by Galois theory, for they have trivialGalois group as there is only one root to move around.

EXERCISE 31. Let f : X 99K Y be a dominant rational map of irreducible affinevarieties which induces a purely inseparable field extension k(Y )/k(X). Provethat there is an open-dense subset V ⊂ Y such that f defines a homeomorphismf−1V → V . (Hint: show first that it suffices to treat the case when k(X) is obtainedfrom k(Y ) by adjoining the pth root of an element f ∈ k(Y ). Then observe that


if f is regular on the affine open-dense V ⊂ Y , then Y contains as an open densesubset a copy of the locus of (x, t) ∈ V × A1 satisfying tp = f .)

The following Corollary 7.8 suggests that if X is an irreducible affine veriety,then the transcendence degree of k(X)/k may be understood as the dimension ofX. We shall come back to this and then improve upon the corollary below.

COROLLARY 7.8. Let X be an irreducible affine variety X and denote by r thetranscendence degree of k(X)/k. Then there exists an irreducible hypersurface Y inAr+1 and a rational dominant map f : X 99K Y which is purely inseparable in thesense that k(X) is a purely inseparable extension of k(Y ). In particular (in view ofExercise 31 above), there an open-dense subset V ⊂ Y such that f defines a homeo-morphism f−1V → V .

Before we give the proof, we recall some basic facts from field theory. For anyalgebraic extension L/K, the elements of L that are separable over K make upan intermediate extension Lsep/K that is (of course) separable and is such thatL/Lsep is purely inseparable. When L is an algebraic closure of K, then Lsep iscalled a separable algebraic closure of K: it is a separable algebraic extension ofK which is maximal for that property. Then L is as an extension of Lsep obtainedby successive adjunction of p-power roots of elements of Lsep.7 In case L/K is afinitely generated extension, then by the theorem of the primitive element, Lsep/Khas a single generator.

PROOF OF COROLLARY 7.8. Let k[X] be generated by b1, . . . , bm, say, then wemay after renumbering assume that for some r ≤ m, b1, . . . , br are algebraicallyindependent (so that the k-linear field homomorphism k(Ar) = k(x1, . . . , xr) →k(X) which sends xi to bi is injective) and that for i > r, bi is algebraic overk(b1, . . . , bi−1). Then k(b1, . . . , br) ∼= k(Ar) is a purely transcendental extension ofk and k(X) is a finite extension of k(Ar) so that r is the transcendence degree ofk(X)/k. It is clear that the inclusion k(Ar) ⊂ k(X) defines a dominant rationalmap X 99K Ar. The theorem of the primitive element implies that the separableclosure of k(Ar) in k(X) has a single generator b. This b is a root of an irreduciblepolynomial F ∈ k(Ar)[T ]. If we clear denominators, we may assume that thecoefficients of F lie in k[Ar] so that F ∈ k[Ar+1] and then we can take for Y thehypersurface in Ar+1 defined by F . We now have obtained a dominant rationalmap f : X 99K Y such that k(X)/k(Y ) is purely inseparable extension.

Much of the algebraic geometry in the 19th century and early 20th centurywas of a birational nature: birationally equivalent varieties were regarded as notreally different. This sounds rather drastic, but it turns out that many interestingproperties of varieties are an invariant of their birational equivalence class.

Here is an observation which not only illustrates how affine varieties overalgebraically nonclosed fields can arise when dealing with affine k-varieties, butone that also suggests that we ought to enlarge the maximal ideal spectrum. Letf : X → Y be a dominant morphism of irreducible affine varieties. This implies

7In case L/K is a finite normal extension (i.e., a finite extension with the property that everyf ∈ K[x] that is the minimal polynomial of some element of L has all its roots in L), then the fixedpoint set of the Galois group of L/K is a subextension Linsep/L that is purely inseparable, whereasL/Linsep separable. Then the natural map Lsep ⊗K Linsep → L is an isomorphism of K-algebras.


that f∗ : k[Y ] → k[X] is injective and that f(X) contains an open-dense subset ofY . Then we may ask whether there exists something like a general fiber: is therean open-dense subset V ⊂ Y such that the fibers f−1(y), y ∈ V all “look the same”?The question is too vague for a clear answer and for most naive ways of making thisprecise, the answer will be no. For instance, we could simply refuse to specify onesuch V by allowing it to be arbitrarily small, but if we then want to implement thisidea by taking the (projective) limit lim←−V open-dense in Y

f−1V , then we end up withthe empty set unless Y is a singleton. However, its algebraic counterpart, whichamounts to making all the nonzero elements of k[Y ] in k[X] invertible,

lim−→V open-dense in Y

k[f−1V ] = (k[Y ] r 0)−1k[X] = k(Y )⊗k[Y ] k[X]

(these equalities follow from the fact that k[X] = k[Y ] ⊗k[Y ] k[X]), is nontrivial.It is in fact a reduced finitely generated k(Y )-algebra and this hints that an ade-quate geometric description requires that we include more points. First of all, wewould like to regard the maximal ideal spectrum of k(Y ) ⊗k[Y ] k[X] as an affinevariety over the (algebraically nonclosed) field k(Y ) so that every regular functionon X which comes from Y is now treated as a scalar (and will be invertible whennonzero).8 And secondly, in order to give this a geometric content, we would likethat every irreducible variety Z defines a point ηZ (its generic point) with ‘residuefield’ k(Z), which for a singleton must give us back its unique element with thefield k. For we then can extend f to the points defined by closed irreducible subsetsZ ⊂ X by putting f(ηZ) := η

f(Z). Then as a set, the generic fiber of f is the fiber of

this extension over ηY , i.e., the set of ηZ for which f |Z : Z → Y is dominant. Suchconsiderations directly lead to the notion of a scheme that we shall discuss later.

8. Finite morphisms

In this section A is a ring and B is a A-algebra. In other words, we are givena ring homomorphism A → B (that is sometimes denoted by B/A). We adoptthe following standard terminology: we say that B is finite over A if B is a finitelygenerated A-module and we say that B is an extension of A if A → B is injective(so that we may regard A as subring of B). So B/A is called a finite extension ifboth A→ B is injective and B is a finitely generated A-module.

PROPOSITION-DEFINITION 8.1. We say that b ∈ B is integrally dependent on Aif one the following equivalent properties is satisfied.

(i) b is a root of a monic polynomial xn + a1xn−1 + · · ·+ an ∈ A[x],

(ii) A[b] is finitely generated as an A-module,(iii) b is contained in a A-subalgebra C ⊂ B which is finitely generated as an

A-module.

PROOF. (i)⇒ (ii). If b is a root of xn + a1xn−1 + · · ·+ an ∈ A[x], then clearly

A[b] is generated as a A-module by 1, b, b2, . . . , bn−1.(ii)⇒ (iii) is obvious.

8Our notion of affine variety required that we work over an algebraically closed field. This is ofcourse arranged by choosing an algebraic closure L of k(Y ). The maximal ideal spectrum of L ⊗k[Y ]

k[X] is then an affine L-variety, and yields a notion of a general fiber that is even closer to our geometricintuition.

8. FINITE MORPHISMS 35

(iii)⇒ (i). Suppose that C is as in (iii). Choose an epimorphism π : An → C ofA-modules and denote the standard basis of An by (e1, . . . , en). We may (and will)assume that π(e1) = 1B . Since b ∈ C, this extends to an epimorphism A[b]n → C,that we will also denote by π. By assumption, bπ(ei) =

∑nj=1 aijπ(ej) for certain

aij ∈ A. We regard the n × n-matrix σ := (bδij − aij)i,j with entries in A[b] as anA[b]-endomorphism of A[b]n. Then πσ(ei) = 0 for all i, in other words, πσ = 0.

Now Cramer’s rule can be understood as stating that if σ′ is the matrix of co-factors of σ, then σσ′ = det(σ)1n, where we note that det(σ) is a monic polynomialin b with coefficients in A. We thus find that in B,

det(σ) = det(σ)π(e1) = π(det(σ)e1) = π(σσ′(e1)) = (πσ)(σ′(e1)) = 0.

COROLLARY-DEFINITION 8.2. The set elements of B that are integrally dependenton A is an A-subalgebra of B. We call this subalgebra the integral closure of A in B(and denoted it by AB).

PROOF. It is enough to prove that if b, b′ ∈ B are integrally dependent over A,then so is every element of A[b, b′]. Or what amounts to the same: if A[b] and A[b′]are finitely generated A-modules, then so is A[b, b′]. This is clear: if bkn−1

k=0 gener-ates A[b] and b′kn

′−1k=0 generates A[b′], then bkb′k′n,n

′

k=0,k′=0 generates A[b, b′].

DEFINITION 8.3. We say that B is integral over A if every element of B is inte-gral over A (so that B = AB). If in addition the given homomorphism A → B isinjective (so that A may be regarded as a subring of B), then we say that B is anintegral extension of A.

The characterization (iii) of Proposition 8.1 shows that when B is finite over A(resp. a finite extension of A), then B is integral over A (resp. an integral extensionof A). An important class of example appears in algebraic number theory: if L is afinite field extension of Q (also called an algebraic number field), then the integralclosure of Z ⊂ Q defines a subring of L, called the ring of integers of L. This ring isoften denoted by OL.

EXERCISE 32. Prove that ‘being integral over’ is transitive: if B is an A-algebraintegral over A, then any B-algebra that is integral over B is as an A-algebra inte-gral over A.

PROPOSITION 8.4. LetA ⊂ B be an integral extension and supposeB is a domain.Then Frac(B) is an algebraic field extension of Frac(A), which is finite whenever B isa finite extension of A.

PROOF. We first show that Frac(B) = Frac(A)B. Let b ∈ B r 0. It is a rootof a monic polynomial xn + a1x

n−1 + · · · + an = 0 (ai ∈ A) with an 6= 0 and so1/b = −1/an.(b

n−1 + a1bn−2 + · · ·+ an−1) ∈ Frac(A)B.

SinceB is a union of finitely generatedA-modules, Frac(A)B is a union of finitedimensional Frac(A)-vector spaces and hence an algebraic extension of Frac(A).The last assertion follows from the observation that any set of A-module generatorsof B is also a set of Frac(A)-vector space generators of Frac(A)B.

There are two simple ways of producing new integral extensions out of a givenone, namely reduction and localization: Suppose A → B is integral. Then forevery ideal J ⊂ B, J ∩ A is (clearly) an ideal of A and A/J ∩ A ⊂ B/J is anintegral extension. And if S ⊂ A is a multiplicative subset, then the induced ring


homomorphism S−1A → S−1B is integral. Both appear in the proof of the ‘Goingup theorem’ below. For this we will also need:

LEMMA 8.5 (Nakayama’s Lemma). Let R be a local ring with maximal ideal mand M a finitely generated R-module. Then a finite subset S ⊂ M generates Mas a R-module if (and only if) the image of S in M/mM generates the latter as aR/m-vector space. In particular (take S = ∅), mM = M implies M = 0.

PROOF. The special case S = ∅ is in fact the general case, for we reduce to itby passing to M/RS, for our assumptions then say that then M = mM and wemust show that M = 0. Let π : Rn → M be an epimorphism of R-modules anddenote the standard basis of Rn by (e1, . . . , en). By assumption there exist rij ∈ msuch that π(ei) =

∑sj=1 rijπ(ej). So if σ := (δij − rij)i,j ∈ EndR(Rn), then πσ = 0.

Notice that det(σ) ∈ 1+m. Since 1+m consists of invertible elements, Cramer’s ruleshows that σ is invertible. So π = π(σσ−1) = (πσ)σ−1 = 0 and hence M = 0.

PROPOSITION 8.6 (Going up). Let A ⊂ B be an integral extension and let p ⊂ Abe a prime ideal of A. Then the going up property holds: p is of the form q∩A, whereq is a prime ideal of B. If also is given is a prime ideal q′ of B with the property thatp ⊃ q′ ∩ A, then we can take q ⊃ q′. Moreover the incomparability property holds:two distinct prime ideals of B having the same intersection with A cannot obey aninclusion relation.

PROOF. The localization A → Ap yields a local ring with maximal ideal pAp

and the prime ideals of Ap correspond (by taking the preimage in A) to the primeideals of A that contain p. The localization ApB (as a A-module) is by the obser-vation above, an integral extension of Ap. If we find a prime ideal q of ApB withq∩Ap = pAp, then the preimage q of q in B is a prime ideal of B with the propertythat q ∩ A is the preimage of pAp in A and so this is just p. Hence for the firstassertion there is no loss in generality in assuming that A is a local ring and p is itsunique maximal ideal mA.

We claim that mAB 6= B. Suppose this is not so. Then write 1 ∈ B as an mA-linear combination of a finite set elements of B. Denote by B′ the A-subalgebraof B generated by this finite set. Since B is an integral extension of A, B′ is finiteover A. Since 1 ∈ mAB

′, we have B′ = mAB′, and it then follows from Nakayama’s

Lemma 8.5 that B′ = 0. It follows that 1 = 0 so that A is the zero ring. Thiscontradicts our assumption that A has a maximal ideal.

Since mAB 6= B, we can take for q any maximal ideal of B which contains theideal mAB: then qB ∩A is a maximal ideal of A, hence equals mA.

For the refinement and the incomparability property we can, simply by passingto A/(q′∩A) ⊂ B/q′ (which is still an integral extension by the observation above),assume that q′ = 0. This reduces the refinement to the case already treated and forthe incomparability property we apply this reduction to the case q′ = q: then thisamounts showing that for any nonzero prime ideal q′′ of B, q′′ ∩ A is nonzero. Tosee this, let b ∈ q′′r0. Then b is a root of a monic polynomial: bn+a1b

n−1 + · · ·+an−1b + an = 0 (ai ∈ A), where we can of course assume that an 6= 0 (otherwisedivide by b). We then find that 0 6= an ∈ Bb ∩A ⊂ q′′ ∩A.

REMARK 8.7. Let us agree to call a prime chain (of length n) in a ring R astrictly ascending sequence p0 ( p1 ( · · · ( pn of prime ideals in R. The goingup property may then be restated as saying that for an integral extension A ⊂ B,


any prime chain in A is the intersection with A of a prime chain in B, where weeven may prescribe the first member of the latter in advance. The incomparabilityproperty says that the intersection a prime chain in B with A is a prime chain in A.

DEFINITION 8.8. We say that a morphism of affine varieties f : X → Y is finiteif the k-algebra homomorphism f∗ : k[Y ]→ k[X] is finite, i.e., makes k[X] a finitelygenerated k[Y ]-module (so k[X] is then integral over k[Y ]).

So f∗ : k[Y ]→ k[X] is a finite extension if and only if f is finite and dominant.

EXERCISE 33. Prove that if Y is an affine variety, then the disjoint union of itsirreducible components is finite over Y .

Propositions 8.4 and 8.6 give in the algebro-geometric setting:

COROLLARY 8.9. Let f : X → Y be a finite, dominant morphism of affine vari-eties. Then f is surjective and every closed irreducible subset P ⊂ Y is the image ofa closed irreducible subset Q ⊂ X. If furthermore is given a closed irreducible subsetQ′ ⊂ X with P ⊂ f(Q′), then we may choose Q ⊂ Q′. If in addition X is irreducible,then so is Y and f∗ : k(Y )→ k(X) is a finite algebraic extension of fields.

PROOF. To see that f is surjective, let p ∈ Y . Then there exists by Proposition8.6 a prime ideal q ⊂ k[X] such that mp = (f∗)−1q. So if q ∈ Z(q), then mp =(f∗)−1q ⊂ (f∗)−1mq and hence f(q) = p. Similarly, if P ⊂ Y is irreducible, thenp := I(P ) is a prime ideal and so (f∗)−1q = p for some prime ideal q. ThenQ := Z(q) is irreducible. Note that f∗ induces the morphism k[P ] = k[Y ]/p →k[X]/q = k[Q]. This is a finite extension and so by what we just proved, f inducesa surjective morphism Q→ P . In other words, f(Q) = P . The other two assertionsare easy consequences of Proposition 8.6.

EXERCISE 34. Let f : X → Y be a finite morphism of affine varieties. Provethat f is closed and that every fiber f−1(y) is finite (possibly empty). Assumingthat in addition that f is surjective, prove that if Y ′ ⊂ Y is closed or a principalopen subset of Y , then the restriction f : f−1Y ′ → Y ′ is also a finite morphism.

THEOREM 8.10 (Noether normalization). Let K be a field and m an integer≥ 0. Every K-algebra A with m generators is a finite (hence integral) extension of apolynomial algebra over K with ≤ m generators: there exist an integer 0 ≤ r ≤ mand an injection K[x1, . . . , xr] → A such that A is finite over K[x1, . . . , xr].

The proof will be with induction on m. The following lemma provides theinduction step.

LEMMA 8.11. Let φ : K[x1, . . . , xm] → A be an epimorphism of K-algebraswhich is not an isomorphism. Then there exists a surjective K-algebra homomorphismφ′ : K[x1, . . . , xm] → A such that φ′(xm) = φ(xm) and φ′(xm) is integral overφ′(K[x1, . . . , xm−1]).

PROOF. We define for any positive integer s a K-algebra automorphism σs ofK[x1, . . . , xm] by σs(xi) := xi + xs

m−i

m when i ≤ m − 1 and σs(xm) = xm. This isindeed an automorphism with inverse given by σ−1

s (xi) = xi − xsm−i

m for i < m− 1and σs(xm) = xm. So if I = (i1, . . . , im) ∈ Zm≥0, then

σs(xi11 · · ·ximm ) = (x1 + xs

m−1

m )i1 · · · (xm−1 + xsm)im−1ximm .


When viewed as an element of K[x1, . . . , xm−1][xm], this is a monic polynomial inxm of degree pI(s) := i1s

m−1 + i2sm−2 + · · ·+ im. Now give Zm≥0 the lexicographic

ordering. Then I > J implies pI(s) > pJ(s) for s large enough. Choose a nonzerof ∈ ker(φ). If I ∈ Zm≥0 is the largest multi-exponent of a monomial that appearsin f with nonzero coefficient, then for s large enough, σs(f) is a constant times amonic polynomial in xm of degree pI(s) with coefficients in K[x1, . . . , xm−1]. Inother words, the image of xm in K[x1, . . . , xm]/(σs(f)) is integral over the imageof K[x1, . . . , xm−1] in K[x1, . . . , xm]/(σs(f)). This is then certainly true for theirimages in K[x1, . . . , xm]/σs(ker(φ)) = K[x1, . . . , xm]/ ker(φσ−1

s ). This is equivalentto the image of σ−1

s (xm) = xm in K[x1, . . . , xm]/ ker(φ) ∼= A being integral overφσ−1

s (K[x1, . . . , xm−1]). Hence φ′ := φσ−1s is as desired.

PROOF OF NOETHER NORMALIZATION. Let φ : K[x1, . . . , xm] → A be an epi-morphism of K-algebras. When φ is an isomorphism, there is nothing to show.Otherwise, Lemma 8.11 tells us that there exists a K-algebra homomorphism φ′ :K[x1, . . . , xm] → A such that φ′(xm) is integral over A′ := φ′(K[x1, . . . , xm−1]).Since the K-algebra A′ has ≤ m− 1 generators, it is by induction a finite extensionof some polynomial algebra K[x1, . . . , xr] with r ≤ m− 1. Hence so is A.

REMARK 8.12. If A is a domain, then according to Proposition 8.4, Frac(A) willbe a finite extension of K(x1, . . . , xr) and so r must be the transcendence degree ofFrac(A)/K. In particular, r is an invariant of A.

COROLLARY 8.13. For every affine variety X there exists an integer r ≥ 0 and afinite surjective morphism f : X → Ar.

This corollary gives us a better grasp on the geometry of X, especially whenX is irreducible, for it shows that X can be ‘spread’ in a finite-to-one manner overaffine r-space. Proposition 8.4 has a kind of converse, also due to Noether, whichwe state without proof.

*THEOREM 8.14 (Emmy Noether). Let A be a domain that contains a fieldover which it is finitely generated as an algebra. Then for any finite field extensionL/Frac(A), the integral closure AL of A in L is finite over A.

If we take L = Frac(A), then AL is called the normalization of A. If A equalsits normalization, then we say that A is normal. We carry this terminology to thegeometric setting by saying that an irreducible affine variety X is normal whenk[X] is. The affine space An is normal. More generally:

LEMMA 8.15. Any unique factorization domain is normal.

PROOF. Let A be a UFD. Any b ∈ Frac(A) integral over A obeys an equationbd + a1b

d + · · · + ad = 0 with ai ∈ A. Write b = r/s with r, s ∈ A such that r ands are relatively prime. The identity rd + a1rs

d−1 + · · · + adsd = 0 shows that any

prime divisor which divides s must divide rd and hence also r. As there is no suchprime, it follows that s is a unit so that b ∈ A.

Proposition 8.14 has a remarkable geometric interpretation: let be given anirreducible affine variety Y and a finite field extension L/k(Y ). Then Proposi-tion 8.14 asserts that AL is a finitely generated k[Y ]-module. It is also a do-main (because it is contained in a field) and so it defines an irreducible varietyYL := Spec(AL). Since A ⊂ AL is an integral extension, we have a finite surjective


morphism YL → Y . This morphism induces the given field extension L/k(Y ). Soevery finite field extension of k(Y ) is canonically realized by a finite morphism ofirreducible affine varieties!

If L is an algebraic closure of k(Y ), then this does not apply, for L/k(Y ) willnot be finite, unless Y is a singleton. But L can be written as a monotone union offinite field extensions: L = ∪∞i=1Li with Li ⊂ Li+1 and Li+1/Li finite. This yieldsa sequence of finite surjective morphisms

Y YL1 YL2

YL3 · · ·

of which the projective limit can be understood as a “pro-affine variety” (a point ofthis limit is given by a sequence (yi ∈ YLi)∞i=1 such that yi+1 maps to yi for all i). Itsalgebra of regular functions is ∪∞i=1k[Y ]Li = k[Y ]L (which is usually not a finitelygenerated k-algebra) and its function field is L.

Of special interest is the case of a finite normal9 field extension.

NORMAL EXTENSIONS. We recall that an algebraic field extension L/K is normal if anirreducible polynomial in K[x] that has one root in L has all its roots in L, i.e., factors inL[x] into polynomials of degree one. A Galois extension L/K is the same thing as a normalseparable extension (this property can be used as a definition). But a purely inseparableextension L/K is also normal, for then these irreducible (monic) polynomials are of theform (x − b)q, with b ∈ L and q the smallest power such that bq ∈ K (if q 6= 1, then thecharacteristic p of k must be positive and q will be a power of p). Clearly an algebraic closureK of K is normal.

Part of Galois theory still works for normal extensions. If L/K is normal, then all theK-linear field embeddings L → K have the same image and so this image is invariantunder the full Galois group of K/K. The latter then restricts to the group of K-linear fieldautomorphisms of L (the Galois group of L/K) and this group permutes the roots of aminimal polynomial in K[x] of any element of L transitively.

For an arbitrary algebraic field extension F/K, one defines its normal closure in K asthe smallest normal extension of K in K that admits a K-linear embedding of F into it. Itis obtained as the subfield of K generated by the roots of all the irreducible polynomials ofK[x] that have a root in F . When F is finite over K, then so is its normal closure in K.

We begin with the corresponding result in commutative algebra. This has alsoimportant applications in algebraic number theory.

PROPOSITION 8.16. Let A be a normal domain and L/Frac(A) be a finite normalextension with Galois group G. Then G leaves invariant the integral closure AL of Ain L, and for every prime ideal p ⊂ A, G acts transitively on the set of prime idealsq ⊂ AL that lie over p (i.e., with q ∩A = p).

For the proof we need:

LEMMA 8.17 (The prime avoidance lemma). Any ideal of a ring that is containedin a finite union of prime ideals is contained in one of them.

PROOF. Let R be a ring, q1, . . . , qn prime ideals in R and I ⊂ R an ideal con-tained in ∪ni=1qi. We prove with induction on n that I ⊂ qi for some i. The casen = 1 being trivial, we may assume that n > 1 and that for every i = 1, . . . , n, I

9As the statement of Proposition 8.16 illustrates, this adjective is a bit overused in mathematics: anormal field extension should not be confused with the normality of a ring.


is not contained in ∪j 6=iqj . Choose ai ∈ I r ∪j 6=iqj . So then ai ∈ qi. Considera := a1a2 · · · an−1 + an. Then a ∈ I and hence a ∈ qi for some i. If i < n,then an = a − a1a2 · · · an−1 ∈ qi and we get a contradiction. If i = n, thena1a2 · · · an−1 = a − an ∈ qn and hence aj ∈ qn for some j ≤ n − 1. This is also acontradiction.

PROOF OF PROPOSITION 8.16. That any g ∈ G leaves AL invariant is clear, forg fixes the coefficients of an equation of integral dependence over A.

Let q and q′ be prime ideals of AL that lie over p. We show that q′ ⊂ ∪g∈Ggq.This suffices, for then the prime avoidance lemma implies that q′ ⊂ gq for someg ∈ G. As both q′ and gq lie over p, we must have q′ = gq by incomparability.

Let b ∈ q′ be nonzero. Let f(x) = xn + a1xn−1 + · · · + an ∈ Frac(A)[x] be a

minimum polynomial for b. Since L/Frac(A) is a normal extension, f completelyfactors in L with roots in Gb: f(x) = (x − g1(b)) · · · (x − gn(b)) for certain gi ∈ G.Since gi(b) ∈ AL we have g1(b) · · · gn(b) ∈ AL. On the other hand, g1(b) · · · gn(b) =(−1)nan ∈ Frac(A) and since A is normal it follows that g1(b) · · · gn(b) ∈ A. One ofthe factors is b and so g1(b) · · · gn(b) ∈ A ∩ (b) ⊂ A ∩ q′ = p ⊂ q. Since q is a primeideal, some factor gib lies in q and hence b ∈ ∪g∈Ggq.

We translate this into geometry:

COROLLARY 8.18. Let Y be a normal variety and L/k(Y ) a normal finite exten-sion with Galois group G. Then G acts naturally on YL in such a manner that forany closed irreducible Z ⊂ Y , G acts transitively on the irreducible components of thepreimage of Z in YL, In particular any fiber of YL → Y is a G-orbit.

This also leads for normal domains to a supplement of the going up property:

COROLLARY 8.19 (Going down). Let A ⊂ B be a finite extension with B a do-main and A normal. Given prime ideals p in A and q′ in B such that p ⊂ q′ ∩A, thenthere exists a prime ideal q in B with q ⊂ q′ and q ∩A = p.

PROOF. Put K := Frac(A) and let L be the normal closure of Frac(B) in analgebraic closure of Frac(B). Since B is integral over A, we have B ⊂ AL. Now Lis a finite normal extension of Frac(B) (with Galois group G, say), and this bringsus in the situation of Proposition 8.16 above. Since Frac(B) is finite over K, L isfinite over K and so AL is by 8.14 finite over A (and hence also over B).

Put p′ := q′ ∩ A so that p′ ⊃ p. According to Proposition 8.6 we can find ina prime ideal q′ in AL which meets B in q′. The same proposition tells us thatthere exist nested prime ideals p′ ⊃ p in AL which meet A in p′ ⊃ p. Since q′

and p′ both meet A in p′, there exists according to Proposition 8.16 a g ∈ G suchthat gp′ = q′. Upon replacing p′ ⊃ p by g′p′ ⊃ gp, we may then assume thatp′ = q′. Now q := p ∩ B is as desired, for it meets A in p and is contained inp′ ∩B = q′ ∩B = q.

REMARK 8.20. We can rephrase this in the spirit of Remark 8.7 by saying thatany prime chain in A is the intersection of prime chain in B for which the lastmember has been prescribed in advance.

9. Dimension

One way to define the dimension of a topological space X is with induction:agree that the empty set has dimension −1 and that X has dimension ≤ n if it

9. DIMENSION 41

admits a basis of open subsets such that the boundary of every basis element hasdimension ≤ n − 1. This is close in spirit to the definition that we shall use here(which is however adapted to the Zariski topology; as you will find in Exercise 35,it is useless for Hausdorff spaces).

DEFINITION 9.1. Let X be a nonempty topological space. We say that the Krulldimension of X is at least d if there exists an irreducible chain of length d in X, thatis, a strictly descending chain of closed irreducible subsets X0 ) X1 ) · · · ) Xd

of X. The Krull dimension of X is the supremum of the d for which an irreduciblechain of length d exists and we then write dimX = d. We stipulate that the Krulldimension of the empty set is −1.

LEMMA 9.2. For a locally closed subset Z of a topological space X we havedimZ ≤ dimX.

PROOF. Our hypothesis implies that if Y is closed in Z, then its closure Y in Xhas the property that Y ∩ Z = Y . We also know that if Y ⊂ Z is irreducible, thenso is Y . So if we have an irreducible chain of length d in Z, then the closures of themembers of this chain yield an irreducible chain of length d in X. This proves thatdimZ ≤ dimX.

EXERCISE 35. What is the Krull dimension of a nonempty Hausdorff space?

EXERCISE 36. Let U be an open subset of the space X. Prove that for anirreducible chain Y • in X of length d with U ∩ Y d 6= ∅, U ∩ Y • is an irreduciblechain of length d in U . Conclude that if U is an open covering of X, then dimX =supU∈U dimU .

EXERCISE 37. Suppose that X is a noetherian space. Prove that the dimensionof X is the maximum of the dimensions of its irreducible components. Prove alsothat if all the singletons (= one element subsets) in X are closed, then dim(X) = 0if and only if X is finite.

It is straightforward to translate this notion into algebra:

DEFINITION 9.3. The Krull dimension dim(R) of a ringR is the supremum of theintegers d for which there exists an prime chain of length d in R, where we stipulatethat the zero ring (i.e., the ring which has no prime ideals) has Krull dimension −1.

It is clear that for a closed subset X ⊂ An, dim k[X] = dimX. Since anyprime ideal of a ring R contains the ideal

√(0) of nilpotents, R and its ‘reduction’

Rred := R/√

(0) have the same Krull dimension. So the Krull dimension of a finitelygenerated k-algebra A is that of the affine variety Spm(A).

Remark 8.7 shows immediately:

LEMMA 9.4. The Krull dimension is invariant under integral extension: if B isintegral over A, then A and B have the same Krull dimension.

REMARK 9.5. For a domain A the zero ideal (0) is a prime ideal and so dim(A) = 0if and only if A (0) is maximal, i.e., A is a field. We say that a noetherian domain A is aDedekind domain if dim(A) ≤ 1, in other words, if every nonzero prime ideal is maximal. Forinstance, a unique factorization domain (such as Z and K[X] with K a field) is a Dedekinddomain. The importance of this notion comes from the fact that a converse holds on thelevel of ideals: any ideal of a Dedekind domain is uniquely written a product of primeideals. Lemma 9.4 shows that any finite extension of a Dedekind domain (such as the ring ofintegers of an algebraic number field and a finite extension of K[X]) is a Dedekind domain.


The Krull dimension was easy to define, but seems difficult to compute in con-crete cases. How can we be certain that a given prime chain has maximal possiblelength? It is not even clear how to tell whether the Krull dimension of a given ringis finite. We will settle this in a satisfactory manner for a domain B containing afield K over which it is a finitely generated: we show that a length of a prime chainin B is bounded by the transcendence degree Frac(B)/K and that we have equalitywhen the prime chain is maximal (so that the length of any maximal prime chainis the Krull dimension).

THEOREM 9.6. LetK be a field and B a finitely generatedK-algebra without zerodivisors. Then the Krull dimension of B equals the transcendence degree of Frac(B)/Kand every maximal prime chain in B (i.e., one that cannot be extended to a longerprime chain) has length dimB.

PROOF. We prove both assertions with induction on the transcendence degreeof Frac(B)/K. By Noether normalization there exists an integer r ≥ 0 and a K-algebra monomorphism K[x1, . . . , xr] → B such that B is finite over K[x1, . . . , xr].We put A := K[x1, . . . , xr]. Then Frac(B) is a finite extension of Frac(A) =K(x1, . . . , xr) and so the transcendence degree of Frac(B)/K is r. In A we havethe length r prime chain (0) ( (x1) ( (x1, x2) ( · · · ( (x1, x2, . . . , xr). By Remark8.7 this is the intersection of A with a prime chain in B and so the Krull dimensionof B is at least r.

To prove the remaining assertions, let q• := ((0) = q0 ( q1 ( · · · ( qm) be aprime chain in B. We prove that its length m is at most r with equality when q• isa maximal prime chain. By the incomparability property of ‘going up’ (Proposition8.6), p• := q• ∩ A will be a prime chain in A, also of length m. The idea is toshow that p1 defines a closed subset of ArK of dimension ≤ m − 1. Choose anirreducible f ∈ p1. After renumbering the coordinates, we may assume that f doesnot lie in K[x1, . . . , xr−1]. So if we write f =

∑Ni=0 aix

ir with ai ∈ K[x1, . . . , xr−1]

and aN 6= 0, then N ≥ 1. Since f is irreducible, A/(f) is a domain. The imageof xr in Frac(A/(f)) is a root of the monic polynomial tN +

∑N−1i=0 (ai/a0)ti ∈

K(x1, . . . , xr−1)[t], and so Frac(A/(f)) is a finite extension of K(x1, . . . , xr−1). Inparticular, Frac(A/(f)) has transcendence degree r − 1 over K. By our inductioninduction hypothesis, A/(f) has then Krull dimension r − 1. Since the image ofp1 ( · · · ( pm in A/(f) is a prime chain of length m − 1 (it is strictly ascending,because it is so in A/p1), it follows that m− 1 ≤ r − 1. Hence m ≤ r.

If q• is a maximal prime chain in B, then by ‘going down’ (Corollary 8.19),we find a prime ideal q ⊂ q1 such that q ∩ A = (f), and the maximality of q•then implies that q = q1 and hence that (f) = p1. Since Frac(B/q1) is a finiteextension of Frac(A/(f)) (which in turn is a finite extension of K(x1, . . . , xr−1)),it has transcendence degree r − 1 over k. As q1 ( q2 · · · ( qm defines a maximalprime chain in B/q1, it follows from our induction hypothesis that m − 1 = r − 1and so m = r.

COROLLARY 9.7. In the situation of Theorem 9.6, let m be a maximal ideal of B.Then the Krull dimension of the localization Bm = (B rm)−1B is that of B.

PROOF. Any prime chain in Bm is a prime chain in B contained in m. So byTheorem 9.6, the Krull dimension of Bm is finite. If such a chain is maximal forthis property, then it will end with m and will also be maximal in B (for there is no

10. NONSINGULAR POINTS 43

prime ideal strictly containing m) and again by Theorem 9.6 its length is then theKrull dimension of B.

COROLLARY 9.8. Let X be an irreducible affine variety of dimension d. Then everymaximal irreducible chain in X has length d. Moreover, every nonempty open affineU ⊂ X has dimension d and for every p ∈ X, the Krull dimension of OX,p is d.

REMARK 9.9. If S ⊂ R is a multiplicative system, then the preimage of a primeideal q of S−1R under the ring homomorphism R → S−1R is a prime ideal q of Rwhich does not meet S and we have q = S−1q. This sets up a bijection betweenthe prime ideals of S−1R and those of R not meeting S. If we take S = R− p, withp a prime ideal, then this implies that dimRp is the supremum of the prime chainsin R which end with p. Since the prime chains in R which begin with p correspondto prime chains in R/p, it follows that dimRp + dimR/p is the supremum of theprime chains in R having p as a member. When R is a domain finitely generatedover a field, then this is dimR by Theorem 9.6.

REMARK 9.10. An affine variety C of dimension 1 is called a curve. When Cis irreducible this amounts to k(C) being of transcendence degree one. It followsfrom Remark 9.5 that this is also equivalent to: k[C] is a Dedekind domain 6= k.

EXERCISE 38. Prove that a hypersurface in An has dimension n− 1.

EXERCISE 39. Let X be an irreducible affine variety and Y ⊂ X a closed ir-reducible subset. Prove that dimX − dimY is equal to the Krull dimension ofk[X]I(Y ).

EXERCISE 40. Prove that when X and Y are irreducible affine varieties, thendim(X×Y ) = dimX+dimY . (Hint: Embed each factor as a closed subset of someaffine space. You may also want to use the fact that the equality to be proven holdsin case X = Am and Y = An.)

10. Nonsingular points

In this section we focus on the local properties of an affine variety X = Spm(A)(so A := k[X] is here a reduced finitely generated k-algebra) at a point p. Thereforea central role will be played by the local algebra OX,p = Amp whose maximal idealis mX,p = (A−mp)

−1mp.

If k = C and X ⊂ Cn is a closed subset of dimension d, then we hope thatthere is a nonempty open subset of X where X is ‘smooth’, i.e., where X looks likea complex submanifold of complex dimension d. Our goal is to define smoothnessin algebraic terms (so that it make sense for our field k) and then to show that theset of smooth points of a variety is open and dense in that variety.

Our point of departure is the implicit function theorem. One version statesthat if U ⊂ Rn is an open neighborhood of p ∈ Rn and fi : U → R, i =1, . . . n − d are differentiable functions zero in p such that the total differentialsat p, df1(p), . . . , dfn−d(p) are linearly independent in p (this is equivalent to: theJacobian matrix of (∂fj/∂xi)(p))i,j has maximal rank n − d), then the commonzero set of f1, . . . , fn−d is a submanifold of dimension d at p whose tangent spaceat p is the common zero set of df1(p), . . . , dfn−d(p). In fact, one shows that thissolution set is near p the graph of a map: we can express n − d of the coordinates


as differentiable functions in the d remaining ones. Conversely, any submanifold ofRn at p of dimension d is locally thus obtained.

We begin with the observation that for any ring R, partial differentiation of apolynomial f ∈ R[x1, . . . , xn] (where the elements of R are treated as constants)is well-defined and produces another polynomial. The same goes for a fractionφ = f/g in R[x1, . . . , xn][g−1]: a partial derivative of φ is a rational function (inthis case with denominator g2). We then define the total differential of a rationalfunction φ ∈ R(x1, . . . , xn) as usual:

dφ :=

n∑i=1

∂φ

∂xi(x)dxi,

where for now, we do not worry about interpreting the symbols dxi: we think of dφsimply as a regular map from an open subset of An to a k-vector space of dimensionn with basis dx1, . . . , dxn, leaving its intrinsic characterization for later. However,caution is called for when R is a field of positive characteristic:

EXERCISE 41. Prove that f ∈ k[x] has zero derivative, if and only if f is constantor (when char(k) = p > 0) a pth power of some g ∈ k[x].

Generalize this to: given f ∈ k[x1, . . . , xn], then df = 0 if and only if f isconstant or (when char(k) = p > 0) a pth power of some g ∈ k[x1, . . . , xn].

We should also be aware of the failure of the inverse function theorem:

EXAMPLE 10.1. Let C ⊂ A2 be the curve defined by y2 = x3 + x. By anyreasonable definition of smoothness we should view the origin o := (0, 0) as asmooth point of C. Indeed, when k = C, the projection f : C → A1, (x, y) 7→ y,would be a local-analytic isomorphism at o. But the map is not locally invertiblewithin our category: the inverse requires us to find a rational function x = u(y)which solves the equation y2 = x3 + x and it is easy to verify that none exists. (Wecan solve for x formally: x = u(y) = y2 − y6 + 3y10 + · · · , where it is important tonote that the coefficients are all integers so that this works for every characteristic.)In fact, the situation is worse: no open neighborhood U of o in C is isomorphic toan open subset V of A1. The reason is that this would imply that k(C) = k(U) isisomorphic to k(V ) = k(x) and one can show that this is not so.

Somewhat related to this is an issue illustrated by the following example.

EXAMPLE 10.2. Consider the curve C ⊂ A2 defined by xy = x3 + y3. Thepolynomial x3 +y3−xy is irreducible in k[x, y], so that k[C] is without zero divisorsandC ′ is irreducible. Hence the local ringOC,o ⊂ k(C) is also without zero divisors.But C seems to have two branches at o which apparently can only be recognizedformally: one such branch is given by y = u(x) = x2 +x5 + 3x8 + · · · and the otherby interchanging the roles of x and y: x = v(y) = y2 + y5 + 3y8 + · · · . If we useξ := x− v(y) and η := y − u(x) as new formal coordinates, then C is simply givenat 0 by the reducible equation ξη = 0.

These examples make it clear that for a local understanding of a variety X ato, the local ring OX,o still carries too much global information. One way to get rid


of this overload is by passing formal to power series. This is accomplished by whatis known as formal completion10.

10.3. FORMAL COMPLETION. Let R be a ring and I ( R a proper ideal. Forevery R-module M , the descending sequence of submodules M ⊃ IM ⊃ I2M ⊃· · · ⊃ InM ⊃ · · · gives rise to a sequence of surjective R-homomorphisms

0 = M/M M/IM M/I2M M/I3M · · ·M/InM ← · · ·

from which we can form the R-module MI := lim←−nM/IMn, called the I-adic com-pletion of M . So any a ∈ MI is uniquely given by a sequence (αn ∈ M/InM)n≥0

whose terms are compatible in the sense that αn is the reduction of αn+1 for alln. In this way MI can be regarded as an R-submodule of

∏n≥0(M/InM). The

natural R-homomorphisms M → M/InM combine to define a R-homomorphismM → MI . Its kernel is ∩∞n=0I

nM and this turns out to be trivial in many cases ofinterest. If we do this for the ring R, we get a ring RI and R → RI is then a ringhomomorphism.

The R-module structure on MI extends naturally to a RI -module structure:for any r = (ρn ∈ R/In)∞n=0 ∈ RI we define ra simply as given by the se-quence (ρnαn)n≥0 (note that ρnαn is indeed the reduction of ρn+1αn+1). AnyR-homomorphism φ : M → N of R-modules sends M/InM to N/InN , and the re-sulting homomorphisms M/InM → N/InN are compatible in the sense that theydetermine a map φI : MI → NI . This is in fact a RI -homomorphism. We have thusdefined a functor from the category of R-modules to the category of RI -modules.It is easily verified that if φ is an epimorphism, then a compatible sequence in(N/InN)n≥0 is the image of one in (M/InM)n≥0 so that φI is an epimorphism aswell. This need not be true for monomorphisms, but we will see that this is so inthe noetherian setting.

EXAMPLE 10.4. Take the ring k[x1, . . . , xn]. Its completion with respect to themaximal ideal (x1, . . . , xn) is just the ring of formal power series k[[x1, . . . , xn]].We get the same result if we do this for the localization OAn,o of k[x1, . . . , xn] at(x1, . . . , xn). We will find that for (C, o) in Example 10.1 resp. 10.2 the completionof OC,o with respect to the maximal ideal is isomorphic to k[[x]] resp. k[[x, y]/(xy).

EXAMPLE 10.5. Take the ring Z. Its completion with respect to the ideal (n),n an integer ≥ 2, yields the ring of n-adic integers Z(n): an element of Z(n) isgiven by a sequence (ρi ∈ Z/(ni))∞i=1 with the property that ρi is the image of ρi+1

under the reduction Z/(ni+1) → Z/(ni). We get the same result if we do this forthe localization Z(n) = a/b : a ∈ Z, b ∈ Z − (n). It follows from the Chineseremainder theorem that Z(n) =

∏p|n Z(p).

10.6. ADIC TOPOLOGIES. We can understand MI and RI as completions withregard to a topology onM andR. This often helps to clearify their dependence on I(which is weaker than one might be inclined to think). For this we endow every R-module M with a topology, the I-adic topology, of which a basis is the collection of

10Another approach would be to allow ‘algebraic’ functions of the type that we encountered inthe two examples above, but then we would have to address the question what the domain of such afunction should be. This can not be achieved by refining the Zariski topology. Rather, this forces us togeneralize the very notion of a topology, leading up to what is called the etale topos. Despite its ratherabstract nature this is closer to our geometric intuition than the Zariski topology.


additive translates of the submodules InM , i.e., the collection of subsets a+ InM ,a ∈ M , n ≥ 0. This is a topology indeed: given two basic open subsets a + InM ,a′ + In

′M , then for any element c in their intersection, the basic open subset c +

Imaxn,n′M is also in their intersection. So a sequence (an ∈ M)n≥1 converges toa ∈ M precisely when for every integer k ≥ 0, an ∈ a + IkM for n large enough.The fact that our basis is translation invariant implies that with this topology, M isa topological abelian group ((a, b) ∈ M ×M 7→ a − b ∈ M is continuous). If weendow R also with the I-adic topology and R×M with the product topology, thenthe map (r, a) ∈ R ×M → ra ∈M which gives the action by R is also continuous.It is clear that any R-module homomorphism is continuous for the I-adic topology.

If J ⊂ I is an ideal with J ⊃ Ir for some r ≥ 0, then the J -adic topologyon M is the same as the J -adic topology. Also, for any n0 ≥ 0, the collectionIn+n0Mn≥0 is a neighborhood basis of 0 in M and hence also defines the I-adictopology. Note that the topology on M comes from one on M/ ∩n≥0 I

nM in thesense that the open subsets of M are pre-images of open subsets of M/ ∩n≥0 I

nM .When M is Hausdorff, then its topology is even metrizable: if φ : Z+ → (0,∞) is any

function with φ(n+ 1) ≤ φ(n) and limn→∞ φ(n) = 0 (one often takes φ(n) = u−n for someu > 1), then a metric δ on M is defined by

δ(a, a′) := infφ(n) : a− a′ ∈ InM.

This metric is nonarchimedean in the sense that δ(a, a′′) ≤ maxδ(a, a′), δ(a′, a′′). A se-quence (an ∈ M)∞n=0 is then a Cauchy sequence if and only if for every integer k ≥ 0 allbut finitely many terms lie in the same coset of IkM in M ; in other words, there existsan index nk ≥ 0 such that am − an ∈ IkM for all m,n ≥ nk. This makes it clear thatthe notion of Cauchy sequence is independent of the choice of φ. Such a Cauchy sequencedefines a compatible sequence of cosets (αn ∈ M/InM)n≥0 and hence an element of MI .Recall that a metric space is said to be complete if every Cauchy sequence in that space con-verges. A standard construction produces a completion of every metric space M : its pointsare represented by Cauchy sequences in M , with the understanding that two such sequencesrepresent the same point if the distance between the two nth terms goes to zero as n→∞.In the present situation we thus recover MI . Note that the homomorphism M → MI is acontinuous injection with image dense in MI .

EXERCISE 42. Let I and J be ideals of a ring R and m a positive integer withJm ⊂ I. Prove that the J -adic topology is finer than the I-adic topology and thatthere is a natural continuous ring homomorphism RJ → RI . Conclude that whenR is noetherian, RI can be identified with R√I .

If M is an R-module, then the inclusion M ′ ⊂ M of any submodule is con-tinuous for the I-adic topology. The Artin-Rees lemma says among other thingsthat when R is noetherian, this is in fact a closed embedding (so that M ′ has theinduced topology).

*LEMMA 10.7 (Artin-Rees). Let R be a noetherian ring, I ⊂ R an ideal, M afinitely generated R-module and M ′ ⊂ M an R-submodule. Then there exists aninteger n0 ≥ 0 such that for all n ≥ 0:

M ′ ∩ In+n0M = In(M ′ ∩ In0M).

The proof (which is ingeneous, but not difficult) can be found in any standardtext book on commutative algebra (e.g., [1]). The lemma implies that for everyn ≥ 0 there exists a n′ ≥ 0 such that M ′∩ In′M ⊂ InM ′ (we can take n′ = n+n0).


So the inclusion M ′ ⊂ M is not merely continuous, but M ′ inherits its I-adictopology from that of M . We will use the Artin-Rees lemma via this property only.A special case is when M ′ := ∩n≥0I

nM : it then follows that M ′ = In+n0M ∩M ′ =In(M ′ ∩ In0M) = InM ′. By taking n = 1, Nakayama’s lemma 8.5 then yields thatM ′ = 0, provided that R is a local ring. We record this as:

COROLLARY 10.8. If R is a noetherian local ring, then any finitely generatedR-module M is Hausdorff for the I-adic topology: ∩n≥0I

nM = 0.

COROLLARY 10.9. In the situation of Lemma 10.7, the homomorphism M ′I → MI

induced by the inclusion M ′ ⊂M is a closed embedding and MI/M′I can be identified

with the I-adic completion of M/M ′. In case R is also a local ring so that by Corollary10.8 we may regard M resp. M ′ as a submodule of MI resp. M ′I , then M ∩M ′I = M ′.

Note that the first part of this corollary says that when R is noetherian, I-adiccompletion is an exact functor on the category of finitely generated R-modules.

PROOF OF COROLLARY 10.9. Observe that we have epimorphisms

M ′/In+n0M ′ M ′/(M ′ ∩ In+n0M) = M ′/In(M ′ ∩ In0M)M ′/InM ′,

where we note that the middle term is the image of M ′ in M/In+n0M . So ifj : M → MI denotes the obvious map, then after taking the projective limitswe obtain continuous epimorphisms M ′I j(M ′) M ′I . Their composite is theidentity and so this is in fact a homeomorphism followed by its inverse. In partic-ular, M ′I → MI is a closed embedding. We next show that the RI -epimorphismMI → M/M ′I induced by the projection M →M/M ′ has kernel M ′. It is clear thatthis kernel contains M ′. For the converse, we observe that the kernel of M/InM →(M/M ′)/In(M/M ′) ∼= M/(M ′ + InM) is (M ′ + InM)/InM ∼= M ′/(M ′ ∩ InM),which we may identify for n ≥ n0 with M ′/In(M ′ ∩ In0M). So the kernel ofMI → M/M ′I is represented by compatible sequences in (M ′/In(M ′∩In0M))n≥n0

.Such sequences represent elements of M ′I .

The last assertion follows from the identity j−1j(M ′) = M ′ + ∩n≥0InM and

the fact that ∩n≥0InM = 0 by Corollary 10.8.

Let R be a local ring with maximal ideal m and residue field κ. We use the roofsymbol for completion with respect to m.

Then m is a finitely generated R-module. Since the ring R acts on m/m2 viaR/m = κ, m/m2 is a vector space over κ. If R is noetherian, then m/m2 is finitelygenerated as a R-module, in other words, finite dimensional as a κ-vector space.

DEFINITION 10.10. The Zariski cotangent space T ∗(R) of R is the κ-vector spacem/m2 and its κ-dual, T (R) := Homκ(m/m2, κ) (which is also equal to HomR(m, κ)),is called the Zariski tangent space T (R) of R. The embedding dimension embdim(R)is the dimension of m/m2 as a vector space over κ.

If X is an affine variety and p ∈ X, then we define the Zariski cotangent spaceT ∗pX, the Zariski tangent space TpX and the embedding dimension embdimpX ofX at p to be that of OX,p.

For instance, the embedding dimension of An at any point p ∈ An is n. Thisfollows from the fact that the map dp : f ∈ mAn,p 7→ df(p) ∈ kn defines an iso-morphism of k-vector spaces mAn,p/m

2An,p

∼= kn. We note in passing that we here


have a way of understanding the total differential at p ∈ An in more intrinsic termsas the map dp : OAn,p → mAn,p/m

2An,p which assigns to f ∈ OAn,p the image of

f − f(p) ∈ mAn,p in mAn,p/m2An,p. Thus, a differential of f at p can be understood

as a k-linear function df(p) : TpAn → k and (dp(xi) = dxi(p))ni=1 is a basis of

mAn,p/m2An,p = T ∗pAn.

Let us observe that since embedding dimension and Zariski tangent space of alocal ring R are defined in terms of m/m2, these notions only depend on R.

EXERCISE 43. Let (R′,m′) and (R,m) be local rings with residue fields κ resp. κ′

and let φ : R′ → R be a ring homomorphism with the property that φ−1m = m′ (wethen say that φ is a local homomorphism). Prove that φ induces a field embeddingκ′ → κ and a linear map of κ-vector spaces T (φ) : T (R)→ κ⊗κ′ T (R′).

An application of Nakayama’s lemma to the R-module m yields:

COROLLARY 10.11. The embedding dimension of a noetherian local ring R is thesmallest number of generators of its maximal ideal. The embedding dimension is zeroif and only if R is a field.

DEFINITION 10.12. A noetherian local ring R is said to be regular if its Krulldimension equals its embedding dimension. A point p of an affine variety X iscalled regular if its local ring OX,p is so; otherwise it is called singular. The corre-sponding subsets of X are called the regular locus resp. singular locus of X and willbe denoted Xreg resp. Xsing. An affine variety without singular points is said to benonsingular.

We shall see that the regularity of a local ring OX,p indeed amounts to X being‘like a manifold’ at p. We begin with a formal version of the implicit functiontheorem.

LEMMA 10.13. Let p ∈ An and let f1, . . . , fn−d ∈ mAn,p be such that the differen-tials df1(p), . . . , dfn−d(p) are linearly independent. Then p := (f1, . . . , fn−d) ⊂ OAn,pis a prime ideal and OAn,p/p is a regular local ring of dimension d whose comple-tion with respect to its maximal ideal is isomorphic to k[[x1, . . . , xd]] as a completelocal k-algebra. Moreover, there exists an affine neighborhood U of p in An on whichf1, . . . , fn−d admit representatives f1, . . . , fn−d ∈ k[U ] which generate in k[U ] a primeideal P and then Z(P ) = ∩iZ(fi) ⊂ U is an irreducible affine variety having p asa regular point of dimension d with Zariski tangent space equal to the kernel of thelinear surjection (df1(p), . . . , dfn−d(p)) : TpAn → kn−d.

PROOF. Let us abbreviate OAn,p by O and its maximal ideal mAn,p by m. Ex-tend f1, . . . , fn−d to a system of regular functions f1, . . . , fn ∈ m such that thedf1(p), . . . , dfn(p) are linearly independent. This means that their images in m/m2

are linearly independent over k. After an affine-linear transformation, we then may(and will) assume that p is the origin o of An and that fi ≡ xi (mod m2). Hencethe monomials of degree r in f1, · · · , fn map to a k-basis of mr/mr+1. With induc-tion on r it then follows that the monomials of degree ≤ r in f1, · · · , fn make up ak-basis of O/mr+1. This amounts to the assertion that the map

yi ∈ k[[y1, . . . , yn]] 7→ fi ∈ O

defines an isomorphism k[[y1, . . . , yn]] ∼= O of complete local rings (a ring isomor-phism that is also a homeomorphism). The restriction of its inverse to O is a topo-logical embedding of O in k[[y1, . . . , yn]] (sending fi to yi). The ideal generated


by (y1, . . . , yn−d) in k[[y1, . . . , yn]] is the closure p of the image of p. The quotientring is the domain k[[yn−d+1, . . . , yn]] and so this is clearly a prime ideal. Accordingto Corollary 10.9, the preimage of p in O is p (hence p is a prime ideal) and theembedding

O/p → k[[y1, . . . , yn]]/(y1, . . . , yn−d) = k[[yn−d+1, . . . , yn]]

realizes the m-adic completion of O/p. This argument applied applied to the idealpi ⊂ OAn,p generated by f1, . . . , fi shows that pi a prime ideal for all i = 0, . . . , n−1.So we have prime chain (0) = p0 ( p1 ( · · · ( pn in O of length n. Accordingto Corollary 9.7, O has Krull dimension n, and so this prime chain is maximal.Theorem 9.6 then implies that O/p = O/pn−d has Krull dimension d. This is alsothe embedding dimension of O/p, for fn−d+1, . . . , fn map to a k-basis of m/(m2 +(f1, . . . , fn−d)). So O/p is a regular local ring of dimension d.

Let g′ ∈ k[x1, . . . , xn] be a common denominator for f1, . . . , fn−d with g(o) 6= 0and denote by P ′ ⊂ k[x1, . . . , xn][1/g′] the preimage of p under the localizationmap k[x1, . . . , xn][1/g′] → O. This is a prime ideal which contains (the images of)f1, . . . , fn−d and has the property that its localization at o ∈ An is p. Choose afinite set of generators φ1, . . . , φr of P ′. We may write in O, φi =

∑n−dj=1 uijfj with

uij ∈ O. If g ∈ (g′) with g(o) 6= 0 is a common denominator for the uij , then putU = Ang . Then P := P ′[1/g] is a prime ideal in k[U ]. It is generated by the imagesof φ1, . . . , φr and hence also by the images f1, . . . , fn−d of f1, . . . , fn−d in k[U ]. So(U ; f1, . . . , fn−d) is as desired.

THEOREM 10.14. Let X ⊂ An be locally closed and let p ∈ X. Then the local ringOX,p is regular of dimension d if and only there exist regular functions f1, . . . , fn−don a principal neighborhood U of p in An such that these functions generate the idealin k[U ] defining X ∩ U and df1, . . . , dfn−d are linearly independent in every point ofU . In that case X ∩ U is regular and for every q ∈ X ∩ U the Zariski tangent spaceTqX is the kernel of the linear map (df1(q), . . . , dfn−d(q)) : TqAn → kn−d.

PROOF. Suppose that OX,p is regular of dimension d. Let IX,p ⊂ OAn,p be theideal of regular functions at p vanishing on a neighborhood of p in X, in otherwords, the kernel of OAn,p → OX,p. The latter is a surjective homomorphism oflocal rings and so the preimage of mX,p resp. m2

X,p is IX,p + mAn,p = mAn,p resp.IX,p + m2

An,p. It follows that the quotient mAn,p/(IX,p + m2An,p)

∼= mX,p/m2X,p has

dimension d. So the image (IX,p+m2An,p)/m

2An,p of IX,p in the n-dimensional vector

space mAn,p/m2An,p must have dimension n − d. Choose f1, . . . , fn−d ∈ IX,p such

df1(p), . . . , dfn−d(p) are linearly independent. We show among other things thatthese functions generate IX,p (this will in fact be the key step).

According to Lemma 10.13, the ideal pi ⊂ OAn,p generated by f1, . . . , fi isprime and so we have a prime chain

(0) = p0 ( p1 ( · · · ( pn−d ⊆ IX,p.Since dimOX,p = d, there also exists a prime chain of length d containing IX,p:

IX,p ⊆ q0 ( q1 ( · · · ( qd ⊆ OAn,p.

As OAn,p has dimension n, these two prime chains cannot make up a prime chainof length n+ 1 and so pn−d = IX,p = q0.

In particular, f1, . . . , fn−d generate IX,p. Let U ′ be an affine neighborhood ofp in An on which the fi’s are regular, generate a prime ideal in k[U ′] and are such


that their common zero set is X ∩ U ′. Since the df1(p), . . . , dfn−d(p) are linearlyindependent, there exist n − d indices 1 ≤ ν1 < ν2 < · · · < νn−d ≤ n such thatδ := det((∂fi/∂xνj )i,j) ∈ k[U ′] is nonzero in p. Then U := Uδ ⊂ U ′ has all theasserted properties (with the last property following from Lemma 10.13).

The converse says that if U , p and f1, . . . , fn−d are as in the theorem, thenthe functions f1, . . . , fn−d generate a prime ideal Ip in OAn,p such that OAn,p/Ip isregular of dimension d. This follows Lemma 10.13.

PROPOSITION 10.15. The regular points of an affine variety X form an open-dense subset Xreg of X.

For the proof we will assume Proposition 10.16 below, which we will leaveunproved for now. (Note that it tells us only something new in case k has positivecharacteristic.)

*PROPOSITION 10.16. Every finitely generated field extension L/k (with k asin these notes, i.e., algebraically closed) is separably generated, by which we meanthat there exists an intermediate extension k ⊂ K ⊂ L such that K/k is purelytranscendental (i.e., of the form k(x1, . . . , xr)) and L/K is a finite separable extension(which by the theorem of the primitive element is then obtained by adjoining the rootof an irreducible, separable polynomial in K[x]).

This improves upon Corollary 7.8, as this implies that every irreducible affinevariety is birationally equivalent to a hypersurface.

PROOF OF PROPOSITION 10.15. Without loss of generality we may assume thatX is irreducible. Since we already know that Xreg is open, it remains to see thatit is nonempty. It thus becomes an issue which only depends on k(X). Hence itsuffices to treat the case of a hypersurface in An so that I(X) is generated by anirreducible polynomial f ∈ k[x1, . . . , xn]. In view of Lemma Lemma 10.13 it thensuffices to show that df is not identically zero on X. Suppose otherwise, i.e., thateach partial derivative ∂f/∂xi vanishes on X. Then each ∂f/∂xi must be multipleof f and since the degree of ∂f/∂xi is less than that of f , this implies that it isidentically zero. But then we know from Exercise 41 that the characteristic p of kmust then be positive (so ≥ 2) and that f is of the form gp. This contradicts thefact that f is irreducible.

EXERCISE 44. Let X be a nonsingular variety. Prove that X is connected if andonly if it is irreducible.

REMARK 10.17. This enables us to find for an affine variety X of dimensiond (with downward induction) a descending chain of closed subsets X = Xd ⊃Xd−1 ⊃ · · · ⊃ X0 such that dimXi ≤ i and all the (finitely many) connectedcomponents of XirXi−1 are nonsingular subvarieties of dimension i (such a chainis called a stratification of X): if Xi has been defined, then we take for Xi−1

the union of the singular locus Xireg and the irreducible components of dimension

≤ i − 1. Then dimXi−1 ≤ i − 1 and every connected component of Xi r Xi−1

is an nonempty open subset of some Xireg and hence a nonsingular subvariety of

dimension i.

10.18. DIFFERENTIALS AND DERIVATIONS. The differential that we defined earlier has anintrinsic, coordinate free description that turns out to be quite useful. Let us begin with theobservation that the formation of the total differential of a polynomial, φ ∈ k[x1, . . . , xn] 7→


dφ :=∑ni=1(∂φ/∂xi)(p)dxi is a k-linear map which satisfies the Leibniz rule: d(φψ) =

φdψ + ψdφ. This property is formalized with the following definition. Fix a ring R (the basering) and an R-algebra A.

DEFINITION 10.19. Let M be a A-module. An R-derivation of A with values in M isan R-module homomorphism D : A → M which satisfies the Leibniz rule: D(a1a2) =a1D(a2) + a2D(a2) for all a1, a2 ∈ A.

The last condition usually prevents D from being an A-module homomorphism. Let usobserve that (by taking a1 = a2 = 1) we must have D(1) = 0. Since D is R-linear, it thenfollows that for every r ∈ R, D(r) = rD(1) = 0. Note also that if b ∈ A happens to beinvertible in A, then 0 = D(1) = D(b/b) = 1/bD(b) + bD(1/b) so that D(1/b) = −D(b)/b2

and hence D(a/b) =(D(a)b− aD(b)

)/b2 for every a ∈ A.

Given a1, . . . , an ∈ A, then the values of D on a1, . . . , an determine its values onthe subalgebra A′ by the ai’s, for if φ : R[x1, . . . , xn] → A denotes the corresponding R-homomorphism and f ∈ R[x1, . . . , xn], then

Dφ(f) =

n∑i=1

φ( ∂f∂xi

)Dai.

If we combine this with the formula for D(1/a), we see that not just determines D on the R-subalgebra A′ of A generated by the ai’s, but also on the biggest localization of A′ containedin A. In particular, if we are given a field extension L/K, then a K-derivation of L withvalues in some L-vector space is determined by its values on a set of generators of L as afield extension of K.

Observe that the set of R-derivations of A in M form an R-module: if D1 and D2 areR-derivations of A with values in M , and a1, a2 ∈ A, then a1D1 + a2D2 is also one. Wedenote this module by DerR(A,M).

EXERCISE 45. Prove that if D1, D2 ∈ DerR(A,A), then [D1, D2] := D1D2 − D2D1 ∈DerR(A,A). What do we get for R = k and A = k[x1, . . . , xn]?

It is immediate from the definition that for every A-module homomorphism φ : M → Nthe composition of a D as above with φ is an R-derivation of A with values in N . We cannow construct a universal R-derivation of A, d : A → ΩA/R (where ΩA/R must of coursebe an R-module) with the property that every D as above is obtained by composing d witha unique homomorphism of A-modules D : ΩA/R → N . The construction that is forcedupon us starts with the free A-module A(A) which has A itself as a generating set—letus denote the generator associated to a ∈ A by d(a)—which we then divide out by the A-submodule ofA(A) generated by the expressions d(ra)−rd(a), d(a1+a2)−d(a1)−d(a2) andd(a1a2)−a1d(a2)−a2d(a2), with r ∈ R and a, a1, a2 ∈ A. The quotientA-module is denotedΩA/R and the composite of d with the quotient map by d : A → ΩA/R. The latter is an R-derivation of A by construction. Given an R-derivation D : A→M , then the map which as-signs to d(a) the value Da extends (obviously) as an A-module homomorphism A(A) →M .It has the above submodule in its kernel and hence determines an A-module homomorphismof D : ΩA/R →M . This has clearly the property that D = Dd. In other words, composition

with d defines an isomorphism of A-modules HomA(ΩA/R,M)∼=−→ DerR(A,M). We call

ΩA/R the module of Kahler differentials. We shall see that the map d : A → ΩA/R can bethought of as an algebraic version of the formation of the (total) differential.

The universal derivation of a finitely generated R-algebra A can be constructed in amore direct manner as follows. We first do the case when A is a polynomial algebra P :=R[x1, . . . , xn]. For any R-derivation D : P →M we have Df =

∑ni=1(∂f/∂xi)Dxi and this

yields (Dx1, . . . , Dxn) ∈ Mn. Conversely, for any n-tuple (m1, . . . ,mn) ∈ Mn, we have anR-derivation D : P → M defined by Df =

∑ni=1(∂f/∂xi)mi. So Dxi can be prescribed


arbitrarily as an element of M . But to give an element of Mn is to give a P -homomorphismPn,M) and hence ΩP/R is the free P -module generated by dx1, . . . , dxn. Thus the universalR-derivation d : P → ΩP/R, which is given by f 7→

∑ni=1(∂f/∂xi)dxi, may be regarded as

the intrinsic way of forming the total differential.Next consider a quotient A := P/I of P , where I ⊂ P is an ideal. If M is an A-module

and D′ : A → M is an R-derivation, then its composite with the projection π : P → A,D = D′π : P → M , is an R-derivation of P with the property that Df = 0 for every f ∈ I.Conversely, every R-derivation D : P → M in an A-module M which is zero on I factorsthrough an an R-derivation D′ : A → M . Note that for any R-derivation D : P → M , itsrestriction to I2 is zero, for if f, g ∈ I, then D(fg) = fDg+ gDf ∈ IM = 0. Now I/I2 isa module over P/I = A and so we obtain a short exact sequence of A-modules

I/I2 → ΩP/R/IΩP/R → ΩA/R → 0.

It follows from our computation of ΩP/R that the middle term is the free A-module gener-ated by dx1, . . . , dxn. So if I is generated by f1, . . . , fm, then ΩA/R can be identified withthe quotient of

∑ni=1Adxi by the A-submodule generated by the A-submodule generated

by the dfj =∑ni=1(∂fj/∂xi)dxi, j = 1, . . . ,m.

Note that if R is a noetherian ring, then so is A (by the Hilbert basis theorem) and sinceΩA/R is a finitely generated A-module, it is noetherian as an A-module. This applies forinstance to the case when R = k and A = k[X] for some affine variety X. We then writeΩ(X) for Ωk[X]/k.

EXERCISE 46. Prove that ΩA/R behaves well with localization: if S ⊂ A is a multi-plicative subset, then every R-derivation with values is some A-module M extends natu-rally to an R-derivation of S−1A with values in S−1M . Prove that we have a natural mapS−1ΩA/R → ΩS−1A/R and that this map is a A-homomorphism.

For an affine variety X and p ∈ X, we write ΩX,p for ΩOX,p/k. The preceding exerciseimplies that ΩX,p is the localization of Ω(X) at p: ΩX,p = (k[X]− pp)

−1Ω(X).

EXERCISE 47. Let X be an affine variety and let p ∈ X. Show that the Zariski tangentspace of X at p can be understood (and indeed, be defined) as the space of k-derivations ofOX,p with values in k, where we regard k as a OX,p-module via OX,p/mX,p ∼= k. Prove thatthis identifies its dual, the Zariski cotangent space, with ΩX,p/mX,pΩX,p.

EXERCISE 48. Show (perhaps with the help of the preceding exercises) that if p ∈ An,then ΩOAn,p/k is the free OAn,p-module generated by dx1, . . . , dxn and that if X is an affinevariety in An that has p a regular point, then ΩX,p is a free OX,p-module of rank dimOX,p.

We must be careful with this construction when dealing with formal power series rings.For instance, as we have seen, the completion of the local k-algebra OA1,0 with respect to itsmaximal ideal is k[[x]] and the embedding of OA1,0 → k[[x]] is given by Taylor expansion.A k-derivation D of k[[x]] with values in some k[[x]]-module is not determined by Dx. Allit determines are the values on the k-subalgebra OA1,0. This issue disappears however if werequire that D is continuous for the (x)-adic topology, for this then means that D commuteswith (infinite) formal series summation: D(

∑∞r=0 crx

r) =∑∞r=0 crrx

r−1Dx.

EXERCISE 49. Prove that the composite d : A → ΩA/R → (ΩA/R)I factors through aderivation dI : AI → (ΩA/R)I and prove that it is universal among all the R-derivations ofA in R-modules that are complete for the I-adic topology (i.e., for which M = MI).

So if p is a regular point of an affine variety X, then ΩX,p is a free OX,p-module of rankdimOX,p.

11. The notion of a variety

We begin with a ‘predefinition’.

11. THE NOTION OF A VARIETY 53

DEFINITION 11.1. A prevariety is a topological space X endowed with a sheafOX of k-valued functions such that X can be covered by finitely many open subsetsU such that (U,OX |U) is an affine variety. Given prevarieties (X,OX) and (Y,OY ),then a morphism of prevarieties f : (X,OX)→ (Y,OY ) is simply a morphism in thecategory of spaces endowed with a sheaf OX of k-valued functions: f is continuousand for every open V ⊂ Y , composition with f takes OY (V ) to OX(f−1V ).

We often designate a prevariety and its underlying topological space by thesame symbol, a habit which rarely leads to confusion. The composite of two mor-phisms is evidently a morphism so that we are dealing here with a category. Theprefix ‘pre’ in prevariety refers to the fact that we have not imposed a separationrequirement which takes the place of the Hausdorff property that one normallyimposes on a manifold (see Example 11.5 below).

Let X be a prevariety. By assumption X is covered by finitely many affine opensubvarieties Uii∈I (I is a finite index set). Suppose κi is an isomorphism of Uionto an affine variety Xi which is given as a closed subset in some Ani . ThenXi,j := κi(Ui ∩ Uj) is an open subset of Xi and κi,j := κjκ

−1i is an isomorphism of

Xi,j onto Xj,i ⊂ Xj . We can recover X from the disjoint union∐i∈I Xi by means

of a gluing process, for if we use κi,j to identify Xi,j with Xj,i for all i, j we getback X. The collection (Ui, κi)i∈I is called an affine atlas for X and κi,j is calleda transition map.

EXERCISE 50. Let (X,OX) be a prevariety.

(i) Prove that X is a noetherian space.(ii) Prove that X contains an open-dense subset which is affine.

(iii) Let Y ⊂ X be locally closed (i.e., the intersection of a closed subset withan open subset). Prove that Y is in natural manner a prevariety in such amanner that the inclusion Y ⊂ X is a morphism of prevarieties.

Much of what we did for affine varieties extends in a straightforward mannerto this more general context. Here are some examples.

Rational functions. A rational function f : X 99K k is defined as before: it isrepresented by a regular function on a subset of X that is open-dense in its set ofclosed points and two such represent the same rational function if they coincide ona nonempty open-dense subset in their common domain of definition.

Function field and dimension. When X is irreducible, the rational functions onX form a field k(X), the function field of X and for an open nonempty affine opensubset U ⊂ X, we have k(X) = k(U) = Frac(O(U)) (but we will see that it is nottrue in general that k(X) = Frac(O(X))). In particular, any nonempty affine opensubset of X has dimension trdegk k(X). According to Exercise 36, this is then alsothe (Krull) dimension of X.

Rational and dominant maps. Similarly, if X and Y are prevarieties, then a ra-tional map f : X 99K Y is represented by morphism from a nonempty open-densesubset of X to Y with the understanding that two such define the same map ifand only if they coincide on a nonempty open-dense subset. If some representativemorphism has dense image in Y , then f is said to be dominant. If in addition both


X and Y are irreducible, then f induces a field extension f∗ : k(Y ) → k(X). Con-versely, a k-linear field embedding k(Y ) → k(X) determines a dominant rationalmap X 99K Y . If U ⊂ X is open and nonempty, then k(U) = k(X) and the inclu-sion is a birational equivalence.

Finite morphisms. A morphism f : X → Y between prevarieties is called finiteif it is locally so over Y , that is, if we can cover Y by open affine subsets V with

the property that f−1V is affine and the restriction f−1Vf−→ V is finite. According

to Exercise 34 a finite morphism between affine varieties is closed. It then followsthat a finite morphism between prevarieties is also closed.

Regular and singular points. Since the notion of a regular point is a local one, itautomatically carries over to this setting. The regular locus of X is an open-densesubset Xreg of X.

The product of two prevarieties. Our discussion of the product of closed subsetsof affine spaces dictates how we should define the product of two prevarieties Xand Y : if (p, q) ∈ X × Y , then let p ∈ U ⊂ X and q ∈ V ⊂ Y be affine openneighborhoods of the components. We require that the topology on U × V bethe Zariski topology so that a basis of neighborhoods of (p, q) consists of the loci(U × V )h where a h ∈ O(U)⊗O(V ) with h(p, q) 6= 0 is nonzero. We of course alsorequire that OX×Y ((U × V )h) = (O(U)⊗O(V ))[1/h].

EXERCISE 51. Prove that this product has the usual categorical characteriza-tion: the two projections X × Y → X and X × Y → Y are morphisms and if Zis a prevariety, then a pair of maps (f : Z → X, g : Z → Y ) defines a morphism(f, g) : Z → X × Y if and only both f and g are morphisms.

The Hausdorff property is not of a local nature: a non-Hausdorff space can verywell be locally Hausdorff. The standard example is the space X obtained from twocopies of R by identifying the complement of 0 in either copy by means of theidentity map. Then X is locally like R, but the images of the two origins cannot beseparated. A topological spaceX is Hausdorff precisely when the diagonal ofX×Xis a closed subset relative to the product topology. As we know, the Zariski topologyis almost never Hausdorff. But on the other hand, the selfproduct of the underlyingspace has not the product topology either and so requiring that the diagonal isclosed is not totally unreasonable a priori. In fact, imposing this condition turnsout to be the appropriate way of avoiding the pathologies that can result from anunfortunate choice of gluing data.

DEFINITION 11.2. A k-prevariety X is called a k-variety if the diagonal is closedin X ×X (where the latter has the Zariski topology as defined above). A subset ofa variety X is called a subvariety if it is open in a closed subset of X. We say thata morphism of varieties f : (X,OX)→ (Y,OY ) is an immersion if it defines an iso-morphism onto a subvariety of Y , that is, is the composite of such an isomorphismand an inclusion.

Strictly speaking, the last part of this definition only makes sense after we haveobserved that a subvariety is in a natural manner a variety that makes the inclusiona morphism. We leave this as an exercise (see also Exercise 50).

The proof of the following assertion is also left as an exercise.

12. CONSTRUCTIBLE SETS 55

PROPOSITION 11.3. The product of two varieties is a variety.

EXAMPLE 11.4. The diagonal in An × An is closed, so An is a variety. Thisimplies that the same is true for any quasi-affine subset of An. Hence a quasi-affineprevariety is in fact a variety.

EXAMPLE 11.5. The simplest example of a prevariety that is not a variety is theobvious generalization of the space described above: let X be obtained from twocopies A1

+ and A1− of A1 by identifying A1

+ r 0 with A1− r 0 by means of the

identity map. If o± ∈ X denotes the image of origin of A1±, then (o+, o−) ∈ X ×X

lies in the closure of the diagonal, but is not contained in the diagonal.

EXAMPLE 11.6. Let f : X → Y be a morphism of varieties. Consider the graphof f , Γf := (x, y) ∈ X × Y : x ∈ U, y = f(x). It is easy to see that Γf isa subvariety of X × Y . The map x ∈ X 7→ (x, f(x)) ∈ Γf and the projectionΓf → X are regular and each others inverse. So they define an isomorphismΓf → X. Notice that via this isomorphism f appears as a projection mapping:(x, y) ∈ Γf 7→ y ∈ Y .

EXERCISE 52. The goal of this exercise is to show that U := A2 r (0, 0) is notaffine. Let Ux ⊂ U resp. Uy ⊂ U be the complement of the x-axis resp. y-axis sothat U = Ux ∪ Uy.

(a) Prove that Ux is affine and that O(Ux) = k[x, y][1/x].(b) Prove that every regular function on U extends to A2 so that O(U) =

k[x, y].(c) Show that U is not affine.

This exercise also leads to an interesting example. Let X be obtained as theobvious generalization of Example 11.5 where A1 is replaced by A2 so that X iscovered by two copies of A2 (appearing as open affine subsets) whose intersection isa copy of A2r(0, 0) (which is not affine). Hence, on a prevariety the intersectionof two affine subsets need not be affine. This cannot happen on a variety and thatis one of the reasons why we like them:

PROPOSITION 11.7. Let X be a variety. Then for any pair U,U ′ of affine opensubsets ofX, U∩U ′ is also an affine open subset ofX andOX(U∩U ′) is as a k-algebragenerated by OX(U)|U ∩ U ′ and OX(U ′)|U ∩ U ′.

PROOF. First note that U × U ′ is an affine open subset of X × X. Since thediagonal ∆(X) of X ×X is closed, its intersection with U ×U ′ is a closed subset ofU × U ′ and hence affine. But the diagonal map sends U ∩ U ′ isomorphically ontothis intersection (the inverse being given by one of the projections) and so U ∩ U ′is affine. Since the diagonal defines a closed embedding U ∩U ′ → U ×U ′ of affinevarieties, the map k[U ]⊗ k[U ′] ∼= k[U ×U ′]→ k[U ∩U ′] is onto. Since the image off ⊗ f ′ ∈ k[U ]⊗ k[U ′] equals f|U∩U ′ · f ′|U∩U ′ , the last assertion follows.

12. Constructible sets

The image of a morphism of varieties need not be a variety as the followingsimple example shows.

EXAMPLE 12.1. Consider the morphism f : A2 → A2, (x1, x2) 7→ (x1, x1x2).A point (y1, y2) ∈ A2 is of the form (x1, x1x2) if and only if y2 is a multiple of y1.


This is the case precisely when y1 6= 0 or when y1 = y2 = 0. So the image of f isthe union of the open subset y1 6= 0 and the singleton (0, 0). This is not a locallyclosed subset, but the union of two such. This turns out to represent the generalsituation and the following definition will help us to express this fact.

DEFINITION 12.2. A subset of variety is called constructible if it can be writtenas the union of finitely many (locally closed) subvarieties.

THEOREM 12.3. Let f : X → Y be a morphism of varieties. Then f takesconstructible subsets of X to constructible subsets of Y . In particular, f(X) is con-structible.

We first show that the theorem follows from the following proposition.

PROPOSITION 12.4. Let f : X → Y be a morphism of varieties withX irreducible.Then f(X) contains a nonempty open subset of its closure.

PROOF THAT PROPOSITION 12.4 IMPLIES THEOREM 12.3. A constructible sub-set is a finite union of irreducible subvarieties and so it is clearly enough to provethat the image of each of these is constructible. In other words, it suffices to showthat the image of a morphism f : X → Y of varieties with X irreducible is con-structible. We prove this with induction on the dimension of X. According toProposition 12.4 the closure of f(X) contains a nonempty open subset U such thatf(X) ⊃ U . It is clear that f−1U is a nonempty open subset of X. If Z := X−f−1U ,then f(X) = U ∪ f(Z). The irreducible components of Z have smaller dimensionthan X and so f(Z) is constructible by induction.

PROOF OF PROPOSITION 12.4. Since Y is covered by finitely many open affinesubsets Vj ⊂ Y we may (upon replacing f by its restriction to f−1Vj) assumewithout loss of generality that Y is affine. For a similar reason we may assumethat X is affine and hence is closed in some An. Then the graph of f identifies Xwith a closed subset of An × Y so that f becomes the restriction of the projectionπ : An×Y → Y to X. In other words, we need to prove that for every closed subsetX ⊂ An × Y , π(X) contains a nonempty open subset of its closure in Y . Since wecan factor πY in an obvious manner into successive line projections (by forgettingthe last coordinate)

An × Y → An−1 × Y → · · · → A1 × Y → Y,

it suffices to do the case n = 1, so that now f factors as X ⊂ A1 × Y π−→ Y . Uponreplacing Y by π(X), we may then also may assume that π|X : X → Y is dominant.When X = A1×Y , there is nothing to show. Otherwise I(X) ⊂ k[A1×Y ] = k[Y ][t]contains a nonzero g ∈ k[Y ][t]. Write g = a0t

N + a1tN−1 + · · ·+ aN with ai ∈ k[Y ]

and a0 nonzero. Since π|X is dominant, we have I(X)∩ k[Y ] = 0 and so N > 0.Hence g/a0 yields for the image of t in k[X][1/a0] = k[Xa0 ] an equation of integraldependence over k[Ya0 ]. This implies that Xa0 is finite over Ya0 and so its image inYa0 is closed by Exercise 34. Since this image is also dense in Ya0 , it follows thatπ(X) contains the open-dense subset Ya0 .

CHAPTER 2

Projective varieties

1. Projective spaces

Two distinct lines in the plane intersect in a single point or are parallel. Inthe last case one would like to say that the lines intersect at infinity so that thestatement becomes simply: two distinct lines in a plane meet in a single point.There are many more examples of geometric configurations for which the specialcases disappear by the simple remedy of adding points at infinity. A satisfactoryapproach to this which makes no a priori distinction between ordinary points andpoints at infinity involves the notion of a projective space.

Given a finite dimensional k-vector space V , then we denote by P(V ) the col-lection of its 1-dimensional linear subspaces. Observe that any linear injectionJ : V → V ′ of vector spaces induces an injection P(J) : P(V ) → P(V ′) (in generalP(J) only makes sense on P(V ) r P(ker(J))). In particular, when J is an isomor-phism isomorphism, then P(J) is a bijection. The following definition makes thisnotion slightly more abstract by suppressing the vector space as part of the data.

DEFINITION 1.1. A projective space of dimension n over k is a set P endowedwith an extra structure that can be given by a pair (V, `), where V is k-vector spaceof dimension n + 1 and ` : P → P(V ) is a bijection, where it is understood thatanother such pair (V ′, `′) defines the same structure if and only if there exists ak-linear isomorphism J : V → V ′ such that `′ = P(J)`′. (We are in fact saying thatthus is defined an equivalence relation on the collection of such pairs and that aprojective structure is given by an equivalence class.)

So for a finite dimensional k-vector space V , the identity map of P(V ) makesP(V ) in a natural manner a projective space. It is called the projective space asso-ciated to V . When V = kn+1 we often write Pn or Pnk and call it simply projectiven-space (over k). The difference between a projectivized vector space and an ab-stract projective space is perhaps elucidated by the following exercise.

EXERCISE 53. Prove that the linear isomorphism φ in Definition 1.1 is uniqueup to scalar multiplication. Conclude that a projective space P determines a vectorspace up to scalar multiplication. Illustrate this by showing that for a 2-dimensionalvector space V we have a canonical isomorphism P(V ) ∼= P(V ∗), but that there isno canonical isomorphism between V and V ∗.

Let P be a projective space of dimension n. We can of course describe itsstructure by a pair (kn+1, `). This gives rise to a ‘coordinate system’ on P as follows:if we denote the coordinates of kn+1 by (T0, . . . , Tn), then every point p ∈ P(V ) isrepresentable as a ratio [p0 : · · · : pn] of n + 1 elements of k that are not all zero:choose a generator p of the line `(p) and let pi = Ti(p). Any other generator is ofthe form λp with λ ∈ k r 0 and indeed, [λp0 : · · · : λpn] = [p0 : · · · : pn]. This is

57

58 2. PROJECTIVE VARIETIES

why [T0 : · · · : Tn] is called a homogeneous coordinate system on P(V ) even thoughan individual Ti is not a function on P(V ) (but the ratios Ti/Tj are, albeit that fori 6= j they are not everywhere defined).

DEFINITION 1.2. Given a projective space P of dimension n over k, then asubset Q of P is said to be linear subspace of dimension d if, for some (and henceany) pair (V, `) as above, there exists a linear subspace VQ ⊂ V of dimension d+ 1such that `(Q) is the collection of 1-dimensional linear subspaces of VQ.

A map j : P → P ′ between two projective spaces over k is said to be linearmorphism if for corresponding structural data (V, `) and (V ′, `′) for P resp. P ′ thereexists a linear injection J : V → V ′ such that `′ = P(J)`.

So a linear subspace has itself the structure of a projective space and its inclu-sion in the ambient projective space is a linear morphism. Conversely, the image ofa linear morphism is linear subspace.

A linear subspace of dimension one resp. two is often called a line resp. a planeand a linear subspace of codimension one (= of dimension one less than the ambi-ent projective space) is called a hyperplane. It is now clear that two distinct linesin a plane intersect in a single point: this simply translates the fact that the inter-section of two distinct linear subspaces of dimension two in a three dimensionalvector space is of dimension one.

We put on a projective space P the structure of a k-variety as follows. A homo-geneous coordinate system [T0, . . . , Tn] for P defines a chart for every i = 0, . . . , n:if PTi ⊂ P is the hyperplane complement defined by Ti 6= 0, then

κi : PTi∼=−→ An, [T0 : · · · : Tn] 7→ (T0/Ti, . . . , Ti/Ti, . . . , Tn/Ti),

is a bijection (chart) with inverse

κ−1i : (a1, . . . , an) ∈ An 7→ [a1 : · · · : ai : 1 : ai+1 : · · · : an] ∈ U.

Clearly, ∪ni=0PTi = P . We show that the collection of charts PTi , κini=0 can serveas an affine atlas for P . The coordinate change for a pair of charts, say for κnκ−1

0

is as follows: the image of PT0∩PTn under κ0 resp. κn is the open subset Anxn resp.

Anx1of An and the transition map is

κnκ−10 : Anxn → Anx1

, (a1, a2, . . . , an) 7→ (1/an, a1/an, . . . , an−1/an),

and hence an isomorphism of affine varieties with inverse κ0κ−1n . An atlas thus

obtained from a homogeneous coordinate system (T0, . . . , Tn) will be called a stan-dard atlas for P ; it gives P the structure of a prevariety (P,OP ): U ⊂ P is openif and only if for i = 0, . . . , n, κi(U ∩ PTi) is open in An and f ∈ OP (U) if andonly if fκ−1

i ∈ O(κi(U)). One can easily check that this structure is in fact that ofa k-variety and that it is independent of the coordinate system. We will not do thishere as we will give in Section 3 a more direct proof of these assertions.

Any hyperplane H ⊂ P can be given as T0 = 0, where [T0 : · · · : Tn] is ahomogeneous coordinate system on P and so its complement U = P r H = PT0

is isomorphic to An. This can also (and more intrinsically) be seen without thehelp of such a coordinate system. Let the projective structure on P be given bythe pair (V, `). Then the hyperplane H corresponds to a hyperplane VH ⊂ V andU corresponds to the set of 1-dimensional linear subspaces of V not contained in

2. THE ZARISKI TOPOLOGY ON A PROJECTIVE SPACE 59

H. If e ∈ V ∗ is a linear form whose zero set is VH , then A = e−1(1) is an affinespace for VH (it has VH as its vector space of translations). Assigning to v ∈ A the1-dimensional linear subspace spanned by v defines a bijection A ∼= U that puts onU a structure of an affine space. This structure is easily checked to be independentof (V, `, φ).

We could also proceed in the opposite direction and start with an affine space A andrealize it as the hyperplane complement of a projective space. For this consider the vectorspace F (A) of affine-linear functions on A and denote by e ∈ F (A) the function on A thatis constant equal to 1. Then e−1(1) is an affine hyperplane in F (A)∗. Any a ∈ A defines alinear form on F (A) by evaluation: f ∈ F (A) 7→ f(a) ∈ k. Note that this form takes thevalue 1 on e so that we get in fact a map A → e−1(1). It is not hard to check that this isan affine-linear isomorphism and so the projective space A := P(F (A)∗) can serve as theprojective completion of A. Paraphrasing the classical Renaissance painters, we might saythat ArA consists of “points at infinity” of A; such a point can be given by an affine line inA with the understanding that parallel lines define the same point at infinity.

2. The Zariski topology on a projective space

We begin with giving a simpler characterization of the Zariski topology on aprojective space. Let P be a projective space of dimension n over k and let [T0 :· · · : Tn] be a homogeneous coordinate system for P . Suppose F ∈ k[X0, . . . , Xn] ishomogeneous of degree d so that F (tT0, . . . , tTn) = tdF (T0, . . . , Tn) for t ∈ k. Theproperty of this being zero only depends on [T0 : · · · : Tn] and hence the zero set ofF defines a subset of P . We shall denote this subset by Z[F ] and its complementP r Z[F ] by PF . We will show in the next section that PF is in fact affine.

PROPOSITION 2.1. The collection PF F , where F runs over the homogeneouspolynomials in k[X0, . . . , Xn], is a basis for the Zariski topology on P . This topologyis independent of the choice of our homogeneous coordinate system [T0, . . . , Tn] and(so) every linear chart is a homeomorphism onto An that identifies the sheaf of regularfunctions on its domain with OAn . If G ∈ k[X0, . . . , Xn] is homogeneous of the samedegree as F , then G/F defines a regular function on PF .

PROOF. We first observe that the obvious equality PF ∩PF ′ = PFF ′ implies thatthe collection PF F is a basis of a topology. The independence of this topology ofthe coordinate choice results from the observation that under a linear substitutiona homogeneous polynomial transforms into a homogeneous polynomial.

Let us verify that this is the Zariski topology defined earlier. First note thatthe domain of each member κi : PTi

∼= An is of the standard atlas is also a basiselement (hence open) for the topology in question. So we must show that each κi isa homeomorphism. If F ∈ k[T0, . . . , Tn] is homogeneous of degree d, then κi(PF ∩PTi) = Anfi , where fi(y1, . . . , yn) := F (y1, . . . , yi, 1, yi+1, . . . , yn) and so κi is open.Conversely, if f ∈ k[y1, . . . , yn] is nonzero of degree d, then its ‘homogenization’F (T0, . . . , Tn) := T di f(T1/T0, . . . Ti/Ti . . . Tn/T0) is homogeneous of degree d andκ−1i (Anf ) = PF ∩ PTi . So κi is also continuous.

For the last statement first observe that G/F indeed defines a function on PF(think of it as regular function on An+1

F that is constant under scalar multiplica-tion). Its pull-back under κi is gi(y1, . . . , yn)/fi(y1, . . . yn), where gi(y1, . . . , yn) :=G(y1, . . . , 1, . . . yn), which is indeed regular on Anfi .


EXERCISE 54. Let 0 6= F ∈ k[X0, . . . , Xn] be homogeneous of degree d. Provethat every function PF → k of the form G/F r, with r ≥ 0 and G homogeneous ofthe same degree as F r, is regular and that conversely, every regular function on PFis of this form.

In order to discuss the projective analogue of the (affine) I ↔ Z correspon-dence, we shall need the following notions from commutative algebra.

DEFINITION 2.2. Let R be a ring. A (nonnegatively) graded R-algebra is an R-algebra A whose underlying additive group comes with a direct sum decompositionA• = ⊕∞k=0Ad into R-submodules such that the product maps Ad × Ae in Ad+e, orequivalently, is such that

∑∞d=0Adt

d is an R-subalgebra of A[t]. An ideal I of suchan algebra is said to be homogeneous if it is the direct sum of its homogeneous partsId := I ∩ Ad. We shall say that an ideal I of A is properly homogeneous1 if it ishomogeneous and contained in the ideal A+ := ⊕d≥1Ad.

IfA is a graded ring, then clearlyA0 is a subring ofA so that we may also regardA as a graded A0-algebra. In fact, A0 is an R-subalgebra of A and A acquires itsR-algebra structure via the one on A0.

If I is a homogeneous ideal of A, then A/I = ⊕∞d=0Ad/Id is again a gradedR-algebra. We will be mostly concerned with the case when A0 = k, so that A+ isthen a maximal ideal (and the only one that is homogeneous).

LEMMA 2.3. If I, J are homogeneous ideals of a graded ring R, then so are I ∩ J ,IJ , I + J and

√I. Moreover, a minimal prime ideal of R is a graded ideal.

PROOF. The proofs of the statements in the first sentence are not difficult andso we omit them. As to the last, it suffices to show that if p ⊂ A is a prime ideal,then the direct sum of its homogeneous parts, p• := ⊕n(p ∩ An) is also a primeideal. Indeed, suppose a, b ∈ A nonzero and such that ab ∈ p•. Let ak resp. bl bethe highest degree part of a resp. b. We prove with induction on k + l that a or bis in p•. Since we have akbl = (ab)k+l ∈ pk+l ⊂ p, it follows that ak ∈ p or bl ∈ p.Let us assume that ak ∈ p. Then ak ∈ p• and so (a − ak)b ∈ p•. By our inductionassumption, then a− ak or b is in p•. It follows that a or b is in p•.

The prime example of a graded k-algebra is furnished by a vector space Vof finite positive dimension (n + 1, say), which we consider as an affine variety,but (in contrast to an affine space) one of which we remember that it comes withthe action of the multiplicative group of k by scalar multiplication. The space ofF ∈ k[V ] that are homogeneous of degree d in the sense that F (tv) = tdF (v)for all v ∈ V and t ∈ k make up a k-linear subspace k[V ]d of finite dimension.This makes k[V ] = ⊕d≥0k[V ]d a graded k-algebra and the decomposition is theone into eigenspaces with respect to the action of scalar multiplication. (A choiceof basis (T0, . . . , Tn) of V ∗ identifies V with An+1 and then k[V ]d becomes thespace of homogeneous polynomials in (T0, . . . , Tn) of degree d.) Note that for anyF ∈ k[V ]d the zero set Z(F ) ⊂ V is invariant under scalar multiplication. Thisis still true for an intersection of such zero sets, in other words, for a properlyhomogeneous ideal I• ⊂ k[V ]+, Z(I) ⊂ V is a closed subset of V that is invariantunder scalar multiplication. Such a closed subset is called an affine cone. The originis called the vertex of that cone. Since k[V ]+ defines the vertex, we always have

1This is not a generally adopted terminology.

2. THE ZARISKI TOPOLOGY ON A PROJECTIVE SPACE 61

0 ∈ Z(I). The intersection of the Z[F ], with F ∈ ∪d≥1Id defines a closed subsetZ[I] of P(V ), whose points correspond to the one-dimensional subspaces of V thatare contained in Z(I).

LEMMA 2.4. For an affine cone C ⊂ V , I(C) is a properly homogeneous radicalideal of k[V ] and hence defines a closed subset P(C) := Z[I(C)] of P(V ).

PROOF. Let F ∈ I(C). Write F =∑d≥0 Fd. We must show that each homoge-

neous component of Fi lies in I(C). As C is invariant under scalar multiplication,the polynomial F (tv) =

∑d≥1 t

dFd(v) (as an element of k[A1 × V ]) vanishes onA1 × C in A1 × V . Clearly the zero set of I(C)[t] is A1 × C ⊂ A1 × V and sincethe quotient is k[V ][t]/I(C)[t] = k[C][t] is reduced, we have I(A1×C) = I(C)[t]. Itfollows that Fd ∈ I(C) for all d.

Conversely, given a closed subset X ⊂ P(V ), let for d ≥ 1, IX,d be the set ofF ∈ k[V ]d for which X ⊂ Z[F ] and put IX,0 = 0. Then IX,d is a k-vector space andIX,d · k[V ]e ⊂ IX,d+e so that IX := ⊕d≥1IX,d is a properly homogeneous ideal ofk[V ]. It is also a radical ideal and we have X = Z[IX ]. So Cone(X) := Z(IX) is thecone in V that as a set is just the union of the 1-dimensional linear subspaces of Vparameterized by X.

COROLLARY 2.5. The maps C 7→ P(C) and X 7→ I(X) set up bijections between(i) the collection of affine cones in V , (ii) the collection of closed subsets of Pn, and(iii) the collection of properly homogeneous radical ideals contained in k[V ]. Thisrestricts to bijections between (i) the collection of irreducible affine cones in V strictlycontaining 0, (ii) the collection of irreducible subsets of P(V ) and (iii) the collectionof homogeneous prime ideals of k[V ] strictly contained in k[V ]+.

PROOF. The first assertion sums up the preceding discussion. The last assertionfollows from the observation that the degenerate cone 0 ⊂ V corresponds to theempty subset of P(V ) and to the homogeneous ideal k[V ]+.

DEFINITION 2.6. The homogeneous coordinate ring of a closed subset X of P(V )is the coordinate ring of the affine cone over X, k[Cone(X)] = k[V ]/IX , endowedwith the grading defined by k[Cone(X)]d = k[T0, . . . Tn]d/IX,d. More generally, ifY is an affine variety, and X is a closed subset of P(V )× Y , then the homogeneouscoordinate ring of X relative to Y of a closed subset is the coordinate ring of thecorresponding closed cone in V × Y over Y , endowed with the grading defined bythe coordinates of V .

EXERCISE 55. Let A be a graded ring.(b) Prove that if I is a prime ideal in the homogeneous sense: if rs ∈ I for

some r ∈ Ak, s ∈ Al implies r ∈ I or s ∈ I, then I is a prime ideal.(c) Prove that the intersection of all homogeneous prime ideals of A• is its

ideal of nilpotents.

Since k[X0, . . . , Xn] is a noetherian ring, any ascending chain of homogeneousideals in this ring stabilizes. This implies that any projective space (and hence anysubset of it) is noetherian. In particular, every subset of a projective space has afinite number of irreducible components whose union is all of that subset.

EXERCISE 56. Let S• be a graded k-algebra that is reduced, finitely generatedand has S0 = k.


(a) Prove that S• is as a graded k-algebra isomorphic to the homogeneouscoordinate ring of a closed subset Y .

(b) Prove that under such an isomorphism, the homogeneous radical idealscontained in the maximal ideal S+ := ⊕d≥1Sd correspond to closed sub-sets of Y under an inclusion reversing bijection: homogeneous idealsstrictly contained in S+ and maximal for that property correspond topoints of Y .

(c) Suppose S• a domain. Show that a fraction F/G ∈ Frac(S•) that is ho-mogeneous of degree zero (F,G ∈ Sd for some d and G 6= 0 defines afunction on Ug.

EXERCISE 57. Let Y be an affine variety.(a) Show that a homogeneous element of the graded ring k[Y ][T0, . . . , Tn]

defines a closed subset of Y × Pn as its zero set.(b) Prove that every closed subset of Y ×Pn is an intersection of finitely many

zero set of homogeneous elements of k[Y ][T0, . . . , Tn].(c) Prove that we have a bijective correspondence between closed subsets of

Y × Pn and the homogeneous radical ideals in k[Y ][T0, . . . , Tn]+.

3. The Segre embeddings

First we show how a product of projective spaces can be realized as a closedsubset of a projective space. This will imply among other things that a projectivespace is a variety. Consider the projective spaces Pm and Pn with their homoge-neous coordinate systems [T0 : · · · : Tm] and [W0 : · · · : Wn]. We also consider aprojective space whose homogeneous coordinate system is the set of matrix coeffi-cients of an (m+1)×(n+1)-matrix [Z00 : · · · : Zij : · · · : Zmn]; this is just Pmn+m+n

with an unusual indexing of its homogeneous coordinates.

PROPOSITION 3.1 (The Segre embedding). The map f : Pm × Pn → Pmn+m+n

defined by Zij = TiWj , i = 0, . . . ,m; j = 0, . . . , n is an isomorphism onto a closedsubset of Pmn+m+n. If m = n, then the diagonal of Pm × Pm is the preimage of thelinear subspace of Pm2+2m defined by Zij = Zji and hence is closed in Pm × Pm.

PROOF. For the first part it is enough to show that for every chart domainPmn+m+nZij

of the standard atlas of Pmn+m+n, f−1Pmn+m+nZij

is open in Pm × Pn andis mapped by f isomorphically onto a closed subset of Pmn+m+n

Zij. For this purpose

we may (simply by renumbering) assume that i = j = 0. So then Pmn+m+nZ00

⊂Pmn+m+n is defined by Z00 6= 0 and is parametrized by the coordinates zij :=Zij/Z00, (i, j) 6= (0, 0). It is clear that f−1Pmn+m+n

Z00is defined by T0W0 6= 0.

This is just PmT0× PnW0

and hence is parametrized by x1 := T1/T0, . . . , xm :=Tm/T0 and y1 := W1/W0, . . . , yn := Wn/W0. In terms of these coordinates,f : f−1Pmn+m+n

Z00→ Pmn+m+n

Z00is given by zij = xiyj , where (i, j) 6= (0, 0) and

where we should read 1 for x0 and y0. So among these are zi0 = xi and z0j = yjand since these generate k[Am × An] = k[x1, . . . , xm, y1, . . . , yn], f indeed restrictsto a closed immersion f−1Pmn+m+n

Z00→ Pmn+m+n

Z00.

In case m = n, we must also show that the condition TiWj = TjWi for 0 ≤i < j ≤ m implies that [T0 : · · · : Tm] = [W0 : · · · : Wm], assuming that not all Tiresp. Wj are zero. Suppose Ti 6= 0. Since Wj = (Wi/Ti).Tj for all j, it follows thatWi 6= 0 and so [W0 : · · · : Wm] = [T0 : · · · : Tm].

4. BLOWING UP AND PROJECTIONS 63

COROLLARY 3.2. A projective space over k is a variety.

PROOF. Proposition 3.1 shows that the diagonal of Pm × Pm is closed.

DEFINITION 3.3. A variety is said to be projective if it is isomorphic to a closedirreducible subset of some projective space. A variety is called quasi-projective if isisomorphic to an open subset of some projective variety.

COROLLARY 3.4. Every irreducible closed (resp. locally closed) subset of Pn is aprojective (resp. quasi-projective) variety. The collection of projective (resp. quasi-projective) varieties is closed under a product.

PROOF. The first statement follows from Proposition 11.3 of Ch. 1 and thesecond from Proposition 3.1.

EXERCISE 58. (a) Prove that the image of the Segre embedding is thecommon zero set of the homogenenous polynomials ZijZkl − ZilZkj .

(b) Show that for every (p, q) ∈ Pm×Pn the image of p×Pn and Pm×qin Pmn+m+n is a linear subspace.

(c) Prove that the map Pn → P(n2+3n)/2 defined by Zij = TiTj , 0 ≤ i ≤ j ≤ nis an isomorphism on a closed subset defined by quadratic equations. Findthese equations for n = 2.

(d) As a special case we find that the quadric hypersurface in P3 defined byZ0Z1 − Z2Z3 = 0 is isomorphic to P1 × P1. Identify in this case the twosystems of lines on this quadric.

EXERCISE 59 (Intrinsic Segre embedding). Let V and W be finite dimensionalk-vector spaces. Describe the Segre embedding for P(V ) × P(W ) intrinsically as amorphism P(V )× P(W )→ P(V ⊗W ).

4. Blowing up and projections

By way of introduction we first explain the blowup of a linear subspace of anaffine space. Fix an integer 1 ≤ c ≤ n and denote by Y ⊂ An the linear codimensionc subspace defined by x1 = · · · = xc = 0. Consider the morphism π : An r Y →Pc−1 defined by π(x1, . . . , xn) = [x1 : · · · : xc]. The graph Γπ of π is the set of((x1, . . . , xn), [T1 : · · · : Tc]) ∈ (An r Y ) × Pc−1 with [T1 : · · · : Tc] = [x1 : · · · : xc].Such points satisfy the equations xiTj = xjTi, 1 ≤ i < j ≤ c. The common zeroset of these equations defines a closed subset of An × Pc−1, called the blowup ofAn along Y and denoted by BlY (An). It is easily seen to be the union of Γπ andY × Pc−1. We claim that Γπ is dense in BlY (An). For this it is of course enoughto show that the closure Γπ of Γπ contains Y × Pc−1. To see this, we note that forinstance BlY (An)∩ (An × Pc−1

T1) is the set of (x1, . . . , xn), [1 : t2 : · · · ; tc]) satisfying

xi = tix1 for i = 2, . . . , c and hence is parametrized by An via

(x1, t2, . . . , tc, xc+1, . . . , xn) 7→ ((x1, t2x1, . . . , tcx1, xc+1, . . . , xn), [1 : t2 : · · · ; tc]).

In terms of this parametrization, Anx1maps onto Γπ∩(Anx1

×Pc−1), whereas Z(x1) ⊂An maps isomorphically onto Y × Pc−1

T1. It follows that Γπ ⊃ Y × Pc−1

T1. Similarly,

Γπ ⊃ Y × Pc−1Ti

for i = 2, . . . , c and hence Γπ ⊃ Y × Pc−1. We call Y × Pc−1

the exceptional divisor of this blowup. The term blowup sometimes also refers toprojection on the first factor, p : BlY (An)→ An, rather than just to its domain.


This construction has a more geometric interpretation. We can think of a pointof Pc−1 as a one dimensional subspace of kc, but here it is better is to think of it as acodimension c− 1 linear subspace of kn = An which contains Y . For then BlY (An)can be regarded as the set of pairs (p, Y ′), where p ∈ An and Y ′ is codimensionc − 1 subspace of An which contains both Y and p and π is then understood asassigning to x ∈ An−Y the linear span of x and Y . The projection onto the secondfactor, BlY (An)→ Pc−1, extends π and is often called the projection away from Y .

REMARK 4.1. The homogeneous coordinate ring of BlY (An) is the graded k[x1, . . . , xn]-algebra k[x1, . . . , xn][T1, . . . , Tc] modulo the ideal generated by the xiTj − xjTi, 1 ≤ i <j ≤ c, with each Ti of degree 1. It admits the following elegant description: consider thehomomorphism of graded k[x1, . . . , xn]-algebras

k[x1, . . . , xn][T1, . . . , Tc]/(xiTj − xjTi, 1 ≤ i < j ≤ c)→∑d≥0

I(Y )dT d, Ti 7→ Txi,

where I(Y ) is the ideal defining Y (so generated by x1, . . . , xc) and I(Y )0 := k[x1, . . . , xn].The right hand side is to be viewed as a k[x1, . . . , xn]-subalgebra of the graded algebrak[x1, . . . , xn][T ], with deg(T ) = 1 (so as a k[x1, . . . , xn]-algebra generated by x1T, . . . , xcT )and can also be written as ⊕d≥0I(Y )d. This is in fact an isomorphism whose inverse is de-fined as follows: a k-basis of the right hand side consists of the monomials xd11 · · ·xdnn T d

with d1 + · · · + dc ≥ d and then the inverse assigns to xd11 · · ·xdnn T d the image in theleft hand side of a monomial xd11 · · ·xdnn T e11 · · ·T ecc , where 0 ≤ ei ≤ di are such that∑i ei = d (check that the image is independent of this choice). The exceptional divi-

sor is defined by the ideal I(Y ) = (x1, . . . , xc). The associated quotient ring is a gradedk[x1, . . . , xn]/I(Y ) = k[Y ]-algebra, which in the first description yields k[Y ][T1, . . . , Tc](this is indeed the homogeneous coordinate ring of Y × Pc−1) and in the second yields∑d≥0 I(Y )d/I(Y )d+1T d ∼= ⊕d≥0I(Y )d/I(Y )d+1. These must of course be isomorphic as

k[Y ]-algebras, but the second description is more canonical in the sense that it identifies theexceptional divisor with the projectivized normal bundle of Y in An. 2

The projection p : BlY (An) → An induces a k[x1, . . . , xn]-algebra homomorphism p∗ :k[x1, . . . , xn]→

∑d≥0 I(Y )dT d that is not the obvious inclusion, but is defined by p∗(xi) =

xiT for i ≤ c and p∗(xi) = xi for i > c. So if H ⊂ An is a hypersurface, and defined byf ∈ k[x1, . . . , xn] say, then p−1H (also called the total transform of H under p) is definedby f(x1T, . . . , xcT, xc+1, . . . , xn), which we view here as an element of

∑d≥0 I(Y )dT d by

writing it as∑d≥0 T

dfd(x1, . . . , xn). The strict transform of H under p is by definition theclosure of H r H ∩ Y in BlY (X) and if m ≥ 0 is such that f0 = · · · = fm−1 = 0 6= fm,then a defining equation for it is (p∗f)str := T−mp∗f =

∑d≥0 T

dfm+d. More generally, ifX ⊂ An is a closed subset, then the ideal generated by the p∗f , f ∈ I(X), defines of courseits total transform p−1(X), whereas the ideal generated by the (p∗f)str, f ∈ I(X), definesthe strict transform of X under p, that is, the closure of X rX ∩ Y in BlY (X).

This blowing up process models a projective analogue that we will discuss next.Let P be a projective space of dimension n andQ ⊂ P a linear subspace of codimen-sion c(= dimP−dimQ). Let us denote by P(P ;Q) the collection of linear subspacesQ′ of P which contain Q as a hyperplane (and so are of dimension dimQ+ 1).

2Nothing stops us now in defining for an arbitrary ring R the ‘blowup’ of an ideal I ⊂ R asthe graded R-algebra ⊕d≥0I

d (where I0 = R) and to regard the graded quotient ⊕d≥0Id/Id+1 as

defining its ‘exceptional divisor’. If Y a closed subset of an affine variety X, then the blowup BlY (X) isdefined by applying this to R = k[X] and I = I(Y ): if I(Y ) has d > 0 generators, then this is a closedsubset of X × Pd−1; for I(Y ) = (0) (so that Y = ∅) we get of course X. When Y is nowhere densein X and f1, . . . , fd generate I(X) in k[X], then BlY (X) is the closure in X × Pd−1 of the graph of[f1 : · · · : fd] : X r Y → Pd−1.

4. BLOWING UP AND PROJECTIONS 65

LEMMA 4.2. The space P(P ;Q) has in a natural manner the structure of a pro-jective space of dimension c − 1 (where dimension −1 means empty). Through ev-ery p ∈ P r Q passes exactly one member of P(P ;Q) and this defines a morphismπQ : P r Q → P(P ;Q). Concretely, if n := dimP and we choose a system of homo-geneous coordinates [T0 : . . . Tn] for P such that Q is given by T0 = · · · = Tc−1, then[T0 : · · · : Tc−1] defines a system of homogeneous coordinates for P(P ;Q) and πQ issimply given by [T0 : · · · : Tn] 7→ [T0 : · · · : Tc−1].

PROOF. Let ` : P ∼= P(V ) be a structural bijection. ThenQ = `−1P(W ) for somelinear subspace VQ ⊂ V and so the Q′ correspond to the linear subspaces VQ′ ⊂ Vwhich contain VQ as a hyperplane. These in turn correspond to the one-dimensionalsubspaces of V/VQ and so we get a bijection P(P ;Q) ∼= P(V/VQ). For anotherchoice of structural bijection (V ′, `′) there must exist a linear isomorphism V ∼= V ′

which then automatically takes VQ onto V ′Q and so induces a linear isomorphismV/VQ ∼= V ′/V ′Q. We thus see that the projective space structure on P(P ;Q) isintrinsically defined. The proof of the last assertion is left to you.

DEFINITION-LEMMA 4.3. The blowup of P alongQ, denoted BlQ P , is the closureof the graph of πQ : P rQ→ P(P ;Q) in P × P(P ;Q) (hence is a projective variety).It enjoys the following properties:

(i) The variety BlQ P is nonsingular and irreducible and the projection on thefirst factor, p1 : BlQ P → P , is an isomorphism over P rQ.

(ii) The preimage over Q is Q × P(P ;Q) and is a nonsingular hypersurface inBlQ P , called the exceptional divisor of the blowup.

(iii) The projection to the second factor defines a locally trivial bundle

p2 : (BlQ P,Q× P(P ;Q))→ P(P ;Q)

of pairs of projective spaces of dimension 1+dimQ and dimQ. To be precise,if U ⊂ P(P ;Q) is a hyperplane complement (hence an affine space), thenthere exists a linear subspace Q′ ⊂ P which contains Q as a hyperplane andan isomorphism p−1

2 U ∼= Q′ × U which is the identity on Q× U and whosesecond component is given by p2.

PROOF. We use a homogeneous coordinate system [T0 : · · · : Tn] for P as above(so that Q is given by T0 = · · · = Tc−1 = 0). If we denote the correspondingcoordinate system for P(P ;Q) by [S0 : · · · : Sc−1], then the graph of πQ in P ×P(P ;Q) is given by the pairs ([T0 : · · · : Tn], [S0 : · · · : Sc−1]) with (T0, . . . , Tc−1) 6=(0, . . . , 0) and [T0 : · · · : Tc−1] = [S0 : · · · : Sc−1]. The last proportionality propertyis equivalent to: TiSj = TjSi for all 0 ≤ i < j < c. Let Γ be the closed subset ofP × P(P ;Q) defined by these equations. We shall eventually see that Γ = BlQ P .As the equations in question are satisfied when (T0, . . . , Tc−1) = (0, . . . , 0), we haveQ × P(P ;Q) ⊂ Γ. On the other hand, for any ([T0 : · · · : Tn], [S0 : · · · : Sc−1]) ∈Γ r (Q × P(P ;Q)), we have Ti 6= 0 for some i < c and so Sj = (Si/Ti).Tj forall j < c. Since not all Sj are zero, we must have Si 6= 0 as well and so then[T0 : · · · : Tc−1] = [S0 : · · · : Sc−1]. Hence Γ r (Q× P(P ;Q)) is the graph of πQ.

Let us now see what the projection Γ→ P(P ;Q) is like over P(P ;Q)S0 in termsof the standard chart (y1, . . . , yc−1) ∈ Am 7→ [1 : y1 : · · · : yc−1] ∈ P(P ;Q)S0

.We let Q′ ⊂ P be defined by T1 = · · · = Tc−1 = 0. It is clear that Q′ containsQ as a hyperplane (defined by T0 = 0). Note that ΓS0

:= Γ ∩ (P × P(P ;Q)S0) is


parametrized by Q′ × P(P ;Q)S0by means of the morphism

([T0 : Tc : Tc+1 : · · · : Tn], [1 : y1 : · · · : yc−1]) ∈ Q′ × P(P ;Q)S0 7→([T0 : T0y1 : · · · : T0yc−1 : Tc : Tc+1 : · · · : Tn], [1 : y1 : · · · : yc−1]) ∈ ΓS0

,

This is an isomorphism (the inverse is obvious) which commutes with the projectionon P(P ;Q)S0

. Since Q × P(P ;Q)S0is defined in ΓS0

by T0 = T1 = · · ·Tc−1 = 0,it follows that its preimage in Q′ × P(P ;Q)S0

is also Q × P(P ;Q)S0. In particular,

Q×P(P ;Q)S0 lies in the closure of the graph of πQ. This remains true if we replaceP(P ;Q)S0 by P(P ;Q)Si , i = 1, . . . , c−1, or by any other hyperplane complement inP(P ;Q). It follows that Γ = BlQ P and that BlQ P enjoys the stated properties.

COROLLARY 4.4. Suppose that in the situation of Definition-Lemma 4.3, Z ⊂ Pis an irreducible and closed subset such that Z ∩Q = ∅. Then πQ|Z : Z → P(P ;Q) isa finite morphism and (so) dimZ < c.

Note that the dimension inequality amounts to the assertion that Z will meetevery linear subspace of P whose codimension is equal to the dimension of Z. Forits proof we shall need the following lemma.

LEMMA 4.5. Let P a projective space, U ⊂ P a hyperplane complement, X bea variety and Z ⊂ U × X a subset that is closed in P × X. Then the projectionπX |Z : Z → X is a finite morphism.

PROOF. Without loss of generality we may assume that X is affine. Since U isalso affine, it follows that Z, being a closed subset of U ×X is affine.

Choose homogeneous coordinates [T0 : . . . ;Tn] for P such that U = PT0 . Soif I ⊂ k[X][T0, . . . , Tn] is the homogeneous ideal defining Z, then the ideal J ⊂k[X][T0, . . . , Tn] generated by I and T0 defines the empty set in P × X and so itsradical is k[X][T0, . . . , Tn]+. In particular, there exists an integer r > 0 such thatT ri ∈ J for i ∈ 1, . . . , n. Write T ri ≡ T0Gi (mod Ir) with G ∈ k[X][T0, . . . , Tn]r−1.We pass to the affine coordinates of U by substituting 1 for T0 and ti for Ti. ThenGi defines a gi ∈ k[X][t1, . . . , tn] = k[X × An] of degree ≤ r − 1 in the t-variablesand we have tri ≡ gi (mod I(Z)). So if we write ti for the image of ti in k[Z], thentri is a k[X]-linear combination of monomials ts11 · · · tsnn with si < r for all i. Thisproves that k[Z] is a finitely generated k[X]-module so that πX |Z : Z → X is afinite morphism.

PROOF OF COROLLARY 4.4. We use the notation of Definition-Lemma 4.3. PutZ := p−1

1 Z. This is closed subset of BlQ(P ) which is disjoint with Q× P(P ;Q) andis mapped by p1 isomorphically onto Z. So it suffices to prove that p2|Z : Z →P(P ;Q) is finite. According to 4.3, p2 : (BlQ(P ), Q × P(P ;Q)) → P(P ;Q) admitsa local trivialization over a hyperplane complement P(P ;Q)Si which identifies ZSiwith a closed subset of Q′ × P(P ;Q)Si , where Q′ ⊂ P is a linear subspace whichcontainsQ as a hyperplane. Our assumption implies that ZSi ⊂ (Q′rQ)×P(P ;Q)Siand so it satisfies the hypotheses of Lemma 4.5 (with U = Q′rQ). So the projectionZSi → P(P ;Q)Si is finite. This implies that the projection Z → P(P ;Q) is finiteand in particular that dimZ = dim Z ≤ dimP(P ;Q) = c− 1.

We have also a kind of converse to Corollary 4.4:

PROPOSITION 4.6. For every closed subset Z of a projective space P there exists alinear subspace in P of codimension dim(Z) + 1 which misses Z.

5. ELIMINATION THEORY AND PROJECTIONS 67

PROOF. We may (and will) assume that Z is irreducible and 6= P . Let i ∈−1, . . . , codim(Z) − 1. We prove with induction on i that Z misses a linear sub-space of dimension i. For i = −1, the empty subspace will do. For i = 0, we musthave Z 6= P and so we can take for our linear subspace any singleton in P − Z.When i > 0, there exists by induction hypothesis a linear subspace Q ⊂ P of di-mension (i − 1) which does not meet Z. By Corollary 4.4, πQ|Z : Z → P(P,Q) isa finite morphism and so dimπQ(Z) = dimZ < dimP − i = dimP(P,Q). Hencethere exist a point in P(P,Q) − πQ(Z). This defines a linear subspace Q′ in P ofdimension i which passes through Q and misses Z.

5. Elimination theory and projections

Within a category of reasonable topological spaces (say, the locally compactHausdorff spaces), the compact ones can be characterized as follows: K is compactif and only if the projection K × X → X is closed for every space X in that cate-gory. In this sense the following theorem states a kind of compactness property forprojective varieties.

THEOREM 5.1. Let P be a projective space. Then for any variety X, the projectionπX : P ×X → X is closed.

We derive this theorem from the main theorem of elimination theory, which westate and prove first.

Given an integer d ≥ 0, let us write Vd for k[T0, T1]d, the k-vector space ofhomogeneous polynomials in k[T0, T1] of degree d. The monomials (T i0T

d−i1 )di=0

form a basis, in particular, dimVd = d+ 1. Given F ∈ Vm and G ∈ Vn, then

uF,G : Vn−1 ⊕ Vm−1 → Vn+m−1, (A,B) 7→ AF +BG

is a linear map between two k-vector spaces of the same dimension m + n. Theresultant R(F,G) of F and G is defined as the determinant of this linear mapwith respect to the monomial bases of the summands of Vn−1 ⊕ Vm−1 and ofVn+m−1. So R(F,G) = 0 if and only if uF,G fails to be injective. Notice that ifF =

∑mi=0 aiT

i0T

m−i1 and G =

∑nj=0 biT

i0T

n−i1 , then the matrix of uF,G with respect

to the monomial bases is

a0 0 0 · · · 0 b0 0 · · · · · · 0a1 a0 0 · · · 0 b1 b0 · · · · · · 0a2 a1 a0 · · · 0 b2 b1 · · · · · · 0· · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·am am−1 ∗ · · · ∗ ∗ ∗ · · · · · · ∗0 am ∗ · · · ∗ ∗ ∗ · · · · · · ∗0 0 am · · · ∗ ∗ ∗ · · · · · · ∗0 0 0 · · · ∗ ∗ ∗ · · · · · · ∗· · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·0 0 · · · · · · am−1 0 0 · · · · · · bn−1

0 0 0 · · · am 0 0 0 · · · bn

from which we see that its determinant R(F,G) is a polynomial in the coefficientsof F and G. So the resultant defines an element of k[Vm × Vn] = k[Vm]⊗ k[Vn].

LEMMA 5.2. R(F,G) = 0 if and only if F and G have a common linear factor.


PROOF. If R(F,G) = 0, then uF,G is not injective, so that there exist a nonzero(A,B) ∈ Vn−1 ⊕ Vm−1 with AF + BG = 0. Suppose that B 6= 0. It is clear thatF divides BG. Since deg(B) = m − 1 < m = degF , it follows that F and G musthave a common factor.

Conversely, if F and G have a common linear factor L: F = LF1, G = LG1,then G1F = F1G and so (G1,−F1) ∈ Vn−1 ⊕ Vm−1 is a nonzero element of thekernel of uF,G.

PROOF OF THEOREM 5.1. Let Z ⊂ P × X be closed. It is clear that πX(Z) isclosed in X if for every open affine subset X ′ ⊂ X, πX(Z) ∩ X ′ is closed in X ′.Since πX(Z)∩X ′ = πX′(Z ∩ (Pn×X ′)) we may (and will) assume that is X affine.We put n := dimP and choose a homogeneous coordinate system [T0 : · · · : Tn] forP . We proceed with induction on n, starting with the crucial case n = 1.

Denote by IZ the homogeneous ideal in the graded algebra k[X][T0, T1] offunctions vanishing on Z. Then Z is the common zero set of the members of IZ(see Exercise 57). For every homogeneous pair F,G ∈ ∪mk[X][T0, T1]m, we canform the resultant R(F,G) ∈ k[X]. We claim that πX(Z) is the common zero setZ(R) ⊂ X of the set of resultants R(F,G) of pairs of homogeneous forms F,Gtaken in ∪mIZ,m, hence is closed in X.

Suppose that y ∈ πX(Z). Then (y, p) ∈ Z for some p ∈ P1 and so p is acommon zero of each pair Fy, Gy, where F,G ∈ ∪mIZ,m and the subscript y refersto substituting y for the first argument. So R(F,G)(y) = 0 and hence y ∈ Z(R).

Next we show that if y /∈ πX(Z), then y /∈ Z(R). Since y×P1 is not containedin Z, there exists an integer m > 0 and a F ∈ IZ,m with Fy 6= 0. Denote byp1, . . . , pr ∈ P1 the distinct zeroes of Fy. We show that there exists a G ∈ IZ,n forsome n such that Gy does not vanish in any pi; this suffices, for this means thatR(Fy, Gy) 6= 0 and so y /∈ Z(R). For any given 1 ≤ i ≤ r, Z

⋃∪j 6=i(y, pj) is

closed in X × P1, so that there will exist a G(i) ∈ ∪mIZ,m with G(i)y zero in all the

pj with j 6= i, but nonzero in pi. Upon replacing each G(i) by some positive powerof it , we may assume that G(1), . . . , G(r) all have the same degree n, say. ThenG := G(1) + · · ·+G(r) ∈ IZ,n and Gy(pi) = G(i)(pi) 6= 0.

Now assume n ≥ 2. Let q = [0 : · · · : 0 : 1] and consider the blowup Pn :=

Blq Pn → Pn. Recall that an element of Pn is the set of pairs in ([T0 : · · · :

Tn], [S0 : · · · : Sn−1]) in Pn × Pn−1 with [T0 : · · · : Tn−1] = [S0 : · · · : Sn−1]. Wehave seen that over the open subset Pn−1

Si⊂ Pn−1 defined by Si 6= 0, the projection

PnSi → Pn−1 is isomorphic to the projection P1×Pn−1Si→ Pn−1

Si. Hence the projection

π1 : Pn ×X → Pn−1 ×X is over Pn−1Si×X like P1 × Pn−1

Si×X → Pn−1

Si×X. So

this projection is closed over Pn−1Si×X. It follows that the projection π1 is closed.

The preimage Z of Z under the projection Pn×X → Pn×X is closed and by whatwe just proved, π1(Z) is closed in Pn−1 ×X. By induction, the image of the latterunder the projection π2 : Pn−1 ×X → X is closed. But this is just πX(Z).

REMARK 5.3. This proof can be adapted to show more, namely that given aclosed and irreducible subset Z ⊂ P × X, then for any x ∈ πX(Z), Zx := p ∈P : (p, x) ∈ Z has dimension ≥ dimZ −dimπX(Z) with equality holding over anopen-dense subset of πX(Z).

Here are two corollaries.

6. THE VERONESE EMBEDDINGS 69

COROLLARY 5.4. Let X be a projective variety. Then any morphism from X to avariety is closed (and hence has closed image).

PROOF. Assume that X is closed in Pn. A morphism f : X → Y to a varietyY can be factored as the obvious isomorphism of X onto the graph Γf of f , theinclusion of this graph in X×Y (which is evidently closed), the inclusion of X×Yin Pn×Y (which is closed sinceX ⊂ Pn is closed) and the projection onto Y (whichis closed by Theorem 5.1). So f is closed.

It is an elementary result from complex function theory (based on Liouville’stheorem) that a holomorphic function on the Riemann sphere is constant. Thisimplies the corresponding assertion for holomorphic functions on complex projec-tive n-space PnC (to see that a holomorphic function on PnC takes the same value onany two distinct points, simply apply the previous remark to its restriction to thecomplex projective line passing through them, viewed as a copy of the Riemannsphere). The following corollary is an algebraic version of this fact.

COROLLARY 5.5. Let X be a projective variety. Then any morphism from X to aquasi-affine variety is constant. In particular, any regular function on X is constant.

PROOF. If f : X → Y is a morphism to a quasi-affine variety Y , then its com-posite with an embedding of Y in some affine space An is given by n regular func-tions on X. So it suffices to prove the special case when Y = A1. By the previouscorollary this image is closed in A1. But if we think of f as taking its values in P1

(via the embedding y ∈ A1 7→ [1 : y] ∈ P1), then we see that f(X) is also closedin P1. So f(X) cannot be all of A1. Since X is irreducible, so is the image and itfollows that f(X) is a singleton. In other words, f is constant.

EXERCISE 60. Let P be a projective space of dimension n.(a) The dual P of P is by definition the collection of hyperplanes in P . Prove

that P has a natural structure of a projective space.(b) Identify the double dual of P with P itself.(c) The incidence locus I ⊂ P × P is the set of pairs (p, q) ∈ P × P with the

property that p lies in the hyperplane Hq defined by q. Prove that I is anonsingular variety of dimension 2n− 1.

(d) Show that we can find homogeneous coordinates [Z0 : · · · : Zn] for P and[W0 : · · · : Wn] for P such that I is given by

∑ni=0 ZiWi = 0.

EXERCISE 61. Let F ∈ k[X0, . . . , Xn]d define a nonsingular hypersurface H inPn. Prove that the map H → Pn which assigns to p ∈ H the projectived tangentspace of H at p is given by [ ∂F∂Z0

: · · · : ∂F∂Zn

]. Prove that the image of this map isclosed in Pn (this image is called the dual of H). What can you say in case d = 2?

6. The Veronese embeddings

Let be given a positive integer d. We index the monomials in Z0, . . . , Zn thatare homogenous of degree d by their exponents: these are the sequences of non-negative integers k = (k0, . . . , kn) of length n + 1 with sum d. They are

(n+dd

)in

number.3 We use this to label the homogeneous coordinates Zk of P(n+dd )−1.

3If we expand∏ni=0(1 − tZi)−1, we see that the coefficient of td is the sum of the monomials in

Z0, . . . , Zn of degree d. So we get the number of such monomials by substituting Zi = 1 for all i: it the


PROPOSITION 6.1 (The Veronese embedding). The map fd : Pn → P(n+dd )−1

defined by Zk = T d00 · · ·T dnn is a closed immersion.

PROOF. It is enough to show that for every chart domain Uk := P(n+dd )−1

Zkof

the standard atlas of the target space, its preimage f−1d Uk is open in Pn and is

mapped by fd isomorphically onto a closed subset of Uk. This preimage is definedby T k00 · · ·T knn 6= 0. Let us renumber the coordinates such that k0, . . . , kr are positiveand kr+1 = · · · = kn = 0. Then f−1

d Uk = PnT0···Tr ⊂ PnT0. So if we use the

standard coordinates (t1, . . . , tn) to identify PnT0with An, then f−1

d Uk is identifiedwith Ant1···tr .

The coordinates on Uk are the functions Zl/Zk with l 6= k. If we write zl−k forthis function, then fd is in terms of these coordinates simply:

fd : Ant1···tr ∼= f−1d Uk → Uk, zl−k = t−k11 · · · t−krr .tl11 · · · tlnn ,

with (l1, . . . , ln) running over all the n-tuples of nonnegative integers with sum≤ d and distinct from (k1, . . . , kr, 0, . . . , 0). Among the components of this map are(t1 . . . tr)

−1 (take li = ki − 1 for i ≤ r and li = 0 for i > r) and ti (take li = ki + 1and lj = kj for j 6= i; this is allowed because then l1 + · · ·+ ln = 1 +k1 + · · ·+kn ≤k0 +k1 + · · ·+kn = d). These generate the coordinate ring k[t1, . . . , tn][1/(t1 . . . tr)]of Ant1···tr and so fd defines a closed immersion of Ant1···tr in Uk.

The following proposition is remarkable for its repercussions in intersectiontheory.

PROPOSITION 6.2. Let H ⊂ Pn be a hypersurface. Then Pn rH is affine and forevery closed irreducible subset Z ⊂ Pn of positive dimension, Z ∩H is nonempty andof dimension ≥ dim(Z)− 1, with equality holding if Z is not contained in H.

PROOF. The hypersurface H is given by a homogeneous polynomial of degreed, say by

∑k ckT

k00 · · ·T knn . This determines a hyperplane H ⊂ P(n+d

d )−1 definedby∑

k ckZk. It is clear that H is the preimage of H under the Veronese morphismand hence the latter identifies Pn r H with a closed subset of the affine spaceP(n+d

d )−1 r H. So Pn rH is affine.For the rest of the argument we may, by passing to the Veronese embedding,

assume that H is a hyperplane. If dim(Z ∩H) ≤ dim(Z) − 2, then by Proposition4.6 there exists a linear subspace Q ⊂ H of dimension dim(H)− (dim(Z∩H)−1 ≤(n− 1)− (dim(Z)− 2)− 1 = n− dim(Z) which avoids Z ∩H. Since this is a linearsubspace of Pn which avoids Z, we thus contradict Corollary 4.4. This provesthat dim(Z ∩ H) ≥ dim(Z) − 1. If Z is not contained in H, then we have alsodim(Z ∩H) ≤ dim(Z)− 1.

REMARK 6.3. A theorem of Lefschetz asserts that if in the situation of Proposi-tion 6.2 above dimZ ≥ 2 (so that dim(Z ∩H) ≥ 1), Z ∩H is connected.

EXERCISE 62. Let d be a positive integer. The universal hypersurface of degree dis the hypersurface of Pn × P(n+d

d )−1 defined by F (X,Z) :=∑

d ZdTd00 T d11 · · ·T dnn .

We denote it by H and let π : H → P(n+dd )−1 be the projection. As of item (c) we

assume that d ≥ 2.

coefficient of td of in (1− t)−(n+1) and hence the value of 1/d!.(d/dt)d(1− t)−(n+1) in t = 0, whichis 1/d!.(n+ 1)(n+ 2) · · · (n+ d) =

(n+dd

).

7. GRASSMANNIANS 71

(a) Prove that H is nonsingular.(b) Prove that projection π is singular at (X,Z) (in the sense that the deriv-

ative of π at (X,Z) is not a surjection) if and only the partial derivativesof FZ ∈ k[X0, . . . , Xn] have X as a common zero.

(c) Prove that the singular set of π is a nonsingular subvariety of Pn×P(n+dd )−1

of codimension n+ 1.(d) Prove that the set of Z ∈ P(n+d

d )−1 over which π has a singular point is ahypersurface. This hypersurface is called the discriminant of π.

(e) For d = 2 we denote the coordinates of P(n+dd )−1 simply by Zij (where it

is understood that Zij = Zji). Prove that the discriminant of π is then thezero set of det(Zij).

7. Grassmannians

Let P be a projective space of dimension n and let d ∈ 0, . . . , n. We wantto show that the collection Grd(P ) of linear d-dimensional subspaces of P is anonsingular projective variety. Let the projective structure on P be defined bythe pair (V, `) so that V is a (n + 1)-dimensional k-vector space and P has beenidentified with P(V ). This identifies Grd(P ) with the collection Grd+1(V ) of linear(d+ 1)-dimensional subspaces of V .

LEMMA 7.1. Let Q ⊂ P be a linear subspace of codimension d + 1. Then thecollection Grd(P )Q of linear d-dimensional subspaces of P contained in P rQ has ina natural manner the structure of an affine space of dimension (n− d)(d− 1).

PROOF. Let Q correspond to the linear subspace VQ ⊂ V of dimension (n +1)− (d+ 1) = n− d. Then the elements of Grd(P )Q correspond to linear subspacesL ⊂ V that complement VQ in the sense that L ⊕ VQ → V is an isomorphism.We claim that the vector space Hom(V/VQ, VQ) acts simply transitively on Grd(P )Q(so that Grd(P )Q becomes an affine space with Hom(V/VQ, VQ) as its group oftranslations). The action is given by letting to σ ∈ Hom(V/VQ, VQ) send L to thegraph of the map L ⊂ V → V/VQ

σ−→ VQ in L ⊕ VQ ∼= V . Indeed, any L′ withV ∼= L′ ⊕ VQ is so obtained for a unique σ.

It can now be shown without much difficulty that Grd(P ) admits a uniquestructure of a variety for which every Grd(P )Q as in this lemma is affine open andits identification with affine space an isomorphism. We will however proceed ina more direct manner to show that Grd(P ) admits the structure of a projectivevariety.

For this we recall that the exterior algebra ∧•V = ⊕p≥0 ∧p V is the quotientof the tensor algebra on V , ⊕∞p=0V

⊗p (here V ⊗0 = k by convention), by the two-sided ideal generated by the ‘squares’ v ⊗ v, v ∈ V . It is customary to denotethe product by the symbol ∧. So we can characterize ∧•V as a (noncommutative)associative k-algebra with unit element by saying that is generated by the k-vectorspace V and is subject to the relations v ∧ v = 0 for all v ∈ V . It is a gradedalgebra (∧pV is the image of V ⊗p) and ‘graded-commutative’ in the sense thatif α ∈ ∧pV and β ∈ ∧qV , then β ∧ α = (−1)pqα ∧ β. If (ε0, . . . , εn) is a basisfor V , then a basis of ∧pV is indexed by the p-element subsets I ⊂ 0, . . . , n: toI = 0 ≤ i1 < i2 < · · · < ip ≤ n is associated to the basis element εI = εi1∧· · ·∧εip(where the convention is that ε∅ = 1 ∈ k = ∧0V ). So dim∧pV =

(n+1p

). Notice


that ∧n+1V is one-dimensional and spanned by ε0 ∧ · · · ∧ εn, whereas ∧pV = 0 forp > n+ 1. We also recall that if V ′ and V ′′ are subspaces of V , then the map

∧•V ′ ⊗ ∧•V ′′ → ∧•V, α⊗ β 7→ α ∧ β,is a linear map of graded vector spaces which is injective (resp. surjective) whenthis is so in degree 1 (i.e., when V ′ ⊕ V ′′ → V is).

We say that α ∈ ∧pV is fully decomposable if there exist linearly independentv1, . . . , vp in V such that α = v1 ∧ · · · ∧ vp. This is equivalent to the existence of ap-dimensional subspace K ⊂ V such that α is a generator of ∧pK.

LEMMA 7.2. Let α ∈ ∧pV be nonzero. Denote by K(α) the set of v ∈ V withv∧α = 0. Then dimK(α) ≤ p and equality holds if and only if α is fully decomposableand spans ∧pK(α).

PROOF. Let ε1, . . . , εr be a basis of K(α) and let V ′ ⊂ V be a subspace supple-mentary to K(α). Then we have a decomposition

∧•V =⊕

I⊂1,...,r

εI ∧ (∧•V ′).

The kernel of εi∧ : ∧•V → ∧•V is the subsum of the εI ∧ (∧•V ′) with i ∈ I. Itfollows that α ⊂ ε1 ∧ · · · ∧ εr ∧ (∧p−rV ′). In particular, r ≤ p with equality holdingif and only if α is a multiple of ε1 ∧ · · · ∧ εp.

If L is a linear subspace of V of dimension d + 1, then ∧d+1L is of dimension1 and will be thought of as a one dimensional subspace of ∧d+1V . We thus havedefined a map δ : Grd(P ) → P(∧d+1V ), [L] 7→ [∧d+1L]. It is called the Pluckerembedding because of:

PROPOSITION 7.3. The map δ : Grd(P ) → P(∧d+1V ) maps Grd(P ) bijectivelyonto a closed subset of P(∧d+1V ).

PROOF. Let α ∈ ∧d+1V be nonzero. According to Lemma 7.2 , [α] is in theimage of δ if and only if K(α) is of dimension d + 1 and if that is the case, thenδ−1[α] has [K(α)] as its unique element. In particular, δ is injective.

The subset Kd+1(V,∧d+2V ) ⊂ Hom(V,∧d+2V ) of linear maps whose kernel isof dimension ≥ d + 1 is (after we have chosen a basis for V ) the common zero setof a system of homogeneous equations in Hom(V,∧d+2V ), namely the (n+1−d)×(n+ 1− d)-minors of the corresponding matrices. Consider the linear map

σ : ∧d+1V → Hom(V,∧d+2V ), α 7→ (v 7→ α ∧ v).

Since σ−1Kd+1(V,∧d+2V ) is given by a set of homogeneous equations it definesa closed subset of P(∧d+1V ). This is just the image of δ, for by Lemma 7.2,σ−1Kd+1(V,∧d+2V ) r 0 is the set of fully decomposable elements of ∧d+1V .

Proposition 7.3 gives Grd(P ) the structure of projective variety. In order tocomplete the construction, let Q ⊂ P be a linear subspace of codimension d. LetVQ ⊂ V correspond to Q and choose a generator β ∈ ∧n−dVQ. Then we have anonzero linear map to the one-dimensional ∧n+1V :

eβ : ∧d+1V → ∧n+1V, α 7→ α ∧ β.Its kernel is a hyperplane whose complement defines principal open subset ofP(∧d+1V ) that we shall denote by P(∧d+1V )Q. Such principal open subsets cover

7. GRASSMANNIANS 73

P(∧d+1V ) (to see this, choose a basis (ε0, . . . , εn) of V and observe that if VQ runsover the codimension d subspaces of V spanned by basis vectors, then P(∧d+1V )Qruns over a collection of principal open subsets defined by the basis (εI)|I|=d+1 of∧d+1V ).

LEMMA 7.4. The preimage of P(∧d+1V )Q under the Plucker embedding δ is theaffine space Grd(P )Q and δ maps this affine space isomorphically onto its image.

PROOF. Let VQ ⊂ V be the linear subspace defining Q and let β be a generatorof ∧n−dVQ as above. If α ∈ ∧d+1V is fully decomposable (and hence generates∧d+1L for a unique (d + 1)-dimensional subspace L ⊂ V ), then L ∩ VQ = 0 ifand only if α ∧ β 6= 0: if L ∩ VQ contains a nonzero vector v then both α and β aredivisible by v and so α ∧ β = 0 and if L ∩ VQ = 0, then we have a decompositionV ∼= L⊕ V and so α ∧ β 6= 0. This implies that δ−1P(∧d+1V )Q = Grd(P )Q.

Let us now express the restriction δ : Grd(P )Q → P(∧d+1V )Q in terms ofcoordinates. Choose a basis (ε0, . . . , εn) for V such that (εd+1, . . . , εn) is a basis forVQ and β = εd+1 ∧ · · · ∧ εn. Then eβ simply assigns to an element α of ∧d+1Vthe coefficient of ε0 ∧ · · · ∧ εd in α. If L0 ⊂ V denotes the span of ε0, . . . , εd,then Grd(P )Q is identified with the affine space Hom(L0, VQ) ∼= A(d+1)×(n−d) of(d+ 1)× (n− d)-matrices via

(aji )0≤i≤d<j≤n 7→ k-span in V of the d+ 1 vectors εi +∑nj=d+1 a

jiεjdi=0 ,

so that δ is given by

(aji )0≤i≤d<j≤n 7→ (ε0 +∑nj=d+1 a

j0εj) ∧ · · · ∧ (ed +

∑nj=d+1 a

jdεj).

The coefficient of εi0 ∧ · · · ∧ εid is a determinant of which each entry is 0, 1 orsome aji and hence is a polynomial in the matrix coefficients aji . It follows thatthis restriction of δ is a morphism. Among the components of δ we find the matrixcoefficients themselves, for aji appears up to sign as the coefficient of ε0 ∧ · · · ∧ εi ∧· · · ∧ εd ∧ εj . Since these generate the coordinate ring of A(d+1)×(n−d), it followsthat δ defines a closed immersion of Grd(P )Q in P(∧d+1V )Q.

COROLLARY 7.5. The Plucker embedding realizes Grd(P ) as a nonsingular irre-ducible subvariety of P(∧d+1V ) of dimension (n−d)(d+1). This structure makes eachsubset Grd(P )Q open and isomorphic to affine (n − d)(d + 1)-space in a way that iscompatible with the one obtained in Lemma 7.1.

PROOF. Every two open subsets of the form Grd(P )Q a have nonempty inter-section and so Grd(P ) is irreducible. The rest follows from the previous corol-lary.

REMARK 7.6. The image of Grd(P ) is a closed orbit of the natural SL(V )-actionon P(∧d+1V ). It lies in the closure of any other SL(V )-orbit.

EXERCISE 63. Let V be a finite dimensional k-vector space. For every linearsubspace W ⊂ V we identify (V/W )∗ with the subspace of V ∗ of linear forms onV that are zero on W . Prove that for every 0 ≤ r ≤ dimV the resulting mapGrr(V )→ GrdimV−r(V

∗) is an isomorphism of projective varieties.

EXERCISE 64. Let V and W be finite dimensional k-vector spaces and let r bea nonnegative integer ≤ mindimV,dimW.


(a) Prove that the subset Homr(V,W ) ⊂ Hom(V,W ) of linear maps of rank ris a (locally closed) subvariety of Hom(V,W ).

(b) Prove that the map Homr(V,W ) → GrdimV−r(V ) resp. Homr(V,W ) →Grr(W ) which assigns to φ ∈ Homr(V,W ) its kernel resp. image is amorphism.

(c) Prove that the resulting morphism Homr(V,W )→ GrdimV−r(V )×Grr(W )is trivial over any product of principal open subsets with fiber the generallinear group GLr(k). Conclude that Homr(V,W ) is nonsingular of codi-mension (dimV − r)(dimW − r).

The Grassmannian of hyperplanes in a projective space is itself a projectivespace (see Exercise 60). So the simplest example not of this type is the Grassman-nian of lines in a 3-dimensional projective space.

Let V be vector space dimension 4. On the 6-dimensional space ∧2V we havea homogeneous polynomial F : ∧2V → k of degree two defined by

F (α) := α ∧ α ∈ ∧4V ∼= k

(the last identification is only given up to scalar and so the same is true for F ). Incoordinates F is quite simple: if e1, . . . , e4 is a basis for V , then (ei ∧ ej)1≤i<j≤4 isbasis for ∧2V . So if we label the homogeneous coordinates of P(∧2V ) accordingly:[T1,2 : · · · : T3,4], then F is given by

F (T1,2, . . . , T3,4) = T1,2T3,4 − T1,3T2,4 + T1,4T2,3.

Notice that F is irreducible. Its partial derivatives are the coordinates themselves(up to sign and order) and so F defines a nonsingular quadric hypersurface ofdimension 4 in a 5-dimensional projective space.

PROPOSITION 7.7. The image of the Plucker embedding of G1(P(V )) in P(∧2V )is the zero set of F .

PROOF. The image of the Plucker embedding is of dimension 4 and so must bea hypersurface. Since the zero set of F is an irreducible hypersurface, it suffices toshow that the Plucker embedding maps to the zero set of F . For this, let α be agenerator of ∧2L for some linear subspace L ⊂ V of dimension 2. If e1, . . . , e4 is abasis of V such that α = e1 ∧ e2, then it is clear that α∧α = 0. This proves that thePlucker embedding maps to the zero set of F .

If the characteristic of k is not 2, then the nonsingular quadric hypersurfaces ofthe same dimension are isomorphic to one another and so this proposition showsthat any nonsingular quadric hypersurface of dimension 4 is isomorphic to theGrassmannian of lines in a three dimensional projective space.

REMARK 7.8. The image of the Plucker embedding Grd(P ) → P(∧d+1V ) is in fact al-ways the common zero set of a collection of quadratic equations, called the Plucker relations.To exhibit these, we first recall that every φ ∈ V ∗ defines a linear ‘inner contraction’ mapιφ : ∧•V → ∧•V of degree −1 characterized by the fact that for v ∈ V , ιφ(v) = φ(v) ∈k = ∧0V and for α ∈ ∧pV, β ∈ ∧•V , ιφ(α ∧ β) = ιφ(α) ∧ β + (−1)pα ∧ ιφ(β). Under thenatural isomorphism End(V, V ) ∼= V ⊗V ∗, the identity of V defines a tensor in V ⊗V ∗. Thewedge-contraction with this tensor defines a linear map BV : ∧•V ⊗∧•V → ∧•V ⊗∧•V ofbidegree (1,−1). Concretely, if (e0, . . . , en) is a basis of V and (e∗0, . . . , e

∗n) is the basis of V ∗

dual to (e0, . . . , en), then

BV (α⊗ β) :=∑nr=0(α ∧ er)⊗ (ιe∗rβ).

8. FANO VARIETIES AND THE GAUSS MAP 75

Notice that if W ⊂ V is a subspace, then BW is just the restriction of BV to ∧•W ⊗ ∧•W .So if α ∈ ∧d+1V is fully decomposable so that α ∈ ∧d+1L for some (d + 1)-dimensionalsubspace L ⊂ V , then BV (α⊗ α) = BL(α⊗ α) = 0. This is the universal Plucker relation.

Conversely, any nonzero α ∈ ∧d+1V for which BV (α ⊗ α) = 0 is fully decomposable.The proof proceeds with induction on n. For n = 0 there is nothing to show. Assumen ≥ 1, let e ∈ V be nonzero and let V ′ ⊂ V be a hyperplane not containing e. If we writeα = α′ + e ∧ α′′ with α′, α′′ ∈ ∧•V ′, then the component of B(α ⊗ α) in ∧•V ′ ⊗ ∧•V ′is BV ′(α′ ⊗ α′) and so α′ is zero or fully decomposable by our induction hypothesis: thereexists a subspace L′ ⊂ V ′ of dimension d + 1 such that α′ ∈ ∧d+1L′. Then the vanishingof the component of B(α ⊗ α) in ∧•V ′ ⊗ e ∧ (∧•V ′) is seen to imply that ιφα′′ = 0 forall φ ∈ (V ′/L′)∗ ⊂ V ′∗. This means that α′′ ∈ ∧dL′. So if we put M := ke + L′, thendimM = d+ 2 and α ∈ ∧d+1M . But then α ∈ ιφ ∧d+2 M for some nonzero φ ∈ M∗. Thenα is a generator of ∧d+1 Ker(φ) and hence fully decomposable.

Let us rephrase this in terms of algebraic geometry: every nonzero linear form ` on∧d+2V ⊗∧dV , determines a quadratic form Q` on ∧d+1V defined by α 7→ `(B(α, α)) whosezero set is a quadratic hypersurface in P(∧d+1V ). This hypersurface contains the Pluckerlocus and the latter is in fact the common zero set of the Q`, with ` running over the linearforms on ∧d+2V ⊗ ∧dV . It can be shown that the Q` generate the full graded ideal definedby the Plucker locus. The quadratic forms Q` are called the Plucker relations.4

8. Fano varieties and the Gauß map

The Fano variety of a projective variety is defined in the following proposition.

PROPOSITION-DEFINITION 8.1. LetX be a closed subvariety of the projective spaceP . If d is an integer between 0 and dimP , then the set of d-linear subspaces of Pwhich are contained in X defines a closed subvariety Fd(X) of Grd(P ), called theFano variety (of d-planes) of X.

PROOF. An open affine chart of Grd(P ) is given by a decomposition V = L⊕Wwith dimL = d + 1 and dimW = n − d and is then parametrized by Hom(L,W )by assigning to A ∈ Hom(L,W ) the graph of A. It suffices to prove that via thisidentification Fd(X) defines a closed subset of Hom(L,W ).

Choose homogeneous coordinates [T0 : · · · : Tn] such that L resp. W is givenby Td+1 = · · · = Tn = 0 resp. T0 = · · ·Td = 0. A linear map A ∈ Hom(L,W )

is then given by A∗Td+i =∑dj=0 a

jiTj , i = 1, . . . , n − d. It defines an element

of Fd(X) if and only for all G ∈ ∪m≥0Im(X), G(T0, . . . , Td, A∗Td+1, . . . A

∗Tn) isidentically zero. This means that the coefficient of every monomial Tm0

0 · · ·Tmdd

in such an expression much vanish. Since this coefficient is a polynomial in thematrix coefficients aji of A, we find that the preimage of Fd(X) in Hom(L,W ) isthe common zero set of a set of polynomials and hence is closed therein.

EXAMPLE 8.2. Consider the case of a quadratic hypersurface X ⊂ P(V ) andassume for simplicity that char(k) 6= 2. So X can be given by a nonzero quadraticform F ∈ k[V ]2. With F is associated a symmetric bilinear form B : V × V → kdefined by B(v, v′) = F (v+v′)−F (v)−F (v′) so that B(v, v) = 2F (v) (so nonzero,because char(k) 6= 2). Let us assume that X is nonsingular. This means thatthe partial derivatives of F have no common zero in P(V ). This translates into:B : V × V → k is nonsingular, that is, the linear map b : v ∈ V 7→ B(-, v) ∈ V ∗is an isomorphism of vector spaces (here we use that char(k) 6= 2). A subspace

4These show up in the algebro-analytic setting of the Sato Grassmannian (for which both d andn− d are infinity) and are then known as the Hirota bilinear relations.


L ⊂ V determines an element of the Fano variety of X precisely when B(v, v) = 0for all v ∈ L. This implies that B is identically zero on L × L. So b maps L to(V/L)∗ ⊂ V ∗. Since b is injective, this implies that dimL ≤ dim(V/L), in otherwords that dimL ≤ 1

2 dimV .This condition is optimal. It not difficult to show that we can find coordinates

(T0, · · · , Tn) such that F = 12

∑ni=0 TiTn−i so that the matrix of B is the unit an-

tidiagonal. If for instance dimX = n − 1 is even, say 2m, then let L resp. L′ bethe linear subspace defined by Tm+1 = · · · = T2m+1 = 0 resp. T0 = · · · = Tm = 0,so that V = L ⊕ L′. Notice that both [L] and [L′] are in Fm(X). The vector spaceHom(L,L′) describes an affine open subset of the Grassmannian of m-planes inP(V ). An element A ∈ Hom(L,L′) is given by A∗Tn−i =

∑mj=0 aijTj , i = 0, . . . ,m.

The correspondingm-plane is contained inX precisely when∑mi,j=0 aijTiTj is iden-

tically zero, i.e., if (aij) is antisymmetric. It follows that [L] ∈ Fm(X) has a neigh-borhood isomorphic to an affine space of dimension 1

2m(m + 1). In particular,Fm(X) is nonsingular.

EXERCISE 65. Let X be a quadratic hypersurface P(V ) of odd dimension 2m+1and assume for simplicity that char(k) 6= 2. Prove that Fm+1(X) = ∅ and thatFm(X) 6= ∅. Prove that Fm(X) is a nonsingular variety and determine its dimen-sion.

EXERCISE 66. Let X ⊂ Pn be a hypersurface of degree d and let 0 ≤ m ≤ n.Prove that the intersection of Fm(X) with a standard affine subset of Grm(Pn) isgiven by

(m+dd

)equations.

EXERCISE 67. Consider the universal hypersurface of degree d in Pn, H ⊂Pn × P(n+d

d )−1.

(a) For every m-plane Q ⊂ Pn, let Yz denote the set of z ∈ P(n+dd )−1 for which

the corresponding hypersurface Hz contains Q. Prove that Yz is a linearsubspace of P(n+d

d )−1 of codimension(m+dd

).

(b) Let Y ⊂ P(n+dd )−1 be the set of z ∈ P(n+d

d )−1 for which Hz contains anm-plane. Prove that Y is a closed subset of P(n+d

d )−1 of codimension atleast

(m+dd

)− (m+ 1)(n−m).

(c) Prove that the family of m-planes contained in a generic hypersurface ofdegree d in Pn is of dimension (m + 1)(n − m) −

(m+dd

)or empty. In

particular, this is a finite set when (m+ 1)(n−m) =(m+dd

).5

Let P = P(V ) be as before and let X be an irreducible closed subset of P of di-mension d. For every nonsingular point p ∈ X, there is precisely one d-dimensionallinear subspace T (X, p) of P which contains p and has the same tangent space atp as X. In other words, it is characterized by the property that the ideals in OP,pdefining X resp. T (X, p) have the same image in OP,p/m2

P,p.

PROPOSITION-DEFINITION 8.3. For every p ∈ Xreg, T (X, p) is the common zeroset of the linear forms defined by the differentials of the homogeneous elements of

5For instance, every cubic surface in P3 (so here n = 3, d = 3 and m = 1) contains a line. If itis nonsingular, then it contains in fact exactly 27 lines. This famous result due to Cayley and Salmonpublished in 1849 is still subject of research.

9. MULTIPLICITIES OF MODULES 77

I(X) evaluated in some representative of p in V r 0. This defines a morphismXreg → Grd(P ), called the Gauß map.6

PROOF. If F ∈ I(X) and G ∈ k[V ] are homogeneous of the same degree msay, with p ∈ XG, then φ := F/G is a regular function on PG which vanishes onXG. The kernel of the linear form dφ(p) : TpP → k contains the tangent spaceTp(X). As such φ generate the ideal defining X at p, Tp(X) is the intersection ofsuch kernels.

We have on VG the identity of differentials d(F/G) = G−1dF − G−2FdG. Weinterpret such a differential as a morphism from VG → V ∗ (the differential at apoint of V is regarded as a linear form on V ). Note that if p ∈ Cone(X)G lies overp, then its value in any p ∈ Cone(X)G is G−1(p)dF (p) ∈ V ∗. The kernel of thislinear form is that of dF (p) ∈ V ∗ and since dF (tp) = tmdF (p), this kernel dependsonly on p. It follows that T (X, p) is the linear subspace defined by the kernels ofthe linear forms dF (p), where F runs over ∪m≥1I(X)m, and that this is also theprojectivized tangent space of Cone(X) at p.

Theorem 10.14 of Ch. 1 tells us that we can choose F1, . . . , Fn−d ∈ I(X) andG ∈ k[V ] such that p ∈ Cone(X)G ⊂ Cone(X)reg and F1, . . . , Fn−d generate theideal defining Cone(X)G in VG such that for every q ∈ Cone(X)G, the tangent spaceof Cone(X)G at q is ∩n−di=1 Ker(dFi(q)). We may (and will) assume that G is homo-geneous. Now dF1 ∧ · · · ∧ dFn−d defines a morphism VG → ∧n−dV ∗ whose imageclearly consists of nonzero fully decomposable elements. In other words, it definesa morphism VG → Grn−d(V

∗). Via the identification Grn−d(V∗) ∼= Grd+1(V ) =

Grd(P ) we regard this as a morphism taking values in Grd(P ). Its restriction toCone(X)G is then just the map which assigns to q the space T (X, q), in other words,factors through the Gauß map XG → Grd(P ). Since Cone(X)G → Grd(P ) is a mor-phism, so will be the induced map XG → Grd(P ) (why?). Hence the Gauß map isa morphism.

REMARK 8.4. The closure of the graph of the Gauss map in X×Grd(P ) is calledthe Nash blowup of X. Its projection to X is clearly an isomorphism over the open-dense subset Xreg and hence birational. A remarkable property of the Nash blowupis that the Zariski tangent space of each of its points contains a distinguished d-dimensional subspace (prescribed by the second projection to Grd(P )) in such amanner that these subspaces extend the tangent bundle ofXreg in a regular manner.

9. Multiplicities of modules

Bezout’s theorem asserts that two distinct irreducible curves C,C ′ in P2 of de-grees d and d′ intersect in dd′ points. Strictly speaking this is only true if C and C ′

intersect as nicely as possible, but the theorem is true as stated if we count eachpoint of intersection with an appropriate multiplicity. There is in fact a generaliza-tion: the common intersection of n hypersurfaces in Pn has cardinality the productof the degrees of these hypersurfaces, provided that this intersection is finite andeach point of intersection is counted with an appropriate multiplicity. One of ouraims is to define these multiplicities. The tools from commutative algebra that weuse for this have an interest in their own right.

6Thus named because it is related to the map that Gauß studied for an oriented surface Σ inEuclidian 3-space E3: it is then the map Σ→ S2 which assigns to p ∈ Σ the unit outward normal vectorof Σ at p.


DEFINITION 9.1. We say that an R-module has length ≥ d if there exist a d-stepfiltration by submodules M = M0 )M1 ) · · · )Md = 0. The length of M is thesupremum of such d (and so may be∞).

EXERCISE 68. Suppose R is a noetherian local ring with maximal ideal m andresidue field K. Prove that the length of a finitely generated R-module M is finiteprecisely when mdM = 0 for some d and is then equal to

∑d−1i=0 dimK(miM/mi+1M).

Prove that if R is a K-algebra, then this is also equal to dimK(M).

In the remainder of this section R is a noetherian ring R and M a finitely gener-ated (and hence noetherian) R-module.

Recall that if p is a prime ideal of R, then Rp is a local ring with maximal idealpRp whose residue field can be identified with the field of fractions of R/p. Wedefine Mp := Rp ⊗RM . So this is a Rp-module.

REMARK 9.2. We can describe Mp and more generally, any localization S−1R ⊗R M ,as follows. Consider the set S−1M of expressions m/s with m ∈ M and s ∈ S with theunderstanding that m/s = m′/s′ if the identity s′′s′m = s′′sm holds in M for some s′′ ∈ S(so we are considering the quotient of S×M by an equivalence relation). Then the followingrules put on S−1M the structure of a R-module:

m/s−m′/s′ := (s′m− sm′)/(ss′), r ·m/s := rm/s.

The map S−1R×M → S−1M , (r/s,m)→ (rm)/s is R-bilinear and hence factors throughan R-homomorphism S−1R ⊗R M → S−1M . On the other hand, the map S−1M →S−1R ⊗R M , m/s 7→ 1/s ⊗R m is also defined: if m/s = m′/s′, then s′′(s′m = sm)for some s′′ ∈ S and so

1/s⊗R m = 1/(ss′s′′)⊗R s′s′′m = 1/(ss′s′′)⊗R ss′′m = 1/s′ ⊗R m.

It is an R-homomorphism and it immediately verified that it is a two-sided inverse of themap above. So S−1R⊗RM → S−1M is an isomorphism.

This description shows in particular that if N ⊂M is a submodule, then S−1N may beregarded as submodule of S−1M (this amounts to: S-localization is an exact functor on thecategory of R-modules).

DEFINITION 9.3. The multiplicity of M at a prime ideal p of R, denoted µp(M),is the length of Mp as an Rp-module.

In an algebro-geometric context we may modify this notation accordingly. Forinstance, if we are given an affine variety X and an irreducible subvariety Y , thenwe may write µY (-) for µp(-) , where it is understood that R = k[X] and p = I(Y ).

REMARK 9.4. LetX be a variety, x ∈ X and I ⊂ OX,x an ideal with√I = mX,x.

So mrX,x ⊂ I ⊂ mX,x for some positive integer r. Then dimk(OX,x/I) is finite (sincedimk(OX,x/mrX,x) is) and according to Exercise 68 equal to the length of OX,x/I asan OX,x-module and hence the multiplicity of OX,x/I at the maximal ideal mX,x.In agreement with our convention, we will denote this multiplicity by µx(OX,x/I).If X is affine and we are given an ideal I ⊂ k[X] whose image in OX,x is I, thenOX,x/I is the localization of k[X]/I at x and so µx(k[X]/I) = µx(OX,x/I) =dimk(OX,x/I). Note that x is then an isolated point of Z(I).

IfX is nonsingular at x of dimension n and I has exactly n generators f1, . . . , fn,then we will see that µp(OX,x/(f1, . . . , fn)) = dimk(OAn,p/(f1, . . . , fn)) can be in-terpreted as the multiplicity of p as a common zero of f1, . . . , fn.

9. MULTIPLICITIES OF MODULES 79

We wish to discuss the graded case parallel to the ungraded case. This meansthat if R is graded: R = ⊕∞i=0Ri, then we assume M to be graded as well, that is,M is endowed with a decomposition as an abelian group M = ⊕i∈ZMi such thatRj sends Mi to Mi+j (we here do not assume that Mi = 0 for i < 0). For example,a homogeneous ideal in R is a graded R-module. In that case we have the notionof graded length of M , which is the same as the definition above, except that weonly allow chains of graded submodules.

CONVENTION 9.5. Given an integer l and a graded module M over a gradedring, then M [l] denotes the same module M , but with its grading shifted over l,meaning that M [l]i := Ml+i.

Let us call a (graded) R-module elementary if it is isomorphic to R/p ((R/p)[l]),for some (homogeneous) prime ideal p (and some l ∈ Z).

Given a (graded) R-module M , then every m ∈ M (m ∈ Ml) defines a homo-morphism or R-modules r ∈ R 7→ rm ∈ M . Its kernel is a (graded) ideal of R, theannihilator Ann(m) of m, so that M contains a copy R/Ann(m) (R/Ann(m)[l]) asa (graded submodule).

LEMMA 9.6. Let M be a finitely generated nonzero (graded) R-module. Thenthe collection of annihilators of nonzero (homogeneous) elements of M contains amaximal element and any such maximal element is a (homogeneous) prime ideal ofR. In particular, M contains an elementary (graded) submodule.

PROOF. We only do the graded case. The first assertion follows from the noe-therian property of R. Let now Ann(m) be a maximal element of the collection(so with m ∈ M homogeneous and nonzero). It suffices to show that this is aprime ideal in the graded sense (see Exercise 55), i.e., to show that if a, b ∈ R arehomogeneous and ab ∈ Ann(m), but b /∈ Ann(m), then a ∈ Ann(m). So bm 6= 0and a ∈ Ann(bm). Since Ann(bm) ⊃ Ann(m), the maximality property of the latterimplies that this must be an equality: Ann(bm) = Ann(m), and so a ∈ Ann(m).

PROPOSITION 9.7. Every finitely generated (graded)R-moduleM can be obtainedas a successive extension of elementary modules in the sense that there exists a finitefiltration by (graded) R-submodules M = M0 ) M1 ) · · · ) Md = 0 such thateach quotient M i−1/M i, i = 1, . . . , d, is elementary.

PROOF. We do the graded case only. Since M is noetherian, the collection ofgraded submodules of M that can be written as a successive extension of elemen-tary modules has a maximal member, M ′, say. We claim that M ′ = M . If M/M ′

were nonzero, then according to Lemma 9.6, it contains an elementary submodule.But then the preimage N of this submodule in M is a successive extension of anextension of elementary modules which strictly contains M ′. This contradicts themaximality of M .

Recall that the annihilator of M , Ann(M), is the set of r ∈ R with rM = 0.It is clearly an ideal of R. We denote by P(M) the set of prime ideals of R whichcontain Ann(M) and are minimal for that property. According to Exercise 14 theseare finite in number and their common intersection equals

√Ann(M) (recall that

here R is noetherian). In the graded setting, Ann(M) is a graded ideal and thenaccording to Lemma 2.3 the members of P(M) are all graded.


PROPOSITION 9.8. In the situation of the preceding proposition, let p(i) be theprime ideal of R such that M i−1/M i ∼= R/p(i). Then P(M) is the set of minimalmembers of the collection p(i)di=1 and for every p ∈ P(M), µp(M) is finite and p

occurs precisely µp(M) times in the sequence (p(1), . . . , p(d)).

PROOF. We first show that√

Ann(M) = p(1) ∩ · · · ∩ p(d). If r ∈ p(1) ∩ · · · ∩p(d), then r maps M i−1 to M i and so rd ∈ Ann(M) and hence r ∈

√Ann(M).

Conversely, if r ∈ R and l ≥ 1 are such that rl ∈ Ann(M), then for all i, rl ∈ p(i)

and hence r ∈ p(i). Since every prime ideal containing√

Ann(M) = p(1)∩· · ·∩p(d)

contains some p(i) it also follows that P(M) is the collection of minimal membersof p(i)di=1.

Fix p ∈ P(M). We then have a filtration Mp = M0p ⊃ · · · ⊃ Md

p = 0(for an inclusion of R-modules induces an inclusion of Rp-modules). We haveM i−1

p /M ip∼= Rp/p

(i)Rp. Either p(i) = p, and then the latter is equal to the residuefield Rp/pRp and hence of length 1. Or p(i) 6= p, and then we cannot have p(i) ⊂ p

by the minimality of p. So there exists an r ∈ p(i) r p. This means that r/1 ∈ p(i)Rp

is invertible so that p(i)Rp = Rp, or equivalently M i−1p /M i

p = 0. Following ourdefinition the first case occurs precisely µp(M) times.

We can of course pass from the graded case to the nongraded case by justforgetting the grading. But more interesting is the following construction, whichwe shall use to pass from a projective setting to an affine one and vice versa.

Let p ⊂ R be a homogeneous prime ideal and let us write mp for the maximalideal pRp of the ring Rp. Given l ∈ Z, let Rp,l denote the set of Rp that arerepresentable as r/s with r ∈ Ri+l and s ∈ Ri r pi for some i and put Rp,• :=⊕l∈ZRp,l. Note that Rp,0 ⊂ Rp,• ⊂ Rp are ring inclusions of which Rp,0 and Rp

are local rings (the maximal ideal mp,0 of Rp,0 is obtained by taking in the previoussentence r ∈ pi), but Rp,• has maximal ideals other than mp ∩Rp,• (see below).

Suppose now that p1 6= R1 and choose s ∈ R1 r p1 so that 1/s ∈ Rp,−1.Then multiplication with sl defines an Rp,0-module isomorphism of Rp,0

∼= Rp,l

(the inverse is given by multiplication with s−l). So Rp,• is the ring of Laurentpolynomials Rp,0[s, s−1] with mp ∩ Rp,• corresponding to mp,0[s, s−1]. So we haveRp,0/mp,0[s, s−1] ∼= Rp,•/mp∩Rp,•). But Rp,•/(mp∩Rp,•) is a subring of the residuefieldRp/mp and generates the latter as a field. It follows thatRp/mp

∼= Rp,0/mp,0(s)is a purely transcendental field extension of Rp,0/mp,0.

This also makes sense for any graded R-module M : Mp,l is the set of fractionsm/s with m ∈Mi+l and s ∈ Ri r pi for some i. Note that this is a Rp,0-module.

COROLLARY 9.9. If M is noetherian, then µp(M) = µmp,0(Mp,0).

PROOF. An iterated extension M = M0 ) M1 ) · · · ) Md = 0 of M by ele-mentary graded R-modules yields an iterated extension of Mp resp. Mp,0 by trivialor by elementary Rp resp. Rp,0-modules. The corollary then follows from the ob-servation that a successive quotient M i−1

p /M ip is nonzero (which is then isomorphic

to the big residue field Rp/mp) if and only M i−1p,0 /M

ip,0 is (which is then isomorphic

to the small residue field Rp,0/mp,0).

We use this observation mainly via the following example.

EXAMPLE 9.10. Let V be a vector space of dimension n + 1, J ⊂ k[V ] a ho-mogeneous ideal and p ∈ P(V ) an isolated point of the closed subset Z[J ] ⊂ P(V )

10. HILBERT FUNCTIONS AND HILBERT POLYNOMIALS 81

defined by J . We take here M := k[V ]/J and take for p the graded ideal Ip ⊂ k[V ]defining p. Then k[V ]Ip,0 can be identified with the local k-algebra OP(V ),p. If Jp ⊂OP(V ),p denotes the ideal corresponding to JIp,0 ⊂ k[V ]Ip,0, then

√Jp = mP(V ),p

and we can identify MIp,0 with OP(V ),p/Jp. According to the above discussionµIp(M) = µp(k[V ]Ip,0/JIp,0) and by Exercise 68 this is just dimk(OP(V ),p/Jp).

10. Hilbert functions and Hilbert polynomials

We shall be dealing with polynomials in Q[z] which take integral values on in-tegers. Such polynomials are called numerical. An example is the binomial functionof degree n ≥ 0: (

z

n

):=

z(z − 1)(z − 2) · · · (z − n+ 1)

n!.

It has the property that its value in any integer i is an integer, for i ≥ n this is anordinary binomial coefficient and hence an integer, for i ≤ −1 this is so up to sign,for then we get (−1)n

(n−1−in

)and for 0 ≤ i ≤ n− 1 it is 0.

Let ∆ : Q[z] → Q[z] denote the difference operator: ∆f(z) := f(z + 1) −f(z). This is a Q-linear map with kernel Q and has the property that it decreasesthe degree of nonconstant polynomials. It clearly sends numerical polynomials tonumerical polynomials and a simple verification shows that it maps

(z

n+1

)to(zn

).

LEMMA 10.1. Every P ∈ Q[z] which takes integral values on sufficiently large in-tegers is in fact numerical and a Z-basis of the abelian group of numerical polynomialsis provided by the binomial functions.

If f : Z→ Z is a function such that for sufficiently large integers ∆f is given by apolynomial, then so is f .

PROOF. The first assertion is proved with induction on the degree d of P . Ifd = 0, then P is constant and the assertion is obvious. Suppose d > 0 and theassertion known for lower values of d. So ∆P (z) =

∑d−1i=0 ci

(zi

)for certain ci ∈ Z.

Then P (z) −∑d−1i=0 ci

(zi+1

)is in the kernel of ∆ and hence is constant. As this

expression takes integral values on large integers, this constant is an integer. Thisproves that P is an integral linear combination of binomial functions.

The proof of the second assertion is similar: let P ∈ Q[z] be such that P (i) =∆f(i) ∈ Z for large i. By the preceding, P (z) =

∑i ai(zi

)for certain ai ∈ Z. So if

we put Q(z) :=∑i ai(zi+1

), then Q is a numerical polynomial with ∆(f −Q)(i) = 0

for large i. This implies that f − Q is constant for large i, say equal to a ∈ Z. Sof(i) = Q(i) + a for large i and hence Q+ a is as required.

We shall see that examples of such functions are furnished by the Hilbert func-tions of graded noetherian modules.

REMARK 10.2. A function f : Z→ Z which is zero for sufficiently negative integers de-termines a Laurent series Lf :=

∑k∈Z f(k)uk ∈ Z((u)). For the function k 7→ max0,

(kn

)

this gives∑k≥n

k(k − 1) · · · (k − n+ 1)

n!uk =

un

n!

dn

dun

∑k≥0

uk =un

n!

dn

dun1

1− u =un

(1− u)n=( u

1− u)n.

So if we also know that for sufficiently large integers f is the restriction of a polynomialfunction, then Lemma 10.1 implies that Lf is a Z-linear combination of powers of u andpowers of u

1−u = 11−u − 1. In particular, Lf ∈ Z[u, 1/(u− 1)].


In the remainder of this section V is a k-vector space of dimension n+1 (we allown = −1). We equip k[V ] with the usual grading (for which each linear form on V hasdegree one) and view it as the homogeneneous coordinate ring of P(V ). A k[V ]-moduleis always assumed to be graded and finitely generated.

Let M be a graded finitely generated k[V ]-module. Then for every i ∈ Z, Mi isa finite dimensional k-vector space and so we may define the Hilbert function of M ,φM : Z→ Z, by φM (i) := dimkMi. For example, the Hilbert function of k[V ] itselfis i 7→

(i+nn

)and so is given by a numerical polynomial of degree n.

The graded ideal Ann(M) defines a closed subset of P(V ) that is called the(projective) support of M and denoted supp(M).

THEOREM 10.3 (Hilbert-Serre). LetM be a graded finitely generated k[V ]-module.Then there exists a unique numerical polynomial PM ∈ Q[z], the Hilbert polynomialof M , such that φM (i) = PM (i) for i sufficiently large. The degree of PM is equal todim supp(M) if we agree that the zero polynomial has the same degree as the dimen-sion of the empty set (namely −1).

PROOF. When V = 0, then M = 0 and there is nothing to show. So assumedimV > 0 and the theorem proved for vector spaces of dimension < dimV .

If N is a graded submodule of M , then Ann(N) ∩ Ann(M/N) has the sameradical as Ann(M) (in fact, Ann(M) ⊃ Ann(N) ∩ Ann(M/N) ⊃ Ann(M)2) andso supp(M) = supp(N) ∪ supp(M/N). It is clear that dimkMi = dimkNi +dimk(Mi/Ni) and so we have φM = φN + φM/N . It follows that if the theoremholds for N and M/N , then it holds for M . As M is a successive extension of ele-mentary modules, it suffices to do the case M = A[l], where A = k[V ]/p with p agraded prime ideal. But φ(A[l](i) = φA(i + l) and since the degree of a polynomialdoes not change after the substitution z 7→ z+ l, it is even enough to do the case A.

Then supp(A) = Z[p] is the closed irreducible subset of P(V ) defined by thegraded ideal p. In case p = k[V ]+, the theorem holds trivially: we have dimk Ai = 0for i > 0 (so that we may take PA to be identically zero) and supp(A) = ∅. Supposetherefore that p 6= k[V ]+, so that there exists a T ∈ k[V ]1 = V ∗ that is not in p1.Denote by V ′ ⊂ V its zero hyperplane. Since k[V ]/p is a domain, multiplication byT induces an injection A → A (increasing the degree by one) with cokernel A′ :=A/TA and so φA′(i) = φA(i)− φA(i− 1) = ∆φA(i− 1). Since Ann(A′) = p + (T ),we have supp(A′) = supp(A) ∩ P(V ′). According to Proposition 6.2 we then havedim supp(A′) = dim supp(A) − 1. Since A′ is a quotient of k[V ′], our inductionhypothesis tells us that there exists a polynomial PA′ of degree dim supp(A′) suchthat φA′ and PA′ coincide on large integers. Since ∆φA(i − 1) = PA′(i) for largei, Lemma 10.1 implies that there exists a polynomial PA of degree one higher thanthat of PA′ (so of degree dim supp(A)) which coincides with φA for sufficientlylarge integers.

REMARK 10.4. For M as in this theorem we may also form the Laurent series LM (u) :=∑i dim(Mi)u

i (this is usually called the Poincare series of M). It follows from Remark 10.2and Theorem 10.3 that LM (u) ∈ Z[u][(u− 1)−1].

It follows from Lemma 10.1 that when PM is nonzero, then its leading term hasthe form cdz

d/d!, where d is the dimension of supp(M) and cd is a positive integer.This observation leads to a notion of degree (that is not to be confused with thedegree of PM):


DEFINITION 10.5. If d = dim supp(M), then the (projective) degree deg(M)is d! times the leading coefficient of its Hilbert polynomial (an integer, which westipulate to be zero in case supp(M) = ∅). For a closed subset Y ⊂ P(V ), the Hilbertpolynomial PY resp. the degree deg(Y ) of Y is that of k[V ]/I(Y ) as a k[V ]-module.

REMARK 10.6. Observe that if Y ⊂ P(V ) is nonempty, then deg(Y ) > 0. Forthen IY,d 6= k[V ]d for every d ≥ 0 and so the Hilbert function of the homogeneouscoordinate ring k[Cone(Y )] = k[V ]/I(Y ) is positive on all nonnegative integers.This implies that PY is nonzero with positive leading coefficient. We also note thatsince deg(Y ) only depends on the dimensions of the graded pieces of k[Cone(Y )],it is for this notion irrelevant whether Y happens to lie in a lower dimensionalprojective space Y ⊂ P(V ′) ⊂ P(V ) for some V ′ ⊂ V . For example, the degree of asingleton y ⊂ P(V ) is the degree of the k[T ]-module k[T ] and hence equal to 1.

One can show that there exists a nonempty open subset of linear subspacesQ ⊂ P(V ) of codimension equal to dimY which meet Y in exactly deg(Y ) points.This characterization is in fact the classical way of defining the degree of Y .

EXERCISE 69. Suppose that M is not of finite length. Prove that there is aunique integer d ≥ 0 such that i 7→ ∆dφM (i) is a nonzero constant for i sufficientlylarge. Show that d is the dimension of the support of M and that the constant is itsdegree.

EXERCISE 70. Compute the Hilbert polynomial and the degree of

(a) the image of the d-fold Veronese embedding of Pn in P(n+dn )−1,

(b) the image of the Segre embedding of Pm × Pn in Pmn+m+n.

EXERCISE 71. Let Y ⊂ Pm and Z ⊂ Pn be closed and consider Y ×Z as a closedsubset of Pmn+m+n via the Segre embedding. Prove that the Hilbert function resp.polynomial of Y ×Z is the product of the Hilbert functions resp. polynomials of thefactors.

We may now supplement Theorem 10.3 as follows. Let M be as in that theo-rem: a finitely generated graded k[V ]-module. Recall that P(M) denotes the set ofminimal prime ideals containing Ann(M). For every p ∈ P(M) not equal to k[V ]+,the associated closed subset Z[p] ⊂ P(V ) is an irreducible component of supp(M)and all irreducible components of supp(M) are so obtained (for p = k[V ]+ we getthe empty set, but we also have deg(k[V ]/k[V ]+) = 0). Denote by Po(M) the set ofp ∈ P(M) that define an irreducible component of supp(M) of the same dimensionas supp(M).

PROPOSITION 10.7. Let M be a finitely generated graded k[V ]-module. Then

deg(M) =∑

p∈Po(M)

µp(M) deg(Z[p]).

PROOF. We write M as an iterated extension by elementary modules: M =M0 ) M1 ) · · · ) Md = 0 with M i−1/M i ∼= k[V ]/pi[li]. Then φM (i) =∑di=1 φk[V ]/pi(i+li). Now φk[V ]/pi is a polynomial of degree equal to the dimension

of supp(k[V ]/pi) = Z[pi] ⊂ P(V ). This degree does not change if we replace thevariable i by i + li. In view of Proposition 9.8 we only get a contribution to theleading coefficient of φM when pi ∈ Po(M) and for any given p ∈ Po(M) thishappens exactly µp(Mp) times. The proposition follows.


REMARK 10.8. Note the special case when M has finite support: then Po(M) =P(M) \ k[V ]+ and this set is in bijective correspondence with the points ofsupp(M). For p ∈ Po(M), Z[p] ⊂ P(V ) is just a singleton p and so deg(Z[p]) = 1.Furthermore, µp(M) is the length of Mp as a k[V ]-module and this is by Remark9.10 equal to dimkMp, whereMp := Mp,o is a OP(V ),p = k[V ]p,o-module of finitelength. So the above formula then says that deg(M) =

∑p∈supp(M) dimk(Mp).

EXERCISE 72. Let Y ⊂ P(V ) be closed. Prove that if Y1, . . . , Yr are the distinctirreducible components of Y of maximal dimension (= dimY ), then deg(Y ) =∑ri=1 deg(Yi).

We can now state and prove a result of Bezout type.

PROPOSITION 10.9. Let M be a graded k[V ]-module and F ∈ k[V ]d with F /∈Ann(M). Then deg(M/FM) = ddeg(M).

PROOF. Our assumption implies that the sequence

0→M(−d)·F−→M →M/FM → 0

is exact. This shows that PM/FM (z) = PM (z)− PM (z − d). Put m := dim supp(M)

so that is we write PM (z) =∑mi=0 aiz

i/i!, then am = deg(M). Since we havezi/i! − (z − d)i/i! = dzi−1/(i − 1)!+ lower order terms, we find that PM/FM (z) =

damzm−1/(m− 1)!+ lower order terms. So deg(M/FM) = dam = ddeg(M).

Note the special case for which M = k[V ] and F is a generator of the idealdefining a hypersurface H ⊂ P(V ). Then PM (z) =

(n+zn

)and so the degree of M

(which is also the degree of P(V )) is 1 and hence the degree of H is d, just as wewould expect. We can now state:

THEOREM 10.10 (Theorem of Bezout). Let Hi ⊂ P(V ) be a hypersurface ofdegree di > 0 (i = 1, . . . n), and assume that H1 ∩ · · · ∩ Hn is finite. Each Hi

determines at p ∈ H1 ∩ · · · ∩Hn a principal ideal in OP(V ),p; denote by Ip ⊂ OP(V ),p

the sum of these ideals. Then

d1d2 · · · dn =∑

p∈H1∩···∩Hn

dimk(OP(V ),p/Ip).

Here dimk(OP(V ),p/Ip) should be interpreted as the intersection multiplicitythe hypersurfaces H1, . . . ,Hn at p. So the theorem can be paraphrased as sayingthat H1, . . . ,Hn meet in d1d2 · · · dn points, provided we count each such point withits intersection multiplicity.

We shall need the following result which we state without proof.

*PROPOSITION 10.11. Let F1, . . . , Fr be r ≤ n + 1 homogeneous elements ofk[V ] such that dim(k[V ]/(F1, . . . , Fr)) = n + 1 − r . Then the image of Fr ink[V ]/(F1, . . . , Fr−1) is not a zero divisor.

PROOF OF THEOREM 10.10. Let Fi ∈ k[V ]di defineHi. Denote byM i the k[V ]-module k[V ]/(F1, . . . , Fi) (so that M0 = k[V ]). Then according to Propositions10.9 and 10.11 we have deg(M i) = di deg(M i−1). Since degM0 = 1, it followsthat deg(Mn) = d1d2 · · · dn. The support of Mn is H1 ∩ · · · ∩Hn and hence finite.Its degree is then also computed as

∑p∈Po(Mn) µp(Mn). But according to Remark

10.8 this is just∑p∈H1∩···∩Hn dimk(OP(V ),p/Ip).


EXAMPLE 10.12. Assume char(k) 6= 2. We compute the intersection multiplic-ities of the conics C and C ′ in P2 whose affine equations are x2 + y2 − 2y = 0and x2 − y = 0. There are three points of intersection: (0, 0), (−1, 1) and (1, 1)(so none at infinity). The intersection multiplicity at (0, 0) is the dimension ofOA2,(0,0)/(x

2 + y2−2y, x2− y) as a k-vector space. But OA2,(0,0)/(x2 + y2−2y, x2−

y) = OA1,0/(x4 − x2) = k[x]/(x2) (for (x2 − 1) is invertible in OA1,0). Clearly

dimk(k[x]/(x2)) = 2 and so this is also the intersection multiplicity at (0, 0). Theintersection multiplicities at (−1, 1) and (1, 1) are easily calculated to be 1 and thusthe identity 2 + 1 + 1 = 2 · 2 illustrates the Bezout theorem.

REMARK 10.13. If Y ⊂ Pn is closed, then PY (0) can be shown to be an invariantof Y in the sense that it is independent of the projective embedding. In many ways,it behaves like an Euler characteristic. (It is in fact the Euler characteristic of OYin a sense that will become clear once we know about sheaf cohomology.) Forexample, PY×Z(0) = PY (0)PZ(0).

We have seen that for a hypersurface Y ⊂ Pn of degree d > 0, PY (z) =(z+nn

)−(

z−d+nn

)and so PY (0) = 1 −

(−d+nn

)= 1 − (−1)n

(d−1n

). For n = 2 (so that Y is a

curve), we get PY (0) = 1− 12 (d−1)(d−2). The number 1−PY (0) = 1

2 (d−1)(d−2)is then called the arithmetic genus of the curve. If the curve is nonsingular andk = C, then we may regard it as a topological surface (a Riemann surface) and g isthen just the genus of this surface (and so PY (0) is half its Euler characteristic).

CHAPTER 3

Schemes

1. Presheaves and sheaves

The notion of a prevariety and its generalizations are best understood in themore general context of ringed spaces. This involves the even more basic notion ofa sheaf.

DEFINITION 1.1. Let X be a topological space. An abelian presheaf F on Xconsists of giving for every open subset U ⊂ X an abelian group F(U) (whoseelements are called sections of F over U) and for every inclusion of open subsetsU ⊂ U ′ a homomorphism of groups F(U ′) → F(U) (called the restriction map)such that:

(i) for the identity U = U we get the identity in F(U),(ii) if U ⊂ U ′ ⊂ U ′′ are open sets, then the homomorphism F(U ′′) → F(U)

is equal to the composite F(U ′′)→ F(U ′)→ F(U) and

So in categorical language, a presheaf is a contravariant functor F : O(X) →Ab from the category O(X) of open subsets of X (whose morphisms are inclusionsof open subsets of X) to the category of abelian groups Ab. The terminology ‘sec-tion over U’ and ‘restriction’ is suggestive, although can sometimes be a bit mislead-ing. This is mirrored by the notation: if we are given an inclusion of open subsetsU ′ ⊂ U , then the image of s ∈ F(U) under the restriction map F(U) → F(U ′) isoften denoted s|U ′ (the obvious categorical notation for the restriction map wouldbe F(U ′ ⊂ U), but this is rarely used; more common is resU,U ′).

In case the groups F(U) come with the structure of a ring and the restrictionmaps are ring homomorphisms, then we say that F is a presheaf of rings (in otherwords, the contravariant functor takes values in the category Ring of rings). Like-wise, we have the notion of a presheaf of modules over a fixed ring R. In somesituations no structure on F(U) is imposed at all (the target category is then theone of sets), but we will mostly deal with presheaves that are at least abelian.

EXAMPLE 1.2 (The constant presheaf). Given an abelian group G and a topo-logical space X, then we have an abelian presheaf defined on X if we take for everynonempty open U ⊂ X the group G and for each inclusion between two such setsthe identity map of G.

EXAMPLE 1.3. Given a topological space X and an abelian topological groupG, then assigning to every nonempty open U ⊂ X the group of continuous mapsU → G (with the obvious restriction maps) is an abelian presheaf CX,G. Of specialinterest areG = R andG = C, in which we get a presheaf of R-algebras (resp. of C-algebras). Another case of interest is when G has been given the discrete topology:then this assigns to U the space of locally constant maps U → G.

87

88 3. SCHEMES

EXAMPLE 1.4. For a smooth manifold M , we have defined the presheaf EMof R-algebras which assigns to every nonempty open U ⊂ M the R-algebra ofdifferentiable functions U → R.

EXAMPLE 1.5. On a complex manifold M , we have defined the presheaf OanM of

C-algebras which assigns to every nonempty open U ⊂M the C-algebra Oan(U) ofholomorphic functions U → C.

EXAMPLE 1.6. On a k-prevarietyX, we have defined the presheafOX of regularfunctions.

DEFINITION 1.7. Given a presheaf F on X and a point p ∈ X, then a germ ofa section of F at p, is a section of F on an unspecified neighborhood of p, withthe understanding that two such sections represent the same germ if they coincideon a neighborhood of p contained in their common domain of definition. The setof germs of sections of F at p is called the stalk of F at p and denoted by Fp.(In categorical language, this is expressed as an inductive limit Fp = lim−→U3p F(U),where one observes that the collection of neighborhoods of p in X form a projectivesystem: the intersection of two neighborhoods of p is another one.)

For instance in the case of Example 1.5, when M is an open subset of C, thenthe stalk Oan

M,p can be identified with the ring of convergent power series in z − p.Let us observe that an s ∈ F(U) determines an element sp ∈ Fp for every

p ∈ U . We will also write s+(p) for sp, where we view s+ as a map U →∏p∈U Fp.

The last three examples differ from the first in that they satisfy two additionalproperties:

DEFINITION 1.8. An abelian presheaf F on X is called an abelian sheaf if forevery open U ⊂ X and open covering Uii∈I of U ,

(iv) every system of sections si ∈ F(Ui)i∈I that is compatible in the sensethat for each pair (i, j) in I, si|Ui ∩ Uj = sj |Ui ∩ Uj comes from somesection s ∈ F(U) in the sense s|Ui = si for all i ∈ I, and

(v) two sections s, s′ ∈ F(U) coincide if s|Ui = s′|Ui for all i ∈ I.

In other words, the sequence

0→ F(U)→∏i∈IF(Ui)→

∏(i,j)∈I2

F(Ui ∩ Uj)

in which the first map is given by s 7→ (s|Ui)i∈I and the second by (si)i∈I 7→(si|Ui ∩ Uj − sj |Ui ∩ Uj)i,j is exact. This implies that F(∅) is the trivial group 0.The argument relies on a categorical notion of product1, but for our purposes wemay just as well stipulate that this is so.

A space X endowed with with a sheaf of rings is also called a ringed space.Examples are 1.3, 1.4, 1.5 and 1.6. A constant presheaf is usually not a sheaf: ifU1, U2 are disjoint open subsets and we take gi ∈ F(Ui) = G distinct for i = 1, 2,

1If we are given a product of abelian groups with index set I,∏i∈I Gi, then for every subset

J ⊂ I, this product maps isomorphically to the product (∏i∈J Gi) × (

∏i∈I−J Gi). By taking J = I,

we see that a product with empty index set must be the trivial group. Similarly, every union of sets∪i∈ITi, can be written as (∪i∈JTi) ∪ (∪i∈I−JTi), from which deduce (by taking J = I), that a unionwith empty index set must be empty. If we then apply the above exact sequence to U = ∅ by taking anopen covering with empty index set, we find that F(∅) = 0.

1. PRESHEAVES AND SHEAVES 89

then clearly these elements cannot be the restriction of a single g ∈ F(U1∪U2) = G.There is however a simple modification which is a sheaf, namely the constant sheafGX which assigns to U ⊂ X the space of continuous maps from to U to G, wherewe endow G with the discrete topology (so these are the maps which are locallyconstant). This turns it into a special case of Example 1.3.

1.9. SHEAFIFICATION. This modification is a special case of a general construc-tion which produces a sheaf F+ out of a presheaf F with the same stalks: a sections of F+ over an open subset U is by definition a map from U to the disjoint unionof the stalks Fp, p ∈ U , with the property that locally this map is of the form s+: wemay cover U by open subsets Ui for which there exist si ∈ F(Ui) such that sp = si,pfor all p ∈ Ui. It is straightforward to verify that this is well-defined, defines a sheafand that F+ = F in case F happens to be a sheaf.

The sheaf F+ has a universal property. In order to explain this, we need thenotion of a homomorphism of presheaves. Let C be one of our categories. If Fand G are C-valued presheaves on X, then a homomorphism of presheaves φ : F →G is what is called in category theory a natural transformation of functors: forevery open U ⊂ X, we are given C-homomorphism φ(U) : F(U) → G(U) suchthat this collection is compatible with restriction: if U ′ ⊂ U and s ∈ F(U), thenφ(U)(s)|U ′ = φ(U ′)(s|U ′).

For example, the formation of F+ defines a homomorphism of presheaves F →F+. It is universal in the sense that if φ : F → G is a homomorphism of presheavesand G is a sheaf, then there is unique sheaf homomorphism φ+ : F+ → G such thatφ is the composite of F → F+ and φ+.

In general a homomorphism of presheaves φ : F → G induces a sheaf homo-morphism φ+ : F+ → G+.

EXERCISE 73 (Local nature of a sheaf). Let X be a topological space and Ua basis of open subsets of X. Prove that an abelian sheaf F on the space X isdetermined by its restriction to that basis, that is, by the collection F(U), and therestriction maps F(U ′)→ F(U), for the pairs (U,U ′) ∈ U × U with U ⊂ U ′.

Prove also a converse: suppose that the basis U is closed under finite intersec-tions and assume that for every U ∈ U is given an abelian group F(U) and for everyinclusion U ⊂ U ′ of members of U a homomorphism F(U ′) → F(U) such that theproperties (i) through (iii) are satisfied. Then F generates a sheaf F+ on X.

EXERCISE 74 (Direct image of a sheaf). Suppose we are given a continuousmap f : X → Y between topological spaces and an abelian presheaf F on X. Thena presheaf f∗F on Y is defined by assigning to an open V ⊂ Y the group F(f−1V ).Prove that if F is a sheaf, then so is f∗F . Is it true that (f∗F)+ = f∗(F+)?

EXAMPLE 1.10. Let f : S1 → S1 be defined by f(z) = z2 and let ZS1 be theconstant sheaf on S1 with stalks Z. Then f∗ZS1 is a sheaf on S1 which is locallylike the constant sheaf Z2. But (f∗ZS1)(S1) = ZS1(S1) = Z and so is of rank lessthan 2. Other examples in the same spirit are gotten as the sheaf on C r 0 thatis the direct image of the constant sheaf on C r 0 with stalk Z under the mapz ∈ C r 0 7→ zn ∈ C r 0 (n ∈ Z r 0) or of the direct image of the constantsheaf on C with stalk Z under the map z ∈ C 7→ ez ∈ Cr 0.

1.11. KERNELS AND COKERNELS. We begin with some obvious definitions:given a presheaf F on X, then a subpresheaf of F is simply a presheaf F ′ on X

90 3. SCHEMES

with the property that F ′(U) ⊂ F(U) for every open subset U of X. If F ′ happensto be a sheaf, then we call it a subsheaf of F .

Suppose and F and G are abelian presheaves over X and let φ : F → Gbe a homomorphism of presheaves. So for every open U ⊂ X, we then have ahomomorphism φ(U) : F(U) → G(U). By assigning to U , Ker(φ)(U), Imφ(U),Coker(φ(U)) we have defined three presheaves, of first two being subpresheaves ofF and G respectively. If Ker(φ) is the zero sheaf, then clearly F can be thought of asubpresheaf of G. Suppose now that F and G are in fact sheaves. Then Ker(φ)(U)is easily seen to be a sheaf (and indeed denoted by Ker(φ)). But the other two neednot be (see Example 1.12 below) and so we define the image sheaf Im(φ) resp. thecokernel sheaf Coker(φ) as the sheaf associated to the corresponding presheaves.Since this process does not affect the stalks, we have that for every x ∈ X, Im(φ)xresp. Coker(φ)x is the image resp. the cokernel of φx : Fx → Gx. Moreover, Im(φ)xis a subsheaf of G. The possible failure of the image presheaf being a sheaf is ex-pressed by the fact that the image of φ(U) : F(U)→ G(U) is contained in Im(φ)(U)but need not equal it and a similar statement can be made for the cokernel sheaf.

We finally note that the homomorphisms from F to G form an abelian group;we denote that group by Hom(F ,G).

If F is a sheaf on X, then F(X) is called its group of global sections. Oneoften denotes this group by Γ(X,F) or H0(X,F) instead. The reason for the latternotation will become clear later.

An interesting twist to the notion of a constant sheaf is that of a local system(and indeed called by some algebraic topologists a system of twisted coefficients):this is a locally constant sheaf. For instance, if G is an abelian group, then a localsystem of type G on X is an abelian sheaf F on X with the property that X can becovered by open subsets U such that F|U is isomorphic to the sheaf associated tothe constant presheaf G on U . Examples were given in 1.10. Here is another one,which illustrates some other phenomena as well.

EXAMPLE 1.12. Let X be C r 0 with its usual Hausdorff topology. On X wedefine a local system L of 2-dimensional complex vector spaces by letting L(U)be the space of C-valued functions that are of the form f + c with f : U → Ccontinuous and satisfying ef(z) = z. In other words, f is a branch of the logarithmfunction. But recall that branch need not defined and that

The constant functions define a constant subsheaf CX ⊂ L. The quotient sheafL/CX is also isomorphic to CX , for it admits as a global generating section thefunction “log” (note that the multivaluedness of this function disappears if we workmodulo constants). But the section “log” cannot be lifted to X: the only globalsections of L(X) are the constants. So the exact sequence of abelian sheaves

0→ CX → LX → LX/CX → 0

evaluated on X yields a sequence of abelian groups

0→ Γ(X,CX)→ Γ(X,LX)→ Γ(X,LX/CX)→ 0.

We observed that the last sequence can be identified with 0 → C = C 0−→ C → 0.So it fails to be exact at Γ(X,CX). This also shows that the image presheaf ofL → LX/CX and the cokernel presheaf of CX → L are not sheaves.

1.13. THE PREIMAGE OF A SHEAF. Here is a relative version of the sheafificationprocess. Suppose we are given a continuous map f : X → Y between topological

1. PRESHEAVES AND SHEAVES 91

spaces and an abelian sheaf G on Y . We want to define a sheaf f−1G on X with theproperty that the stalk of f−1G at x ∈ X is Gf(x). If U ⊂ X is open and s assignsto x ∈ X, sx ∈ Gf(x), then we stipulate that s defines a section of f−1F if for eachx ∈ U there exists a neighborhood Vx of y in Y and a section t ∈ G(Vy) such thatfor x′ in a neighborhood of x in f−1Vx ∩ U , we have sx′ = tf(x). This is indeed asheaf, called the sheaf pull-back of G. If Y is a single point y, so that G is givenby a single group G = Gy, then this reproduces the sheaf of locally constant mapsfrom open subsets of X to G.

If Y ⊂ X is a subspace, then we often write Γ(Y,F) for the group of globalsections of i−1F , where i : Y ⊂ X denotes the inclusion. Notice that when Y isopen, this is just F(Y ).

EXERCISE 75 (Adjunction). Suppose that f : X → Y is a continuous map andlet F resp. G be an abelian sheaf on X resp. Y .

(a) Prove that we may identify Hom(G, f∗F) with Hom(f−1G,F), in categor-ical language, that the functor f−1 : AbY → AbX , G 7→ f−1G, is the leftadjoint of the functor f∗ : AbX → AbY defined by F 7→ f∗F and thelatter is the right adjoint of the former.

(b) Deduce the existence of natural sheaf homomorphisms G → f∗f−1G and

f−1f∗F → F .(c) Give examples to show that neither of these needs to be an isomorphism.

(Hint for the second case: look at Example 1.10.)(d) Suppose that f is a homeomorphism. Show that under the identification

(a) a sheaf isomorphism G → f∗F yields a sheaf isomorphism f−1G → Fand vice versa. (When such an isomorphism exists, we say that (X,F)and (Y,G) are isomorphic.)

A space X endowed with a sheaf of rings OX is called a ringed space; the sheafin question is then often referred to as its structure sheaf. This notion becomes muchmore useful if we also have a notion of morphism of ringed spaces: if (X,OX) and(Y,OY ) are ringed spaces, then by a morphism from (X,OX) to (Y,OY ) we mean apair consisting of a continuous map f : X → Y and a sheaf homomorphism of ringsOY → f∗OX . According to Exercise 75 the latter is equivalent to giving a sheafhomomorphism of rings f−1OY → OX . (We often abuse notation by denoting thispair by the single symbol f .)

REMARK 1.14. Via Exercise 73 we can (at least formally) characterize certainextra structures on spaces in a uniform manner. For example, given a topologicalm-manifoldM , then a Ck-differentiable structure onM is simply given by a subsheaf ofthe sheaf R-valued continuous functions on M which has the property that locallyit is isomorphic to the sheaf of Ck-differentiable functions on Rm. It is a sheafof R-algebras. A Ck-map between two Ck-manifolds is then simply a morphismbetween the corresponding ringed spaces (or rather, spaces endowed with a sheafof R-algebras). This is more conceptual (and perhaps also more concise) thanthe definition based on an atlas. A holomorphic structure on M (which turns Minto a complex manifold) and a holomorphic map between complex manifolds canbe defined in a similar manner. An affine variety is naturally a ringed space andindeed, a k-prevariety, is a ringed space locally isomorphic as a ringed space to anaffine k-variety. This is why such sheaves are often referred to as structure sheaves.We shall define in the same spirit a structure sheaf on the spectrum of a ring.

92 3. SCHEMES

2. The spectrum of a ring as a locally ringed space

Let R be a ring. The (prime ideal) spectrum Spec(R) is the set of its primeideals. As a rule, we denote for a prime ideal p of R the corresponding element ofSpec(R) by [p] and the prime ideal associated to x ∈ Spec(R) by px ⊂ R.

For s ∈ R, Z(s) ⊂ Spec(R) denotes the set of [p] ∈ Spec(R) with s ∈ p andSpec(R)s := Spec(R)− Z(s). The collection Spec(R)ss∈R, is basis for a topologyon Spec(R), the Zariski topology. We therefore refer to any subset of Spec(R) of theform Spec(R)s as a principal open subset of Spec(R). So the closed subsets of thistopology are of the form Z(S) = ∩s∈SZ(s), where S is any subset of R (but notethat Z(S) does not change if we replace S by the radical of the ideal generated byS). The same argument as used in Lemma 4.7 shows that Spec(R) is quasicompact.

The spectrum of a ring R is a singleton precisely when R has just one primeideal. Since

√(0) is the intersection of all the prime ideals of R, it follows that this

prime ideal equals√

(0) and must be a maximal ideal. (So if R is noetherian it willhave finite length.) In particular,

√(0) = (0) precisely when R is a field.

The spectrum of a local ring has just one closed point (namely the one definedby its maximal ideal), but can have plenty of nonclosed points, of course.

A ring homomorphism φ : R′ → R determines a map

Spec(φ) : Spec(R)→ Spec(R′), [p] 7→ [φ−1p].

It has the property that Spec(φ)−1(Spec(R)r′) = Spec(R)φ(r′) so that Spec(φ) iscontinuous. The ring homomorphism R′ → R → Rp factors through R′p′ (by Exer-cise 11) and so we have an associated homomorphism of local rings φp : R′p′ → Rp.We have φ−1

p (pRp) = p′Rp′ and so this homomorphism is local in the sense of:

DEFINITION 2.1. A homomorphism ψ : A′ → A of local rings is said to be alocal homomorphism if ψ−1mA = mA′ (so that it defines an embedding of residuefields A′/mA′ → A/m).

Hence a homomorphism of local rings is local precisely if the induced map onspectra sends the closed point to the closed point. For instance, the inclusion of thelocal ring k[[t]] in its quotient field k((t)) is not local, for the maximal ideal of k((t))is the zero ideal, but the zero ideal is not equal to the maximal ideal tk[[t]] of k[[t]].Thus the singleton Spec(k((t))) maps to the nonclosed point of Spec(k[[t]]) definedby the zero ideal.

For every subset Y ⊂ Spec(R), we may consider the set I(Y ) of r ∈ R which‘vanish’ on Y , that is for which Z(r) ⊃ Y . This is clearly an ideal in R. In thiscontext, a Nullstellensatz comes easily as it is basically the content of Lemma 2.16:

PROPOSITION 2.2. Given a ring R, then for every ideal J ⊂ R, I(Z(J)) =√J .

PROOF. The property r ∈ I(Z(J)) means Z(r) ⊃ Z(J). This in turn meansthat every prime ideal which contains J must contain r. In other words, I(Z(J)) isthe intersection of all the prime ideals that contain J . According to Exercise 14 thisintersection is precisely

√J and so r ∈

√J .

PROPOSITION 2.3. Let R be a ring.(i) The closure of the singleton [p] consists of the [q] ∈ Spec(R) with q ⊃ p.

(ii) A singleton [p] is closed if and only if p is a maximal ideal (so that Spec(R)contains Spm(R) as the set of its closed points).

2. THE SPECTRUM OF A RING AS A LOCALLY RINGED SPACE 93

(iii) A closed subset of Spec(R) is irreducible if and only if it is the closure of asingleton.

(iv) A subset of Spec(R) is an irreducible component of Spec(R) if and only if itis of the form [p] with p a minimal prime ideal of R.

PROOF. The proofs of (i) and (ii) are left to you as an exercise. As for (iii)and (iv), let C ⊂ Spec(R) be a closed irreducible subset. We first prove that I(C)is prime. For suppose that r1r2 ∈ I(C) for certain r1, r2 ∈ R. Then Z(r1r2) =Z(r1) ∪ Z(r2) contains C and since C is irreducible it follows that Z(r1) ⊃ C orZ(r2) ⊃ C or equivalently, that r1 ∈ I(C) or r2 ∈ I(C). By definition, x ∈ C ifand only if px ⊃ I(C) and so C is the closure of the singleton xI(C). On theother hand, a singleton is irreducible and hence so is its closure. Now (iii) and (iv)follow.

EXERCISE 76. Let φ : R′ → R be a ring homomorphism. Prove that if I ′ ⊂ R′

is a proper ideal, then the natural map Spec(R′/I ′) → Spec(R′) is injective withimage the closed subset of Spec(R′) defined by I ′. Show that the preimage of thisclosed subset under Spec(φ) : Spec(R) → Spec(R′) is the closed subset of Spec(R)defined by the ideal of R generated by φ(I).

REMARK 2.4. The ring Z has as its prime ideals the zero ideal (0) and the idealpZ, where p is a prime number. We denote the corresponding points of Spec(Z) byx0 resp. xp. The only nonclosed point of Spec(Z) is x0, which contains every xp init closure.

Given any ring R, then we have a natural ring homomorphism φ : Z → R andhence a continuous map f := Spec(φ) : Spec(R) → Spec(Z). Let us see what itsfibers are. Let n ≥ 0 be such that ker(φ) = Zn. If p ⊂ R is a prime ideal, thenφ−1p is a prime ideal containing Zn. When n = 0 (so that φ is injective), then thiscan be any prime ideal, otherwise this must be of the form Zp with p dividing n. Inother words, in the first case f is onto and in the second case the image of f is afinite iset of closed points. A prime ideal p ⊂ R with the property that φ−1(p) = Zpcorresponds to a prime ideal of R/pR = Fp⊗R and so the closed fiber f−1(xp) canbe identified with Spec(Fp ⊗ R) (even as topological spaces). The fiber f−1(x0) isnonempty only when φ is injective. We can then localize with respect to the imageof Z r 0 and find that a point of f−1(x0) is given by a prime ideal of Q ⊗ R.Conversely, given a prime ideal of Q ⊗ R, then its inverse image in R under thelocalization homomorphism R→ Q⊗R is a prime ideal of R whose preimage in Zis trivial. Hence f−1(x0) = Spec(Q⊗R), even as topological spaces.

EXAMPLE 2.5 (The affine line). We define the affine line over a ring R asSpec(R[x]). If R is the algebraically closed field k, then its closed points are inbijection with the points of k (any prime ideal is of the form (x− a), with a ∈ k).

When R = R, we get as closed point also the complex conjugate pairs (definedby a prime ideal of the form (x2 + bx+ c) with b2 < 4c). In particular, x2 + 1 has azero in A1

R (namely at the point defined by the prime ideal (x2 + 1)).Next we turn to A1

Z = Spec(Z[x]). From the remark above we see that we havenatural map f : A1

Z → Spec(Z) with f−1(x0) = Spec(Q[x]) = A1Q and f−1(xp) =

Spec(Fp[x]) = A1Fp . The latter is closed in A1

Z. We treat these cases separately.Since Q[x] is a principal ideal domain, every prime ideal of Q[x] is either the

zero ideal or is generated by an irreducible polynomial P ∈ Q[x] of positive degree.

94 3. SCHEMES

In the first case, the prime ideal defines the (generic) point of Spec(Q[t]), whoseclosure is dense in Spec(Q[t]). In the second case we can make P unique by takingit monic. Galois theory tells us that if Q is an algebraic closure of Q, then the rootsof P in Q make up an orbit under the Galois group Aut(Q) and that conversely,any Aut(Q)-orbit in Q is so obtained. It follows the closed points of A1

Q correspondbijectively to the Aut(Q)-orbits in Q. The nonclosed point of A1

Q comes from thezero ideal in Z[x], and so is, when viewed as a point of A1

Z, dense in A1Z.

The case A1Fp can be treated similarly. Fix an algebraic closure Fp of Fp. Since

Fp[x] is a principal ideal domain, every nonzero prime ideal of Fp[x] is generated byan irreducible monic polynomial f ∈ Fp[x] of positive degree r, say. Such a primeideal is then maximal and so defines a finite extension of Fp of the same degreer. Since f has coefficients in Fp, its set of roots is invariant under the Frobeniusautomorphism Φ : a ∈ Fp 7→ ap ∈ Fp. If a is a root of f , then a,Φ(a), . . . ,Φr−1(a)

are pairwise distinct, whereas Φr(a) = a. So∏r−1i=0 (x − Φi(a)) divides f , hence

equals f . Conversely, for any a ∈ Fp which generates a field extension of Fp ofdegree r (equivalently, for which Φr(a) = a), the minimal polynomial for a isirreducible and so generates a maximal ideal in Fp[x]. It follows that the closedpoints of A1

Fp are in bijective correspondence with the Φ-orbits in Fp. There is oneother (nonclosed) point in A1

Fp and its closure equals A1Fp .

Let R be a ring and write X for Spec(R). When s ∈ R is nilpotent, then s iscontained in every prime of R so that Xs = ∅. We then also have that R[1/s] isthe zero ring and since a prime ideal must always be a proper ideal, it follows thatSpec(R[1/s]) = ∅. The following proposition identifies principal open subsets withspectra of simple localizations and shows that inclusions between them come fromobvious ring homomorphisms.

PROPOSITION 2.6. Given s ∈ R, denote by js : R → R[1/s] the natural map.Then Spec(js) : SpecR[1/s]→ Spec(R) = X is a homeomorphism onto Xs. We haveXs ⊂ Xs′ if and only if s ∈

√(s′) and the inclusion is then induced by the associated

homomorphism R[1/s′]→ R[1/s].

PROOF. When R[1/s] is the zero ring there is nothing to prove and so let usassume that this is not the case. Then s is not nilpotent. If p is a prime ideal R[1/r],then p = j−1

s p is a prime ideal with s /∈ p and we have p = p[1/s]. On the otherhand, for a prime ideal p of R with s /∈ p, p[1/s] is a prime ideal of R[1/r] andthe preimage of j−1

s p[1/s] = p. In other words, Spec(js) is injective with imageXs. This map is continuous. It is also open: a nonempty principal open subset ofSpecR[1/s] is defined by a nonnilpotent s′ ∈ R[1/s]. If s′′ ∈ R denotes the productof s and the numerator of s′, then R[1/s][1/s′] = R[1/s′′] and so this principal openset is also a principal open subset of Spec(R).

We have Xs ⊂ Xs′ if and only if Z(s) ⊃ Z(s′) and by Proposition 2.2 thismeans that sn = s′r for some r ∈ R and some integer n > 0. Hence R[1/s′] mapsto R[1/(s′r)] = R[1/s] and this map of course defines the inclusion Xs ⊂ Xs′ .

Henceforth we will use Spec(js) to identify Xs with SpecR[1/s].The following theorem is of fundamental conceptual importance. It interprets

in a geometric manner the localizations of R and the natural maps between them.

2. THE SPECTRUM OF A RING AS A LOCALLY RINGED SPACE 95

THEOREM-DEFINITION 2.1. Let R be a ring. There is a sheaf of rings OSpec(R)

on Spec(R) characterized by the property that it assigns to any principal open setSpec(R)s the ring R[1/s], and the restriction map defined by an inclusion of suchsubsets, Spec(R)s ⊂ Spec(R)s′ , is the associated homomorphism R[1/s′] → R[1/s].Its stalk at the point defined by the prime ideal p ⊂ R is the local ring Rp. We callOSpec(R) the structure sheaf of Spec(R).

PROOF. For the existence and uniqueness of OSpec(R), it suffices, in view ofExercise 73, to prove the following:

Let be given a collection of nonempty principal open subsets Spec(R)sαα∈A ofSpec(R) whose union is a principal open subset Spec(R)s and let uα ∈ R[1/sα]α∈Abe a collection with the property that for every pair α, β ∈ A, uα and uβ become equalin R[1/(sαsβ)]. Then the members of this collection come from a unique elementu ∈ R[1/s].

The verification of this property is modeled after the proof of Proposition 5.2.Since Spec(R)s = SpecR[1/s] is quasicompact, there is a finite subset K ⊂ A suchthat Spec(R)s = ∪α∈KSpec(R)sα . For κ ∈ K, we write uκ = rκ/s

Nκκ . We can take

Nκ larger if we want to and so we may assume that they are all equal to some N ′.For κ, λ ∈ K, rκ/sN

′

κ and rλ/sN′

λ become equal in R[1/(sκsλ)], and so rκsN′

λ −rλs

N ′

κ is annihilated by (sκsλ)Nκ,λ for some Nκ,λ ≥ 0. We can take Nκ,λ largerif we want to, and so we may assume that they are all equal to some N ′′. Nowput rκ := rκs

N ′′

κ and N := N ′ + N ′′. Then uκ = rκ/sN and rκs

Nλ = rλs

Nκ for all

κ, λ ∈ K.Since Spec(R)s = ∪κ∈KU(sκ), we have

Z(s) = ∩κ∈KZ(sκ) = ∩κ∈KZ(sNκ ) = Z(∑κ∈K

RsNκ ).

Then sM of s will lie in the ideal∑κ∈K Rs

Nκ for some M > 0 by Proposition 2.2:

sM =∑κ∈K

tκsNκ

for certain tκ ∈ R. We put r :=∑κ∈K tκrκ ∈ R. Then for every λ ∈ K,

rsNλ =∑κ∈K

tκrκsNλ =

∑κ∈K

tκrλsNκ = rλs

M

This proves that u := r/sM ∈ R[1/s] maps to uκ in R[1/sκ] for κ ∈ K. This doesnot prove yet that u has this property for an arbitrary α ∈ A. Still we could inthe above argument replace K by K ∪ α and thus come up with a u′ ∈ R[1/s]which is such that it not only maps to uκ in R[1/sκ] for κ ∈ K, but also to uαin R[1/sα]. It therefore remains to show u = u′ (which then also would establishuniqueness). The fact that u − u′ ∈ R[1/s] maps to the zero element of R[1/sκ]means that snκκ (u − u′) = 0 for some nκ > 0. Since some power sn of s lies in theideal

∑κ∈K Rs

nκ , it follows that sn(u− u′) = 0. This means that u = u′ in R[1/s].It remains to prove the assertion about the stalks. An element of this stalk at [p]

is represented by a fraction r/s ∈ R[1/s] with s /∈ p, with the understanding thattwo such fractions r/s ∈ R[1/s] and r′/s′ ∈ R[1/s′] are considered equal if theycoincide in some R[1/s′′] with s′′ /∈ p, where s and s′ divide s′′. This means thatrs′− r′s is annihilated by some power of s′′. This implies that they define the sameelement of Rp. In other words, the stalk at [p] embeds in Rp. On the other hand,

96 3. SCHEMES

any element of Rp thus appears as it is represented by a fraction r/s with s /∈ p(and so nonnilpotent).

REMARK 2.7. If in the previous proposition we take s = 1, we find that R =Γ(Spec(R),OSpec(R)) and so no information is lost when we pass from R to theassociated ringed space.

EXERCISE 77. Determine the points and the stalks of the spectrum of the localrings K[[t]], K[t](t) (with K a field) and Z(p) (the fractions whose denominator isnot divisible by p).

Proposition 2.1 associated to a ring R a ringed space. We expect a ring homo-morphism φ : R′ → R to induce a morphism of ringed spaces. This is indeed thecase.

PROPOSITION 2.8. LetR andR′ be rings and putX := Spec(R), X ′ := Spec(R′).Every ring homomorphism φ : R′ → R defines a continuous map f = Spec(φ) :

X → X ′ and gives rise to a natural sheaf homomorphism of rings φ : OX′ → f∗OXwhose value on X ′, φ(X ′) : R′ = Γ(X ′,OX′)′ → Γ(X ′, f∗OX) = Γ(X,OX) = R, isφ. If we think of φ as a sheaf homomorphism f−1OX′ → OX , then its stalk at x = [p]is given by the local sheaf homomorphism R′φ−1p → Rp induced by φ.

Conversely, any pair (f, φ) with f : X → X ′ a continuous map and φ : OX′ →f∗OX a sheaf homomorphism φ : OX′ → f∗OX such that the induced map on stalksis local, comes from the ring homomorphism R′ → R defined by φ(X ′).

PROOF. For the first assertion, we explain how φ is defined on a principal openU(r′) ⊂ X ′, i.e., we give the homomorphism Γ(U(r′),OX′) → Γ(U(r′), f∗OX).This will determine φ as a sheaf homomorpism. We have Γ(U(r′),OX′) = R′[1/r′].Put r := φ(r′). We have already seen that f−1U(r′) = U(r). The ring homomor-phism R′[1/r′]→ R[1/r] induced by φ can be understood as a ring homomorphism

Γ(U(r′),OX′) = R′[1/r′]→ R[1/r] = Γ(U(r),OX) = Γ(U(r′), f∗OX)

(when r is nilpotent, U(r) = ∅ and the right hand side is the zero ring). We thushave defined a homomorphism φ : OX′ → f∗OX of sheaves of rings. It is clear thatthis homomorphism is on the stalks as asserted.

Suppose now given (f, φ) is as in the proposition and let φ : R′ → R be thering homomorphism that is the value of φ on X ′. The sheaf property implies thatfor every x = [p] ∈ X we have a commutative diagram

R′ = Γ(X ′,OX′)φ−−−−→ Γ(X ′, f∗OX) = Ry y

R′p′ = OX′,x′φp−−−−→ OX,x = Rp

where p′ is the prime ideal of R′ associated to x′ := f(x). The bottom map is localand so the preimage of pRp is p′R′p′ . The commutativity of the diagram implies thatthen φ−1p = p′. In other words, Spec(φ) and f take the same value on x and definethe same map on the stalk at this point. Hence φ induces the pair (f, φ).

3. THE NOTION OF A SCHEME 97

3. The notion of a scheme

Let us say that a locally ringed space is a ringed space whose stalks are localrings and let us agree that a morphism of locally ringed spaces is a morphism ofringed spaces that are locally ringed and which has the additional property that itgives local homomorphisms on the stalks. This defines the category of locally ringedspaces. To any ring R we have associated a locally ringed space (which we denoteSpec(R) by abuse of notation) and we found that ring homomorphisms R′ → Rare in bijective correspondence with the morphisms Spec(R)→ Spec(R′) as locallyringed spaces. Thus the spec construction identifies the dual of the category ofrings as a full subcategory of the category of locally ringed spaces. We enlarge thiscategory by taking the full subcategory whose objects are locally like the spectrumof a ring and thus arrive at the central notions of algebraic geometry:

DEFINITION 3.1. An affine scheme is a ringed space that is isomorphic to Spec(R)for some ring R. A quasi-affine scheme is a ringed space that is isomorphic to anopen subset to Spec(R) for some ring R. A scheme is a ringed space that is locally anaffine scheme (hence this is always a locally ringed space). A morphism of schemesis a morphism between such objects in the category of locally ringed spaces.

We denote the category of schemes by Sch and its subcategory of affine schemesby Affsch. The following corollary essentially restates Proposition 2.8:

COROLLARY 3.2. The passage to spec defines a contravariant functor which iden-tifies the opposite of the category of rings, Ring, with the category Affsch of affineschemes.

We often denote a scheme by a single symbol (e.g., X), but sometimes alsouse that symbol for its underlying topological space. But if the distinction must bemade, we denote that space by the absolute value sign: a scheme X is then a space|X| with structure sheaf OX .

REMARK 3.3. We have seen that the closed points of an affine scheme Spec(R)bijectively correspond to the maximal ideals of R. In particular, Spec(R) has aclosed point. But an arbitrary scheme need not have a closed point (see [8] Exerc.3.27 for an example).

Base schemes. It is often useful to fix a ‘base scheme’ S and to consider theovercategory Sch/S as defined above. Its objects are S-schemes, that is, schemesX endowed with a morphism X → S. According to the definition, an S-morphismX/S → X ′/S, is then a morphism X → X ′ whose composite with the given mor-phism X ′ → S yields the given morphism X → S. But when S = Spec(R), weusually say that X is an R-scheme (rather than a Spec(R)-scheme) and we writeX/R accordingly. This simply means that OX is a sheaf of R-algebras and that theR-morphisms f : X/R→ Y/R are those with the property that the induced homo-morphisms on stalks are R-algebra homomorphisms. The category of R-schemes isof course denoted Sch/R. Since Z is an initial object for Ring, Spec(Z) is a finalobject for Affsch. It then also follows that it is a final object for Sch: every schemeX is one over Z: Sch = Sch/Z (to see this, cover X by affine open subsets Ui andobserve that the morphisms Ui → Spec(Z) and Uj → Spec(Z) will have the samerestriction to Ui ∩ Uj).

We also observe that the category of k-prevarieties is a full subcategory of thecategory Sch/k of k-schemes.

98 3. SCHEMES

The proj construction. We describe a way of producing a scheme out of agraded ring, such that the relation between the algebra and the geometry gener-alizes the way a projective variety defines a homogeneous coordinate ring. Westart out with a graded ring A = ⊕∞k=0Ak (so AkAl ⊂ Ak+l; in particular A is aA0-algebra). This has a distinguished ideal, A+ := ⊕∞k=1Ak. We define a schemeProj(A) whose underlying topological space |Proj(A)| has as its set of points thehomogeneous prime ideals strictly contained in A+ (briefly: strict homogeneousprime ideals). In particular, if A is reduced to A0 (so that An = 0 for n > 0),then Proj(A) = ∅. Given s ∈ Ad with d > 0, then the strict homogeneous primeideals of A which contain s define a subset Zs ⊂ |Proj(A)|. We denote its comple-ment by Proj(A)s. We have Proj(A)s ∩ Proj(A)s′ = Proj(A)ss′ as usual and so thesubsets Proj(A)s are a basis for a topology on |Proj(A)|. We give |Proj(A)| thistopology. We put Fk := Adk/s

k ⊂ A[1/s], which we interpret for k = 0 as the im-age of A0 in A[1/s]. We have Fk ⊂ Fk+1, FkFl ⊂ Fk+l and so the monotone unionA[1/s]0 := ∪k≥0Fk is a filtered ring (and we would then write Fk = Fk(A[1/s]0));it is the subring of A[1/s] of degree zero.

We claim that the topological space underlying Proj(A)s can be identified with|Spec(A[1/s]0)|. For if p ⊂ A is a strict homogeneous prime ideal which does notcontain s, then ∪k≥0pdk/s

k is a (filtered) prime ideal of A[1/s]0. Conversely, if p′

is a prime ideal of A[1/s]0, then denote by pk ⊂ Ak the set of a ∈ Ak for whichaAdl−k/s

l ⊂ p′ for l large enough (if this inclusion holds for some l, then it holdsalso for any larger l). This is a homogeneous ideal which does contain s. It is also aprime ideal: if a ∈ Ak and a′ ∈ Ak′ are such that aa′Adm−k−k′/sm ⊂ p′ for some m,then by enlarging m we may assume that we can write m = l + l′ such that k ≤ dl

and k′ ≤ dl′. If neither aAdl−k/sl ⊂ p′ nor a′Adl−k′/sl′ ⊂ p′, then choose r ∈ Adl−k

and r′ ∈ Adl−k′ such that neither ar/sl nor a′r′/sl′

is in p′. But since their productis in p′ we get a contradiction. You can check that the two constructions are eachothers inverse.

Given s′ ∈ Ad′ , then we have an inclusion of principal open sets Proj(A)s ⊂Proj(A)s′ if and only s′ divides some positive power of s, and then such an inclusionis covered by an obvious ring homomorphism A[1/s′]0 → A[1/s]0. So the mapProj(A)s 7→ A[1/s]0 defines a sheaf of rings on |Proj(A)| with the property that itsrestriction to Proj(A)s equals Spec(A[1/s]0) as a locally ringed space. This definesa scheme structure on |Proj(A)|; we denote that scheme by Proj(A) and usuallyrefer to it as the ‘proj’ ofA. The ring homorphismA0 → A[1/s]0 defines a morphismof schemes Proj(A)s → Spec(A0). These morphisms compatible (we get the samemorphism if we restrict the morphism Proj(A)s′ → Spec(A0) to Proj(A)s and thusdefine together define a morphism Proj(A)→ Spec(A0).

Notice that if p is a strict homogeneous prime ideal of A, then the stalk ofOProj(A) at the point x ∈ Proj(A) associated to p is the filtered subring Ap,0 =∪k≥1Ak/(Ak − pk) of degree zero elements of Ap. The image of x ∈ Proj(A) inSpec(A0) is defined by the prime ideal p0 and so the homomorphisms on stalks isthe local homomorphism (A0)p0 → Ap,0.

PROPOSITION-DEFINITION 3.4. We have thus defined an A0-scheme Proj(A). Thepoints of Proj(A) are in bijective correspondence with the strict homogeneous primeideals of A. Every homogeneous element s ∈ ∪∞n≥1An defines a principal open affine


subscheme Proj(A)s that we may identify with Spec(A[1/s]0). If p is a strict homo-geneous prime ideal of A, then the stalk of OProj(A) at the point x associated to p isAp,0 = ∪k≥1Ak/(Ak − pk) and the image of x in Spec(A0) is defined by p0 ⊂ A0.

An A0-scheme X/A0 is said to be projective (over A0) if it is isomorphic toProj(A) for some finitely generated A0-algebra. Such a scheme is quasi-compact.

PROOF. We already established all the assertions, but the last one. If (si ∈Adi)

ki=1 generates A as an A0-algebra, then Proj(A) is covered by the affine open

subsets Proj(A)si and hence is quasi-compact.

In case R is a ring and we take for A the graded ring R[T0, . . . , Tn] (withdeg(Ti) = 1 for all i), then the associated projective scheme Proj(R[T0, . . . , Tn])is called projective n-space over R and is denoted PnR. In case R = k, its set ofclosed points reproduces the projective space as defined in Chapter 2. We shalllater discuss these projective schemes in more detail.

The functor of points. This is a functorial way of thinking of a scheme, whichcan be extremely useful. First note that a scheme X defines a contravariant functorhX = MorSch( , X) from the category of schemes Sch to the category Set of sets.This functor depends covariantly on X: a morphism f : X → Y induces a naturaltransformation h∗(f) : hX ⇒ hY . General category theory (the Yoneda Lemma)tells us that nothing is lost this way. The restriction of this functor to affine schemescan be thought of as a covariant functor Ring→ Set (for Affsch can be regarded asthe dual of Ring). We call this the functor of points defined by X. Mumford denotesit by h(o)

X , and so shall we when we have to, but we will often we abuse notationa bit by simply using X. So it assigns to a ring R the set X(R) of morphismsSpec(R) → X and if φ : R → R′ is a ring homomorphism, then compositionwith Spec(φ) : Spec(R′) → Spec(R) defines the map X(φ) : X(R) → X(R′). Amorphism f : X → Y of schemes induces a natural transformation from the functorassociated to X to the functor associated to Y and it is now logical to denote thisnatural transformation also by f . This just expresses the fact that for every ringhomomorphism the square

X(R)f(R)−−−−→ Y (R)

X(φ)

y Y (φ)

yX(R′)

f(R′)−−−−→ Y (R′)

commutes.In caseX is affine, i.e., whenX = Spec(A) for some ring A, then the associated

functor is essentially hA, the (contravariant) functor which assigns to a ring Rthe set of ring homomorphisms Hom(R,A). According to the Yoneda lemma thisembeds the category of affine schemes Affsch (which is the opposite of the categoryof rings) in the category of functors Ring → Set. (At first this looks striking untilone realizes that X has a special A-valued point defined by the identity Spec(A) =X.) Again nothing is lost by this passage:

100 3. SCHEMES

PROPOSITION 3.5. Let X and Y be schemes. Then the map which assigns to amorphism f : X → Y the natural transformation h(o)(f) : h

(o)X ⇒ h

(o)Y defines a

bijection between MorSch(X,Y ) and Nat(h(o)X , h

(o)Y ).2

PROOF. We checked this for preschemes. The result for general schemes fol-lows from a rather straightforward glueing argument (using the fact that both Xand Y admit an open covering by affine schemes).

It is remarkable that the not so straightforward notion of a scheme (whichinvolves the category of locally ringed spaces) amounts to a certain type functorfrom Ring to Set.

EXAMPLE 3.6. We first take for X the line over Z, A1Z = Spec(Z[x]) (which we

now prefer to denote A1, thereby abandoning our convention that this stands forA1k). Then R 7→ A1(R) is just the forgetful functor Ring → Set: an element of

A1(R) is by definition given by a ring homomorphism φ : Z[x] → R and this inturn is given by φ(x) ∈ R; conversely, any r ∈ R determines a ring homomorphismφr : Z[x] → R that takes x to r. Likewise, if An stands for SpecZ[x1, . . . , xn] (towhich we might refer as absolute affine n-space), then

φ ∈ An(R) = MorRing(Z[x1, . . . , xn], R) 7→ (φ(x1), . . . , φ(xn)) ∈ Rn

identifies An(R) with Rn with the inverse assigning to r = (r1, . . . , rn) the ringhomomorphism φr : Z[x1, . . . , xn] → R which sends xi to ri. Taking this onestep further: if X is the spectrum of a quotient ring A of Z[x1, . . . , xn], so that X =Spec(Z[x1, . . . , xn]/(f1, . . . , fk)) for certain elements f1, . . . , fk of Z[x1, . . . , xn], thenan element of X(A) is given by a ring homomorphism A → R, which amountsto giving a ring homomorphism φ : Z[x1, . . . , xn] → R with the property thatfi(φ(x1), . . . , φ(xn)) = 0 for all i. In other words, X(R) can be identified withthe solution set in Rn of the system of equations fi(r1, . . . , rn) = 0, i = 1, . . . , k.For a ring homomorphism φ : R → R′ the induced map X(φ) : X(R) → X(R′)is obtained by applying φ to each coordinate. So this just corresponds to the moreintuitive notion idea of an algebraic set, but now repackaged in functorial form.

When R = k, then X(k) is the underlying set of an affine variety, but bewarethat we have not endowed X(k) with a topology, let alone with a sheaf of regularfunctions.

If we fix a scheme with associated functor S : Ring → Set, then the categoryof schemes over S, Sch/S, can be described as schemes for which the associatedfunctor X : Ring→ Set is endowed with a natural transformation X → S.

The functor of points approach is a natural one in the theory of algebraic groupssince it affords the following simple definition:

DEFINITION 3.7. An algebraic group scheme is a scheme X for which we aregiven a factorization of the associated functorX : Ring→ Set through the categoryof groups: X : Ring→ Grp→ Set.

This just means that for every ring R, the set of R-valued points X(R) has beenendowed with a group structure in such a manner that for any ring homomorphismφ : R→ R′ the associated map X(φ) : X(R)→ X(R′) is a group homomorphism.

2In other words, the functor of points defines an fully faithful functor from Sch to the functorcategory Func(Ring,Set).


EXERCISE 78. Prove that there is defined a group scheme Ga which assigns toevery ring R the abelian group underlying R (so this really an additional structureon the affine line SpecZ[x]). Similarly, prove that there is defined a group schemeGm (also denoted GL1) which assigns to every ring R the (multiplicative) groupR× on invertible elements (so this really an additional structure on the puncturedaffine line Z[x, x−1]). More generally, show that we have a group scheme GLnwhich assigns to R the group GLn(R) of n × n-matrices with entries in R anddeterminant in R× (such matrices have an inverse with entries in R).

Functors Ring → Set arise naturally in what are called moduli problems. Inthis context it becomes important to find criteria for when such a functor is repre-sentable by a scheme.

Special points. Let R be a local ring. Then Spec(R) has just one closed point;if κ := R/mR denotes its residue field, then it is a κ-valued point. Given a schemeX, then an R-valued point of X defines a map Spec(R) → X and so its restrictionto the closed point is going to be a κ-valued point of X: we find a p ∈ X and a localhomomorphism OX,p → R. We will call this an R-valued point over p. Such a localhomomorphism must have the maximal ideal mX,p as kernel so that κ appears asan extension of the residue field κ(p) of p. Conversely, any local homomorphismfrom a stalk of X to a local ring R defines an R-valued point of X.

Of particular interest is the case when we take for R a field, so that R = κ.Then a local homomorphism OX,p → κ is just a field embedding ι : κ(p) → κ.Observe that by our definition, composition of ι with an nontrivial automorphism σof κ defines another point. When κ is algebraically closed, then the absolute Galoisgroup Aut(κ) permutes the κ-valued points over p. Such points have a name:

DEFINITION 3.8. A point of a scheme in the above sense is called a geometricpoint if it takes values in an algebraically closed field.

So when X is an affine k-variety, then a geometric point is given by an irre-ducible subset Y ⊂ X and an embedding of its field of rational functions k(Y ) inan algebraically closed fieldK. Note thatK may be larger than an algebraic closureof k(Y ). For instance, if Y → Y is a dominant morphism of irreducible k-varieties,then K could be an algebraic closure of k(Y ).

EXERCISE 79. Let f ∈ Q[x] be nonzero and let X = Spec(k[x]/(f)). Fix analgebraic closure Q of Q and describe X(Q).

EXERCISE 80. Prove that a noetherian ring is a DVR if and only if it has onlytwo prime ideals, one of which is the zero ideal (so that it is without zero divisors)and the other principal.

EXERCISE 81. Let X be an irreducible variety over the algebraically closed fieldk of dimension n and let D ⊂ X be a subvariety of dimension n − 1 with theproperty that D is not contained in the singular part of X. Prove that this definesa Z-valuation vD on k(X) for which the associated DVR is the local ring OX,xDof the (possibly nonclosed) point xD ∈ Spec(X) defined by D. (For f ∈ k(X)×,vD(f) is to be understood as ‘the order of vanishing’ of f along D.) (This appliesin particular to the germ of a nonsingular curve: to be precise, the local ring of ak-variety X that is nonsingular of dimension 1 at the closed point x ∈ X is a DVR.)

102 3. SCHEMES

EXERCISE 82. Prove that the completion of a DVR with respect to its maximalideal is still a DVR. Describe this completion for κ[x](x) and Z(p).

Since every prescheme determines and is determined by a functor Ring→ Set,the question arises whether we can characterize all the functors Ring → Set thusobtained. This turns out to be a highly nontrivial (but important) issue.

4. Formation of products

Given a ring R, then the under category R/Ring has a coproduct: if we aregiven (commutative) R-algebras A and B (in other words, we are given ring ho-momorphisms φ : R→ A and ψ : R→ B), then we can form the R-algebra A⊗RB.It fits in the commutative diagram

A −−−−→ A⊗R B

φ

x xR

ψ−−−−→ B

which makes A ⊗R B in fact the coproduct: if a ring C has both the structure of aA-algebra and a B-algebra, but in such a manner that the resulting two R-algebrastructures coincide, then C is in fact a A⊗R B-algebra (this is trivial).

Since the category Affsch of affine schemes is dual to the category Ring ofrings, it has therefore arbitrary fibered products: if f : X → S and g : Y → S aremorphisms of affine schemes, then the fibered product X ×S Y (= X

∏S Y ) exists

and is simply defined as Spec(A ⊗R B), where A, B resp. R is the ring of globalsections of the structure sheaf of X, Y resp. S. The tensor product commutes withlocalization: first, if we localize in S, by taking h ∈ S, then A[h−1]⊗R[h−1]B[h−1] =

(A ⊗R B)[h−1]. This proves that f−1Sh ×Sh g−1Sh is a principal open of X ×S Y .A similar argument shows that if X ′ resp. Y ′ is an arbitrary principal open of Xresp. Y , then X ′ ×S Y ′ is a principal open of X ×S Y . This quickly leads to theexistence of arbitrary fibered products for the full category of schemes:

THEOREM 4.1. The category of schemes, Sch, admits arbitrary fibered products.

PROOF. Let f : X → S and g : Y → S are morphisms of schemes. First notethat it suffices to treat the case when S is affine: S = Spec(R). Then cover X resp.Y by affine open subsets Ui = Spec(Ai)i∈I and Vj = Spec(Bj)j∈J . We canform Ui ×S Vj = Spec(Ai ⊗R Bj). Then (Ui ∩ Ui′) ×S (Vj ∩ Vj′) can be identifiedwith open an affine open of both Ui ×S Vj and Ui′ ×S Vj′ . These identificationsyield the scheme X ×S Y (see Hartshorne [7], Ch. II Thm. 3.3) for a detaileddiscussion).

Notice that for S = Spec(k) and X and Y k-varieties gives gives us back theproduct defined in 6.1 of Chapter 1.

Base change and the fibers of a morphism. The formation of a fiber productis sometimes also called a base change. The use of this terminology is a only apsychological device, as is encourages us to think a bit differently about a fiberedproduct, namely given a morphism S → S, we get a functor Sch/S → Sch/S

simply by associating to every S-scheme X/S the S-scheme S ×S X/S.If we take S to be a point s of S, then we get the notion of the fiber of a

morphism f : X → S: the fiber f−1(s) (also written as X ⊗ κ(s) or X(s) when f is

5. ELEMENTARY PROPERTIES OF SCHEMES 103

understood) is Spec(κ(s))×SX. This is naturally a scheme over the field κ(s). Thisobservation leads one to think of X/S as a family schemes parametrized by S in analgebraic manner (which may be sometimes helpful). Note that in the generalitywe consider here this makes sense for any A-valued point of S (where A is a ring):the fiber is then the A-scheme Spec(A)×SX. If we take this to be a geometric point(meaning that A is an algebraically closed field), then we call this a geometric fiberof X/S.

Extension of scalars. If S is affine: S = Spec(R), so that A is an R-algebra,then the formation of Spec(A) ×Spec(R) X is sometimes called extension of scalars(and in the literature you will often see this abbreviated as A×RX or A⊗RX). If Ris a domain, then (0) ⊂ R is a prime ideal and the corresponding point of Spec(R)(see also Section 5 below), called the generic point of S, has the field of fractions,Frac(R), as its residue field. The fiber over this point is called the generic fiber ofX/S. A general point of Spec(R) is by definition given by an embedding of R inalgebraically closed field K. The corresponding fiber is then called a general fiberof X/S.

Here is another important case. Let K ⊃ Ko be a field extension. The inducedmap Spec(K) → Spec(Ko) is a trivial one (the underlying spaces are singletons).But now treat this as a base change. Suppose Xo/Ko is an Ko-scheme. ThenX := K ×Ko Xo will be a scheme over K. For instance, we could take Xo =Spec(Ko[x]/(f)). If we assume f ∈ Ko[x] irreducible, then Ko[x]/(f) is a field and|Xo| is a singleton. But X = Spec(K[x]/(f)), where f is now viewed as an elementof K[x] and its points correspond to the distinct irreducible factors of f in K[x]. Wewill later take closer look at this situation.

Since every scheme X is one over Z, Remark 2.4 extends to schemes: a closedpoint of Spec(Z) is given by a prime p and has residue field Fp and its uniquenonclosed point is defined by the prime ideal (0) and has residue field Q, and sothe fibers X ⊗ Fp and X ⊗Q over these point are schemes over Fp resp. Q (whichmay be empty, of course).

REMARK 4.2. Suppose X/S and Y/S are schemes over a scheme S and let Rbe a ring. Then an element of (X ×S Y )(R) is by definition given by a morphismSpec(R) → X ×S Y . The latter is by definition given by morphisms Spec(R) → Xand Spec(R) → Y such that the composites Spec(R) → X → S and Spec(R) →Y → S coincide. This amounts to giving a point of X(R)×S(R) Y (R). We thus seethat the set (X×S Y )(R) can be identified with the set X(R)×S(R)Y (R). The sameargument works of course if we replace Spec(R) by an arbitrary scheme.

EXERCISE 83. Prove that if X/S and Y/S are schemes over a scheme S, thenthere is a natural map |X ×S Y | → |X| ×|S| |Y |. Show that this map is surjective,but need not be injective. (Hint and permission: you may use Exercise 85 of thenext section.) Work this out for the following two simple cases: for X = Y = A1

k,S = Spec(k) and for the case where L/K is genuine finite separable field extensionand X = Y = Spec(L), S = Spec(K).

5. Elementary properties of schemes

Integral schemes. We say that a scheme is connected resp. irreducible if theunderlying topological space has that property. In particular, an irreducible schemeis connected.

104 3. SCHEMES

We say that a scheme X is reduced if its structure sheaf takes values in thecategory of reduced rings, that is, for every open U ⊂ X, OX(U) is reduced. If aring R is reduced, then so are it localizations. Hence the spectrum of a reducedring is reduced.

PROPOSITION-DEFINITION 5.1. A scheme X is irreducible and reduced if and onlyif for every nonempty open U ⊂ X, OX(U) is an integral domain. We then say thatX is an integral scheme.

PROOF. An integral domain is reduced. If X is reducible, then it containsnonempty pairwise disjoint open subsets U0, U1. Let U := U0 ∪ U1. If we choosefi ∈ OX(U) so that it takes the value 1 on Ui and zero on U1−i, then f0f1 = 0 andso OX(U) is not a domain.

If on the other hand a nonempty open subset U is irreducible and OX(U) isreduced, then OX(U) must be a domain: if f0, f1 ∈ O(U) are such that f0f1 = 0,then U = Z(f0) ∪ Z(f1), and so U = Z(fi) for some i. Since OX(U) is reduced,this implies that fi = 0.

REMARK 5.2. An integral scheme X contains a special point whose local ringis a field. If U ⊂ X is affine open and nonempty, then U is dense in X and OX(U)is an integral domain so that (0) is a prime ideal. This prime ideal defines a pointηU ∈ U whose residue field is Frac(OX(U)). That point is independent of thechoice of U : if U ′ is nonempty affine open subset of U , then O(U) → O(U ′) is alocalization map and since O(U) is an integral domain, it is injective and inducesan isomorphism of fraction fields. In particular, the inclusion U ′ ⊂ U maps ηU ′ toηU . We call this the generic point of X and denote it by ηX . This argument alsoshows that for U as above, Frac(OX(U))→ OX,ηX is an isomorphism of fields. Wecall this field the function field of X and sometimes denote it K(X). It is the localring of the generic point of X.

Subschemes and immersions. If (X,OX) is a scheme, then for every opensubset U ⊂ X, (U,OX |U) has of course the structure of a scheme. We say that amorphism f : Y → X of schemes is an open immersion if the image of f is an opensubset U of X and f factors through an isomorphism of Y onto U . In other words,|f | : |Y | → |X| is an open embedding and for every y ∈ Y , fy : OX,f(x) → OY,yis an isomorphism. For a closed subset of |X| the situation is a bit different in thesense that a closed subset may have many distinct structures as a scheme.

Before we give the relevant definition, we recall that a subspace Y of a spaceX is called locally closed if for every p ∈ Y there exists an open neighborhood Up ofp in X such that Y ∩ Up is a closed subset of Up. This is equivalent to: there existsan open subset UY of X which contains Y and is such that Y is closed in UY (justtake U = ∪p∈Y Up).

DEFINITION 5.3. A morphism of schemes f : Y → X is called an immersionif it maps |Y | homeomorphically onto a locally closed subset of |X| and the sheafhomomorphism f−1OX → OY is surjective (which we recall means that is surjectiveon the stalks: for every y ∈ Y , fy : OX,f(y) → OY,y is onto). If |f | is an inclusion(so that |Y | is a locally closed subspace of |X|), then we say that Y is subscheme ofX (we then may write Y ⊂ X). When |Y | is a closed subset of |X|, we say that fis a closed submersion (resp. that Y is a closed subscheme of X).

For an affine scheme this notion has a simple interpretation.

5. ELEMENTARY PROPERTIES OF SCHEMES 105

LEMMA 5.4. Let X is be an affine scheme: X = Spec(A). Then every idealI ⊂ A defines a closed subscheme and every surjection of rings A→ B defines a closedimmersion.

PROOF. Clearly an ideal I ⊂ R defines a closed subset Z(I) of |X|. This closedsubset underlies the scheme Y := Spec(R/I). More precisely, the ring homomor-phism R → R/I induces a scheme morphism Y → X whose underlying map oftopological spaces is the inclusion of Z(I) ⊂ |X|. Moreover, a point y ∈ |Y | ⊂ |X|is defined by a prime ideal p ofRwhich contains I and the homomorphism on stalksOX,y → OY,y is just the natural ring homomorphism Rp → Rp/Ip ∼= (R/I)p, wherep ⊂ R/I denotes the image of p. This ring homomorphism is clearly surjective.

The second assertion is left to you.

We will see (Corollary 7.5) that the converse also holds: any closed subschemeof an affine scheme Spec(R) is defined by some ideal of R.

EXAMPLE 5.5. Let k be an algebraically closed field (as always). For t ∈ k, theideal It ⊂ k[z] generated by z(z + t) (with t ∈ k) defines a closed subscheme ofA1k. It consists of two copies of Spec(k) (two points) when t 6= 0, but for t = 0 they

coalesce: we get a single point with structure sheaf (stalk) given by k[z]/(z2).

PROPOSITION 5.6. Let f : X → Y be a morphism. Then the graph Γf : X →X × Y (which is associated to the identity 1X : X → X and f) is an immersion. Inparticular, f factors as as the composite of an immersion and a projection:

f : XΓf−−→ X × Y → Y.

PROOF. Given x ∈ X, then choose first an affine open neighborhood V ⊂ Yof f(x) and then an affine open neighborhood U of x in f−1V so that U × V is anopen affine neighborhood of Γf (x). Then Γf |U maps to U × V and the resultingmorphism U → U × V is just the graph of the morphism fU,V : U → V that is therestriction of f . It now suffices to see that fU,V is a closed immersion. Write U =Spec(A) and V = Spec(B) so that fU,V is given by a ring homomophism φ : B → A.Then ΓfU,V is associated to the ring homomorphism a ⊗ b ∈ A ⊗ B 7→ aφ(b) ∈ A,which is clearly surjective (take b = 1). So fU,V is a closed immersion.

EXERCISE 84. LetX be affine: X = Spec(R). Show thatXred := Spec(R/√

(0))is a reduced affine scheme that is a closed subscheme of X with |Xred| = |X|. Provethat this ‘reduction process’ generalizes to an arbitrary scheme: every scheme Zcontains a closed subscheme Zred that is reduced and has the same underlyingpoint set.

EXERCISE 85. Let X be a scheme. Every closed integral subscheme Y ⊂ X hasa generic point ηY . Identify ηY with its image in X so that we have defined a mapη from the collection of closed integral subschemes of X to the points of X. Provethat this is bijection and that under this bijection we have Y ⊂ Y ′ ⇔ ηY ∈ ηY ′ .

Finiteness properties. We proved in Lemma 4.7 that the maximal ideal spec-trum of a ring is quasi-compact. The same proof shows that this is true for theprime ideal spectrum. So every open covering of an affine scheme admits a finitesubcovering.

PROPOSITION-DEFINITION 5.7. For a scheme X the following are equivalent:

106 3. SCHEMES

(i) X admits an open affine covering by the spectra of noetherian rings,(ii) every open affine subset of X is the spectrum of a noetherian ring.

When these equivalent conditions are fulfilled, we say that X is locally noetherian.

PROOF. The only nontrivial assertion is that (ii) implies (i). So assume (i)and let U ⊂ X be open affine: U = Spec(R). Let x ∈ U . We claim that thereexists a s ∈ O(U), such that x ∈ Xs = Spec(R[1/s]) with R[1/s] is noetherian.Since the localization of a noetherian ring is noetherian, a principal open subsetof the spectrum of a noetherian ring is one, too. Now by assumption x ∈ U iscontained in an open subset U ′ of X with U ′ the spectrum some noetherian ring.By passing to a principal open neighborhood of x in U ′ contained in U , we mayassume that U ′ ⊂ U . Another such passage also ensures that U ′ is principal in U :U ′ = Xs = Spec(R[1/s]) for some s ∈ O(U) and so R[1/s] is noetherian.

It follows that we can cover U by principal open subsets that are spectra ofnoetherian rings. Since U is affine, it is quasi-compact and so a finite number ofthese will do: Spec(R) = ∪kα=1 Spec(R[1/sα]) for certain sα ∈ R. Write φα : R →R[1/sα] for the natural map.

Claim: Given an ideal I ⊂ R, denote by Iα ⊂ R[1/sα] the ideal generated byφα(I). Then I = ∩αφ−1

α Iα.Proof. We check the nontrivial inclusion ⊃. If r ∈ ∩αφ−1

α Iα, then φα(r) =rα/s

mαα for some rα ∈ I and mα ≥ 0. This means that there exists a nα ≥ 0 such

that snαα (rsmαα −rα) = 0. In particular, smα+nαα r ∈ I. Since the U(sα) = U(smα+nα

α )cover U , there exist tα ∈ R such that

∑α tαs

mα+nαα = 1. It follows that

r =∑αtαs

mα+nαα r =

∑αtαsα ∈ I.

We can now complete the proof. If I• = (I1 ⊂ I2 ⊂ · · · ) is an monotonesequence of ideals in R, then each sequence (I•)α becomes stationary (for R[1/sα]is noetherian). Hence the same is true for the sequence ∩αφ−1

α (I•)α = I•.

DEFINITION 5.8. We say that a scheme is noetherian if it is locally noetherianand quasi-compact (in other words, if it has an open covering by a finite numberspectra of noetherian rings).

DEFINITION 5.9. We say that a morphism of schemes, f : X → S islocally of finite type if each point of S has an open affine neighborhood V =

Spec(R) with the property that f−1V is covered by affine open subsets that arespectra of finitely generated R-algebras,

quasi-compact if each point of S has an open affine neighborhood V with theproperty that f−1V is quasi-compact,

of finite type if f is locally of finite type and quasi-compact (so that each pointof S has an open affine neighborhood V = Spec(R) with the property that f−1V iscovered by finitely many affine open subsets that are spectra of finitely generatedR-algebras),

affine if each point of S has an open affine neighborhood V such that f−1V isaffine,

finite if each point of S has an open affine neighborhood V = Spec(R) suchthat f−1V is affine and the spectrum of a finitely generated R-algebra.

The notions ‘of finite type’ and ‘finite morphism’ clearly generalize their name-sakes for k-varieties.

6. SEPARATED AND PROPER MORPHISMS 107

As the above definition illustrates, any property P for schemes admits a simplegeneralization to S-schemes: a morphism f : X → S is said to satisfy P if we cancover S by affine open subsets V such that P holds for f−1V (so that the originalproperty is recovered by taking S = Spec(Z)).

EXAMPLE 5.10. If we take for X a reduced k-scheme X/k, then to say that X/kis of finite type means that X is covered by a finite number of open subsets that arethe prime ideal spectra associated to affine k-varieties. This essentially means thatX is a k-prevariety.

6. Separated and proper morphisms

Separation property. LetX/S be an S-scheme. Its self-product in the categoryof S-schemes is the fibered product X ×S X. The two identity maps from X tothe factors determine a morphism ∆ : X → X ×S X, called for good reasons thediagonal morphism, for the composite with any of the two projections X×SX ⇒ Xis the identity. Were we working in the topological category (so that X → X ×S Xwould be a subspace of the product space X × X), then the requirement that ∆is a closed is equivalent to each fiber of X/S being a Hausdorff space. Althoughthe topological space underlying a scheme is almost never a Hausdorff space, thischaracterization renders in this setting almost the same service:

DEFINITION 6.1. We say that X/S is separated if the diagonal morphism is aclosed immersion. We say that X is separated if it so as a scheme of Spec(Z) (i.e.,when ∆ : X → X ×X is a closed immersion).

This property can in fact be tested on the underlying topological spaces. Wefirst observe

LEMMA 6.2. Suppose X/S is affine, i.e., suppose that every point of S has an openaffine neighborhood whose preimage in X is affine. Then X/S is separated.

PROOF. It suffices to prove this locally over S and so we may as well assumethat S is affine: S = Spec(R) so that also X is affine: X = Spec(A), where A is aR-algebra. Then ∆ comes from the ring homomorphism A⊗R A→ A, a1 ⊗R a2 7→a1a2. This is a surjection and so ∆ is a closed immersion.

COROLLARY 6.3. The S-scheme X/S is separated if and only if the image of thediagonal |∆| : |X| → |X ×S X| is closed.

PROOF. Suppose the map |∆| : |X| → |X ×S X| has a closed image. Then it isa homeomorphism onto its image, for a continuous inverse is given by restriction ofa projection |X ×S X| → |X| to the image of |∆|. It remains to see that OX×SX →∆∗OX is onto. If V is an affine open subset of S and U is one of X such thatf(U) ⊂ V , then the restriction U → V defined by f determines the affine schemeU ×V U . We have a natural morphism U ×V U → X ×S X. This morphism isan open immersion and so we may identify U ×V U with an open affine subset ofX ×S X. According to Lemma 6.2, the morphism OU×V U → ∆∗OU is surjective.Since such open subsets cover the diagonal of X ×S X, the corollary follows.

REMARKS 6.4 (Elementary properties of the separation condition). It is clearfrom the definition that the property of f : X → S being separated is local on S: ifwe are given an open covering Vαα of S, then f is separated if and only if everyf−1Vα → Vα is separated.

108 3. SCHEMES

It is also clear that if f : X → S is separated, then so is the restriction of f toany open subset of X.

Since an open immersion is affine, it is separated by Lemma 6.2. The same istrue for a closed immersion X → Y : just check that X ×Y X can be identifiedwith X and that the diagonal morphism is then the identity morphism. The latteris certainly a closed immersion.

The following properties amount to an exercise in topology; for more details,you may consult [6] Prop. 1.3.26 (or alternatively, invoke the valuative criterionTheorem 6.17 discussed below).

Closed under composition: if f : X → Y and g : Y → Z are separated,then so is gf : X → Z.

Fullness: if X/S and Y/S are separated, then so is every S-morphism fromX to Y .

Closed under taking products: if we are given a base scheme S and twoseparated S-morphisms X → Y and X ′ → Y ′, then so is the associatedproduct S-morphism X ×S X ′ → Y ×S Y ′.

Closed under base change: if X/S is separated and S → S is a morphism,then S ×S X is separated over S.

So if we fix a base scheme S, then the first three properties say that the schemesthat are separated over S make up a full subcategory of Sch/S that is closed underS-products. The last property implies that the fibers of separated morphism areseparated.

The following proposition generalizes Proposition 11.7. The proof is essentiallythe same.

PROPOSITION 6.5. If X is a separated scheme, and U,U ′ are affine open subsets,then U∩U ′ is affine andOX(U∩U ′) is generated byOX(U)|U∩U ′ andOX(U ′)|U∩U ′.

PROOF. First observe that U ×U ′ is an affine open subset of X×X and that itspreimage under ∆ : X → X ×X is U ∩ U ′. Since ∆ is a closed embedding, U ∩ U ′must be affine with OX(U ∩U ′) a quotient of OX×X(U ×U ′) = OX(U)⊗OX(U ′).This means that OX(U ∩U ′) is generated by OX(U)|U ∩U ′ and OX(U ′)|U ∩U ′.

REMARK 6.6 (The graph of a morphism). Let S be a scheme and let f : X/S →Y/S be a morphism of S-schemes. The identity map of X and f combine to definean S-morphism Γf : X → X×SY , the graph of f . The two S-morphisms X×SY →X → Y and the identity of Y combine to define an S-morphism f×S1Y : X×SY →Y ×S Y . We thus obtain a commutative diagram

XΓf−−−−→ X ×S Y

f

y yf×S1YY

∆Y−−−−→ Y ×S Y.This is in fact a fiber product: if we are given S-morphisms Z → X×SY and Z → Ysuch the composites Z → X ×S Y → Y ×S Y and Z → Y → Y ×S Y coincide,then there is unique morphism Z → X which makes everything commute, namelythe composite Z → X ×S Y → X. In particular, if Y/S is separated, so that ∆Y isa closed immersion, then Γf is a closed immersion (for this is a property preserved


under base change, as you can easily check) so that we may identify X with aclosed subscheme of X ×S Y .

REMARK 6.7. Following up on Example 5.10, we can now identify the categoryof k-varieties with the full subcategory of Sch/k of k-schemes that are of finite typeand are separated over k.

Valuation rings. We now want to discuss the valuative criterion of separated-ness. This necessitates a short excursion into commutative algebra, for a centralrole is played here by the notion of a valuation ring. This is a generalization of thatof a discrete valuation ring. For more on this, see [1], Ch. 5.

DEFINITION 6.8. A valuation ring is an integral domain R with the propertythat every element of Frac(R)−R has its inverse in R. We say that a valuation ringis discrete (abbreviated as a DVR) if it is noetherian, but is not a field.

We will write K for Frac(R). So such a ring R contains a large part (‘more thanhalf’) of K.

LEMMA 6.9. A a valuation ring R is a local ring. MoreoverAlmost PID: Any finitely generated ideal in R is principal.Ideal total ordering: The ideals of R are totally ordered by inclusion: if I

and J are ideals of R, then I ⊂ J or J ⊂ I.Normality: R is integrally closed in its field of fractions

PROOF. Let a, b ∈ R. Then one of ab−1 or a−1b is in R. In other words, wehave a ∈ (b) or b ∈ (a). It follows (with induction) that any finitely generated idealin R is generated by one element.

Let I and J be ideals of R such that I 6⊂ J . Choose a ∈ I such that a /∈ J . Thenfor every b ∈ J , we cannot have a ∈ (b) and so b ∈ (a). In other words, J ⊂ (a) ⊂ I.In particular, R has only one maximal ideal, so is a local ring.

Let x ∈ K be integral over R. Then xn + a1xn−1 + · · · + an = 0 for certain

n and ai ∈ R. If x ∈ R we are done. Otherwise x−1 ∈ R, and so we havex = −(a1x

−1 + a2x−2 + · · ·+ anx

−(n−1)) ∈ R[x−1] = R.

Our definition of a DVR is equivalent to a more familiar one:

LEMMA 6.10. A DVR has the property that every nonzero ideal is a power of itsmaximal ideal. Conversely, any integral domain that is not a field and of which everynonzero ideal is a power of a the maximal ideal is a DVR.

PROOF. Let R be a DVR. The assumption that R is not a field implies that itsmaximal ideal m is nonzero. Since R in noetherian, m is finitely generated andhence admits a generator π ∈ m. If r ∈ R is arbitrary, then r ∈ (π) = m or π ∈ (r)(and hence m ⊂ (r)). Since m is maximal, the last possibility can only occur whenwe have (r) = m or (r) = R. We conclude that either r is a unit or r is divisible byπ: r = r1π for some r1 ∈ R. In the last case we continue with r1 and find that eitherr1 is a unit or that r1 = r2π for some r2 ∈ R. This produces a strictly ascendingsequence (r) ( (r1) ( (r2) ( · · · and so this process will terminate, because R isnoetherian. It follows that r = uπn with u ∈ R× and n ≥ 0. In particular, everyideal of R is of the form mn with n a nonnegative integer.

The proof of the second assertion is left to you.

110 3. SCHEMES

Let R be a DVR and let π be a generator of its maximal ideal (this is oftencalled a uniformizer). Then its field of fractions K := Frac(R) is simply obtainedby inverting π so that every a ∈ K× = K r 0 is uniquely written as uπn withu ∈ R× and n ∈ Z. We denote this exponent by v(a). We now have obtained a Z-valued valuation, that is, a surjective homomorphism v : K× → Z with the propertythat the nonarchimedean inequality holds: v(a + b) ≥ minv(a), v(b) (assuminga + b 6= 0). Notice that v has kernel R× and so defines an isomorphism of groupsK×/R× ∼= Z. A field endowed with such a valuation is called a local field. Given alocal field (K, v), then the union of 0 and the subset of K where this valuation is≥ 0 is then automatically a DVR whose field of fractions is K.

REMARK 6.11. For a valuation ring R we can also define a valuation, but onethat takes values in a totally ordered abelian group, the group in question beingΓ := K×/R× (for which we shall write the group law additively) and the totalorder being defined by aR× ≤ bR× ⇔ (a) ⊃ (b). This makes Γ a totally orderedgroup in the sense that γ ≤ γ′ implies γ+γ′′ ≤ γ′+γ′′ for all γ′′ ∈ Γ. The resultinggroup homomorphism v : K× → Γ satisfies the nonarchimedean inequality, for ifa, b, a + b ∈ K× and for instance v(a) ≤ v(b) so that b ∈ (a), then a + b ∈ (a) andso v(a+ b) ≥ v(a) = minv(a), v(b).

EXAMPLES 6.12 (DVR’s). A standard example of a DVR is the local ring of asmooth k-variety C of dimension one (a curve) at a closed point x: the associatedfield is Frac(OC,x) and if f ∈ Frac(OC,x), then the valuation v(f) is the orderof vanishing of a rational function at x (this is negative when it has a pole in x).Another is the C-algebra of convergent power series Cz. But the residue field andthe field of fractions may have different characteristic, witness the example Z(p),with p a prime, where they are Fp and Q respectively. More generally, suppose thatR is a noetherian domain and p ⊂ R is a nonzero prime ideal that is minimal for thisproperty. Then Rp has no other prime ideals than (0) and pRp and so by Exercise80 to say that this is a DVR is to say that pRp has a single generator. Exercise 81shows why DVR’s are also important in classical algebraic geometry.

EXAMPLE 6.13 (A valuation ring that is not a DVR). The ring of Laurent seriesRo := k[[t]] is a DVR with fraction field Ko := k((t)). The valuation νo : K×o → Zassigns to any nonzero element its order: if a =

∑n≥no ajt

j with ano 6= 0, thenν(a) = no so that Ro is the union of 0 and the part where νo ≥ 0. The maximalideal mo = (t) is the union of 0 and the part where νo ≥ 0.

Now suppose k is of characteristic zero. Then the multiplicative group R×o isdivisible: for every positive integer n and every u ∈ R×o there exist a unit un ∈ R×osuch that (un)n = u (first extract an nth root from the constant term, which ispossible since k is algebraically closed. This allows you to assume that the constantterm of u is 1 so that the Taylor series of (1 + a1t + a2t

2 + · · · )1/n will producethe desired root). So if we adjoin to Ro for every n > 1 an element t1/n satisfying(t1/n)n = t, then we obtain a ring R such that every element has an nth root. (Wemust do this in way that if n = mm′ then (t1/n)m = t1/m

′.)

Then R is a valuation ring having K := R[t−1] as its field of fractions. In fact,any element of K can be written as a Laurent series in some t1/n. We thus see thatνo can be extended to a map ν : K× → Q and is such that R is the union of 0 andthe part where ν ≥ 0. This implies that R is a valuation ring. Note also that themaximal ideal m of R is the union of 0 and the part where ν > 0. But any f ∈ m


has a square root in K, which is then of course also in m. In other words, m = m2.Were R noetherian, then it would follow that m = 0, which is clearly not the case.So R is not noetherian and hence not a DVR. One can show that K is an algebraicclosure of Ko and that R is the integral closure of Ro in K.

EXAMPLE 6.14 (Not a valuation ring). Let C ⊂ A2k be the cuspidal curve defined

by x3 = y2 and let o := (0, 0) ∈ C. The the local ring OC,o =(k[x, y]/(x3−y2)

)(x,y)

is not integrally closed in its field of fractions and hence is not a valuation ring. Tosee this, observe that t := y/x satisfies the equation t2 − x = 0, hence is integralover OC,o.

Observe that the spectrum of a DVR R has just two points: the closed point odefined by the ideal m and with local ring R and residue field κ = R/m and thegeneric point η defined by the zero ideal (0) with local ring equal its residue fieldK = Frac(R). So if X is a scheme, then any x ∈ X(R) determines two points withvalues in a field: one in X(κ) and one in X(K). If we think of x as a morphismSpec(R)→ X, then these are x(o) and x(η) respectively.

Let K be a field. We consider the collection of subrings of K that are local. Onthis collection we define a partial ordering as follows: given local subrings A,B ofK, we say that B dominates A (and write A ≤ B) if A ⊂ B and the inclusion islocal: mB∩A = mA. For instance in Example 6.13, for the local ringsRn := Ro[t

1/n]in K we have Rn ≤ Rm if and only of n divides m.

PROPOSITION 6.15. A local subring of K which is maximal relative to the partialorder ≤ is a valuation ring having K as its field of fractions. In particular, (by Zorn’slemma) any local subring of K is contained in a valuation ring in K having K as itsfield of fractions.

PROOF. Let R ⊂ K be a local subring that is maximal for ≤ and let x ∈ K −R.We must show that x−1 ∈ R. We first observe that R must be integrally closed(otherwise its integral closure yields a local ring which strictly dominates R). Wenext claim that mR[x] = R[x]. For otherwise mR[x] is contained in a maximal idealm of R[x]. Then the localization R := R[x]m is a local ring contained in K whichcontainsR. The homomorphismR→ R local: since the maximal ideal mR does notcontain 1, its intersection with R is a proper ideal in R and since this ideal containsm, it must be equal to m. But then the maximality of R implies that R = R. Thiscontradicts our assumption that x /∈ R.

Since mR[x] = R[x], there exist n ≥ 0 and r0, . . . , rn ∈ m such that

1 = r0 + r1x+ · · · rnxn.

We can arrange that r0 = 0, simply by subtracting from both members r0 and thendividing by the unit 1 − r0. Assuming this to be the case, then we get an equationof integral dependence for x−1: x−n = r1x

−(n−1) + · · · + rn. Since R is integrallyclosed, it follows that x−1 ∈ R.

In algebraic geometry valuation rings often come up via:

COROLLARY 6.16. Let X be a scheme and let p, q ∈ X be such that p lies in theclosure of q. Then there exists a valuation ring R and a morphism Spec(R) → Xwhich sends the closed point of Spec(R) to p and the generic point of Spec(R) to q.

112 3. SCHEMES

PROOF. Let Q ⊂ X be the closed integral subscheme that has q as its genericpoint. So κ(q) is the function field ofQ. Since p ∈ Q,OQ,p is a local ring inside κ(q).According to Proposition 6.15, this ring is dominated by a valuation ring R ⊂ κ(q)having κ(q) as its field of fractions. The resulting morphism Spec(R) → Q ⊂ X isthen as desired: it sends the generic point to q (and defines the identity on theirresidue fields) and the closed point defined to p (this is the local homomorphismOX,p → OQ,p ⊂ R on the stalks).

In the situation of the corollary above one says that p is a specialization of q.

Valuative approach to the separation property. If R is a valuation ring withfield of fractions K, then for every scheme X we have a natural map X(R) →X(K). This need not be injective or surjective. If we are given some x ∈ X(K),then we call any x ∈ X(R) mapping to x a lift of x. The main result of thissubsection is:

THEOREM 6.17 (Valuative characterization of being separated). Let f : X → Sbe a morphism of schemes. If f is separated if and only if ∆ is quasi-compact and forevery valuation ringR with field of fractionsK, the natural mapX(R)→ S(R)×S(K)

X(K) is injective.

In other words, if we are given a K-valued point x ∈ X(K) whose imagef(x) ∈ Y (K) has been lifted to an R-valued point y ∈ Y (R), then there is at mostone lift x ∈ X(R) of x which maps to y. Exercise 86 shows that the propertyof ∆ being quasi-compact (in which case one also says that f is quasi-separated)should be regarded as a finiteness condition and Exercise 87 should help to makethe valuative criterion more useful.

EXERCISE 86. Let f : X → S be a morphism of schemes. Prove that if ∆ : X →X ×S X → X is quasi-compact, then for any two open affine subschemes U,U ′ ofX, the restriction f |U ∩ U ′ : U ∩ U ′ → S is quasi-compact.

EXERCISE 87. Let f : X → S be a morphism of schemes.(a) Prove that if f is of finite type and S is locally noetherian, then S can be

covered by open subsets V such that f−1V is noetherian.(b) Prove that if S can be covered by open subsets V with f−1V is noetherian,

then ∆ : X → X ×S X is quasi-compact.(We may thus conclude that if f is of finite type and S is locally noetherian,

then f is quasi-separated.)

We shall need:

LEMMA 6.18. Let f : X → Y be a quasi-compact morphism of schemes. Thenevery po ∈ Y in the closure of f(X) lies in the closure of f(q) for some q ∈ X. In fact,there exist a valuation ring R with field of fractions K, a morphism φ : Spec(K)→ Xand an extension Spec(R) → Y of fφ which maps the closed point to po (so po thenlies in the closure of the fφ-image of generic point of Spec(R)).

PROOF. Every point of Y lies in an affine open subset and so without loss ofgenerality we may assume that Y is affine. Then X is quasi-compact and hence aunion of a finitely many open affines U1, . . . , Un. We denote by Yi the closure off(Ui) in Y , endowed with its reduced structure. If we put Xi := (Ui)red, then fmaps Xi to Yi so that f restricts to a dominant morphism fi : Xi → Yi between


reduced affine schemes. The closure of f(X) in Y is the union of the closures ofthe (finitely many) fi(Xi) (for the closure of a finite union of subsets of a subspaceis the union of their closures). So without loss of generality we assume that bothX and Y are affine and reduced: X = Spec(A) and Y = Spec(B) with A and Breduced, and that f : X → Y is dominant: the associated ring homomorphismφ : B → A is injective (otherwise ker(φ) would define a proper closed subschemeof Y to which f maps).

Let now po ∈ Y be defined by a prime ideal po ⊂ B. By Zorn’s lemma hereexists a minimal prime ideal p in B contained in po. This defines a point p ∈ Ywhich contains po in its closure. So this closure is a closed integral subscheme Pof Y that contains po and is maximal for this property. Notice that the residue fieldκ(p) of p is equal to the localization Bp. The scheme theoretic fiber f−1p is equal toSpec(κ(p) ⊗B A) by definition. It is nonempty: since localization is exact, the ringhomomorphism φp : κ(p) → κ(p) ⊗B A is injective. In particular, κ(p) ⊗B A is notthe zero ring.

Choose q ∈ f−1p and put K := κ(q). This contains κ(p) as a subfield and κ(p)contains OP,po as a local ring. According to Corollary 6.16 there exists a valuationring R of K which dominates OP,po . Then this defines a morphism φ : Spec(R) →X which sends the generic point to q and for which fφ : Spec(R) → X sends theclosed point to po (for mP,po ⊂ mR).

PROOF OF THEOREM 6.17. Suppose first that f is separated. Then ∆ is a closedimmersion and hence quasi-compact. Let now R be a valuation ring with fieldof fractions K and let φ1, φ2 ∈ X(R) have the same image S(R) ×S(K) X(K).That is, we have morphisms φ1, φ2 : Spec(R) → X which become equal aftercomposing with f : X → S or after restriction to Spec(K). This means that wehave a morphism (φ1, φ2) : Spec(R) → X ×S X which maps the generic point tothe diagonal. Since the diagonal is closed, (φ1, φ2) maps to the diagonal also andhence φ1 = φ2.

Suppose now that f possesses the valuative property and is such that ∆ : X →X ×S X is quasi-compact. We must show that ∆ : X → X ×S X has closed image.Since ∆ is quasi-compact, it suffices by (the first part of) Lemma 6.18 to show thatwhen p ∈ ∆(X) and po ∈ X ×S X lies in the closure of p, then po ∈ ∆(X).

According to Corollary 6.16 there exists a valuation ring R and a morphismφ : Spec(R)→ X ×S X which sends the generic point η to p and the closed point oto po. Since φ(η) lands in the image of ∆, we must have π1φ(η) = π2φ(η) (whereπi : X ×S X → X denotes projection on the ith factor). If we denote this commonimage by q, then ∆(q) = p. Since fπ1 = fπ2, π1φ and π2φ define two elements ofX(R) with the same image in S(R)×S(K) X(K). Our assumption tells us that theymust then coincide: π1φ = π2φ. If we denote this last morphism by φ : Spec(R)→X, then we must have ∆φ = φ (by the universal property of the fiber product) andso po = φ(o) = ∆(φ(o)) ∈ ∆(X).

Proper morphisms. A continuous map f : X → S between topological spacesis said to be proper if for every compact subset K ⊂ S, its pre-image f−1K isalso compact. In particular, such a map has compact fibers. This notion is themost useful for the category of locally compact Hausdorff spaces (which containsall spaces homeomorphic to a closed subset of some Rn) and if we restrict to thatcategory, then there is an alternate definition: f : X → S is proper if and only if

114 3. SCHEMES

it is universally closed, which means that for every locally compact Hausdorff spaceS and continous map S → S, the natural map S ×S X → S is closed: the imageof every closed subset of S ×S X in S is closed. We take this as our cue for thedefinition of proper in the category of schemes:

DEFINITION 6.19. A morphism of schemes f : X → S is proper when it isseparated, of finite type and universally closed in the category of schemes. We saythat a scheme X is proper, if it is so over Z.

REMARK 6.20. Some authors use an apparently weaker condition for univer-sally closed and only ask that for any scheme T , the morphism 1T × f : T ×X →T × S is closed. The difference is indeed only apparent. For if g : T → S is a mor-phism, then according to Proposition 5.6 its graph identifies T with a locally closedsubspace of T × S. Via this identification, the restriction of 1T × f to this subspacegives us back the base change morphism T×SX → T . But if 1T×f : T×X → T×Sis closed, then its restriction over any locally closed subspace of T × S is closed aswell.

REMARKS 6.21 (Elementary properties of the properness condition). Most ofthe remarks we made for the separation condition extend easily to properness. Theproperty of f : X → S being proper is local on S: if we are given an open coveringVαα of S, then f is proper if and only if every f−1Vα → Vα is proper. A closedimmersion X → Y is proper.

Closed under composition: if f : X → Y and g : Y → Z are proper, thenso is gf : X → Z.

Fullness: if X/S and Y/S are proper, then so is every S-morphism fromX → Y .

Closed under taking products: if we are given a base scheme S and twoproper S-morphisms X → Y and X ′ → Y ′, then the associated productS-morphism X ×S X ′ → Y ×S Y ′ is also proper.

Closed under base change: if X/S is proper and S → S is a morphism,then S ×S X is proper over S.

So the first three properties say that if we fix a base scheme S, then the schemesthat are proper over S make up a full subcategory of Sch/S that is closed underS-products. The last property implies that any scheme theoretic fiber of a propermorphism is proper over its base ring. Note that the first property implies that whenf : X → S is proper and Y ⊂ X is a closed subscheme of X, then f |Y : Y → S isalso proper.

Integral affine schemes are rarely proper, witness:

PROPOSITION 6.22. Let X be an integral scheme proper over the algebraicallyclosed field k. Then any f ∈ Γ(X,OX) must be constant. In particular, X cannot beaffine unless it is a copy of Spec(k).

PROOF. Since X is integral, the first assertion follows if we show that f isconstant on any open affine subscheme U ⊂ X. Define a closed subscheme Cfof A1

k ×k X as the subset defined by the ideal tf − 1. So if U = Spec(A), withA a reduced k-algebra, then A1

k ×k U = SpecR[t], f |U ∈ A, and the restrictionof Cf to A1

k ×k U is Spec(A[t]/(tf |U − 1). The image of Cf under the projectionπ : A1

k ×k X → A1k will not contain the origin of A1

k (the closed point defined by


the maximal ideal (t)). Since X is proper over k, this projection is closed, and soπ(Cf ) cannot contain the generic point. In particular, (f |U)∗ : k[t] → A is notinjective: its kernel contains a nonzero polynomial. Then f |U will have its imagein the (finite) zero set of this polynomial. But A is an integral domain, and so f |Uis then constant.

If X is affine, then we can take X = U . It then follows the finitely generatedk-algebra A consists of constants. Since A is a finitely generated field extension ofk and k is algebraically closed, it follows that A = k.

THEOREM 6.23 (Valuative characterization of being proper). Let f : X → S be amorphism of schemes that is of finite type and for which the diagonal ∆ : X → X×SXis quasi-compact. Then f is separated if and only if for every valuation ring R withfield of fractions K, the natural map X(R)→ S(R)×S(K) X(K) is bijective.

PROOF. We first assume that f is proper and show that for R a valuation ring,X(R) → S(R)×S(K) X(K) is bijection. Since f is separated, we know this map isinjective and so we only need to show it is onto.

So let be given morphism φ : Spec(K) → X and a lift ψ : Spec(R) → Spec(S)of fφ. We make a base change over ψ: let Xψ := Spec(R) ×S X and denote byfψ : Xψ → Spec(R) the projection. We can now think of φ as defining a section offψ over the generic point η ∈ Spec(R). Put q := φ(η) and denote by Q the integralclosed subscheme of Xψ which has q as its generic point. From fψ(q) = η we getthat κ(q) ⊃ K and from φ(η) = q we get the opposite inclusion: K ⊃ κ(q). Soκ(q) = K. Since f is universally closed, fψ(Q) is closed. Hence it must containthe closed point o ∈ Spec(R): there exists a qo ∈ Q such that fψ(qo) = o. In otherwords, K ⊃ OQ,qo ⊃ R. In view of the maximality of R, this implies thatOQ,qo = Rand hence we thus have defined a morphism Spec(R) → Q whose composite withQ ⊂ Xψ → X equals φ and whose composite with Q ⊂ Xψ → X → S (which isthe composite Q→ Spec(R)→ S) equals ψ.

We now assume that f has the valuative property and prove that f is universallyclosed. Let S → S be a morphism and let f : X := S ×S X → S be the associatedmorphism. We must show that for any closed subset Z ⊂ X, f(Z) is closed. Sincef is quasi-compact, so is f |Z. Let po ∈ S be in the closure of f(Z). Accordingto Lemma 6.18 we can find a valuation ring R with field of fractions K and amorphism φ : Spec(K) → X such that fφ : Spec(K) → S lifts to a morphismψ : Spec(R)→ S. Then the composites Spec(K)→ X → X and Spec(R)→ S → Sdefine an element of S(R)×S(K)X(K). By assumption that element originates froma unique morphism ψ : Spec(R) → X. Then the pair (ψ, ψ) defines a morphismφ : Spec(R)→ S×SX = X. Since φ maps the generic point η to Z and Z is closed,φ(o) is also a point of Z. We have f(φ(o)) = po and so po ∈ f(Z).

Our main example of a proper scheme will be absolute projective n-space,Pn := PnZ and the schemes PnS := S ×Spec(Z) Pn obtained by base change. We recallthat Pn is obtained as the proj construction applied to the graded ring Z[T0, . . . , Tn].It is covered by n + 1 affine open pieces U0, . . . , Un, where Ui = UTi is defined byTi 6= 0 and is identified with the spectrum of the ring

Z[T0

Ti, . . . , Ti−1

Ti, Ti+1

Ti, . . . , TnTi ].

In particular, Pn is of finite type.

116 3. SCHEMES

The following lemma links the schematic to the conventional definition ofprojective space. Its proof is modeled after the proof of following more clas-sical fact: any meromorphic map φ from the punctured disk 0 < |z| < 1 toPn(C) extends holomorphically across 0. This is shown as follows: represent φas [φ0(z) : · · · : φn(z)] with each φi meromorphic. If k ∈ Z is the smallest integersuch that each zkφi is holomorphic at 0 (so that not all vanish there), then it is clearthat [φ0(z) : · · · : φn(z)] = [zkφ0(z) : · · · : zkφn(z)] is holomorphic at 0.

LEMMA 6.24. LetR be a valuation ring with field of fractionsK. Then the naturalmap Pn(R)→ Pn(K) is a bijection and Pn(K) is naturally identified with the classicalprojective space Pnclss(K) of nonzero vectors in Kn+1 given up to a nonzero scalar.

PROOF. We first define a map Pn(R)→ Pnclss(K) and prove it is a bijection. Wethen observe that this also shows that a similarly defined map Pn(K)→ Pnclss(K) isa bijection and that the resulting bijection Pn(R) ∼= Pn(K) is the obvious map.

An element of Ui(R) is given by a ring homomorphism

φ : Z[T0

Ti, . . . , Ti−1

Ti, Ti+1

Ti, . . . , TnTi ]→ R.

By assigning to φ its values on T0/Ti, . . . , Ti−1/Ti, Ti+1/Ti, . . . , Tn/Ti, we get anidentification of Ui(R) with Rn, but for reasons that, if not clear now, will be clearin a moment, we prefer to assign to φ the element

[φ(T0

Ti) : · · · : φ(Ti−1

Ti) : 1 : φ(Ti+1

Ti) : · · · : φ(TnTi )] ∈ Pnclss(K).

This defines a map ei : Ui(R) → Pnclss(K) which is easily seen to be injectivewith image the set of [a0 : · · · : an] for which ai 6= 0 and aj/ai ∈ R for all j.The embeddings ei and ej coincide on Ui(R) ∩ Uj(R) and so the ei’s assemble toproduce a map e : Pn(R)→ Pnclss(K). This map is still injective. It is also surjective:given a p ∈ Pnclss(K), then represent p by some (a0, . . . , an) ∈ Kn+1 r 0. Aftermultiplication by a common denominator, we can arrange that each ai is in R. Theideal in R generated by the a0, . . . , an is generated by one of these elements, say byai, so that ai divides each aj . This implies that p ∈ e(Ui).

We conclude that e is a bijection. The field K is also an example of a valua-tion ring (one which equals its field of fractions) and so we also have a bijectionPn(K) ∼= Pnclss(K). The resulting bijection Pn(R) ∼= Pn(K) is indeed induced by theobvious map.

EXERCISE 88. Let C be a curve over Spec(k) and let p ∈ C be a smooth point ofC. Prove that every morphism f : C r p → Pnk extends uniquely to a morphismC → Pnk . (Hint: observe that OC,p is a valuation ring.)

COROLLARY 6.25. The absolute projective space Pn = PnZ is proper (over Z) andhence for every scheme S, PnS := PnZ × S is proper over S.

PROOF. We may conclude from either Exercise 86 or Exercise 87 (Z is noe-therian) that the diagonal Pn → Pn ×Spec(Z) Pn is quasi-compact. By the valuativecriterion, it then suffices to show that for every valuation ring R with field of frac-tions K, the natural map

Pn(R)→ Pn(K)×Spec(Z)(K) Spec(Z)(R)

is a bijection. But for any ring A, Spec(Z)(A) is a singleton (Z is an initial object ofRing) and so the right hand side is simply Pn(K). The desired property therefore

7. QUASI-COHERENT SHEAVES 117

amounts to the assertion that Pn(R) → Pn(K) is a bijection, and this is part ofLemma 6.24.

DEFINITION 6.26. A morphismX → S is called projective (resp. quasi-projective)if it factors through a closed S-immersion (resp. a S-immersion) in PnS for some n.

THEOREM 6.27. A projective morphism is proper.

PROOF. Suppose f factors through a closed S embedding i : X → PnS . Both iand the projection PnS → S are proper, hence so is f .

7. Quasi-coherent sheaves

For a scheme X we introduce here an important class of OX -modules. (Werefer to the appendix for a discussion of modules over a sheaf.) If M is an OX -module, and x ∈ X, we write Mx for its stalk at x (an OX,x-module). The κ(x)-vector space κ(x) ⊗OX,x Mx

∼= Mx/mX,xMx is called the fiber of M at x and isoften denotedM(x).

PROPOSITION-DEFINITION 7.1. LetR be a ring and writeX for its spectrum. Thenfor every R-module M there is a sheaf of OX -modules M on X (the sheafificationof M) characterized by the property that for a principal open set Xs, M(Xs) =M [1/s], and the restriction map defined by an inclusion Xs ⊂ Xs′ is the associatedhomomorphism M [1/s′] → M [1/s]. Its stalk at x ∈ X is the localization Mpx andM(X) = M .

The R-module M and its sheafificationM have the same support.Sheafification is compatible with the tensor product: if M and N are R-modules,

then M⊗OX N is the sheafification of M ⊗R N and with arbitrary direct sums: ifMii is a collection of R-modules, then ⊕iMi is the sheafification of ⊕iMi. It definesa functor ModR →ModOX which is exact and fully faithful.

PROOF. We omit the proof of the first part as it is similar to the one of 2.1.The proofs of the second and third statements then follow from the fact that

the natural R[1/s]-homomorphisms (M ⊗R N)[1/s]→M [1/s]⊗R[1/s] N [1/s] resp.(⊕iMi)[1/s]→ ⊕iMi[1/s] are isomorphisms.

Sheafification is exact: if 0 → M → M ′ → M ′′ → 0 is an exact sequence ofR-modules, then for every prime ideal p of R, the sequence 0 → Mp → M ′p →M ′′p → 0 is still exact (localization is exact) and this is just the sequence of stalks ofsheafified sequence at x = [p].

Sheafification is fully faithful: we must show that any sheaf homomorphismφ : M → N comes from a unique R-homomorphism Φ : M → N . There is nouniquess issue, for we must Φ = φ(X) : M = M(X) → N (X) = N . But thenexistence is also clear: on a principal open subset Xs, φ(Xs) : M [1/s] → N [1/s]must be induced by Φ.

The following supplements Proposition 2.8. The proof is straightforward andtherefore omitted.

PROPOSITION 7.2. Let φ : R′ → R be ring homomorphism and denote by f : X →X ′ the associated morphism of affine schemes. Sheafification is compatible with therestriction functor ModR → ModR′ : for an R-module M , f∗M is the sheafificationof M regarded as a R′-module. It is also compatible with its left adjoint, the inductionfunctor: for an R′-module M ′, f∗M′ is the sheafification of R⊗R′ M ′.

118 3. SCHEMES

We will be mostly concerned with sheaves that arise locally from sheafifyingmodules.

PROPOSITION-DEFINITION 7.3. Let X be scheme. For an OX -module F the fol-lowing are equivalent:

(i) X admits a covering by affine open subsets U such that F|U is the sheafifi-cation of the OX(U)-module F(U).

(ii) for every affine open affine subset U of X, F|U is the sheafification of theOX(U)-module F(U).

When these equivalent conditions are fulfilled, we say that F is quasi-coherent.In particular, for an affine scheme Spec(R), the sheafification functor defines an

equivalence between the category of R-modules and the category of quasi-coherentOSpec(R)-modules and hence the global section functor sends a short exact sequence ofquasi-coherent OSpec(R)-modules to a short exact sequence of R-modules.

PROOF. We assume (i) and prove (ii). We do this in steps. Let U = Spec(R) ⊂X be open and affine and put M := F(U). We must show that F|U is the sheafifi-cation of the R-module M .

Step 1: There exist a finite set si ∈ Rri=1 such that the Usi cover U and F|Usiis the sheafification of the O(Usi)-module Mi := F(Usi).

Given a point x ∈ U , let U(x) = Spec(Rx) by an affine open subset containing xsuch that F|U(x) is the sheafification of the Rx-module M(x) := F(U(x)). Choosea principal open neighborhood U ′(x) of x contained in U∩U(x), so that there existsan s′x ∈ Rx and an sx ∈ R such that U ′(x) = U(x)s′x = Usx . Then F|U ′(x) is thesheafification of F(U ′(x)) = M(x)[1/s′x]. Since U is quasi-compact, finitely manyof the U ′(x)’s suffice to cover U : U = ∪ri=1Usxi . Then si := sxiri=1 is as desired.

We have a natural sheaf homomorphism M → F|U . To prove that this is anisomorphism, it suffices to show that this induces an isomorphismM(U ′)→ F(U ′)for U ′ running over a basis of open subsets of U . So it suffices to show that for anarbitrary s ∈ O(U), the ring homomorphismM [1/s]→ F(Xs). We prove injectivityand surjectivity separately.

Step 2: The homomorphism M [1/s]→ F(Us) is injective.Since sheafification is compatible with localization, we have that F(U(ssi)) =

Mi[1/s]. Now let a ∈ M = F(U) be such that its restriction to F(Us) is zero.We must show that sNa = 0 for some N ≥ 0. The image of a in F(Us ∩ Usi) =F(Ussi) = Mi[1/s] is of course also zero. So if ai ∈Mi = F(Usi) is the image of a,then sNiai = 0 for some integer Ni ≥ 0. Then N = maxiNi has the property thatsNai = 0 in Mi = F(Usi) for all i. This means that sNa = 0 in F(U) = M .

Step 3: The homomorphism M [1/s]→ F(Us) is surjective.Let b ∈ F(Us). Then its image in F(Us ∩ Usi) = Mi[1/s] is of the form bi/s

Ni

for some bi ∈ Mi and some Ni ≥ 0. We multiply numerator and denominator by apositive power of s so that we can take Ni independent of i: Ni = N . Now bi andbj have the same image in F(Us ∩ Usi ∩ Usj ) = F(Usisj )[1/s] and so there existsan integer Nij ≥ 0 such that bi|Usi ∩ Usj − bj |Usi ∩ Usj is killed by sNij . So if welet N ′ := maxi,j Nij , then sN

′bi and sN

′bj have the same restriction to Usi ∩ Usj

for all i, j. This implies that there exists a b ∈ F(U) = M having the property thatb|U(si) = sN

′bi for all i. Then b/sN+N ′ ∈M [1/s] restricts to b|F(Us ∩ Usi) for all i

and hence restricts to b ∈ F(Us).


COROLLARY 7.4. Let f : X → Y be a morphism of schemes. Then(a) the kernel, cokernel and image of a homomorphism between quasi-coherentOX -modules is also quasi-coherent, in other words, the category Qcoh(X) ofquasi-coherent OX -modules is abelian subcategory of the category ModOXof OX -modules,

(b) the tensor product over OX of two quasi-coherent OX -modules is quasi-coherent,

(c) f∗ takes QcohY to QcohX ,(d) if f is quasi-compact and quasi-separated (note that this is the case when X

is noetherian), then f∗ takes QcohX to QcohY .

PROOF. The first two assertions are of a local nature, hence follow from Propo-sition 7.3 and the fact that ModR is abelian. The same is true for (c) (this is localon X).

For (d) we may assume that Y is affine: Y = Spec(S). If X is also affine:X = Spec(R), then we are done of course, for any quasi-coherent sheaf M on Xis the sheafification of an R-module M and f∗M is then the sheafification of Mviewed as a S-module via the ring homomorphism S → R.

We now turn to the general case. Since f is quasi-compact, this implies thatX is covered by finitely many affine open subsets Ui)Ni=1. Since f is also quasi-separated, Ui ∩ Uj is covered by finitely many affine open subsets Uijk

Nijk=1. Let

now F be quasi-coherent on X. The sheaf property of F implies that we have anexact sequence of OY -modules

0→ f∗F → ⊕if∗(F|Ui)→ ⊕i,j,kf∗(F|Uijk).

By the case treated above, f∗(F|Ui) and f∗(F|Uijk) are quasi-coherentOY -modules.Hence, f∗F , being the kernel of a OY -homomorphism between two such modules,is also quasi-coherent.

COROLLARY 7.5. Let f : Y → X be a morphism of schemes that is quasi-compactand quasi-separated. Then ker(OX → f∗OY ) is a quasi-coherent ideal sheaf in OXand hence defines a closed subscheme of X whose underlying subset is the closure ofthe image of f . We call this the scheme theoretic image of f . In particular anyclosed subscheme of a scheme X is defined by a quasi-coherent ideal sheaf in OX andconversely any quasi-coherent ideal sheaf in OX defines a closed subscheme of X.

PROOF. Since f is quasi-compact and quasi-separated, part (d) of Corollary 7.4implies that f∗OY is quasi-coherent. By part (a) of that corollary, the ideal sheafker(OX → f∗OY ) is then quasi-coherent as well. It is clear that a quasi-coherentideal sheaf in OX defines a closed subscheme. Since f∗OY is zero on any opensubset of X which does not meet the image of Y , the underlying subset is theclosure of the image of f .

The second statement follows from the fact that the inclusion of a closed sub-scheme ofX inX is separated (and hence quasi-compact and quasi-separated).

We use this corollary to prove an ‘almost converse’ to Theorem 6.27 in case thebase is noetherian. It is known an Chow’s Lemma and helps us to extend propertiesof projective morphisms to proper morphisms.

PROPOSITION 7.6 (Chow’s Lemma). Let f : X → S be a proper morphismwith S noetherian. Then there exists a projective morphism π : X ′ → X that is an

120 3. SCHEMES

isomorphism over an open-dense subset of X (so that π will be surjective)3 and hasthe property that fπ : X ′ → S is also projective.

PROOF. It is not hard to see that it suffices to prove this for each irreduciblecomponent of X separately and so we will assume that X is irreducible.

Since f is of finite type, we can cover X by a finite number of nonempty affineopen subsets Uiri=1 such that f(Ui) is contained in an affine open subset Vi of Sand OX(Ui) a finitely generated OS(Vi)-algebra. Since each Ui is nonempty (hencedense in X), so is U := ∩iUi. Choose a finite set of OS(Vi)-generators of OX(Ui) sothat we have a closed immersion Ui → AkiVi and hence a locally closed immersionUi → AkiVi ⊂ PkiS . We write Pi for PkiS and P for P1 ×S · · · ×S Pr. Note that P isprojective over S. The diagonal morphism j : U → X ×S P is an immersion so thatwe can regard U as a locally closed subscheme of X ×S P . We then take to be X ′

the closure of U as a subscheme (here we invoke Corollary 7.5) and π : X → X ′

the projection on the first factor.Clearly π is projective. Let πi : X ′ → Pi denote the projection. We claim that

π−1i Ui = π−1Ui and that the projection of this open subset of X ′ onto Ui is an

open immersion. For this we note that the projection (π, πi) : X ′ → X ×S Pi isprojective, hence closed. Its restriction to the open-dense U ⊂ X ′ is the diagonalU → U ×S U ⊂ X ×S Pi and so the image of X ′ lies in the closure of U ×S U . Thisis also the closure of Ui ×S Ui. Since X/S is separated, this implies our claim.

We show that this implies the proposition. Since the Ui cover X, the π−1Ui willcover X ′. It then follows from our claim that the projection (π1, . . . , πn) : X ′ → Pis an immersion. Since X is proper over S, this projection is also closed and henceX ′ is thus realized as a closed subscheme of P . In particular, the projection fπ :X ′ → S is projective. It also follows that π−1U is the diagonally embedded U inX ×S P so that π is an isomorphism over U .

PROPOSITION-DEFINITION 7.7. Let X be a noetherian scheme. For an OX -module F the following are equivalent:

(i) X admits a covering by affine open subsets U such that F|U is the sheafifi-cation of a finitely generated OX(U)-module F(U).

(ii) for every affine open affine subset U of X, F|U is the sheafification of thefinitely generated OX(U)-module F(U).

When these equivalent conditions are fulfilled, we say that F is coherent4.In particular, for a noetherian ring R the sheafification functor defines an equiva-

lence between the category of finitely generatedR-modules and the category of coherentOSpec(R)-modules. Moreover, any closed subscheme of Spec(R) is defined by a uniqueideal of R. Furthermore:

(a) the category Coh(X) of coherent sheaves is abelian subcategory of the cate-gory ModOX of OX -modules,

(b) the tensor product of two coherent OX -modules is a coherent OX -module,

3Such a morphism is often called a projective modification of X.4This definition is fine when the scheme X is noetherian (and we will only consider coherent

modules on such schemes), but for a general ringed space (X,O) it takes the following form: anO-module F is coherent if we can cover X by open subsets U for which F|U and the kernel of any O|U -homomorphism Or|U → F|U are finitely generated O|U -modules. This generalization is for instancethe one that is customary in the setting of complex-analytic spaces (the ring of holomorphic functionson a nonempty open subset of Cn, n ≥ 1, is not noetherian).


(c) if f : X → Y is a morphism between noetherian schemes, then f∗ takesCohY to CohX ,

(d) any closed subscheme of X is defined by a coherent sheaf of ideals.

PROOF. We first prove that (i) implies (ii), the other assertions follow as inthe quasi-coherent case. Let U ⊂ X be open and affine and put R := OX(U).Since X is noetherian, then by Proposition 5.7, so is R. According to the previousproposition, F|U is the sheafification of the R-module M := F(U) and (in thesituation of the proof of that proposition), we can assume that each M [1/si] is anoetherian R[1/si]-module. Then the argument given in Proposition 5.7 shows thatM must be noetherian as well.

The proofs of the other assertions straightforward.

REMARK 7.8. A direct image of a coherent module on a projective scheme isin general not coherent. The simplest example is perhaps the direct image of OA1

k

under the projection map A1k → Spec(k): this is the k-vector space Γ(A1

k,OA1k) =

k[t], which is not a finite dimensional. We will see however that this is the case fora projective morphism.

The following results show among other things that on a noetherian schemequasi-coherent modules are well approximated by coherent ones.

PROPOSITION 7.9. Let X be a noetherian scheme, U ⊂ X an open subset and Ga quasi-coherent OX -module whose restriction to U is coherent. Then there exists acoherent submodule F ⊂ G with F|U = G|U .

PROOF. First consider the case when X = Spec(R) is affine. Then G is thesheafification of an R-module N . Cover U by finitely many principal open affinesXf1 , . . . , Xfr . Since G|U is coherent, N [1/fi] is a finitely generated R[1/fi]-module.Let Si ⊂ N be a finite subset whose image in N [1/fi] generates the latter as aR[1/fi]-module. Then the R-submodule M ⊂ N generated by the finite set ∪iSiyields after sheafification a coherent subsheaf F ⊂ G with the desired properties.

In general we proceed as follows. Cover X with finitely many affine opensubsets U1, . . . , Un and put U0 := U , Vk := U0 ∪ U1 ∪ · · · ∪ Uk and Gk := G|Vk.Let 1 ≤ k ≤ n and assume with induction that we have constructed a coherentsubmodule Fk−1 of Gk−1 which on U coincides with G (this is an empty statementfor k = 1). The difference Zk := Vk−Vk−1 is a closed subset of Vk that is containedin the affine open Uk and is disjoint with U . Denoting the inclusion Vk−1 ⊂ Vkby j, then we have natural homomorphisms of quasi-coherent OVk -modules Gk →j∗Gk−1 and j∗Fk−1 → j∗Gk−1. Let F ′k ⊂ Gk be the preimage of the image ofj∗Fk−1 → j∗Gk−1 in Gk. This is a quasi-coherent OVk -submodule of Gk whoserestriction to Vk−1 = Vk − Zk equals Fk−1 (and hence is coherent). By the affinecase we already treated, there exists a coherent submodule F ′′k ⊂ F ′k|Uk whichextends F ′k|Uk − Zk. We then complete the induction step by letting Fk ⊂ G|Vk bethe coherent submodule characterized by Fk|Uk = F ′′k and Fk|Vk−1 = Fk−1.

COROLLARY 7.10. Let X be a noetherian scheme. Then

(i) if U ⊂ X is open, then every coherent OU -module extends to a coherentOX -module and

(ii) every quasi-coherent OX -module is the union of its coherent submodules.

122 3. SCHEMES

PROOF. If F is a coherent OU -module, then assertion (i) follows by applyingthe above proposition to its direct image G := j∗F under j : U ⊂ X.

For (ii) we must show that if G is a quasi-coherent OX -module, then any localsection s ∈ Γ(U,G) is contained in a coherent OX -submodule. This follows byapplying (i) to the OU -submodule OUs ⊂ G|U , which is indeed coherent.

8. When is a scheme affine?

The main theorem of this short section is a characterization for a noetherianscheme to be affine in terms of cohomology (for sheaf cohomology we refer to theappendix).

THEOREM 8.1 (Serre). For a noetherian scheme X the following are equivalent:(i) X is affine,

(ii) every quasi-coherent sheaf on X is Γ-acyclic: Hk(X,F) = 0 when k 6= 0,(iii) for every quasi-coherent ideal sheaf I ⊂ OX , H1(X, I) = 0.

The implication (ii) ⇒ (iii) is trivial and so it suffices to establish (i) ⇒ (ii)and (iii)⇒ (i). For the former we shall need the following general result on sheafcohomology.

LEMMA 8.2 (G. Kempf). Let F be an abelian sheaf on a space X and let U be abasis of open subsets. Suppose that for some positive integer n we have Hk(X,F) = 0for 0 < k < n (so this condition is empty for n = 1). Then for every a ∈ Hn(X,F)we can cover X by U ∈ U with the property that a dies in Hn(X, jU∗j

−1U F) (where

jU : U ⊂ X so that jU∗j−1U F(V ) = F(U ∩ V )).

PROOF. Embed F in a flasque sheaf I (for instance the sheaf defined by I(U) =∏x∈U Fx) and put G := I/F . Then the long exact sequence associated to this

quotient yields an exact sequence

0→ Γ(X,F)→ Γ(X, I)→ Γ(X,G)→ H1(X,F)→ 0

and isomorphisms Hk(X,G) ∼= Hk+1(X,F) for k ≥ 1. For U ∈ U, put GU :=jU∗j

−1U I/jU∗j

−1U F . This embeds in jU∗j

−1U G (jU∗ is left exact). Since jU∗j−1

U I isalso flasque, we have likewise an exact sequence

0→ Γ(X, jU∗j−1U F)→ Γ(X, jU∗j

−1U I)→ Γ(X,GU )→ H1(X, jU∗j

−1U F)→ 0

and Hk(X,GU ) ∼= Hk+1(X, jU∗j−1U F) for k ≥ 1.

We prove the assertion with induction on n and so we first treat the casen = 1. The first exact sequence shows that we can lift a ∈ H1(X,F) to an el-ement a′ ∈ Γ(X,G). We can cover X by principal open subsets U ∈ U suchthat a′|U lifts to a section a′′ ∈ Γ(U, I) = Γ(X, jU∗j

−1U I). Then the image of

a in H1(X, jU∗j−1F) is the image of a′′ under the composite Γ(X, jU∗j

−1U I) →

Γ(X,GU )→ H1(X, jU∗j−1F) and hence zero.

Now assume n > 1. We have Hk−1(X,G) ∼= Hk(X,F) = 0 for k > 1, so that Gsatisfies our hypothesis with n replaced by n− 1. By our induction assumption, wecan then cover X by U ∈ U such that Hn−1(X, jU∗j

−1U G) = 0. Obviously GU |U =

G|U and so Hn(X, jU∗j−1U F) ∼= Hn−1(X, jU∗j

−1U GU ) = Hn−1(X, jU∗j

−1U G) = 0.

PROOF OF (i)⇒ (ii). We prove (ii) with induction: assume that we have es-tablished that any quasi-coherent sheaf on an affine scheme has zero cohomologyin degrees 1, . . . , n− 1 (this assertion being empty for n = 1). Then the hypotheses

8. WHEN IS A SCHEME AFFINE? 123

of Lemma 8.2 hold for the quasi-coherent sheaf F on X. Since X is noetherian, wefind that for any a ∈ Hn(X,F) we can cover X by a finite number of affine opensubsets Uiri=1 such that a dies in Hn(Ui,F), i = 1, . . . , r. Since the Ui’s coverX, the sheaf homomorphism F → ⊕ri=1jUi∗j

−1UiF is a monomorphism of quasi-

coherent sheaves. So its cokernel, denoted G, is also quasi-coherent. Since X isaffine, say equal to Spec(R), this comes from an exact sequence of R-modules

0→ Γ(X,F)→ ⊕ri=1Γ(X, jUi∗j−1UiF)→ Γ(X,G)→ 0.

Feeding this in the long exact sequence for cohomology shows that then H1(X,F)→⊕ri=1 H1(X, jUi∗j

−1UiF) is injective. So if n = 1, then a = 0. For n > 1 we find that

a lies in the image of Hn−1(X,G) → Hn(X,F). But then it must be zero, since byour induction assumption Hn−1(X,G) = 0.

For the proof that (iii)⇒ (i) we need the following criterion for affineness:

LEMMA 8.3. A scheme X is affine if and only if there exist f1, . . . , fr ∈ Γ(X,OX)such the corresponding open subschemes Xfi ⊂ X are affine and cover X.

PROOF. Let X be a scheme and put A := Γ(X,OX). Then there is an obviousmorphism X → Spec(A), which of course is an isomorphism precisely when X isaffine. We only prove the nontrivial direction and assume that A contains f1, . . . , fras in the lemma. Then each fi defines a principal open subset SpecA[1/fi]) ⊂Spec(A) whose preimage in X is Xfi . So it will be enough to show that for every

f ∈ A, the restriction Xfφ−→ SpecA[1/f ]) is an isomorphism. In other words, we

must show that the natural ring homomorphism A[f ] → OX(Xf ) is an isomor-phism. For this we will use the observation that for every i, Xf ∩ Xfi is a prin-cipal open subset of the affine Xfi so that OX(Xf ∩ Xfi) = O(Xfi)[1/fi], wherefi := f |Xfi .

To prove injectivity, suppose g ∈ A is such that g|Xf = 0. Then for every i,g|Xf ∩ Xfi = 0 and so exists an integer ni ≥ 0 such that fnig|Xfi = 0. So if welet n := maxni, then fng|Xfi for all i and hence fng is zero in Γ(X,OX) = A. Inother words, g has zero image in A[1/f ].

To prove surjectivity, let g ∈ OX(Xf ). Then g|Xfi ∩Xf ∈ O(Xfi)[1/fi] has theproperty for some an integer ni ≥ 0, fnig extends to a section of OX |Xfi . So if welet n := maxni, then fng extends to a section h ∈ A of OX and g is the image ofh/fn ∈ A[1/f ].

PROOF OF (iii)⇒ (i). We will show that the hypotheses of Lemma 8.3 are sat-isfied. Let x ∈ X be a closed point and let Ux be an open affine neighborhood ofx. Let Ix ⊂ OX denote the ideal sheaf that defines the closed subscheme X − Uxand let J x ⊂ Ix be the kernel of the natural surjection Ix → ix∗κ(x). SinceH1(X,J x) = 0, the evaluation map Γ(X, Ix) → κ(x) is surjective and so thereexists a fx ∈ Γ(X, Ix) whose image in κ(x) is 1. Notice that Xfx can be understoodas a principal open subset of Ux and is therefore an affine neighborhood of x in X.

Since X is noetherian a finite number of such subsets will cover X. Now applyLemma 8.3.

We turn our attention to higher direct images of quasi-coherent sheaves. In theappendix we show that a for morphism (X,O) → (X ′,O′) of ringed spaces, theright derived functors of ‘ringed direct image functor’ ModO → ModO′ composed

124 3. SCHEMES

the forgetful functor ModO′ → AbX′ produce the right derived functors of the‘ordinary direct image functor AbX → AbX′ .

PROPOSITION 8.4. Let f : X → Y be a morphism of schemes with X noetherian.Then Rkf∗ sends Qcoh(X) to Qcoh(Y ). In fact, if V ⊂ Y is open affine, then for everyquasi-coherent sheaf F on X, Rkf∗(F)|V is the sheafification of the OX(V )-moduleHk(f−1V,F).

PROOF. It suffices to prove the last assertion and so we assume that Y =Spec(R) is affine. For a quasi-coherent sheaf F on X, we denote by T k(F) thesheafification of the R-module Hk(X,F). It is clear that then T 0(F) = f∗F . Itis also clear that the T k ’s are part of a δ-package and so it remains to see thatit each T k is effaceable. For this let jν : Uν ⊂ Xnν=1 be a finite affine opencovering of X. For every affine open subset V ⊂ Y , we have T k(jν∗j

∗νF)(V ) =

Hk(f−1V, jν∗j∗νF) = Hk(f−1V ∩ Uν ,F) and since f−1V ∩ Uν is affine, Serre’s cri-

terion for affineness implies that this is zero when k > 0. So if we put F ′ :=⊕νjν∗j∗νF), then T k(F ′) = 0 for k > 0. We have a natural OY -homomorphismF → jν∗j

∗νF ′ of quasi-coherent sheaves which combine to a monomorphism F →

F ′. So T k is effacable.

EXERCISE 89. Let f : X → Y be a quasi-compact morphism of schemes withX locally noetherian. Prove that f is affine if and only if for every quasi-coherentsheaf, R1f∗(F) = 0. Show that in either case, Rkf∗(F) = 0 for all k > 0.

9. Divisors and invertible sheaves

We fix a scheme X. An OX -module L is called invertible if we can cover Xby open subsets U such that L|U ∼= OU . So for every x ∈ X, the fiber L(x) =κ(x)⊗OX,x Lx = Lx/mX,xLx is then a κ(x)-vector space of dimension one. We notethat the OX -dual, defined as L∨ := HomOX (L,OX) is also invertible and so is itstensor product L ⊗ L′ with another invertible sheaf L′. We also write Ln for resp.L⊗n, OX , (L∨)⊗(−n) when resp. n > 0, n = 0, n < 0. The tensor product turns theset isomorphism classes of invertible sheaves on X into a group (with the inverserepresented by the dual), the Picard group Pic(X) of X. The group structure iswritten addively so that [Ln] = n[L]. It turns out that the Picard group often hasthe structure of a group scheme.

Observe that for any morphism of schemes f : X → Y , f∗ takes an invertiblesheaf L on Y to one on X. This is verified on affine open subsets U = Spec(R) ⊂ Xthat map to an affine open subset V = Spec(S) of Y on which L is trivial and thisit is a simple consequence of the fact that for the associated ring homomorphismφ : S → R, the map r ∈ R 7→ r ⊗ 1 ∈ R ⊗S S is an isomorphism with inverser ⊗ s ∈ R ⊗S S 7→ rφ(s) ∈ R. This pull-back is compatible with ⊗ and takesisomorphic invertible sheaves to isomorphic invertible sheaves. Hence f induces agroup homomorphism:

Pic(f) : Pic(Y )→ Pic(X).

We thus have defined a contravariant functor Pic : Sch → Ab. This notion of a Pi-card group makes sense for any ringed space and in the appendix we prove that inthat generality Pic(X) can be identified with H1(X,O×X). Here we describe Pic(X)(at least part of it) in terms of Cartier divisors.

9. DIVISORS AND INVERTIBLE SHEAVES 125

Cartier divisors. For a ring R, we have defined its fraction ring Frac(R) (2.14)as the localization with respect to the multiplicative set S(R) of all nonzero divisors.It has the property thatR→ Frac(R) is injective. We note that the property of being(not) a zero divisor can be tested ‘locally’: For every prime ideal p of R, we havethe localization map R→ Rp. Clearly, a zero divisor of R will map to a zero divisorof Rp or to zero. On the other hand, for every prime ideal p of R, S(R) maps toS(Rp): if s ∈ S(R) kills a/b ∈ Rp, then sac = 0 in R for some c ∈ R − p. Since s isnot a zero divisor, we then have ac = 0, which means that a/b = 0.

We phrase this in scheme language: if U ⊂ X is affine, then s ∈ Γ(U,OX) isnot a zero divisor if and only if its image in each stalk OX,x with x ∈ U is not a zerodivisor. It is now clear that we have defined on X a presheaf of rings which assignsto an arbitrary open subset U ⊂ X, the localization of Γ(U,OX) with respect to themultiplicative system of s ∈ Γ(U,OX) which map to a nonzero divisor in OX,x forall x ∈ U . We denote the associated sheaf5 by KX . It is characterized by the prop-erty that for an affine open subset U ⊂ X, KX(U) = Frac(OX(U)). The naturalsheaf homomorphism OX → KX is a monomorphism and we will therefore thinkof OX as a subsheaf of KX . We denote by K×X ⊂ KX the subsheaf of invertibleelements. Notice that when X is an integral scheme, then KX is the constant sheafthat assigns to any nonempty open subset of X the function field K(X) of X.

A quasi-coherent OX -submodule I ⊂ KX is called a fractional ideal if we cancoverX by affine open subsets U with the property that fI ⊂ OX for some nonzerodivisor f ∈ Γ(U,OX). Note that a fractional ideal is automatically coherent whenX is locally noetherian. Invertible fractional ideals arise as follows. Let L be aninvertible OX -module. Any section s ∈ Γ(X,L) defines a subscheme Z(s) ⊂ Xwhose defining ideal sheaf IZ(s) is locally principal: it is characterized by the prop-erty that if U ⊂ X is an open affine subset and φ : L|U ∼= OX |U is an isomorphism,then IZ(s)|U = OUφ(s). This is indeed independent of the choice of φ, for any twosuch differ by a factor in O(U)×, hence generate the same principal ideal sheaf. Wecan also view s as defining a sheaf homomorphism OX → L which maps 1 to s.This is an embedding precisely when for every (U, φ) as above, φ(1) is not a zerodivisor in O(U). In that case, we can reconstruct L from Z := Z(s) as a fractionalideal I−1

Z ⊂ KX . This submodule is characterized by the property that for everylocal generator f ∈ IZ(U) of IZ , I−1

Z |U := OUf−1. Then the sheaf homomorphismOX → L defined by s extends to an isomorphism I−1

Z∼= L. This observation is the

point of departure for the discussion that follows.

DEFINITION 9.1. A Cartier divisor on X is an element of the (additively written)abelian group Γ(X,K×X/O

×X). We say that a Cartier divisor is principal if it lies in

the subgroup that is the image of Γ(X,K×X)→ Γ(X,K×X/O×X). The quotient group

is denoted CaCl(X). Two Cartier divisors are said to be linearly equivalent if theyhave the same image in CaCl(X), i.e., if their difference (really their quotient, butwe agreed to write the group law additively) is principal.

We will see that often a Cartier divisor has the simple geometric meaning asformal linear combinations of integral subschemes of codimension one that can

5Because of this roundabout definition there are a number of subtle issues here (see S.L. Kleiman:Misconceptions about KX). For instance, for x ∈ X, the natural map KX,x → Frac(OX,x) is alwaysinjective, but need not be surjective. In particular, KX need not be quasi-coherent.

126 3. SCHEMES

locally be given by a single equation. This explains why we write the group lawadditively. In any case, we note that a Cartier divisor H on X can always be givenby a collection of pairs (Uα, fα)α, where U = (Uα)α is an affine covering of X andfα ∈ K(U)× is such that the restrictions of fα and fβ in Uα∩Uβ have a quotient thatlies in O(Uα ∩ Uβ)× (in a sense only the zeroes and the poles of the fα’s matter).But beware that we are not claiming that any open affine covering of X will do forthis purpose: indeed, quite often an affine variety has Cartier divisors that are notprincipal.

EXAMPLE 9.2. Consider the quadric Q = Spec(k[x, y, z]/(xy − z2 + z)). Theline ` ⊂ Q defined by the ideal (y, z) (the ‘x-axis’) defines a Cartier divisor: wecover Q by the open subsets U0 defined by z 6= 0 and U1 defined by z 6= 1 (sothat U0 ∩ U1 is defined by z2 − z 6= 0, or equivalently, xy 6= 0) and we take on U0

the regular function y|U0 and on U1 the regular function constant 1. Then theirquotient y/1 = y will be indeed nonzero on U0 ∩ U1. But this Cartier divisor is notprincipal: there is no rational function on f onQwith the property that both f/y|U0

and f |U1 are invertible. For this would imply that f is a regular function onQwhichgenerates the ideal sheaf I` inOQ defining `. But then a representative f ∈ k[x, y, z]of f must generate the ideal (y, z) in the coordinate ring k[x, y, z]/(xy− z2 + z). Inother words, (f , xy−z2+z) = (y, z) in k[x, y, z]. This is easily seen to be impossible.

EXAMPLE 9.3. This example shows that not every curve on a surface defines aCartier divisor. Consider the quadric cone C = Spec(k[x, y, z]/(xy − z2)) and let` ⊂ C be the line defined by the ideal (y, z) (the ‘x-axis’). Then the zero sets x, y, zrestricted to C are resp. the y-axis with multiplicity 2, the x-axis with multiplicity2, the union of the x-axis and the y-axis. It follows that no regular function onC can vanish on ` with multiplicity 1. This remains true after localization at thevertex o ∈ C: if f, g ∈ k[x, y, z] are such that g(o) 6= 0, then the image of f/g inOC,o can not vanish of order one on the x-axis and not anywhere else near o (forthen f would have the same property). So ` can not be defined by a Cartier divisor.On the other hand, the principal divisor defined by y|Q yields twice `.

PROPOSITION 9.4. A Cartier divisor H on X determines an invertible fractionalideal L(H) ⊂ KX characterized by the property that if H is represented on an opensubset U ⊂ X by f ∈ K(U)×, then L(H)|U := OUf−1. Conversely, any invertiblefractional ideal L ⊂ KX defines a Cartier divisor D(L) characterized by the propertythat given a local isomorphism φ : OU ∼= L|U , then φ(1)−1 ∈ KX(U) representsD(L)|U .

This sets up a bijection between the group of Cartier divisors onX and the group ofinvertible fractional ideals of KX , where the group structure on the latter correspondsto the (tensor) product: given invertible subsheaves L ⊂ KX and L′ ⊂ KX , then wecan form L⊗OXL′ ⊂ KX . Under this correspondence the principal divisors correspondto the invertible fractional ideals of KX that are trivial (i.e., isomorphic to OX).

PROOF. If H is represented on an affine open subset U ′ ⊂ X by f ′ ∈ K(U)×,then the images of f and f ′ in K(U ∩ U ′)× have a quotient in O(U ∩ U ′)× and sothey generate the same subsheaf of KU∩U ′ . This proves that L(H) is well-defined.It is clearly invertible.

Of the other assertions we only verify the last one (the others are staightfor-ward). If we are given a OX -monomorphism φ : OX → KX , then φ(1) defines


a global section of K×X and the associated divisor is then the principal divisor de-fined by φ(1). Conversely, if the divisor for L ⊂ KX is principal, say given byf ∈ Γ(X,K×X), then L = OXf ⊂ KX and so multiplication by f−1 defines anisomorphism L ∼= OX .

REMARK 9.5. The following may help to explain the inversion convention inProposition 9.4. Let L be an invertible sheaf on X and s ∈ Γ(X,L) a section.We observed that s defines a principal ideal sheaf IZ ⊂ OZ and that under someassumption, L can be identified with the invertible fractional ideal I−1

Z ⊂ KX .If U ⊂ X and φ : OU ∼= L|U is an isomorphism, then s|U = fφ(1) for somef ∈ OX(U) and so f generates IZ |U . Hence f−1 is a generator of I−1

Z |U and byour convention, the Cartier divisor associated to I−1

Z is on U represented by thepair (U, f).

Proposition 9.4 immediately implies:

COROLLARY 9.6. The group of Cartier divisors modulo its subgroup of principaldivisors (denoted CaCl(X)) can be identified with the subgroup of Pic(X) of isomor-phism classes of invertible sheaves that are embeddable in KX .

We can obtain this embedding also in terms of the cohomological descriptionof the Picard group: the short exact sequence

0→ O×X → K×X → K

×X/O

×X → 0

gives rise to the long exact sequence

0→ H0(X,O×X)→ H0(X,K×X)→ H0(X,K×X/O×X)→ H1(X,O×X)→ H1(X,K×X)

and we thus see that CaCl(X) = Coker(H0(X,K×X) → H0(X,K×X/O×X)) maps in-

jectively to H1(X,O×X) ∼= Pic(X) (with the cokernel contained in H1(X,K×X)).

COROLLARY 9.7. If X is an integral scheme, then the map CaCl(X)→ Pic(X) isan isomorphism.

PROOF. Then K×X is a constant sheaf (with value the units of the function fieldof X) on an irreducible space and hence flasque. So H1(X,K×X) = 0. Now feed thisinto the exact sequence above.

Weil divisors. As some of the examples discussed below will show, the Picardgroup of a scheme can be quite interesting, but is in general hard to determine. Forcomputations it is helpful to think of a Cartier divisor as a formal linear combina-tion of hypersurfaces in the sense that if the Cartier divisor is given by the system(Uα, fα), then we would want to define a linear combination

∑i riPi with Pi an

integral subscheme of codimension one and ri ∈ Z such that fα has along Pi ∩ Uαa zero of order ri (when this ri < 0 is to be understood as a pole of order −ri)and fα has no other zeros or poles. It is then reasonable to assume that each Piwhich may possibly occur is at least generically given by a single equation and thatwe can speak of order of vanishing. This just amounts to insisting that for sucha P = Pi, the generic point ηP defined by P , OX,ηP is a DVR (whose valuationFrac(OX,ηP ) → Z we denote by vP ): when P ∩ Uα 6= ∅, then fα ∈ Frac(O(Uα)has a nonzero image in Frac(OX,ηP ) and so the valuation on Frac(OX,ηP )× has awell-defined value vP (fα) on fα. This value only depends on the Cartier divisor,for if we also have P ∩ Uβ 6= ∅, then P ∩ Uα ∩ Uβ 6= ∅ (for P is irreducible) and

128 3. SCHEMES

since fβ/fα is a unit in O(Uα ∩ Uβ), it is also a unit of Frac(OX,ηP )× and hencevP (fβ) = vP (fα). This motivates the following definition.

DEFINITION 9.8. A closed integral subscheme P of X (or its associated pointηP ∈ X) whose local ring OX,ηP is a DVR is called a prime divisor of X. A Weildivisor on X is a finite formal Z-linear combination of prime divisors D =

∑P nPP

(the sum is over prime divisors, but for only finitely many we have nP 6= 0) andwe then denote by OX(D) ⊂ KX the subsheaf of f ∈ KX with vP (f) ≥ −nP forall P . We say that the Weil divisor is effective if all the coefficients are ≥ 0 (so thatOX(D) ⊃ OX).

EXAMPLE 9.9. Here are a few simple examples. Let K be a field.(a) Any nonconstant irreducible polynomial f ∈ K[x1, . . . , xn] defines a prime

divisor in Ank and every prime divisor on Ank is so obtained.(b) Any irreducible homogeneous polynomial F ∈ K[T0, . . . , Tn] of positive

degree defines a prime divisor on PnK and every prime divisor on PnK is so obtained.(c) Any closed point on a nonsingular curve C defined over K defines a prime

divisor and every prime divisor on C is so obtained.(d) Suppose K/Q is a number field and denote by OK its ring of integers. Then

the prime divisors of Spec(OK) are defined by the prime ideals of OK .

The above program works best if we impose a few extra conditions on X. Con-sider for instance the affine cuspidal curve C := Spec(k[x, y]/(y2 − x3)). As wehave seen, the local ring at the origin OC,o is not a DVR, yet x|C and y|C have azero at o. So if we focus on local rings that are DVR’s such a zero goes undetected.The simplest way out is to assume that there are no such points. This we can doby stipulating every closed integral subscheme of X that is not an irreducible com-ponent of X is contained in a prime divisor. This is equivalent to: the closure ofevery x ∈ X is an irreducible component of X or is contained in a prime divisor D.This implies that OX,x is a field or contained in a DVR and this implies that OX,xhas no zero divisors (hence is reduced). So X is reduced and every point of X iscontained in exactly one irreducible component. In other words, each connectedcomponent of X is integral. It is then hardly a restriction to assume that X itself isintegral, for we are then just asking that X be connected.

Another natural condition to impose is that for every f ∈ K(X)× only a finitenumber valuations are nonzero on it. We will see that this is automatic when X isnoetherian. So in the rest of this section we assume the

DVR domination property: X is noetherian and every closed integral sub-scheme of X distinct from X is contained in a prime divisor (so that X isintegral).

All the schemes described in the examples in 9.9 have this property. Anotherinteresting case is that of (the spectrum of) a Dedekind domain: this is a noether-ian domain with the property that every maximal ideal is a DVR. These occur inalgebraic geometry as affine normal curves over a field F and in algebraic numbertheory as rings of integers in number fields. More generally, any integral noether-ian scheme all of whose closed points are prime divisors satisfies these hypotheses.This includes the case of a nonsingular curve over a field.

EXERCISE 90. Prove that an irreducible variety over k of dimension n whosesingular locus is of dimension ≤ n− 2 satisfies the DVR domination property.


PROPOSITION-DEFINITION 9.10. Let X be a scheme having the DVR dominationproperty.

(i) Then for every f ∈ K(X)×, the sum (f) :=∑P vP (f)P (with D running

over the prime divisors) is finite and hence defines a Weil divisor. This defines a grouphomomorphism from K(X)× to the group of Weil divisors.

The image of this homomorphism is called the group of principal Weil divisors.Its cokernel is called the divisor class group of X and will be denoted Cl(X).

(ii) Any Cartier divisor H on X determines a Weil divisor (H) on X characterizedby the property that if H is on an open affine U represented by f ∈ KX(U), then(H)|U = (f). This defines a homomorphism from the group of Cartier divisors tothe group of Weil divisors, and this induces a monomorphism Pic(X) ∼= CaCl(X) →Cl(X).

PROOF. Since X is noetherian, it is for (i) enough to show that for any affineopen U ⊂ X, (f)|U is finite. In other words, we may assume that X = Spec(R).We first assume that f is regular: f ∈ R. A prime divisor for U is given a by aprime ideal p with the property that Rp is a DVR. The only a prime ideal containedin such a p is the zero ideal. We have vp(f) > 0 if and only f ∈ p. In that case, pappears in the primary decomposition of

√(f) and since this involves only a finite

number of primes, the sum (f) :=∑D vD(f)D is finite. For an arbitrary nonzero

f = f1/f0 ∈ Frac(R), (f) = (f1)− (f0) and so this sum is also finite. At the same itis clear that (fg) = (f) + (g).

If the Cartier divisor H on X is on the affine open U represented by f andg, then f/g ∈ O(U)× and so (f) = (g). This implies that (H) is well-defined.This also implies that H 7→ (H) is a homomorphism of groups. If (H) is principal(as a Weil divisor), then H is principal (as a Cartier divisor), and so the inducedhomomorphism Pic(X) ∼= CaCl(X) → Cl(X) will be injective.

REMARK 9.11. In order that Pic(X) ∼= CaCl(X) → Cl(X) be also surjective, wewant that every prime divisor P represents a Cartier divisor. This amounts to: theideal sheaf IP defining P is locally principal. Then for every Weil divisor D, OX(D)is an invertible subsheaf of KX and this defines the inverse Cl(X) → Pic(X). Thisis so when the local rings of X are all UFD’s. An important example in algebraicnumber theory is the spectrum of the ring of integers of a number field K/Q (byabuse of language this is then often called the divisor class group of K). A theoremof Auslander-Buchsbaum asserts that every regular local ring is a UFD. This impliesthat for instance that for a nonsingular irreducible variety X, we have Pic(X) ∼=Cl(X) (but this is not hard to prove independently of the Auslander-Buchsbaumtheorem).

REMARK 9.12. We elaborate on Remark 9.5 a bit. Suppose X satisfies the DVRdomination property. Let L be an invertible sheaf on X and s ∈ Γ(X,L) a nonzerosection. Then s defines a principal ideal sheaf IZ(s) ⊂ OX and an isomorphism ofL onto the invertible fractional ideal I−1

Z(s) ⊂ KX . If U ⊂ X is open, φ : OU ∼= L|Uan isomorphism and f ∈ OX(U) such that s|U = fφ(1), then the Cartier divisorassociated to I−1

Z(s) is on U represented by (U, f). The Weil divisor attached to thisCartier divisor is then on U represented by

∑P vP (f)P . Clearly, vP (f) ≥ 0 and can

be understood as the order of vanishing of s along P . So the Weil divisor attachedthis situation and has then all its coefficients ≥ 0 and can be understood as the

130 3. SCHEMES

‘zero divisor’ of the section s. We write (s) for that divisor. Since IZ(s) is the sheafof g ∈ OX with

∑P vP (g)P ≥ (s), (s) and Z(s) determine each other.

This generalizes to the case of a ‘rational section’ of L, that is, a nonzero ele-ment s of the generic fiber Lη: then s determines an embedding L → KX such thats ∈ Lη maps to 1 ∈ KX,η = K(X). The Weil divisor of the image (an invertiblefractional ideal) is the divisor (s) of s.

EXAMPLES 9.13. We revisit some of the Examples 9.9 (so K is a field).(a) Since K[x1, . . . , xn] is a UFD, we have Pic(AnK) = Cl(AnK) = 0.(b) A prime divisor P on Pnk is given by a nonzero homogeneous irreducible

polynomial F ∈ K[T0, . . . , Tn]. This polynomial is unique up a scalar in K× andso has has a well-defined degree, n, say. One such prime divisor is the hyperplaneH in PnK defined by the ideal T0. Now F/Tn0 is a rational function on PnK and itsdivisor is P − nH. It follows that Pic(PnK) = Cl(PnK) is generated by the class ofH. On the other hand, for n > 0, nH is not principal (otherwise we would get arational function on PnK without poles and we know that such a function must beconstant). It follows that the ‘degree’ defines an isomorphism Pic(PnK) ∼= Z.

(c) We now treat of the cuspidal curve C := Spec(K[x, y]/(x3−x2). This curveis parametrized by the morphism f : C := A1

K → C, f∗x = t2, f∗y = t3 and thusidentifies Γ(C,OC) with the subalgebra K[t2, t3] of K[t]. This is the algebra of allthe polynomials in t without linear term. Similarly, if we regard OC,o as a subringOC,o, then the linear terms ct form a supplementary one-dimensional subspace andany u ∈ O×

C,ois uniquely written u(1 + λt) with u ∈ O×C,o and λ ∈ K. In fact, the

map u 7→ λ defines a group homomorphism O×C,o→ Ga(K) with kernel O×C,o.

Now let L be an invertible sheaf on C. Then f∗L is trivial and so admits anowhere zero section s. Note that s is unique up to scalar and that Γ(C, f∗L) =K[t]s. The latter contains Γ(C,L) as a K[t2, t3]-submodule. A generator s for thefree rank one OC,o-module Lo will also be a generator for f∗L as a OA1

K ,0-module

and so will be of the form us with u ∈ O×A1K ,0

. Since s is unique up to an element of

O×C,o, the image of u in OC,o/O×C,o only depends on L. We identified this quotient

with Ga(K×). Notice that (1 +λt)s can then be understood as a section of L whichhas a zero only in f(−λ−1) (assuming λ 6= 0).

This gives us a complete invariant of L. If λ ∈ Kr0, then we let Lλ ⊂ OC bethe OC -submodule which over Cr−λ−1 is generated by 1 +λt and over Crois generated by 1. This is clearly an invertible OC -module. We have thus obtaineda group isomorphism Ga(K) ∼= Pic(C).

(d) We compute the Picard variety of the nodal curve C over a field K definedby Spec(K[x, y]/((x − y)3 + xy)) (the node is at o = (0, 0)). Put C := P1 r 1.Then f : C → C, f(t) := (t(t − 1)−3, t2(t − 1)−3) parametrizes C. Notice thatf(0) = f(∞) = o. Thus f∗ identifies Γ(C,OC) with the subalgebra of K[t][1/(t−1)]of functions which take the same value in t = 0 and t = ∞. Now let L be aninvertible sheaf on C. Since C ∼= A1

K has trivial Picard group, f∗L is trivial and soadmits a nowhere zero section s. Because s(0) and s(∞) are both generators of theone-dimensional K-vector space L(o) = κ(o) ⊗OC,o Lo there exists a λ ∈ K× withs(∞) = λs(0). Since s := (λt − 1)/(t − 1)s has the property that s(0) = s(0) ands(∞) = s(∞) we find that s defines a section of L. It has a zero in t = 1/λ. Thisidentifies L with the sub OC -module of OC which over C r 1/λ is generated by


(λt− 1)/(t− 1) and over C r 0,∞ is generated by 1. This identifies Pic(C) withthe multiplicative group Gm(K).

Divisors on a nonsingular curve. Let C be a connected nonsingular curveover k. Then a Weil divisor D on C can be given as a formal linear combinationof closed points of C: D =

∑p∈C(k) np(p) with np nonzero for only finitely many

p ∈ C(k). We then define the degree deg(D) of D as the sum∑p np. In order to

better understand this notion we need the following proposition.

PROPOSITION-DEFINITION 9.14. Let π : C → C ′ a nonconstant proper morphismof connected nonsingular curves over k. Then π∗OC is locally free of finite rank; thisrank is called the degree of π and denoted deg(π). If we define the local degreedegp(π) of π at p ∈ C as the k-dimension of OC,p/mC′,π(p)OC,p, then

deg(π) =∑

p∈π−1(p′)

degp(π).

PROOF. Since π is proper, π(C) is closed in C ′. Since π(C) is connected, butnot a singleton, it follows that π is onto and that any fiber over a closed point ofC ′ is finite. In particular, π is dominant: π∗ embeds k(C ′) in k(C). For p ∈ C wehave OC,p ⊂ k(C), OC′,f(p) ⊂ k(C ′) and π∗ maps OC′,f(p) to OC,p. So π∗ restrictsto a monomorphism π∗p : OC′,f(p) → OC,p. Since OC,p is an integral domain, thisimplies that OC,p is a torsion free OC′,f(p)-module.

The fact that π is proper further implies that π∗OC is a coherent OC′ -module(we will prove a general result of this kind in 11.5). Let p′ ∈ C ′ and choose a neigh-borhood U ′ of p′ and f1, . . . , fd ∈ Γ(U ′, π∗OC) such that their images in the fiber(π∗OC)(p′) = ⊕p∈π−1(p′)OC,p/mC′,p′OC,p make up a k-basis. Then their imagesin the stalk (π∗OC)p′ = ⊕p∈π−1(p′)OC,p generate that stalk as a OC′,p′ -module byNakayama’s Lemma. As we have seen, each summand of this stalk is torsion free.Since a torsion free finitely generated module over a DVR is free, (f1, . . . , fd) willmap to aOC′,p′ -basis of (π∗OC)p′ . So the homomorphism of coherentOU ′ -modulesOdU ′ → π∗OC |U ′ defined by the f1, . . . , fd has its kernel and cokernel zero in p′ andhence is an isomorphism on a neighborhood of p′. This proves that π∗OC is locallyfree. The identity is obtained by taking dimensions.

Now suppose that in the above situation we are given a Weil divisor D′ =∑p′ np′(p

′) on C ′. Associated to D′ is the invertible sheaf OC(D′) of f ∈ KC′with vp(f) ≥ np′ for all p′. Now π∗OC(D′) is also an invertible sheaf and as astraightforward local computation shows, it comes with a divisor denoted π∗D′

that it is equal to∑p∈C(k) degp(π)nf(p). So if we define the degree of the divisor D′

by deg(D′) =∑p′∈C′(k) np′ , then the degree of π∗D′ is equal to deg(π) deg(D′).

PROPOSITION 9.15. If C is a connected nonsingular curve proper over k, then thedegree of a Weil divisor on C only depends on its equivalence class; in fact the degreefactors through a surjective group homomorphism deg : Pic(C)→ Z.

Moreover the map p ∈ C(k) 7→ (p) ∈ Pic(C) is injective, unless C ∼= P1k.

PROOF. For the first assertion we must show that a principal divisor D on Chas zero degree. By definition D is given by a nonzero f ∈ K(C). We can think ofsuch an element as defining a morphism f : C → P1

k. To be precise, let U0 resp. U1

be the open subset where f resp. 1/f is regular (a closed point p ∈ C(k) is in U0

132 3. SCHEMES

resp. U1 if and only if vp(f) is ≥ 0 resp. ≤ 0). Then f |U0 resp. f |U1 is defined byf∗(T1/T0) = f resp. f∗(T0/T1) = 1/f . The morphism f is proper (for C is properover k and nonconstant so that Proposition 9.14 applies. Since D is obtained as f∗

applied tot the degree zero divisor (0)− (∞)) is has degree zero.For the second assertion we must show that if p, q ∈ C(k) are distinct and

(p) − (q) is principal, then C ∼= P1k. For such a divisor (p) − (q) the preceding

yields a morphism f : C → P := P1k of degree one. This means that f is an

isomorphism.

REMARK 9.16. It is clear that the image of C(k) in Pic(C) will generate Pic(C)as a group. If we let Pic0(C) denote the kernel of deg : Pic(C)→ Z, then it is alsoeasy to see that the image of (p, q) ∈ C(k)× C(k) 7→ (p)− (q) ∈ Pic0(C) generatesthe latter as a group. When k = C, C determines a compact connected Riemannsurface, of genus g say, and then Pic0(C) can be identified with the Jacobian of C.This is an abelian variety of dimension g: as a complex manifold it is a copy of thereal torus of dimension 2g, endowed with a translation invariant complex structure,i.e., of the form Cg/L, where L ⊂ Cg is the Z-span of some R-basis of Cg and admitsan embedding a complex projective space. One can show that something like thisis true in general: the abelian group Pic0(C) is naturally the set of k-valued pointsof a connected projective group scheme over k that is characterized by the propertythat C(k)× C(k)→ Pic0(C) comes from a morphism of schemes.

10. The projective setting

Projective sheafification. Let A = ⊕∞k=0Ak be grading ring and write R for itssubring A0. We recall that the proj construction associates to A a scheme Proj(A)(which we shall here often abbreviate by P ) over Spec(R). The points of P are inbijection with the homogeneous prime ideals p strictly contained in A+. For s ∈ Ad(d > 0), Z(s) ⊂ P denotes the set of points associated to homogeneous primeideals containing s. Then the points of Ps = P − Z(s) are also associated to thehomogeneous prime ideals of A[1/s]0 = ∪k≥0Adk/s

k (this is an increasing union,for Adk/sk = sAdk/s

k+1 ⊂ Ad(k+1)/sk+1) and a scheme structure on P is charac-

terized by the property that any such Ps is open and identified with Spec(A[1/s]0).The stalk of OP at the point associated to p is then (Ap)0.

REMARKS 10.1. This notion leads us to make a number of simple observations.First, if si ∈ Adii is a set of homogeneous elements of positive degree (di > 0)

which generates A as a R-algebra, then the associated open subsets Psii coverP : if p ( A+ is a strict homogeneous prime ideal, then there exists a s ∈ A+ suchthat s /∈ p and since s can be written as a polynomial in the si’s, there must exist asi such that si /∈ p and so the point of P defined by p lies in Psi .

Let us also note that for every positive integer m, Adk/sk ⊂ Adkm/skm, and so

Proj(A) does not change if we replace A by its subalgebra A(m) := ⊕∞k=0Amk.Finally, a surjective homomorphism φ : A→ B of graded algebras gives rise to

a closed immersion Proj(B)→ Proj(A). To see this, let s ∈ Ad, d > 0 be as above.Then φ induces a graded ring epimorphism A[1/s] → B[1/φ(s)]. This restricts to aring epimorphism A[1/s]0 → B[1/φ(s)]0, which amounts to a closed immersion ofaffine schemes Proj(B)φ(s) → Proj(A)s. Since the Proj(B)φ(s) cover Proj(B), ourclaim follows.

A graded A-module gives rise to a quasi-coherent sheaf on P :

10. THE PROJECTIVE SETTING 133

PROPOSITION-DEFINITION 10.2. A graded A-module M defines a quasi-coherentsheaf MP on P characterized by the property that for any d ≥ 0 and s ∈ Ad,Γ(Ps,MP ) = M [1/s]0 = ∪k∈ZMdk/s

k. It has the property that the stalk of MP

at the point associated to a strict homogeneous prime ideal p ( A+ equals (Mp)0.We call this the projective sheafification of M . The sheaf is coherent when A isnoetherian and M is finitely generated.

PROOF. The proof is straightforward.

Notice that any element of M0 gives rise to a section of MP : we have anobvious R-homomorphism

M0 → Γ(P,MP ).

In general this is neither injective nor surjective. For instance, if we take M =A/A+, then M = M0. For every s ∈ Ad, d > 0, we have sM = 0 and so M [1/s] = 0.HenceMP = 0. If we take A = R[T ] (with T of degree 1 and R in degree 0) andM = (T ), then M0 = 0, but it is not hard to check that Γ(P,MP ) = Γ(P,OP ) =R. The reason is that the passage from Spec(A) to Proj(A) goes via the unionof the Spec(A[1/s]) taken over all homogeneous s ∈ A+. This union is equal toSpec(A) r o, where o is the ‘vertex point’ defined by A+, and hence ignores anypossible uggliness of Spec(A) at o. Similarly,MP only depends on the sheafificationof M restricted to Spec(A) r o.

Given a graded A-module M and an integer n ∈ Z, then we have defined anew graded A-module M(n) by shifting degrees: M(n)k := Mk+n. The associatedOP -moduleM(n)P is characterized by the property that for any d ≥ 0 and s ∈ Ad,it assigns to Ps the A[1/s]0-module

M [1/s]n = ∪k∈ZMdk+n/sk so thatMP (n)[p] = (Mp)n.

We have of course a natural R-homomorphism Mn → Γ(P,M(n)P ). So if weform the quasi-coherent OP -moduleM(•)P := ⊕n∈ZM(n)P , then we have naturalhomomorphism M → ⊕n∈ZΓ(P,M(n)P ) = Γ(P,M(•)P ) of graded R-modules. Ifwe apply this to A we find an associated quasi-coherent sheaf O(n)P and a naturalgraded homomorphism A→ Γ(P,O(•)P ). But there is more structure present, for itis immediate from the definitions that multiplication defines for every pair n, r ∈ Za sheaf homomorphism

O(n)P ⊗OP M(r)P →M(n+ r)P .

so that O(•)P is in fact a sheaf of graded OP -algebras, and M(•)P is a gradedmodule over this sheaf. In particular, Γ(P,M(•)P ) = ⊕n∈ZΓ(P,M(n)P ) is a graded⊕n∈ZΓ(P,O(n)P )-module. This is compatible with the natural graded maps

A→ Γ(P,O(•)P ), M → Γ(P,M(•)P )

in the sense that the former is a ring homomorphism via which the latter becomesan A-module homomorphism.

PROPOSITION 10.3. Suppose A is generated as a R-algebra by A1. Then forevery n and r, theOP -moduleO(n)P is invertible and theOP -module homomorphismO(n)P ⊗OP M(r)P →M(n+ r)P is an isomorphism.

PROOF. Our assumption implies that P is covered by the open subsets Ps withs ∈ A1. So it suffices to show that O(n)P ⊗OP M(r)P →M(n+ r)P is an isomor-phism on an open subset Ps with s ∈ A1. For r = 0 this amounts to the assertion

134 3. SCHEMES

that multiplication by sn defines an isomorphism of Γ(Ps,MP ) = M [1/s]0 ontoΓ(Ps,M(n)P ) = M [1/s]n, which is obvious. If we apply this toMP = OP we findthat multiplication by sn defines an isomorphism OP |Ps ∼= O(n)P |Ps. The generalcase then follows immediately.

Under the assumption of this proposition (i.e., A is generated as a R-algebraby A1), O(1)P is called the Serre twisting sheaf. Proposition 10.3 then allows us todefine for any quasi-coherent OP -module F , its Serre twist F(n) := O(n)P ⊗P F(so that we can write MP (n) for M(n)P and in particular OP (n) for O(n)P ).We can now also go in the opposite direction and form F(•) := ⊕n∈ZF(n) andF := Γ(P,F(•)) (so that Fn := Γ(P,F(n))). This is a graded module over thegraded algebra Γ(P,OP (•)) and hence, via the graded ring homomorphism A →Γ(P,OP (•)), also one over the graded algebra A. This has an associated quasi-coherent sheaf FP . A natural sheaf homomorphism FP → F is then defined asfollows: for s ∈ A1, an element of Γ(Ps,FP ) can be given by a fraction a/sm witha ∈ Fm = Γ(P,F(m)) for some m with the understanding that b/sn defines thesame element if smb− sna is annihilated by some power of s. Now a|Ps is a sectionof F(m)|Ps and s−m|Ps is a section of OP (−m)|Ps and so their product defines asection of

(OP (−m) ⊗OP F(m)

)|Ps ∼= F|Ps (by 10.3). This defines a natural map

Γ(Ps,FP )→ Γ(Ps,F). Its sheafification defines sheaf homomorphism FP → F onPs and it is easily checked that the sheaf homomorphisms thus defined agree onoverlaps and thus produce a sheaf homomorphism FP → F .

PROPOSITION 10.4. Suppose A is generated as a R-algebra by a finite subset ofA1. Then the OP -homomorphism FP → F is an isomorphism. In particular, everyquasi-coherent sheaf on P comes from a graded A-module.

PROOF. Let s1, . . . , sr ∈ A1 generate A as a R-algebra. Then P = ∪Psi and soP is quasi-compact.

Since both FP and F are quasi-coherent, it is enough to verify that for everys ∈ A1, the map Γ(Ps,FP ) = F [1/s]0 → Γ(Ps,F) is injective and surjective. Theproof proceeds exactly as that of Step 2 resp. Step 3 of Proposition 7.3. We do thisthis only for injectivity and leave the surjectivity to you.

An element of the kernel is represented by a fraction a/sn with a ∈ Γ(P,F(n))and a|Ps = 0. We must show that then a is killed by some positive power of s. Theimage of a in Γ(Ps ∩ Psi ,F(n)) = Γ(Psi ,F(n))[1/s] is zero and hence the image ofa in Γ(Psi ,F(n)) is killed by smi for some mi ≥ 0. So if we take m = maximi, thensma|Psi = 0 for all i. This means that sma = 0.

PROPOSITION 10.5. Let A be the graded R-algebra R[T0, . . . , Tr] (with deg(Ti) =1) so that P = PrR. Then the R-algebra homomorphism A → Γ(P,OP (•)) is anisomorphism. (So for n ≥ 0, An ∼= Γ(PrR,OP (n)) is a free R-module with basis themonomials in T0, . . . , Tr of degree n, whereas Γ(PrR,OP (n)) = 0 for n < 0.)

PROOF. For r = 0 there is nothing so show and so we assume r > 0. LetF ∈ Γ(P,OP (n)). Then F |PTi can be written as FiTn−nii ∈ A[1/Ti]n for someFi ∈ Ani (so ni ≥ 0). Since FiTn−nii and FjT

n−njj have the same restriction to

PTi ∩ PTj = PTiTj we find that they are equal in A[1/TiTj ]. Now assume i 6=j. The left hand side lies in A[1/Ti] and the right hand side in A[1/Tj ]. SinceA[1/Ti] ∩ A[1/Tj ] = A it follows that FiTn−nii = FjT

n−njj ∈ An. So n ≥ 0 and

FiTn−nii is independent of i. Hence F is uniquely represented by F0T

n−n00 .

11. COHOMOLOGY OF COHERENT MODULES ON A PROJECTIVE SCHEME 135

Let us take R = k so that P = Prk. We observed that the isomorphism classof the invertible sheaf OP (1) generates the infinite cyclic group Pic(P ). The othergenerator is then of course OP (−1) and can be distinguished from the former bythe property that it has no nonzero sections. In particular, for any automorphismσ of P there exists an isomorphism σ∗OP (1) ∼= OP (1) (which will be unique up toan element of Γ(P,O×P ) = k×). The obvious k-linear isomorphism Γ(P,OP (1)) ∼=Γ(P, σ∗OP (1)) can then be composed with an isomorphism as above to producea k-linear automorphism σ∗ of Γ(P,OP (1)) that is unique up multiplication bya scalar. Proposition 10.5 shows that T0, . . . , Tr is a basis of Γ(POP (1)). If wethen write σ∗(Ti) =

∑i,j σi,jTj , then we recover σ as the projectivization of the

adjoint of (σi,j) ∈ GLr+1(k). In particular, we find that σ is given by an element ofPGLr+1(k). We conclude:

COROLLARY 10.6. The natural group homomorphism PGLr+1(k)→ Autk(Prk) isan isomorphism.

The preceding discussion also enables us to describe the closed subschemes ofP = PrR in terms of A = R[T0, . . . , Tr] as projective schemes: a closed subschemei : Y ⊂ P is by definition given by a quasi-coherent ideal sheaf I ⊂ OP so thatOP /I = i∗OY . Since OP (n) is invertible, I(n) → OP (n) is an injection and so wemay then form I := ⊕n≥1In with In := Γ(P, I(n)). Clearly In ⊂ Γ(P,OP (n)) andby Proposition 10.5 the latter is just An. It follows that I is a strict homogeneousideal of A. According to Proposition 10.4 we recover I from I by means of projec-tive sheafification. Let us now form the graded R-algebra B := A/I (so that wehave B0 = R) and put Y ′ := Proj(B). If Tj denotes the image of Tj in B1, thenΓ(Y ′

Tj,OY ′) is by definition B[T−1

j ]0. Since localization is exact, we can also ob-

tain this by taking A[T−1j ]0 and reduce it modulo the ideal I[T−1

j ]0. But the formeris Γ(PTj ,OP ) and the latter is Γ(PTj , I) and so the quotient is Γ(PTj , i∗OY ) (forPTj is affine). Since the PTj cover P , we conclude that Y may be identified withProj(B) and that via this isomorphism the sheaf OP (n)/I(n) can be understood asthe Serre twist OY (n). But we are not claiming that Γ(Y,OY (n)) can be identifiedwith Bn = An/In: the exact sequence

0→ Γ(P, I(n))→ Γ(P,OP (n))→ Γ(Y,OY (n))→ H1(P, I(n))→ . . .

merely tells us that we have an injection Bn → Γ(Y,OY (n)).We record part of this discussion as

PROPOSITION 10.7. Every closed subscheme Y of PrR is given by a homogeneousideal in I ⊂ R[T0, . . . , Tr]

+ (so that Y can be identified with Proj(R[T0, . . . , Tr]/I)).In particular, Y is projective over R.

11. Cohomology of coherent modules on a projective scheme

Let R be a noetherian ring, r an integer ≥ 1 and and put P := PrR. We haveseen that H0(P,OP (n) is zero for n < 0 and equal to R[T0, . . . , Tr]n for n > 0. Thefollowing proposition gives all the cohomology of each OP (n).

PROPOSITION 11.1. The R-module Hr(P,OP (−r−1)) is free of rank one and thenatural homomorphism of R-modules

HomOP (OP (n),OP (−r − 1))→ HomR(Hr(P,OP (n),Hr(P,OP (−r − 1))

136 3. SCHEMES

which assigns to φ ∈ Hom(OP (n),OP (−r− 1)) the induced R-homomorphism Hr(φ)is an isomorphism (hence the choice of a generator of Hr(P,OP (−r − 1)) identi-fies HomR(Hr(P,OP (n)), R) with the R-module HomOP (OP (n),OP (−r − 1)) ∼=H0(P,OP (−n− r − 1) ∼= R[T0, . . . , T − r]−n−r−1).

The cohomology of OP (n) is zero in degrees 6= 0, r.

We will see that the proof will at the same time show that

Hk(Ar+1R r 0,OAr+1

R r0) = ⊕n∈ZHk(P,OP (n)).

PROOF. Note that OP (n) is the degree n-part of the graded OP -algebra A :=

⊕n∈ZOP (n). It is clear that F is quasi-coherent. We make explicit the Cech complexfor F relative to the standard open covering U := PTiri=0: Γ(PTi ,A) can as agraded R-module be identified with the localization R[T0, . . . , Tr][1/Ti]. Likewise,if 0 ≤ i0 < i1 < · · · < ik ≤ r, then Γ(PTi0 ∩ · · · ∩ PTik ,A) = Γ(PTi0 ···Tik ,A) =

R[Ti0 , . . . , Tr][1/(T0 · · ·Tik)] and so

Ck(U,A) = ⊕0≤i0<i1<···<ik≤rR[T0, . . . , Tr][1/(T0 · · ·Tik)].

We observe that this is also the Cech complex for the structure sheaf of Ar+1R r 0

relative the open affine covering (Ar+1R )Tiri=0. So we have a graded isomorphism

Hr(P,A) = Hr(U,A) = Coker(Cr−1(U,A)→ Cr(U,A))

= Coker(⊕

i

R[T0, . . . , Tr][1/(T0 · · · Ti · · ·Tr)]→ R[T0, . . . , Tr][1/(T0 · · ·Tr)])

= R[T0, T−10 , . . . Tr, T

−1r ]/

(∑i

R[T0, . . . , Tr][T−10 , . . . , T−1

i , . . . , T−1r ])

of modules over H0(P,A) = R[T0, . . . , Tr]. Let us write Dr for this last moduleso that Hr(P,OP (n)) is the degree n part of Dr. Since we may identify Dr with(T0 · · ·Tr)−1R[1/T0, . . . , 1/Tr] as an R-module, we find that Hr(P,OP (n)) is zerounless n = −r − 1 − k with k ≥ 0, in which case we get (T0 · · ·Tr)−1 times theR-module spanned by the monomials in 1/T0, . . . , 1/Tr of degree k. In particular,Hr(P,OP (−r−1)) is obtained as the free R-module generated by (T0 · · ·Tr)−1. Re-call that H0(P,OP (k)) can be identified with the free R-module generated by themonomials in T0, . . . , Tr of degree k. For any such monomial M , (T0 · · ·Tr)−1M−1

is a basis element of Hr(P,OP (−r − 1 − k)). Multiplication by a degree k mono-mial Mo sends the basis element (T0 · · ·Tr)−1M−1 of Hr(P,OP (−r − 1 − k)) tozero unless M = Mo in which case we get the basis element (T0 · · ·Tr)−1 ofHr(P,OP (−r − 1)).

It remains to show the vanishing of the intermediate cohomology. We do thiswith induction on r. Observe that for r = 1 this assertion is empty. Consider thehyperplane i : P ′ ⊂ P defined by Tr = 0. Then the ideal sheaf IP ′ ⊂ OP is justOP (−1) and so we have for every n ∈ Z a short exact sequence

0→ A(−1)→ A→ i∗A′ → 0,

where A′ = ⊕n∈ZOP ′(n). We have Hk(P, i∗A′) = Hk(P ′,A′), and by inductionassumption, Hk(P ′,A′) = 0 for 0 < k < r−1. The long exact cohomology sequencebegins with

0→ R[T0, . . . , Tr]Tr·−−→ R[T0, . . . , Tr]→ R[T0, . . . , Tr−1]

11. COHOMOLOGY OF COHERENT MODULES ON A PROJECTIVE SCHEME 137

and ends with

DrTr·−−→ Dr → Dr−1 → 0.

It is clear from the definitions that the last map of the former is onto and thefirst map of the latter is into. So if we take 0 < k < r, then it follows thatmultiplication by Tr induces an isomorphism Hk(P,A(−1)) ∼= Hk(P,A). If weignore the grading, then A(−1) = A. So this implies that Hk(P,A) is in fact aR[T0, . . . , Tr][1/Tr]-module so that Hk(P,A) = Hk(P,A)[1/Tr]. But the latter isjust Hk(PTr ,A) (localization is exact) and hence zero, since is PTr is affine.

THEOREM 11.2 (Serre). Let R be a noetherian ring, X ⊂ PrR a closed subschemeand F a coherent OX -module. Then

(i) for n sufficiently large, F(n) is generated as a OX -module by its globalsections,

(ii) for every k and n ∈ Z, Hk(X,F(n)) is a finitely generated R-module. It iszero when k 6= 0 and n is sufficiently large.

We first make the simple observation that there is no loss in assuming that Xcoincides with P := PrR. The reason is that since i : X ⊂ P is closed immer-sion, i∗F will be a coherent OP -module and (i∗F)(n) = i∗(F(n)). Furthermore,Hi(X,F(n)) = Hi(P, i∗F(n)) for all i (this is because i∗ applied to flasque sheafon X is a flasque sheaf on P ). We first prove part (ii), derive a Corollary and thencontinue with the proof of the other assertions.

PROOF OF PART (I). Since F is coherent and PTi is affine, F|PTi is the sheafi-fication of finitely generated Γ(PTi ,OP )-module. In particular, F|PTi is generatedas a Γ(PTi ,OP )-module by a finite number of its sections, sijj , say. The usualargument shows that there exists an integer nij ≥ 0 such that Tniji sij extends toan element sij ∈ Γ(P,F(nij)). Let n0 := maxnij . For n ≥ n0, the collectionTn−niji sij ∈ Γ(P,F(n))i,j has the property that for i = 0, . . . , r, its restriction toPTi generates F(n)|PTi . Hence this collection generates F(n).

COROLLARY 11.3. For every coherent sheaf F on P = PrR with R noetherian thereexists a surjective OP -homomorphism OP (−n)` F for certain n ≥ 0 and ` ≥ 0. Inparticular, F admits a left resolution P• → F by such OP -modules.

PROOF. According to Assertions (i) (case i = 0) and (ii), there exists a pos-itive integer n such that F(n) is generated as a OP -module by a finite numberof global sections, say by s1, . . . , s`. This means that the the OP -homomorphism(f1, . . . , f`) ∈ O`P 7→

∑i fisi ∈ F(n) is onto. If we twist this with OP (−n) we get a

OP -homomorphism OP (−n)` → F that will still be surjective.

COROLLARY 11.4. Let X be a projective scheme over a noetherian ring R. Thenevery coherent OX -module F admits a left resolution P• → F by locally free OX -modules.

PROOF. Let i : X → P = PrR be closed submersion. Then there exists anepimorphism P0 → i∗F with P0 locally free. This induces an epimorphism i∗P0 →F and i∗P0 is locally free as a OX -module. Take P0 := i∗P0 and continue withker(P0 → F).

138 3. SCHEMES

PROOF OF PART (II) OF THEOREM 11.2. By Theorem 11.1, the assertions holdfor any F of the form OP (m) and hence also feo any F that is a finite direct sumof such modules. The fact that Hk(P,F(n)) = 0 for all n ∈ Z and all k > r, allowsus to prove the corollary with downward induction on k. According to Corollary11.3 there exists an epimorphism φ : P F , with P a finite direct sum of copiesof O(m)’s. Then K := ker(φ) is coherent OP -module. The kernel of P(n) F(n)is K(n) and we have an associated long exact sequence

· · ·Hk(P,K(n))→ Hk(P,P(n))→ Hk(P,F(n))→ Hk+1(P,K(n))→ · · ·

Our induction hypothesis says that for l > k, Hl(P,K(n)) is a finitely generated R-module that is zero for n sufficiently large. The exact sequence above then showsthat Hk(P,P(n)) has the same property.

Here is an important consequence of part (ii) (case i = 0) of Theorem 11.2 andChow’s lemma:

COROLLARY 11.5. Let f : X → S be a proper morphism of noetherian schemes.Then f∗ maps Coh(X) to Coh(S).

PROOF. According to Chow’s lemma 7.6 there exists a projective modificationπ : X ′ → X such that f := fπ is projective. Let F be a coherent OX -module. Wehave a natural in adjoint homomorphism F → π∗π

∗F . Since π is surjective, thishomomorphism is injective and so f∗F → f∗π∗π

∗F = (πf)∗π∗F is also injective.

But π∗F is a coherent OX′ -module and Theorem 11.2-(i) tells us that that forevery affine open subset V = Spec(R) ⊂ S, Γ(f−1V, π∗F) is a finitely generatedR-module. It contains Γ(f−1V,F) as a R-submodule and since R is noetherian, thissubmodule must also be finitely generated. So f∗F is a coherent OS-module.

Euler characteristics. Serre’s theorem allows us to define the Euler character-istic for a coherent sheaf F on a projective scheme X over k as

χ(F) :=∑i

dimk Hk(X,F).

This is indeed a finite sum. If i : X → Prk is a closed immersion, then i∗F is acoherent OPrk -module with the same cohomology as F and so χ(F) = χ(i∗F). It istherefore here no restriction to assume that X is a projective space.

EXAMPLE 11.6. Proposition 11.1 enables to compute χ(OPrk(n)). According tothis proposition, OPrk(n) has for n ≥ 0 only cohomology in degree zero and thatcohomology is then k[T0, . . . , Tr]n. So χ(OPrk(n)) =

(r+nr

)when n > 0. When

n ≤ −r − 1, OPrk(n)) has only cohomology in degree r and this is k-dual tok[T0, . . . , Tr]−n−r−1. Hence χ(OPrk(n)) = (−1)r

(−n−1r

)=(n+rr

)when n ≤ −r − 1.

For −r ≤ n ≤ −1, OPrk(n) has no cohomology and so χ(OPrk(n)) = 0, which alsohappens to be

(n+rr

). We thus find that χ(OPrk(n)) =

(n+rr

)for all n ∈ Z.

A short exact sequence 0 → F ′ → F → F ′′ → 0 of coherent sheaves givesa long exact sequence for cohomology and from this we see that χ is additivein the sense that χ(F) = χ(F ′) + χ(F ′′). The same argument shows that when0 → Pn → Pn−1 → · · · → P0 → F is an exact sequence of coherent sheaves, thenχ(F) =

∑k(−1)kχ(Pk). If X is a projective space over k, then by Corollary 11.3

we can always take the Pk to be locally free (we will later see that we then may

12. AMPLE AND VERY AMPLE SHEAVES 139

take Pk = 0 for k > dimX).

These Euler characteristics enter in the famous Hirzebruch-Riemann-Roch the-orem. They also lead to a somewhat different approach to the Hilbert-Serre poly-nomials introduced earlier. The support of a quasi-coherent sheaf F , supp(F), isthe closure of the set of x ∈ P with Fx 6= 0.

PROPOSITION 11.7. Let P be a projective space over k. Then for every coherentOP -module F , the function PF : n ∈ Z 7→ χ(F(n)) is given by a numerical polyno-mial in Q[T ] of degree ≤ dim(supp(F)), called the Hilbert polynomial of F . This isalso the polynomial we associated in Theorem 10.3 to the graded module Γ(P,F(•)).

PROOF. We prove this with induction on dim(supp(F)). When F = 0, the sup-port is empty and we get the zero polynomial. Let therefore F have nonempty sup-port of dimension r and assume the proposition proved for coherent OP -moduleswhose support has dimension < r. Choose a hyperplane i : Q ⊂ P that is notcontained in supp(F) unless supp(F) = P . This ensures that dim(supp(i∗F)) < r.The defining ideal IQ ⊂ IP can be identified with OP (−1), so that we a short exactsequence 0 → OP (−1) → OP → i∗OQ → 0. By the right exactness of the tensorproduct. Tensoring with F yields the exact sequence F(−1) → F → i∗i

∗F → 0.The first map need not be injective, but its kernel K will have support of dimension< r. From the exact sequences

0→ K(n)→ F(n− 1)→ F(n)→ i∗i∗F(n)→ 0

we read off the formula PF (n) − PF (n − 1) = Pi∗F (n) − PK(n). Our inductionhypotheses tell us that Pi∗F −PK is given by a numerical polynomial of degree < r.This implies that PF is given by a numerical polynomial of degree ≤ r.

Part (ii) of Theorem 11.2 tells us that for n sufficiently large, we have χ(F(n)) =dimk(H0(P,P(n))). This implies the last assertion.

12. Ample and very ample sheaves

Let S be a scheme. We have seen that the homogeneous coordinates T0, . . . , Trof P = PrS can be understood as sections of OP (1) and that they in fact generateOP (1). So if X is a scheme and f : X → P is morphism, then Ti gives rise toa section f∗Ti of invertible sheaf f∗OP (1). Since T0, . . . , Tr generate OP (1) asa OP -module, f∗T0, . . . , f

∗Tr generate f∗OP (1) as a OX -module. The followingproposition shows that these sections completely determine f once X is given asan S-scheme:

PROPOSITION 12.1. Let X/S be a S-scheme, L an invertible OX -module thatcomes with r + 1 generating sections s0, . . . , sr. Then we have a morphism f : X →PrS =: P and an isomorphism F : f∗OP (1) ∼= L which takes f∗Ti to si. The pair(f, F ) is unique.

PROOF. Without loss of generality we may assume that S is affine, say S =Spec(R). Our assumption says that the open subsets Ui := X − |Z(si)| cover X.Since si|Ui is a generating section, we have sj |Ui = f ijsi|Ui for some f ij ∈ OX(Ui).An S-morphism Ui → PTi is given by a homomorphism R[T0/Ti, . . . , Tn/Ti] →OX(Ui) of R-algebras. We define fi : Ui → PTi to be the morphism that sendsTj/Ti to f ij . Since si|Ui is a generating section of L|Ui and f∗Ti|Ui is a generatingsection of f∗O(1) we have defined over Ui a unique isomorphism of trivialized

140 3. SCHEMES

invertible sheaves Fi : f∗OP (1)|Ui ∼= L|Ui that takes f∗Ti|Ui to si|Ui and hencef∗Tj |Ui = f ijf

∗Ti to f ijsi|Ui = sj |Ui. The pairs (fi, Fi) and (fj , Fj) coincide onUi ∩ Uj and hence such pairs define a pair (f, F ) as asserted. The uniquenessassertion is left to you.

The following proposition addresses the question of when the f found in Propo-sition 12.1 is a closed immersion. We state and prove this only in case X is properover S = Spec(k), although the proposition below holds in a setting where X andS are noetherian and f is proper.

PROPOSITION 12.2. In the situation of the previous proposition, assume that R =k and that X is proper over k. Then f is a closed immersion (so that X is in factprojective over k) if and only if:

Two-point separation: For any two distinct closed points x, x′ ∈ X the im-ages of the si in L(x)⊕ L(x′) span the latter as a k-vector space.6

Infinitesimal separation: For every closed point x ∈ X, the images of the siin L(2x) := LX,x/m2

X,xLX,x span the latter as a k-vector space.

PROOF. We prove that these properties imply that f is a closed immersion, theconverse is left as an exercise. The separation property implies that f is injectiveon closed points and hence is injective. Since f is a morphism between proper k-schemes, f is also proper and hence is closed. It follows that f is a homeomorphismonto its image. So it remains to see that f∗x : OP,f(x) → f∗OX,x is onto for all x ∈ X.Again its suffices to check this for closed points only (why?).

By corollary 11.5, f∗OX is a coherent OP -module so that for every x ∈ X,f∗x : OP,f(x) → OX,x is a homomorphism of finitely generated OP,f(x)-modules. Bymeans of a local generator of Lx we can identify f∗x with F ∗x : OP (1)f(x) → Lx. Wehave si,x = F ∗x (Ti) and so if s′i ∈ OX,x corresponds to si,x, then s′i is in the image off∗x . Our assumption regarding infinitesimal separation amounts to the projectionsof the s′i in OX,x/m2

X,x spanning the latter over k. Lemma 12.3 below then impliesthat f∗x : OP,f(x) → OX,x is onto.

LEMMA 12.3. Let φ : A → B a local homomorphism of noetherian rings that isfinite (i.e., is such that B is a finitely generated A-module). If the composite of φ withB → B/m2

B is onto, then so is φ.

PROOF. By assumption the ideal in B generated by φ(mA) = φ(A) ∩ mB mapssurjectively to mB/m

2B . Nakayama’s lemma applied to the B-module mB then tells

us that this ideal is all of mB . It follows that B = φ(A) + φ(mA)B, in other words,the A-module Coker(φ) is annihilated by mA. Since Coker(φ) is a finite generatedA-module, another application of Nakayama’s lemma implies that it is trivial.

The preceding discussion leads us to make the following definition.

DEFINITION 12.4. Let f : X → S be a morphism of schemes. An invertiblesheaf L on X is said to be very ample relative to f if we can cover S by affine opensubsets V such that there exists an r ≥ 0 and a V -immersion i : f−1V → PrV andan isomorphism i∗OPrV (1) ∼= L|f−1V .

We have the following companion notion.

6L(x) stands for the fiber of L over x, i.e., the 1-dimensional κ(x)-vector space LX,x/mX,xLX,x.

12. AMPLE AND VERY AMPLE SHEAVES 141

DEFINITION 12.5. We say that an invertible sheaf L on a noetherian scheme Xis ample if for every coherent OX -module F , there exists an integer n0 such thatF ⊗ Ln is generated by its global sections for all n ≥ n0.

REMARKS 12.6. This notion is empty when X is affine, for then even everyquasi-coherent OX -module is generated by its global sections. We also note that ifLm is ample for some positive integer m, then so is L: for given a quasi-coherentOX -module, then there exists an n0 such that F ⊗ Lmn is generated by its globalsections for n ≥ n0. Next we apply then this property to F⊗L−k for k = 0, . . . , n0−1 and find that there exist an nk such that F ⊗ Lmn−k is generated by its globalsections for n ≥ nk. So for n ≥ mmaxk nk, F ⊗ Ln is generated by its globalsections.

PROPOSITION 12.7. Let X be a scheme of finite type over a noetherian ring R.Then an invertible OX -module L is ample if and only if some positive tensor powerLm is very ample over Spec(R).

PROOF. Suppose L is is ample. We first show that there exists a positive integern and a finitely many sections s1, . . . , sr of Ln such that X − Z(si)i is a finitecovering of X by affine open subsets.

Given x ∈ X, let φ : L|U ∼= OU be a trivialization of L on an open affineneighborhood of x in X. Put Y := X−U (a closed subscheme). By our assumption,there exists an nx > 0 such that IY ⊗ Lnx is generated by its sections and so thereexists an s ∈ Γ(X, IY ⊗ Lnx) with x /∈ Z(s). The image s of s in Γ(X,Lnx) hasthen the property that x ∈ X −Z(s) ⊂ U . Since Z(s|U) is defined by the section ofφ(s) of OU , its complement in the affine open U (which is X − Z(s)) is a principalaffine open neighborhood of x in X. Since X is quasi-compact a finite number ofthese principal open subsets will cover X. The corresponding finitely many sectionssi ∈ Γ(X,Lni) are in possibly different powers of L, but we make these equal bytaking for n a common multiple of these and replacing si by sn/nii .

Put Ui := X − Z(si). Since X is of finite type over R, OX(Ui) is a finitelygenerated R-algebra. Choose generators fi0, . . . , firi of OX(Ui) with fi0 = 1. Uponreplacing n by a positive multiple if necessary, we then may assume that each fijsiextends to a section sij of Γ(X,Ln). Then ∩i,jZ(sij) = ∅ so that by Proposition12.1 we have defined an R-morphism

f : X → P := ProjR[(Tij)ij ]

and an isomorphism F ∗ : Ln ∼= f∗OP (1) which takes sij to Tij . The preimage ofthe standard open set PTi0 is the affine open Ui and (f |Ui)∗ takes Tij/Ti0 to fij .

Since the (fij)j generate O(Ui) as a R-algebra, it follows that Uif−→ PTi0 is a closed

embedding. It follows that f maps X onto a closed subscheme of ∪iPTi0 (this unionis open in P ).

Now assume Lm very ample for somem > 0. By Remark 12.6 it suffices to showthat Lm is ample. By definition there exists an R-immersion f : X → P := PrR forsome r > 0 and an isomorphism Lm ∼= f∗OP (1) and so we may as well assume thatX is locally closed in P . If F is a coherent sheaf on X, then according to Corollary7.10 it admits a coherent extension to the closure X. This in turn can be regardedas a coherent OP -module G. By part (ii) of Serre’s theorem, there exists an integern0 such that G(n) is generated by its global sections for n ≥ n0. The same is is thentrue for f∗G ∼= Lm.

142 3. SCHEMES

REMARK 12.8. Let C be a connected proper nonsingular curve. With the helpof the Riemann-Roch theorem on can show that an invertible sheaf on C of positivedegree is ample. In particular, for any p ∈ C(k), the invertible sheaf O(p) hasdegree one and so is ample. But if O(p) is very ample, then we must have C ∼= P1

k.To see this, choose q ∈ C(k) r p. The separation property implies that there is asection of O(p) which is zero in q. In other words, its divisor has all its coefficients≥ 0 with the one in q positive. Since its degree is 1, this divisor must be just (q). So(p)− (q) is principal. But according to Proposition 9.15 this implies that C ∼= P1

k.

For an invertible module on a proper scheme over a noetherian base we havethe following ampleness criterion.

PROPOSITION 12.9. Let X a proper scheme over a noetherian ring R and L aninvertibleOX -module. If L is ample then for every coherentOX -module F , F⊗Ln hasno cohomology for n sufficiently large. Conversely, if for every coherent OX -moduleF , there exist an n > 0 such that H1(X,F ⊗ Ln) = 0, then L is ample.

PROOF. Suppose L is ample. Then Lm is very ample for some m > 0. As aconsequence, we can assume that X is a closed subscheme of PrR for some r andthat Lm = OX(1). Then F(n) = F ⊗ Lmn has no cohomology in nonzero degreefor n ≥ n0, say. Similarly, we find for k = 1, . . . ,mn0 − 1 an integer nk such thatF ⊗ L−k(n) = F ⊗ Lmn−k has no cohomology in nonzero degree for n ≥ nk. Itfollows that F ⊗ Ln has no cohomology in nonzero degree for n ≥ mn0 maxk nk.

Now suppose that for every coherent OX -module F , H1(X,F ⊗ Ln) = 0 forn sufficiently large. We must show that F ⊗ Ln is generated by its sections whenn is sufficiently large. Let x ∈ X be a closed point and denote by i : x ⊂ Xthe inclusion. Then we have an epimorphism from F to the skyscraper sheaf i∗i∗F(its stalk at x is the finite dimensional κ(x)-vector space F(x) = κ(x) ⊗ F) whosekernel is IxF , where Ix ⊂ OX denotes the ideal defining x. We have for everyinteger n ≥ 0 a short exact sequence

0→ IxF ⊗ Ln → F ⊗Ln → i∗i∗F ⊗ Ln → 0.

By assumption, there exists a positive integer nx(F) such that this gives an exact se-quence of sections onX. Since Fx is a finitely generatedOX,x-module, Nakayama’slemma then implies that Γ(X,F ⊗ Lnx(F)) → Fx ⊗ Lnx(F)

x is onto. The coherenceof F implies that the global sections of F ⊗ Lnx(F) will generate this OX -moduleon a neighborhood Ux(F) of x. Put

Ux(F) :=Ux(OX) ∩nx(F)−1⋂k=0

Ux(F ⊗ L−k),

nx(F) := maxnx(OX),

nx(F)−1maxk=0

nx(F ⊗ L−k)

and then observe that for n ≥ nx(F), the global sections of F ⊗ Ln generate thatsheaf over Ux(F). SinceX is quasi-compact, the open covering Ux(F)x∈X admitsa finite subcovering Uxi(F)`i=1. Then for n ≥ max`i=1 nxi(F), F⊗Ln is generatedby its global sections.

13. Blowing up

In this section we fix a noetherian scheme X. Suppose we are given a gradedsheaf of OX -algebras A = ⊕∞d=0Ad with the property that A0 = OX , A1 is a

13. BLOWING UP 143

coherent OX -module which locally generates A by as OX -algebra. In other words,X can be covered by a finite number open affines of the form U = Spec(R) suchthat R is noetherian and the R-algebra A(U) = ⊕∞d=0Γ(U,Ad) is as in Proposition10.4: it is a graded R-algebra with R in degree zero, and generated as a gradedR-algebra by a finite subset of the degree one summand. We then have defined theprojective morphism ProjA(U)→ U and on ProjA(U) an invertible Serre twistingsheaf OProjA(U)(1) that is ample relative the projection. If U ′ ⊂ U is a principalopen subset, then the data we get for U ′ are obtained by the restricting the ones wejust defined and so these morphisms and invertible sheaves glue together to forma projective morphism that we naturally denote as Proj(A)→ X plus an invertibleSerre twisting sheaf OProj(A)(1).

There is one particular case of interest and that is when E := A1 is locally freeof rank r. Then we may form the graded symmetric OX -algebra Sym• E on E . Thelatter means that if U ⊂ X is open and T1, . . . , Tr is a basis of sections of E(U),then Sym• E(U) = E(U)[T1, . . . , Tr]. In more invariant terms, Symd E is the biggestquotient of the d-fold tensor power E ⊗OX E ⊗OX · · · ⊗OX E invariant under thepermutation group Sd. Then Proj(Sym• E) → X is locally like Pr−1

U → U : it is aprojective space bundle that we denote by P(E) → X. So for r = 0, P(E) = ∅ andwhen r = 1, i.e., E invertible, P(E) → X is an isomorphism. Notice that Proj(A)can now be regarded as a closed subscheme of P(A1).

To explain the notation: suppose X = Spec(k) so that E is just an (n + 1)-dimensional k-vector space. Then Sym• E is the algebra of regular functions on thek-dual E∨ of E and so P(E) is the projective space associated to E∨. This is also theprojective space of hyperplanes in E .

We will here focus on a special case, namely the ‘blowup’ of closed subscheme.Let us first do an example which in a way revisits the discussion of Chapter 2,section 4.

EXAMPLE 13.1. LetX := Ank and 1 ≤ r ≤ n−1. Denote by Y ⊂ Ank be the closedsubscheme defined by x1 = · · · = xr = 0 (so this a copy of An−rk ). Consider themorphism X−Y → Pr−1

k defined by (x1, . . . , xn) 7→ [x1 : · · · : xr] and let X ⊂ Pr−1X

be the closure of its graph. If we identify Pr−1X with Proj(Γ(X,OX)[T1, · · · , Tr]),

then X is given by the homogeneous ideal in k[x1, . . . , xn][T1, · · · , Tr] generatedby xiTj − xjTi, 1 ≤ i < j ≤ r. Notice also that on XTr we have r − 1 regularfunctions (ξj := Tj/Tr)

r−1j=1 so that xj = ξjxr for j < r. We may thus identify XTr

with the affine space Spec k[ξ1, . . . , ξr−1, xr, . . . , xn]. The projection X → X is anisomorphism over X − Y , whereas a fiber over a closed point p ∈ Y is all of Pr−1

k .The ideal sheaf I ⊂ OX defining Y is

∑rj=1OXxj . But on XTr we have xj = ξjxr,

and so π∗I|XTr it simply the principal ideal generated by xr. Notice that this idealdefines the preimage Y of Y in XTr . A similar property holds for XTi , i = 1, . . . , r,and as these cover X, it follows that π∗I ⊂ OX is a principal ideal sheaf: π∗I isinvertible and the subscheme Y ⊂ X is a prime divisor.

The case that concerns us here is a far reaching generalization of this situation,where X is a noetherian scheme and Y is a closed subscheme of X. Then Y defines(and is defined by) a coherent ideal sheaf I = IY and we take Ad = Id ⊂ OX , theimage of the d-fold tensor power I ⊗OX · · · ⊗OX I → OX . The preceding construc-tion applies to ⊕∞d=0Id and produces a projective morphism Proj(⊕∞d=0Id) → X.

144 3. SCHEMES

Since Id ⊂ Id+1, it is often helpful to keep track of the degree by means of an extravariable T so that ⊕∞d=0Id can be regarded as a subalgebra of OX [T ]:

⊕∞d=0 Id ∼=∑∞k=0 IdT d ⊂ OX [T ].

We then write BlY (X) for Proj(∑∞d=0 IdT d) and call the morphism π : BlY (X) →

X the blowup of X along (or with center) Y . Notice that Id|X − Y = OX−Y andso∑∞d=0 IdT d|X − Y = OX−Y [T ]. The projection Proj(OX−Y [T ]) → X − Y is an

isomorphism and so π is an isomorphism overX−Y . This blowup has a remarkableproperty with respect to the ideal sheaf I: recall that the OProj(A)-module OProj(A)

is associated to the gradedA-moduleA(1). Since hereA =∑d≥0 IdT d andA(1) =∑

d≥0 Id+1T d, it is clear that A(1) = IA. This identity shows that OBlY (X)(1) maybe identified with the ideal pull-back OBlY (X)π

−1I ⊂ OBlY (X) of I. In particular,this ideal pull-back is invertible. This property characterizes the blowup:

PROPOSITION 13.2. Let f : X ′ → X be a morphism with X ′ noetherian and letY ′ := f−1Y be the closed subscheme defined by the ideal pull-back I ′ := OX′f−1I ⊂OX′ . Then f is uniquely covered by a morphism f : BlY ′(X) → BlY (X). Moreover,I ′ is invertible if and only if f lifts to a morphism X ′ → BlY (X).

PROOF. The homomorphism of OX -modules OX → f∗OX′ takes I to f∗I ′ andso we have a homomorphism of OX -algebras

⊕d≥0IdT d → ⊕d≥0f∗I ′dT d = f∗(⊕d≥0I ′d).

This induces a morphism between proj schemes: f : BlY ′(X′) → BlY (X). If I ′

is invertible, then BlY ′(X′) → X ′ is an isomorphism, and so we then find a lift

X ′ → BlY (X) of f . Conversely, if such a lift X ′ → BlY (X) exists, then I ′ isinvertible, as it is the pull-back of the invertible sheaf π∗I.

The morphism f obtained above is sometimes called the strict transform of f .

REMARK 13.3. If we replace I by Im for some positive integer m, then we geta closed subscheme Y (m) ⊂ X which contains Y , but according to Remark 10.1,BlY (m)(X) can be identified with BlY (X).

The same is true for a closed subscheme Y ⊂ Y ′ ⊂ X for which IY ′ = LIfor some invertible ideal L ⊂ OX : BlY ′(X) ∼= BlY (X). For if U = Spec(R) isan affine open subset of X on which L has a generating section s, and I|U isgiven by the ideal I ⊂ R, then I ′ is given by I ′ = sI. For every f ∈ I, we have∪n≥0I

nf−n = ∪n≥0I′n(sf)−n and hence an isomorphism between the two blow-

ups over U ; this identification is evidently independent of the choice of s.This also leads to the observation that the obvious generalization of a blowup

to one of a fractional ideal I ⊂ KX as ProjX(⊕∞k=0IkT k) → X (where I0 := OX)hardly yields anything new: by definition we may cover X by affine open subsetsU for which there exist a nonzero divisor f ∈ Γ(U,OX) such that fI is an idealin OU and if YU ⊂ U is the corresponding closed subscheme, then BlI(X)|U canbe identified with BlYU U . In fact, if L ⊂ KX is an invertible submodule such thatLI ⊂ OX and Y ⊂ X is the closed subscheme defined by LI, then we may identifyBlI(X) with BlY (X).

Such an L always exists in case X is quasi-projective and integral: then Xsupports an ample invertible OX -module O(1). Since I is a coherent OX -module,so is Hom(I,OX). We have Hom(I,OX)x 6= 0 for all x ∈ X, because there exists

13. BLOWING UP 145

a nonzero fx ∈ OX,x with fIx ⊂ OX,x. Following Part (ii) of Theorem 11.2 thereexists an n such that Hom(I,OX)(n) has a global section which does not vanish ona nonempty open subset. Such a section amounts to a nonzero OX -homomorphismI(−n)→ OX . SinceKX has no zero divisors, this homomorphism must be injectiveand so L := OX(−n) will do.

EXAMPLE 13.4 (Tangent cone). Let X be a scheme and x ∈ X a closed point.We consider the blow-up of x in X. Since this does not affect X r x, we mayjust as well assume that X is the spectrum of a local ring (R,m) (so that R = OX,x).We denote by κ := R/m the residue field. Then BlxX = Proj(

∑k≥0 m

kT k) →Spec(R) (read R for m0). The fiber over x ∈ X is the Proj of the graded κ-algebra

∑k≥0(mk/mk+1)T k. This defines a closed subscheme of the projective

space P(m/m2). If X is (the germ) of a variety over k, then we recognize P(m/m2)as the projectivized Zariski tangent space of X at x. The closed subscheme definedby Proj(

∑k≥0(mk/mk+1)T k) is the called the projectivized tangent cone of X at x.

It is the full projectivized Zariski tangent space precisely when X is nonsingular atx. (The tangent cone of X at x is Spec(

∑k≥0(mk/mk+1)T k); it is a homogeneous

subscheme of the Zariski tangent space Spec(Sym•m/m2).)

Here is a somewhat more concrete model for the blowup π : BlY (X) → Xwhich also makes the connection with Example 13.1. Assume that X is affine, soX = Spec(R) with R noetherian. Then I is defined by a finitely generated idealI ⊂ R. If f0, . . . , fr is a set of generators of I, then [f0 : · · · : fr] defines a morphismX − Y → Prk, but we prefer to view this as defining a section f of PrX/X overX − Y . The scheme theoretic image of f : X − Y → PrX as defined in Corollary7.5 is a closed subscheme f(X − Y ) of PrX whose ideal sheaf is the kernel of theOPrX -homomorphism OPrX → f∗OX−Y .

PROPOSITION 13.5. There is a natural X-isomorphism φ : f(X − Y ) ∼= BlY (X).In particular, if X − Y is irreducible, then so is BlY (X).

PROOF. We define a closed immersion of BlY (X) in PrX with image f(X − Y ).Since the fi generate I, BlY (X)fi is covered by the affine open subsets BlY (X)fiand we will specify this immersion on BlY (X)fi for i = 0, . . . , r as a closed embed-ding in (PrX)Ti with image f(X − Y ) ∩ (PrX)Ti .

Let us do this for i = 0: if (t1, . . . , tr) ∈ ArX 7→ [1 : t1, . . . , tr] ∈ (PrX)T0is

the standard parametrization of a standard open subset of PrX , then we have ahomomorphism of R-algebras φ∗0 : R[t1, . . . , tn]→ R[1/f0] which sends ti to fi/f0.The image of φ∗0 is the subring ∪n≥0I

nf−n0 whose spec is just BlY (X)f0 , and so φdefines a closed embedding of BlY (X)f0 in (PrX)T0

. It is also clear that the idealker(φ∗0) is generated by fi − tif0 (i = 1, . . . , r). This ideal defines the subschemef(X − Y ) ∩ (PrX)T0 in (PrX)T0 and hence this subscheme must be the image of φ0.

Likewise for i = 1, . . . , r, and one checks that these morphisms agree on over-laps. To complete the proof that we get a closed immersion we still need to checkthat the preimage of (PrX)Ti is BlY (X)fi . This can be done directly, or by observingthat the ideal pull-back of I to f(X − Y ) is invertible (so that by Proposition 13.2the inverse is defined).

146 3. SCHEMES

REMARK 13.6. The same argument works for the blowup of a fractional idealI ⊂ KX : a set of OX -generators f0, . . . , fr ∈ Γ(X, I) determines a closed embed-ding of BlI(X) in PrX .

Blowing up on a variety. Here we confine ourselves to the case when X is ak-variety in the sense of this chapter: i.e., a separated integral scheme of finite typeover k. We let Y be a subscheme 6= X so that I 6= 0. Since X is irreducible, so isX−Y and hence is BlY (X) by Proposition 13.5. If U ⊂ X is affine and f ∈ Γ(U, I)nonzero, then ∪k≥0Ikf−k is subring of the integral domain Γ(U,OX)[1/f ] andhence an integral domain. It follows that BlY (X) is integral as well and that theprojection BlY (X)→ X is birational. The following theorem shows that very oftenevery birational projective morphism of varieties is thus obtained.

THEOREM 13.7. A variety X has the property that every birational, projectivemorphism f : X → X is X-isomorphic to the blowup of a fractional ideal on X.When X is quasi-projective, we can assume this in fact to be an ideal sheaf (so that Xis the blowup of a closed subscheme of X).

PROOF. By assumption f factors through a closed X-embedding i : X → PrX .We thus have defined a very ample invertible sheaf OX(1) that comes with r + 1sections T0, . . . , Tr defining the embedding. Moreover, f∗OX(1) is a coherent OX -module. Since X is integral there exists a OX -monomorphism OX(1) → KX . Thisproduces a OX -monomorphism f∗OX(1) → f∗KX . Since f is birational, f∗KX =KX . Thus, f∗OX(1) is realized as a fractional ideal I ⊂ KX . Let U = Spec(R) be anaffine open subset ofX and let (f0, . . . , fk) be set of generating sections of I|U , withfi = Ti for i ≤ r. These define a morphism [f0, . . . , fk] : f−1U → PkU whose imagelies in ∪ri=0(PrU )Ti and whose composite with the projection ∪ri=0(PrU )Ti → PrX isjust i. So f−1U → PkU is a closed U -immersion as well. According to Proposition13.5 and its subsequent Remark 13.6 this also models the blowup of I over U .

If X is quasi-projective, then following Remark 13.3 there exists an invertiblesubsheaf L ⊂ OX such that LI ⊂ OX . Now use that BlLI(X) = BlI(X).

14. Flatness

We first recall the definition of a flat module. Given a ring R , then for anyR-module M , the functor ⊗RM : ModR →ModR is a left exact, but not in generalexact. If it is, then we say that M is a flat R-module. The jth derived functor of⊗RM is denoted by TorRj (M,−) and so this can be expressed by saying that thesefunctors are zero for j > 0.

The flatness property can be made more concrete: it amounts to saying that anyR-linear relation among elements m1, . . .mn of M :

∑ni=1 rimi = 0 is necessarily

of the type that we would have if M were a free R-module: we then can writemi =

∑j rijm

′j and we have

∑i ririj for all j. In particular, a free R-module is flat.

But a flat R-module need not be free. For instance, if R is a domain, then its fieldof fractions Frac(R) is rarely free over R, but is a flat R-module. Here are somebasic properties:

Flatness is local: M is a flat R module if and only if every localization Mp

at a prime ideal p is a flat Rp-module.Transitivity of flatness: LetR→ R′ be a ring homomorphism which makesR′ a flat R-module. Then any flat R′ module is also flat as a R′-module.

14. FLATNESS 147

Base change: Let R → R′ be a ring homomorphism. If M is a flat R mod-ule, then R′ ⊗RM is a flat R′-module.

Flatness over a noetherian ring: LetM be a finitely generated module overa noetherian ring R. Then: M is flat ⇔ M is locally free ⇔ the functionx ∈ Spec(R) 7→ dimκ(x) κ(x)⊗M is locally constant.

Flatness preserved by kernels: The kernel of anR-homomophism betweenflat R-modules is flat.

Flatness preserved by extension: Let M be an R -module and N ⊂ M asubmodule. If N and M/N are exact, then so is M .

Completion is flat: If R is a local noetherian ring, then its m-adic comple-tion is a flat R-module.

Fiber dimension: Let φ : R → R′ be a local homomorphism of noether-ian local rings. Then dim(R′) − dim(R) ≤ dim(R′/mRR

′) and we haveequality if R′ is flat over R.

We refer to [1] (and for the last property to [2]) for proofs. We will explorethe geometric meaning of this notion via the following definition.

DEFINITION 14.1. Let f : X → Y be a morphism of schemes and let F be aOX -module. We say that F is flat over Y if for every x ∈ X, Fx is flat as a OY,f(y)-module. If this is the case for F = OX , then we say that f is a flat morphism.

It is clear that flatness is local property. The properties listed above implyamong other things:

Sheafification of flat modules: Let M be an R-module. Then M is a flatR-module if and only if its sheafificationM is a flat over Spec(R).

Transitivity of flatness: Let Xf−→ Y and Y

g−→ Z be morphisms of schemeswith g flat. Then an OX module F that is is flat over Y is also flat over Z.

Base change: Let f : X → Y a morphism of schemes and F an OX -modulethat is flat over Y and g : Y ′ → Y a morphism. Denoting by g : Y ′×YX →X the associated morphism, then g∗F is flat over Y ′.

Flat coherent modules: Let X be a locally noetherian scheme and F acoherent OX -module. Then F is flat over X ⇔ F is locally free ⇔x ∈ X 7→ dimκ(x) F(x) is locally constant.

Flatness preserved by kernels: The kernel of an homomophism betweenflat quasi-coherent OX -modules is flat.

Flatness preserved by extension: LetM be quasi-coherentOX -module andN ⊂M a submodule. If N andM/N are exact, then so isM.

Flat implies constant fiber dimension: Let f : X → Y a morphism of lo-cally noetherian schemes and x ∈ X. Then dimxX−dimy Y ≤ dimx f

−1(y)with equality when f is flat. Here y := f(x).

We further mention that a morphism of between noetherian schemes is open.

REMARK 14.2. Let f : X → Y be a morphism of noetherian schemes andlet F be a quasi-coherent OX -module. We may then resolve F by f∗-acyclic quasi-coherent OX -modules in a Cechlike manner as follows. Cover X by a finite numberof affine open subsets jν : Uν ⊂ Xnν=1. Then jν∗j

∗νF is a quasi-coherent OX -

module. In case F is flat over Y , then f∗jν∗j∗νF = (f |Uν)∗F|Uν is a flat quasi-

coherent OY -module.

148 3. SCHEMES

Since Uν is affine we have Rkf∗(jν∗j∗νF) = Rk(f |Uν)∗F|Uν = 0 when k > 0.So if we put C0 := ⊕νjν∗j∗νF , then we have an monomorphism F ⊂ C0 of quasi-coherent OX -modules, with C0 f∗-acyclic and f∗C0 a flat OY -module, wheneverF is flat over Y . With induction we thus obtain a resolution F → C• by suchOX -modules.

Let now be given a commutative diagram of morphisms of schemes

X ′g−−−−→ X

f ′y f

yY ′

g−−−−→ Y

We found in Proposition 3.14 that for anyOX -module F on X and any j = 0, 1, . . . ,there exists by a natural homomorphism

φj : g∗Rjf∗F → Rjf ′∗g∗F .

PROPOSITION 14.3. Suppose that the above diagram is cartesian: X ′ = Y ′×Y Xwith X,Y, Y ′ are noetherian and g is flat. When F is a quasi-coherent OX -module,then each φj is an isomorphism.

PROOF. Since this is local issue on Y ′ we may assume that both Y ′ and Y areaffine: Y ′ = Spec(R′) and Y = Spec(R) with R noetherian and R′ a flat R-module.We must then show that the natural map R′ ⊗R Hk(X,F) → Hk(X ′, g∗F) is anisomorphism. Notice that g will be affine and flat by base change.

Let F → C• be a resolution by quasi-coherent Γ-acyclic OX -modules as inRemark 14.2 above. Since g is flat, g∗F → g∗C• is still a resolution and since g isaffine, this resolution is Γ-acyclic. Therefore we have Hk(X,F) = hk(C•(X)) andHk(X ′, g∗F) = hk(g∗C•(X)). But clearly g∗C•(X ′) = R′ ⊗R Ck(X) and since thefunctor R′ ⊗R − commutes with taking cohomology, the assertion follows.

THEOREM 14.4. Let S be a noetherian scheme and F a coherentOPrS -module. Fors ∈ S, denote by Ps ∈ Q[T ] the Hilbert polynomial of i∗sF(s), where is : Prs ⊂ PrS .Then F is flat over S if and only if the Hilbert polynomial defines a locally constantfunction P : |S| → Q[T ] (in which case π∗F(n) is locally free for n 0).

PROOF. Without loss of generality we may assume that S is affine with R noe-therian. We first show that for every s ∈ S the natural map κ(s) ⊗ Γ(PrS ,F(n)) →Γ(Prκ, i∗sF(n)) is an isomorphism when n 0, so that P is in fact computable fromthe graded R-module ⊕nΓ(PrS ,F(n)).

Choose a finite set of generators of mS,s so that have an exact sequence Rk →R→ κ(s)→ 0. This yields an exact sequence

Fk → F → is∗i∗sF → 0

which remains exact after Serre twisting. For n 0, H1 of F(n)k and the image ofF(n)k → F(n) vanish, so that we then have an exact sequence

Γ(PrS ,F(n))k → Γ(PrS ,F(n))→ Γ(Prs, i∗sF(n))→ 0.

The first map is also obtained by tensoring Γ(PrS ,F(n)) with Rk → R and so thenatural map κ(s)⊗ Γ(PrS ,F(n))→ Γ(Prκ, i∗sF(n)) is an isomorphism indeed.

15. SERRE DUALITY 149

In view of the fact that for a finitely generated R-module M , M is flat if andonly if the function s ∈ S 7→ dimκ(s) κ(s) ⊗ M is locally constant, it thereforesuffices to show that F is flat if and only if Γ(PrS ,F(n)) is flat for n 0.

Assume thatF is flat over S. Let 0→ F → C0 → C1 → · · · be an exact sequenceof quasi-coherent flat OPrS -modules that are Γ-exact (as for instance obtained inremark 14.2). This remains so after Serre twisting. For n 0, we have by theSerre vanishing property that 0 → Γ(PrS ,F(n)) → Γ(PrS , C0(n)) → Γ(PrS , C1(n))is also exact so that Γ(PrS ,F(n)), being the kernel of a homomorphism of flat R-modules, is a flat R-module.

Conversely, assume that Γ(PrS ,F(n)) is a flat R-module for n ≥ n0. This meansthat the graded R-module ⊕n≥n0

Γ(PrS ,F(n)) is flat. This implies that F as theassociated coherent sheaf is flat over R.

REMARK 14.5. We may apply the preceding to a closed subscheme X ⊂ PrSwith S is connected and with F = OX . We then find that X is flat over S if andonly if the Hilbert polynomials of the fibers Xs of X → PrS are constant. Sincethe dimension of Xs is the degree of Ps and the degree deg(Xs) of Xs is dim(Xs)!times the leading coefficient of Ps, it follows that then both s 7→ dim(Xs) ands 7→ deg(Xs) are constant functions on S. Since χ(OXs) = Ps(0), is also followsthat χ(OXs) is constant.

REMARK 14.6 (The Hilbert scheme). Fix a positive integer n. If we are givena k-scheme S and a closed subscheme X ⊂ PnS that is flat over S, then for everymorphism f : S′ → S, S ×S′ X is a closed subscheme of PnS′ flat over S′. So wehave a functor HilbPnk : Sch/k → Set which assigns to the k-scheme S the set ofclosed subschemes X ⊂ PnS that are flat over S and assigns to f : S′ → S the mapHilbPnk (S) → HilbPnk (S′) defined by the fiber product construction. Grothendieckshowed that this functor is representable: there exists a k-scheme HilbPnk and aclosed k-scheme XHilb(Pnk ) ⊂ PnHilb(Pnk ) that is flat over Hilb(Pnk ) that is universal:for every X ⊂ PnS as above there exists a unique morphism f : S → Hilb(Pnk ) suchthat X is the preimage of the associated morphism Pnf : PnS → PnHilb(Pnk ) so that wehave a Cartesian diagram

Xf−−−−→ XHilb(Pnk )y y

Sf−−−−→ Hilb(Pnk ).

The preceding implies that over each connected component of Hilb(Pnk ) the Hilbertpolynomial of the fibers is constant. Grothendieck proved that the locus over whichwe have a prescribed Hilbert polynomial is a subscheme of finite type (and is hencea finite union of connected components of Hilb(Pnk )).

15. Serre duality

In this section X denotes a scheme of dimension n that is proper over thealgebraically closed field k. The main result only concerns the case when X isprojective over k. For a coherent OX -module F , Hi(X,F) is a finite dimensionalvector space. So we have defined a functor Hi(X,−) : Coh(X) → Vectk and bydualizing, Hi(X,−)∨ : Coh(X) → Vectk.

150 3. SCHEMES

DEFINITION 15.1. A weak dualizing sheaf for X is a representation of the func-tor Hn(X,−)∨ : Coh(X) → Vectk (recall that n = dim(X)). In other words,this is a pair (ω, t), where ω is a coherent OX -module and t ∈ Hn(X,ω)∨ =Homk(Hn(X,ω), k) (often called the trace map) such that for every coherent OX -module F , the map

HomX(F , ω)(Hn)∨−−−−→ Homk(Hn(X,ω)∨,Hn(X,F)∨) −→ Hn(X,F)∨

(where the last map is evaluation in t) is an isomorphism.We say that the pair (ω, t) is a dualizing sheaf7 if for every coherent OX -module

F , and every i, the composite

ExtiX(F , ω)→ Homk(Hn−i(X,F),Hn(X,ω)) =

= Homk(Hn(X,ω)∨,Hn−i(X,F)∨)→ Hn−i(X,F)∨

(with the first map defined as in the appendix and the last map evaluating in t sothat for i = 0 we get the map appearing above) is an isomorphism.

By the Yoneda lemma a weak dualizing sheaf is unique up to unique isomor-phism8. We will show that it exists when X is projective.

REMARK 15.2. The functors Tj , T ′j : Coh(X) → Vectk defined by Tj(F) :=

Extj(F , ωX) resp. T ′j(F) := Hn−j(X,F)∨ are δ-functors with the first one beinguniversal. (The contravariant nature leads us to use a homological indexing bysubscripts rather than superscripts.) The homomorphisms

Tj(F) = ExtjX(F , ω)→ Homk(Hn−j(X,F),Hn(X,ω))→ Hn−j(X,F)∨ = T ′j(F)

defined above constitute in fact a morphism of δ-functors. To say that (ω, t) is aweak dualizing sheaf means that T0 ⇒ T ′0 is an isomorphism. So to say that such aweak dualizing sheaf is dualizing means that T ′• is also universal.

We first treat the case of a projective space.

PROPOSITION 15.3. For P := Prk, the pair (ωP := OP (−r − 1), tP ), where tP isa generator of the 1-dimensional k-vector space Hr(P,OP (−n − 1))∨, is a dualizingsheaf.

PROOF. Proposition 11.1 shows that (ω = OP (−r − 1), t) has all the desiredproperties for the OP -modules F of the form OP (n) (the dualizing property istrivially fulfilled in degrees 6= 0, r). It then also holds for any OP -module thatis a finite direct sum of such sheaves. In particular, when P = OP (−d)⊕s withd > 0, then both members of the asserted identity vanish unless i = 0 (in whichcase they are naturally isomorphic). As observed in Remark 15.2, F 7→ Ti(F) :=ExtiP (F , ω) and F 7→ T ′i (F) := Hn−i(P,F)∨ are δ-functors. By 11.3 there exists aleft resolution P• → F if F in Coh(P ) by modules P as above (this is of course aright resolution in Coh(P )). The Pi are acyclic for both δ-functors and so by theabstract De Rham theorem, Ti(F) = hi(T0(P•)) and T ′i (F) = hi(T

′0(P•)). Since we

have T0(P•) ∼= T ′0(P•) as chain complexes, the proposition follows.

7Hartshorne calls a dualizing sheaf what we call a weak dualizing sheaf.8Beware that this does not imply that for a given ω, Hn(X,ω)∨ has a canonical element t: we may

just as well replace t by t′ = λt, where λ is a nonzero scalar and then multiplication by λ maps the pair(ω, t) isomorphically onto the pair (ω, t′).

15. SERRE DUALITY 151

THEOREM 15.4. Let i : X ⊂ Prk =: P be a closed subscheme of dimension n andlet (ωP , tP ) be a dualizing sheaf for P . Then ExtjP (i∗OX , ωP ) = 0 for j < r − n andwe have a weak dualizing sheaf (ωX , tX) for X with ωX = i∗ Extr−nP (i∗OX , ωP ). Thisis a dualizing sheaf if also ExtjP (i∗OX , ωP ) = 0 for j > r − n.

PROOF. We first show that ExtjP (i∗OX , ωP ) = 0 for j < r − n.Since OP (q) is invertible, it is equivalent to prove this property for some Serre

twist ExtjP (i∗OX , ωP )(q)). When q 0, then by the Serre’s vanishing theorem 11.2,ExtjP (i∗OX , ωP )(q) is generated by its global sections and is Γ-acyclic. So it sufficesto prove that Γ(P, ExtjP (i∗OX , ωP )(q)) = 0 for q 0. Since OP (q) is locally free,3.18-iv of the appendix implies that

ExtjP (i∗OX , ωP )(q) ∼= ExtjP (i∗OX , ωP (q)) ∼= ExtjP (i∗OX(−q), ωP )).

In particular, ExtjP (i∗OX(−q), ωP ) is Γ-acyclic for q 0. We invoke 3.18-v ofthe appendix and find a natural isomorphism from Γ(P, ExtjP (i∗OX(−q), ωP )) ontoExtjP (i∗OX(−q), ωP ). By Serre duality on P ,

ExtjP (i∗OX(−q), ωP ) ∼= Hr−j(P, i∗OX(−q))∨ ∼= Hr−j(X,OX(−q))∨

and the last vector space is zero when r − j > n, i.e., when j < n− r. 9

We now fix a coherent OX -module F and consider the functor G : Coh(P ) →Vectk which assigns to a coherentOX -module G the k-vector space HomX(F , i∗G) =

HomP (i∗F ,G). We have RjG(G) = ExtjP (i∗F ,G) by definition. We now regard Gas the composite of two functors, to be precise, we write G = GF with F : G 7→HomP (i∗OX ,G). Then RjF (G) = ExtjP (i∗OX ,G) and so if we take G = ωP , then bythe above, these derived functors vanish for j < n − r. Hence by Proposition 2.21there is a natural isomorphism

HomP (i∗F , Extn−rP (i∗OX , ωP )) ∼= Extn−rP (i∗F , ωP ).

The left hand side can be identified with HomX(F , i∗ Extn−jP (i∗OX , ωP )) and theright hand side is by Serre duality on P isomorphic to the k-dual of Hn(P, i∗F) =

Hn(X,F). Thus ωX = i∗ Extn−jP (i∗OX , ωP ) serves as a weak dualizing sheaf.Now assume that we also have that ExtjP (i∗OX , ωP ) = 0 for j > n − r. Then

the stronger part of Proposition 2.21 says that there is a natural isomorphism

ExtjP (i∗F , Extn−rP (i∗OX , ωP )) ∼= Extn−r+jP (i∗F , ωP ), j ≥ 0.

The right hand side can be identified with the k-dual of Hr−j(P, i∗F) = Hr−j(X,F).In order to identify the left hand side, we make the observation that for anyOP -

module G, ExtjP (i∗F ,G) is a i∗OX -module (choose an injective resolution G → I•and note that Hom(i∗F , I•) is a complex of i∗O-modules) so that the natural mapExtjP (i∗FX ,G) → i∗i

∗ ExtjP (i∗FX ,G) is an isomorphism. If G is a OX -module, thenExtjP (i∗F ,G) ∼= ExtjP (i∗F , i∗i∗G) = i∗ ExtjX(F , i∗G) (for i∗ is exact) and similarly,ExtjP (i∗F ,G) = ExtjX(F , i∗G).

If we apply this to the present situation we find that

ExtjP (i∗F , Extn−rP (i∗OX , ωP )) = ExtjP (i∗F , i∗i∗ Extn−rP (i∗OX , ωP )) = ExtjX(F , ωX).

We have thus shown that ExtjX(F , ωX) ∼= Hr−j(X,F)∨.

9You may find it strange that we proved a local property by means of global arguments. A purelylocal (commutative algebra) proof is indeed possible.

152 3. SCHEMES

The Cohen-Macaulay property. The vanishing condition in Theorem 15.4 (ofcertain ExtjP (i∗OX , ωP )) that gives us a dualizing sheaf is remains elusive, as longas we have no way of establishing it. At this point it is also unclear to what extentthis depends on the immersion of X in a projective space. Let us first show thatthis is an issue of a local nature:

PROPOSITION 15.5. Let F be a coherent sheaf on a noetherian scheme Y and lety ∈ Y . Then for every OY -module G, we have ExtjY (F ,G)y = ExtjOY,y (Fy,Gy).

PROOF. For F = OY this is obvious: we have ExtjY (OY ,G) = 0 for j 6= 0

and Ext0Y (OY ,G) = G. For the same reason this is true when F is a free OX -

module of finite rank. In general, choose an affine open neighborhood U of yin Y and a right resolution of F|U by free OU of finite rank: L• → F|U . ThenExtjY (F ,G)|U = ExtjU (F|U,G|U) = hj(HomU L•,G|U)) by 3.18-(ii) and (vi) and soExtjU (L,N )y = hj(HomU (L•,G|U))y = hj HomU (Ly,•,Gy) = ExtjOY,y (Fy,Gy).

CHAPTER 4

Appendix

1. The basics of category theory

In this part of mathematics one quickly runs into foundational aspects of math-ematics and one must be careful about which kind of set theory one adopts (other-wise we will be soon be talking about a nonentity like ‘the set of all sets). We leavethese issues aside. Let us just say that we at least accept the axiom of choice.

Categories. The following definition is modeled after the coarse structure per-ceived in various mathematical areas.

DEFINITION 1.1. A category C consists of(i) a collection (a class, to be precise) called objects Obj(C),

(ii) for any pair X,Y ∈ Obj(C) a (possibly) empty set, called the set of C-morphisms from X to Y and denoted MorC(X,Y ) and for

(iii) any triple X,Y, Z ∈ Obj(C) a map, called composition

(g, f) ∈ MorC(Y, Z)×MorC(X,Y ) 7→ g f ∈ MorC(X,Z).

These data must obey:Associativity: (h g) f = h (g f) whenever that makes sense andIdentity element: for every X ∈ Obj(C) there exists a 1X ∈ MorC(X,X)

such that for all f ∈ MorC(X,Y ) and g ∈ MorC(Y,X), f 1X = f and1X g = g.

A standard argument shows that there is only one element in MorC(X,X) sat-isfying the last property. It is called the identity element of X. We often writef : X → Y instead of f ∈ MorC(X,Y ) and we call X the source and Y the targetof f . We say that f : X → Y is an isomorphism if it has a 2-sided inverse, i.e., ifthere exists a g : Y → X such that g f = 1X and f g = 1Y . Again a standardargument shows that g is then unique; it is called the inverse of f and denoted f−1.Two objects of C are said to be isomorphic if an isomorphism between them exists.

The generality of this notion is both its power and its limitation.

EXAMPLES 1.2. Here are just a few.Set: the category Set of sets and maps between sets,Top: the category of topological spaces and continuous maps,Htp: the category of topological spaces and homotopy classes of maps,Grp: the category of groups and homomorphisms,Ab: its subcategory of abelian groups,Ring: the category of rings and ring homomorphisms,VectK: the category of finite dimensional K-vector spaces and linear maps

(K a field), and more generally,

153

154 4. APPENDIX

ModR: the category ModR ofR-modules andR-homomorphisms (R a ring),AbX : the category of abelian sheaves of on a space X,ModO: the category of sheaves of O-modules, where O is a sheaf of rings

on a space X (to be discussed later).

The definition shows that reversal of the arrows of C produces another category,called the dual of C and denoted C: it has the same objects: Obj(C) := Obj(C),but MorC(X,Y ) := MorC(Y,X) (if we view f ∈ MorC(Y,X) as an element ofMorC(X,X

′), we may denote it by f), the composition in C being the one of C:if f ∈ MorC(X,X ′) and g ∈ MorC(X ′, X ′′), then f g := (g f).

Functors. Categories can be related to each other with the help of the notionof functor.

DEFINITION 1.3. Let C and C′ be categories. A functor from C to C′ is a map Fwhich assigns to each object X of C an object F (X) of C′ and to each morphismf : X → Y a morphism F (f) : F (X) → F (Y ) which sends identity elements toidentity elements: F (1X) = 1F (X) and respects composition: F (gf) = F (g)F (f).This is also called a covariant functor in order to distinguish it from a contravariantfunctor from C to C′, which is simply a functor from C to C′ (so it reverses arrows).

It is clear that functors F : C → C′ and F ′ : C′ → C′′ can be composed toproduce a functor F ′ F : C→ C′′.

An important class of functors form the ones that ‘forget’ structure (and for thisreason often referred to as forgetful functors). For instance, we have the forgetfulfunctors Top → Set (forgets the topology of a space), Grp → Set (forgets thegroup structure of a group). Other examples of functors are π0 : Top → Setand irr : Top → Set (associate to a space its set of connected resp. irreduciblecomponents). We have an evident functor Top → Htp and we have encounteredthe functor Ring → Top which assigns to a ring its prime ideal spectrum. Wealso note that we have a contravariant functor ModR →ModR which assigns to anR-module M , HomR(M,R).

Remarkably often a functor is of the type we describe next. An object A of acategory C determines two functors. The first one is contravariant: it is the functorhA : C → Set which assigns to an object X of C, the set hA(X) := MorC(X,A) andto a morphism ξ : X → X ′ the induced map hA(ξ) : MorC(X ′, A) → MorC(X,A)(given by precomposition with ξ). The other, hA : C→ Set, is covariant and assignsto an object X of C, the set hA(X) := MorC(A,X) and to a morphism ξ : X → X ′

the induced map hA(ξ) : MorC(A,X) → MorC(A,X ′) (given by composition withξ). A functor F that is naturally isomorphic to one of these (i.e., to hA or hA

according to whether F if contravariant or covariant) is said to be representable(we give a more precise definition later).

EXAMPLES 1.4. The forgetful functor Top→ Set is representable by the single-ton space ∗, for assigning to a space X the set of continuous maps ∗ → X (whichis of course the set of all maps) gives us back X as a set. Similarly, the forgetfulfunctor Grp→ Set is representable by Z.

EXERCISE 91. Show that the forgetful functor Ring→ Set is representable.

Our somewhat off-hand definition of a representable functor is incomplete be-cause we did not say what we mean by two functors being naturally isomorphic.This notion is a special case of that of a natural transformation.

1. THE BASICS OF CATEGORY THEORY 155

DEFINITION 1.5. Let F,G : C → C′ be functors between categories. A naturaltransformation from F to G (denoted T : F ⇒ G) assigns to every object X of C aC′-morphism T (X) : F (X) → G(X) with the property that for every C-morphismφ : X → Y the diagram

F (X)F (φ)−−−−→ F (Y )

T (X)

y T (Y )

yG(X)

G(φ)−−−−→ G(Y )

commutes. We say that T is a natural isomorphism (and that F and G are naturallyisomorphic) if every T (X) is an isomorphism; we then write T : F ∼= G.

This is a very fundamental concept and ideally the adjective ‘natural’ when usedin mathematical sense, should always be understood in this way. As the followingexercise shows, a natural isomorphism need not be unique.

EXERCISE 92. Let D : VectK → VectK be the functor which assigns to a K-vector space V its dual V ∗. Construct a natural isomorphism DD ∼= 1VectK

:VectK → VectK . Construct two such when K is not of characteristic 2.

Two categories C and C′ are for all practical purposes the ‘same’ if they areequivalent in the sense of the following definition.

DEFINITION 1.6. A functor F : C → C′ is said to be an equivalence (of cate-gories) is there exist a functor G : C′ → C such that GF ∼= 1C and FG ∼= 1C′ .

In other words, there must exist for every object X of C isomorphisms T (X) :X → GF (X) and T ′(X) : X → FG(X) such that for every C-morphism φ : X → Ywe have GF (φ) T (X) = T (Y ) φ and FG(φ) T ′(X) = T ′(Y ) φ. The followingexample may help to illustrate the subtleties involved here.

EXAMPLE 1.7. Fix a field K and let MatK be the category whose objects arethe nonnegative integers 0, 1, 2, . . . . A morphism m→ n is given by a m×n-matrix(which has m rows and n columns) with entries in K. Composition of morphismsis composition of matrices. This is in fact a subcategory of VectK : we have aninclusion functor F : MatK ⊂ VectK . We claim that this is an equivalence ofcategories. The inverse functor G : VectK → MatK will on objects be clearlythe map which assigns to every finite dimensional K-vector space V its dimensiondimV . But defining it on morphisms requires us to choose an ordered basis ofV , or equivalently, an isomorphism eV : KdimV ∼= V for every V (so here weneed the axiom of choice). Then for a K-linear map φ : V → W we let G(φ)be the matrix of φ in terms of the chosen bases of V and W , in other words, ofe−1W φeV : KdimV → KdimW . Now a natural isomorphism T : 1MatK ⇒ GF is

defined by letting T (n) be the matrix of eKn ∈ GL(n,K) (so had we been careful tochoose each eKn to be the identity, then T (n) would have been the identity matrix)and a natural isomorphism T ′ : 1VectK ⇒ FG is given by letting T ′(V ) := e−1

V .

This example shows that a given functor F : C→ C′ may define an equivalence withoutthere being an obvious choice for its ‘inverse’G. But the idea behind this example generalizeswell to a criterion for F being an equivalence: it is necessary and sufficient that

F is full: for every X,Y ∈ C, F : MorC(X,Y )→ MorC′(F (X), F (Y )) is surjective,F is faithful: for every X,Y ∈ C, F : MorC(X,Y ) → MorC′(F (X), F (Y )) is injec-

tive,

156 4. APPENDIX

F is essentially surjective: every object of C′ is isomorphic to one of the form F (X)with X ∈ C.

The basic example in algebraic geometry is the functor Spec : Ring → Affsch.In this case an inverse functor is also given, for it is the global section functorΓ : Affsch → Ring which assigns to an affine scheme the ring of global sectionsof its structure sheaf. For a ring R resp. an affine scheme X, we found naturalisomorphisms R → Γ Spec(R) resp. X → Spec Γ(X), but it is not always a goodidea to regard these as identities.

Given categories C and C′, then the functors F : C → C′ are the objects of anew category whose morphisms are the natural transformations. We denote thiscategory Func(C,C′), so that MorFunc(C,Set)(F,G) = Nat(F,G).

We encounter an example in the Yoneda lemma that we discuss next.

The Yoneda Lemma. Notice that a C-morphism α : A → A′ defines a naturaltransformation hA ⇒ hA′ (postcomposition with α). In this way we can regardthe map A 7→ hA as a functor h∗ : C → Func(C,Set) from the category C tothe category Func(C,Set) whose objects are the contravariant functors C → Setand whose morphisms are the natural transformations between them. The Yonedalemma implies that this embeds C in Func(C,Set) as a full subcategory. In par-ticular, A ∈ C is determined (up to unique isomorphism) by its associated functorhA.

LEMMA 1.8 (The Yoneda Lemma). Let F : C → Set be a contravariant functorfrom C to Set. Then the map which assigns to any natural transformation T : hA ⇒ Fthe element sT := T (A)(1A) ∈ F (A) defines a bijection between the set Nat(hA, F )of natural transformations hA ⇒ F and F (A). Its inverse assigns to s ∈ F (A) thenatural transformation Ts : hA ⇒ F defined by

Ts(X) : φ ∈ hA(X) = MorC(X,A) 7→ F (φ)(s) ∈ F (X).

In particular (take F = hA′), this yields a bijection Nat(hA, hA′) ∼= hA′(A) =MorC(A,A′) and thus h∗ : C→ Func(C,Set) is fully faithful.

The proof is best left as an exercise.If we replace in the preceding the category C by its opposite, then we find

that A also defines a covariant functor hA, which assigns to an object X of C,the set hA(X) := MorC(A,X) and to a morphism ξ : X → X ′ the induced maphA(ξ) : MorC(A,X) → MorC(A,X ′). The morphism α : A → A′ defines a naturaltransformation hA′ → hA so that A 7→ hA defines a functor h∗ : C → Func(C,Set).In this case the Yoneda lemma asserts that for every functor G : C → Set there isa natural bijection between Nat(hA, G) and G(A).

We say that a contravariant functor F : C → Set is representable if there existsa pair (A, s), where A is an object of C and s ∈ F (A), such that the associatednatural transformation Ts : hA ⇒ F is an isomorphism. In other words, if for everyobject X of C, the map

φ ∈ MorC(X,A) 7→ F (φ)(s) ∈ F (X)

is a bijection. We then call (A, s) a universal element for F .The pair (A, s) is unique up to unique isomorphism. For if F is also represented

by the pair (A′, s′ ∈ F (A′)), then we get a natural isomorphism T−1s′ Ts : hA ∼= hA′

1. THE BASICS OF CATEGORY THEORY 157

and by the Yoneda Lemma this must come from an unique isomorphism A ∼= A′. Itis straightforward to check that it takes s to s′.

Over and under category; initial and final object. Given a category C, thenevery object A of C defines a category denoted C/A: its objects are morphisms withtarget A: f : X → A; a morphism from f : X → A to f ′ : X ′ → A is then simplya C-morphism g : X → X ′ such that f ′g = f . It is often called the category over A.We have a forgetful functor C/A→ C (which tells us to forget about the morphismto A). This is the identity precisely if A is a final object of C, i.e., if for every objectXof C, Mor(X,A) is a singleton. Such an object is unique up to unique isomorphism.

If we reverse all arrows we arrive at the dual notion, the category under A,denoted A/C (whose objects are morphisms A→ X). The forgetful functor A/C→C is the identity precisely if A is an initial object of C: it has the property thatMor(A,X) is a singleton for every object X of C.

EXERCISE 93. Determine of each category listed in 1.2 whether it has a finalresp. initial object and if it exists, describe it. Prove that C/A resp. A/C has a finalobject resp. initial object.

Adjoint functors. The notion of an adjoint functor encodes (and sometimeseven suggests) in a categorical way universal constructions such as the formationof the free group generated by a set, the field of fractions of a ring, the largestgroup that is the quotient of a semigroup etc.

DEFINITION 1.9. Given categories C and C′, then functors F : C → C′ andF ′ : C′ → C are said to make up an adjoint pair if for every object X of C and objectX ′ of C′ we are given a bijection MorC(F ′(X ′), X) ∼= MorC′(X

′, F (X)) which isnatural1 in X and X ′ (covariant in X and contravariant in X ′). We say that F ′ is aleft adjoint of F and that F is a right adjoint of F ′.

Before giving examples, let us note that if we take X = F ′(X ′) and take theimage of 1 ∈ MorC(F ′(X ′), F ′(X ′)) we get a C′-morphism X ′ → FF ′(X ′). Thisdefines a natural transformation η : 1C′ ⇒ FF ′. Similarly, by taking X ′ = F (X),we get a C-morphism F ′F (X) → F giving rise to a natural transformation ε :F ′F ⇒ 1C. We also observe that Yoneda’s lemma implies a right adjoint (or leftadjoint) of a functor is unique up to natural isomorphism.

EXAMPLES 1.10. (ii) A standard example is provided by the forgetful func-tor Forget : Grp → Set. We are looking for left adjoint of it, that is, a functorF : Set → Grp such that for every set X and every group G, Hom(F (X), G) =Map(X,Forget(G)). This problem has a solution indeed: given a set map X → G,we can always extend it (uniquely) to a group homomorphism from the free groupFree(X) generated by X to G (by convention Free(∅) is the trivial group). We thusfind that F := Free does the job. We have indeed a map of setsX → Forget Free(X)and a natural group homomorphism Free Forget(G)→ G.

(ii) The inclusion Ab ⊂ Grp has as left adjoint the abelianization functor Grp→Ab, which assigns to a group G, G/[G,G].

(iii) Let Dom be the category of integral domains whose morphisms are in-jective homomorphisms and let Field ⊂ Dom be its subcategory of fields. Then

1The use of ‘natural’ in this definition is explicated as follows: both (X,X′) 7→ MorC(F ′(X′), X)and (X,X′) 7→ MorC′ (X

′, F (X)) define functors from the product category C× (C′) to the categorySet of sets and we ask here for a natural equivalence between the two.

158 4. APPENDIX

the inclusion has a left adjoint Frac : Dom → Field given by the formation of thefraction field.

(iv) Given a space X, then the (obvious) forgetful functor from the categoryof sheaves on X to the category of presheaves on X has a left adjoint, namelysheafification.

Product and sum. Let C be a category. Given objects A and B of C, then

form the category whose objects are the diagrams A α←− Xβ−→ B in C and where

a morphism from Aα←− X

β−→ B in C to Aα′←− X ′

β′−→ B in C is the given by amorphism f : X → X ′ so that the resulting diagram commutes: α = α′f andβ = β′f . If this category has a final object, then we call it ‘the’ product of A andB and, denoted A u B (the definite article is justified since a final object is uniqueup to unique isomorphism). So A u B comes with morphisms πA : A u B → Aand πB : A u B → B and has the property that whenever we are given a diagram

Aα←− X

β−→ B, then there is a unique morphism f u g : X → A u B such that weget a commutative diagram. If we apply this definition to an over category C/C,then we get in C the notion of a fibered product. So we are then given a diagram

Aα−→ C

β←− B and we form a category whose objects are commutative squares

A ←−−−− X

α

y yC

β←−−−− B

and whose morphisms are defined in an obvious manner. If an final object of thiscategory exists, then we call it the fibered product of the pair (α, β), denoted

AπA/C←−−−− A uC B

α

y yπB/CC

β←−−−− B

In other words, if we are given morphisms f : X → A and X → B such thatαf = βg, then there exists a unique a morphism f u g : X → A uC B over which fand g both factor.

We say that C has finite products if every pair of its objects has a product. Ifthis is the case for every overcategory C/C, then we say that C has finite fiberedproducts. In many categories (such as Set, see Example 1.11) it is more commonto write A×C B instead of AuC B and will in fact use this notation in the categoryof schemes Sch.

Reversal of arrows gives the notion of the sum of A and B, denoted AiA−→

A t B iB−→ B. When applied to the undercategory C/C, we get the notion of the

cofibered sum AiC/A−−−→ A tC B

iC/B←−−− B.

EXAMPLE 1.11. In the category Set, fibered products and cofibered sums exist:the product is the ordinary product of sets and the sum is the disjoint union. The

fibered product of A α−→ Cβ←− B is a subset of A × B and given as the set of

(a, b) ∈ A × B with α(a) = β(b) (and is usually denoted A ×C B). The cofibered

2. ABELIAN CATEGORIES AND DERIVED FUNCTORS 159

sum of Aφ←− C

ψ−→ B is the quotient of A t B and obtained by dividing out by theequivalence relation generated by φ(c) ∼ ψ(c) (and is usually denoted A ∪C B).

In Grp fibered products and cofibered sums also exist: for two groups A, B,A u B = A × B (the direct product) and A t B = A ∗ B (the free product).Furthermore, A uC B is the subgroup A ×C B ⊂ A × B of (a, b) ∈ A × B withα(a) = β(b), and AtC B is the amalgamated product A∗C B (the quotient of A∗Bby the normal divisor generated by the elements φ(c)ψ(c−1), c ∈ C).

EXERCISE 94. Prove the category Ring has fibered products and cofiberedsums. More precisely, show that AuB = A⊗B, AuC B = A⊗C B, AtB = A×Band A tC B is A×B modulo the ideal generated by (φ(c), ψ(−c)).

Monomorphism and epimorphism. The notion of an injective resp. surjectivemap in the category of sets has the following analogue in a categorical setting.

DEFINITION 1.12. We say that a morphism f : A → B in a category C is amonomorphism if for two morphisms g1, g2 ∈ MorC(X,A) are equal if and only iffg1 and fg2 are. Dually, we say that f : A → B is an epimorphism if if for twomorphisms g1, g2 ∈ MorC(B,X) are equal if and only if g1f and g2f are.

In the categories Grp, Ring, Top, ModR these also amount to injection resp.surjection.

EXERCISE 95. Prove that in the category AbX , a morphism of sheaves φ : A →B is a monomorphism resp. epimorphism if and only if for every x ∈ X the homo-morphism of stalks φx : Ax → Bx has this property.

2. Abelian categories and derived functors

Abelian categories. These are categories to which the standard techniquesof homological algebra extend in an immediate manner. But the advantage ofthe categorical approach goes beyond the generalization: it offers also a different,‘higher’ point of view.

The categories C under consideration will all have an object that is both finaland initial. It is called the zero object and will be denoted 0. This determines inevery MorC(A,B) a distinguished element, namely the composite A→ 0→ B; wecall this the zero morphism and denote it also by 0. That enables us to define thenotion of kernel and cokernel: given a morphism f : A→ B in C, then a morphismi : K → A is called a kernel of f if fi = 0 and i is universal for this property:any morphism g : X → A in C such that fg = 0, factors over i be means of aunique morphism g : X → K (g = ig). Notice that the uniqueness property impliesi has to be a monomorphism. We also note that a kernel is unique up to uniqueisomorphism. We denote a kernel of f often by Ker(f) → A. If f happens to be amonomorphism, then it has a kernel, namely the zero map 0→ A. For if g : X → Ais such that gf = 0, then it has that property in common with the zero morphism0 : X → A, and so the monomorphism property implies that we must have g = 0.This just means that g factors (uniquely) as X → 0 → A and so 0 → A may serveas the kernel of f .

Dually, a morphism p : B → C is called a cokernel of f if pf = 0 and p isuniversal for this property: any morphism g : B → X in C such that gf = 0, factorsover p be means of a unique morphism g : C → X (g = gp). We denote a cokernel

160 4. APPENDIX

of f often by B → Coker(f). If f happens to be an epimorphism, then B → 0 is thecokernel of f .

EXERCISE 96. Describe the kernel of f ∈ MorC(A,B) as a representative objectof a functor C → Set and describe the cokernel of f as a representative object ofa functor C→ Set.

EXERCISE 97. Prove that the kernel of A → 0 and the cokernel of 0 → A areboth represented by the identity 1A : A→ A.

We now also assume that any two objects A1, A2 of C have a sum that at thesame time serves as product (this is called in the literature a biproduct). We denoteit by A1 ⊕ A2. This structure turns MorC(A1, A2) into an abelian semigroup whichhas 0 as zero, as follows. The product role of ⊕ makes that A1⊕A2 comes with theprojections p1 : A1 ⊕A2 → A1, p2 : A1 ⊕A2 → A2, whereas its sum role endows itwith morphisms k1 : A1 → A1 ⊕ A2 and k2 : A2 → A1 ⊕ A2. Let us then note that

given fi ∈ MorC(Ai, Bi) (i = 1, 2), then Aifi−→ Bi

ki−→ B1 ⊕ B2 (i = 1, 2) combineto define a morphism

f1 ⊕ f2 : A1 ⊕A2 → B1 ⊕B2.

We have a diagonal ∆ : A→ A⊕A characterized by the property that p1∆ and p1∆are the identity and a ‘codiagonal’ ∇ : A ⊕ A → A characterized by the propertythat ∇k1 and ∇k2 are the identity. We then define the sum of f1, f2 ∈ MorC(A,B)as the composite

f1 + f2 : A∆−−−−→ A⊕A f1⊕f2−−−−→ B ⊕B ∇−−−−→ B.

It is a (not entirely trivial) exercise to check that the sum is commutative, associa-tive and has 0 as its zero. The construction shows that the identity of A⊕B is equalto k1p1 + k2p2.

If we are also given g ∈ MorC(X,A) resp. h ∈ MorC(B, Y ), then you maycheck that (f1 + f2)g = f1g + f2g and h(f1 + f2) = hf1 + hf2. In other words, thecomposition map

MorC(A,B)×MorC(B,C)→ MorC(A,C)

is bilinear.

DEFINITION 2.1. A category C is said to be an abelian category if it is as above,that is, possesses zero object and any two objects have a biproduct, and has inaddition the following properties

(i) the semigroup structure on MorC(A,B) is in a fact a group structure (sothat MorC(A,B) is an (additively written) abelian group,

(ii) every morphism has a kernel and a cokernel,(iii) every morphism f : A → B admits a factorization A → I → B, where

A → I is the cokernel of Ker(f) → A and I → B is the kernel of B →Coker(φ).

The factorization in (iii) is called the formation of the image of f , and I is oftendenoted Im(f). Notice that A → Im(f) is an epimorphism and Im(f) → B is amonomorphism.

Let us observe that the dual of an abelian category is also abelian. We furthernote that for an object A of an abelian category, the functors hA and hA take valuesin the category Ab.


LEMMA 2.2. Let f : A → B be a morphism in an abelian category. Then f is amonomorphism if and only if it has trivial kernel and in that case A → Im(f) is anisomorphism. Dually, f is an epimorphism if and only if it has trivial cokernel and inthat case Im(f)→ B is an isomorphism.

In particular, f is an isomorphism if and only if both its kernel and its cokernelare trivial.

PROOF. With the help Exercise 97 we see that Ker(f) = 0 implies that A→ I isan isomorphism. This also shows that f is monomorphism, as is then the compositeof the isomorphism A ∼= I and the monomorphism I → B (a kernel). We alreadyobserved that a monomorphism has trivial kernel and so the first assertion follows.The proof of the second assertion is similar.

EXAMPLES 2.3. The most basic example is that of the category of abelian groups,Grp. Other examples are ModR and (as we will show later), ModO. But Ring isnot one (why?).

We have now all the ingredients in place for the notions related to exactness.

For instance, if we have a diagram Af−→ B

g−→ C in C with gf = 0, then the com-posite Im(f) → B → C is zero, hence must factor uniquely through a morphismIm(f)→ Ker(g). This is a monomorphism of which we can form the cokernel. Thiscokernel is called the (co)homology of the diagram and denoted h(A −→ B −→ C).We say that the diagram is exact if this is the zero object.

DEFINITION 2.4. A complex in C is a chain of morphisms

A• . . .dk−2

−−−→ Ak−1 dk−1

−−−→ Akdk−→ Ak+1 dk+1

−−−→ . . .

such that dkdk−1 = 0 for all k. The k-th cohomology object hk(A•) is defined asthe cokernel of the natural morphism Im(dk−1)→ Ker(dk). If all of these are zero,then we say that the complex is exact.

Given two complexes A• and B• in C, then a chain morphism φ• : A• → B• isa collection of morphisms φk : Ak → Bk, k ∈ Z, that make a commutative ladder.Two such chain morphisms φ•, ψ• : A• → B• are said to be (chain) homotopic ifthere exist morphisms ki : Ai → Bi−1 such that di−1ki + ki+1di = φi − ψi for alli ∈ Z.

The same proofs as in the category Ab yield the following two lemmas.

LEMMA 2.5. Homotopy is an equivalence relation on the collection of chain mor-phisms which behaves well with respect to composition: a homotopy class of chainmorphisms A• → B• can be composed with a homotopy class of chain morphismsB• → C• to obtain a homotopy class of chain morphisms A• → C•. A chain mor-phism f• : A• → B• induces morphisms hk(f•) : hk(A•)→ hk(B•) for all k ∈ Z andthese morphisms only depend on its homotopy class.

Two chain morphisms f• : A• → B• and g• : B• → C• are said to define ashort exact sequence of chain morphisms, often written as

0→ A•f•−→ B•

g•−→ C• → 0,

when each sequence 0 → Ak → Bk → Ck → 0 is exact (so each fk is then amonomorphism and each gk is an epimorphism).

162 4. APPENDIX

LEMMA 2.6. A short exact sequence of chain morphisms as above generates a longexact sequence

· · · → hk(A•)hk(f•)−−−−→ hk(B•)

hk(g•)−−−−→ hk(C•)δk−→ hk+1(A•) → · · · .

REMARKS 2.7. Often we are given a complex as 0 → A0 d0−→ . . .dk−1

−−−→ Akdk−→

. . . . It is then understood that all the terms Ck with k < 0 are zero.Sometimes we are given a complex with decreasing indexing (the index is then

almost always written as a subscript):

A• . . .dk+1−−−→ Ak

dk−→ Ak−1dk−1−−−→ . . . ,

but one can always pass to an increasing indexing by putting Ak := A−k anddk := d−k.

Derived functors. In what follows F : C → C′ is a functor of abelian cat-egories. It is an exercise to show that F (A ⊕ B) is canonically isomorphic toF (A) ⊕ F (B) (there are natural maps between them in each direction) and thatF takes diagonal to diagonal and codiagonal to codiagonal. It then follows that Fis additive in the sense that for any two objects A,B of C the map MorC(A,B) →MorC′(F (A), F (B)) is a homomorphisms of abelian groups: F (φ1 + φ2) = F (φ1) +F (φ2) for any pair φ1, φ2 ∈ MorC(A,B).

DEFINITION 2.8. We say that F is left exact if it takes an exact sequence ofthe form 0 → A → B → C to a sequence that is still exact (so 0 → F (A) →F (B)→ F (C) is exact); in particular, F takes monomorphisms to monomorphisms.Similarly, we say that F is right exact if it takes an exact sequence of the formA → B → C → 0 to sequence that is still exact (so that F takes epimorphisms toepimorphisms). We say that F is exact if it is both left and right exact.

This definition (and the subsequent discussion) is justified by the fact that manyimportant functors are only half exact, witness the following examples.

EXAMPLES 2.9. (a) Given a ring R and an R-module M , then the functor⊗RM : ModR → ModR, A 7→ A ⊗R M is right exact. It is not always left ex-act: if R = Z, then tensoring the exact sequence 0 → Z ×5−−→ Z → Z/5 with Z/5yields the non-exact sequence 0→ Z/5 ×0−−→ Z/5 ×1−−→ Z/5.

(b) For R and M as above, we can consider HomR(M,−) : ModR → ModRinstead. This functor is left exact, but need not be right exact: taking R = Z andM = Z/5 and applying the functor HomZ(Z/5,−) to the exact sequence Z ×5−−→ Z→Z/5→ 0 yields 0→ 0→ Z/5→ 0 which is evidently not exact.

(c) For a space X we have a functor Γ : AbX → Ab which assigns to an abeliansheaf F its group of sections Γ(X,G) (which we often prefer to write H0(X,F)).This functor is left exact, but not always right exact. For example, lat X be thecircle (with its usual topology). Let E0

S1 be the sheaf of C∞-functions on S1 andE1S1 = E0

S1dθ be the sheaf of C∞-differentials on S1. Then we have a short exactsequence of abelian sheaves

0→ RS1 → E0S1

d−→ E1S1 → 0

(the differential of a function is zero if and only if it is locally constant and anydifferential on an open subset of S1 can locally be integrated to a function). But if


we apply Γ(S1,−) we get

0→ R→ E0(S1)d−→ E1(S1)→ 0,

E i(S1) stands for the functions (i = 0) resp. differentials (i = 1) on S1. Exactnessfails at E1(S1), for the differential dθ can not be integrated to function on S1: θis only defined up to an integral multiple of 2π and is not globally definable as aC∞-function, although dθ is globally defined.

(d) Suppose f : X → Y is a continuous map of topological spaces. Thenf−1 : AbY → AbX is exact, for if 0 → F ′ → F → F ′′ → 0 is an exact sequenceof abelian sheaves on Y , then 0 → f−1F ′ → f−1F → f−1F ′′ → 0 is also exact:this property needs only be tested for stalks, and the stalk sequence at x is just0→ F ′f(x) → Ff(x) → F ′′f(x) → 0.

(e) In the above situation, f∗ : AbX → AbY is left exact, but need not be rightexact. In case Y is a singleton, we recover Example (c).

(f) There are plenty variations and combinations of the previous examples. Forexample, we can start with a ringed space (X,O) and put R := Γ(X,O). Then Γdefines a functor ModO →ModR. This functor is left exact. If we are also given anO-moduleM, then A 7→ HomO(A,O) defines a left exact functor HomO(−,M) :ModO → ModO. And getting a bit ahead of ourselves, we can also consider thefunctor HomO(−,M) : ModO →ModR defined by A 7→ HomO(A,M). This is leftexact as the composite of two left exact functors, namely as

HomO(−,M) : ModOHomO(−,M)−−−−−−−−→ ModO

Γ−−−−→ ModR.

In hindsight, one may say that the desire to thoroughly understand and mea-sure how the half exact functors mentioned in examples (a) and (b) fail to be exactgave rise to the birth of homological algebra. It was observed in the 1950’s thatthe tools developed for this purpose can render the same services in the presentsetting. We go briefly through the relevant notions.

DEFINITION 2.10. We say that an object I of C is injective if the functor hI =MorC(−, I) : C → Ab is exact. We say that C has enough injectives, if for everyobject A of C there exists a monomorphism from A to an injective object.

Dually, an object P of C is said to be projective if the functor MorC(P,−) : C→Ab is exact. In particular, if we are given an epimorphism A → B, then then anymorphism P → B lifts to a morphism P → A. We say that C has enough projectives,if for every object A of C there exists a epimorphism from a projective object to A.

Since MorC(−, I) is always left exact, the injectivity property amounts to: forany monomorphism A→ B, any morphism A→ I extends to a morphism B → I.

Given an object A of C, then a resolution of A is simply a long exact exactsequence which begins with A:

0→ Aα−→ C0 d0−→ C1 d1−→ C2 d2−→ · · ·

We prefer to think of this a morphism of complexes: α : A → C•, where A isidentified with the complex that has A placed in degree zero and zeroes elsewhere.The exactness property then amounts to the statement that α induces an isomor-phism on cohomology: A ∼= h0(C•) and hk(C•) = 0 for k 6= 0. A complex with theproperty that it is exact on all nonzero degrees is alcso called an acyclic complex.

We omit the proof of the following proposition, as it follows a standard pattern.

164 4. APPENDIX

PROPOSITION 2.11. Suppose C has enough injective objects. Then(i) for every object A of C, there exists an injective resolution α : A → I• (by

which we mean that each Ik is injective).(ii) Given a morphism φ : A→ B in C, an arbitrary resolution α : A→ C• and

an injective resolution β : B → J•. Then φ can be extended to a morphismφ• of chain maps:

−−−−→ 0 −−−−→ Aα−−−−→ C0 d0−−−−→ C1 d1−−−−→ C2 d2−−−−→ · · ·y φ

y φ0

y φ1

y φ2

y−−−−→ 0 −−−−→ B

β−−−−→ J0 d0−−−−→ J1 d1−−−−→ J2 d2−−−−→ · · ·and any two such extensions φ•, ψ• are chain homotopic.

In the remainder of this subsection, F : C → C′ is a left exact functor betweenabelian categories and it is assumed that C has enough injectives.

With the help of the previous proposition we can now define a functor RiF :C → C′ as follows. Given an object A, choose an injective resolution A → I•

and put RiF (A) := hi(F (I•)). Before we prove that this is well-defined, we firstdefine RiF on morphisms. Given a morphism φ : A → B in C, choose injectiveresolutions A → I• and B → J• and lift φ to chain morphism φ• : I• → J•.This induces a chain morphism F (φ•) : F (I•) → F (J•) and hence morphismshi(F (I•)) → hi(F (J•)). Given our choice of injective resolutions, the homotopyclass of φ• only depends on φ and hence the same is true for the homotopy class ofF (φ•). We may therefore denote hi(F (I•)) → hi(F (J•)) provisionally by hi(φ, F ).If we have another morphism ψ : B → C and an injective resolution C → K•,then ψ can be lifted to a chain morphism ψ• : J• → K•. So ψ•φ• defines a chain amorphism I• → K•. We thus find that hi(ψφ, F ) = hi(ψ, F )hi(φ, F ).

If we apply this to the case where φ is an isomorphism with inverse ψ, wefind that hi(ψ, F ) is a two sided inverse of hi(φ, F ). In particular, if we takeA = B and φ the identity, then it follows that the object hi(F (I•)) is unique upto unique isomorphism. We denote this object by RiF (A) and we denote the mor-phism RiF (A) → RiF (B) induced by φ ∈ MorC(A,B) by RiF (φ). We have thusconstructed a functor RiF : C→ C′, called the ith right derived functor of F .

PROPOSITION-DEFINITION 2.12. We have R0F = F and when A is injective, then

Ri(A) = 0 for i 6= 0. Moreover any short exact sequence 0→ Aφ−→ B

ψ−→ C → 0 in Cin C gives rise to a long exact sequence in C′:

· · · → RiF (A)RiF (φ)−−−−−→ RiF (B)

RiF (ψ)−−−−−→ RiF (C)δi−→ Ri+1F (A)→ · · · ,

which is functorial: we have a functor from the category of short exact sequences in Cto the category of long exact sequences in C′.

For any resolution A → C• of A we have for every n ≥ 0 a natural morphismhn(F (C•))→ RnF (C).

PROOF. Let A be an arbitrary object of C and let A → I• be an injective res-olution. Since F is left exact, the exactness of 0 → A → I0 → I1 implies the ex-actness of 0 → F (A) → F (I0) → F (I1). This shows that R0F (A) = Ker(F (I0) →F (I1)) ∼= F (A). It is easily checked that under these isomorphisms, R0F (φ) = φfor any morphism in C.


If A is injective, then we can regard A as its own injective resolution (put 0 indegrees 6= 0) and we find that Ri(A) = 0 for i 6= 0.

The long exact sequence follows by choosing our injective resolutions compat-ibly with the short exact sequence:

0 −−−−→ Aφ−−−−→ B

ψ−−−−→ C −−−−→ 0yα yβ yγ0 −−−−→ I•

φ•−−−−→ J•ψ•−−−−→ K• −−−−→ 0

in the sense that α, β, γ are injective resolutions as part of a short sequence ofcomplexes (show that you can do this). Since F is exact on a short sequences ofinjective objects,

0 −−−−→ F (I•)F (φ•)−−−−→ F (J•)

F (ψ•)−−−−→ K• −−−−→ 0

is also a short exact sequence of complexes. The associated long exact cohomologysequence is the exact sequence with the asserted properties.

The last assertion follows from Proposition 2.11.

EXAMPLE 2.13. Let us return to the Examples 2.9-a,b. Let R be a ring andM an R-module. Since R has enough projectives, the right exact functor ⊗RM :ModR → ModR has left derived functors; they are denoted TorRi (M,−) familiarfrom homological algebra. Similarly, since R has enough injectives, the left exactfunctor HomR(M,−) : ModR →ModR has right derived functors; they are denotedExtiR(M,−).

Injective resolutions have in general only a theoretical interest: they serve todefine left derived functors, but in practice they are unfit to do computations with.Proposition 2.14 below implies that for calculations it suffices to have at our dis-posal a large collection of objects on which the Rk for k > 0 vanish.

Derived functors without injective objects. Let C and C′ be abelian cate-gories. We have seen that in case C has enough injective objects, then any left exactfunctor F yields a collection of functors T i = RiF : C′ → C∞i=0 and natural trans-

formations from the category of short exact sequences S : 0 → Aφ−→ B

ψ−→ C → 0in C to the category of long exact exact sequences in C′ that are of the form

· · · → T i(A)T i(φ)−−−→ T i(B)

T i(ψ)−−−−→ T i(C)δiS−→ T i+1(A)→ · · · .

The data needed to write down this sequence are of course just the morphismsδiS : T i(C) → T i+1(A). Let us agree to call this system of data (the T i and the δi)a δ-package (or δ-functor). Notice that T 0 is then automatically left exact. We havean abstract De Rham theorem for such packages:

PROPOSITION 2.14 (Abstract De Rham theorem). Let (T i, δi)i be δ-package for(C,C′) and let be A → C• is a right resolution in C. Then we have for every n ≥ 0 anatural morphism hn(T 0(C•))→ Tn(A) and this is an isomorphism when T i(Cj) =0 for 0 < i ≤ n− j.

PROOF. Denote by Zk the kernel of dk : Ck → Ck+1. We have short exactsequences 0 → A → C0 → Z1 → 0 and 0 → Zk → Ck → Zk+1 → 0 (for k ≥ 0).

166 4. APPENDIX

The associated long exact T -sequences yield for n > 0 a string of maps:

Tn(A)δ←− Tn−1(Z1)

δ←− · · · δ←− T 1(Zn−1)δ←− Coker(T 0(Cn−1)→ T 0(Zn)).

Since T 0 is left exact, we have T 0(Zn) = ker(T 0(Cn) → T 0(Cn+1) and so we canidentify Coker(T 0(Cn−1)→ T 0(Zn)) with hn(T 0(A)). The maps δ in our string areall isomorpisms when T i(Cj) = 0 for 0 < i ≤ n− j.

Note that the preceding proposition applies in particular to right derived func-tors.

REMARK 2.15. The name comes from the fact that its content and proof are inspiredby a key step in the proof of De Rham theorem. The Poincare lemma asserts that on aC∞-manifold M of dimension m, the complex of sheaves of differential forms

0→ RM → E0Md−→ E1M

d−→ · · · d−→ EmM → 0

is exact. We think of this as a resolution RM → E•M of the constant sheaf RM . With the helpof partitions of unity one can show that each EkM is acyclic for the functor Γ : AbM → Ab. Itfollows that RkΓ(M,RM ) (which we usually write as Hk(M,RM )) is computed as the kthcohomology group of the De Rham complex Γ(M, E•M ).

There may exist several δ-packages for given C and C′. (That is for instancethe case, if C has enough injectives and there are several left exact functors C →C′.) The δ-packages for the pair (C,C′) are objects of a category: a morphismfrom on such package (T i, δi)∞i=0 to another such (T ′i, δ′i)∞i=0 consists of naturaltransformations T i ⇒ T ′i (i = 0, 1, . . . ) such that we get for every morphism ofshort exact sequences in C a morphism of long exact exact sequences in C′.

DEFINITION 2.16. We say that the δ-package (T i, δi)∞i=0 for (C,C′) is universal iffor every δ-package (T ′i, δ′i)∞i=0 for (C,C′), every natural transformation T 0 ⇒ T ′0

extends uniquely to morphism of δ-packages from (T i, δi)∞i=0 to (T ′i, δ′i)∞i=0.

In particular, if (T ′i, δ′i)∞i=0 is a package with T ′0 = T0, then such a morphismof δ-packages always exists and is unique. So if (T ′i, δ′i)∞i=0 is also universal, thenthis morphism has a two-sided inverse. Hence there is, up to unique isomorphism,at most one universal δ-package (T i, δi)∞i=0 for a given initial functor T 0.

Here is a useful criterion for a δ-package to be universal.

PROPOSITION 2.17. A δ-package (T i, δi)∞i=0 is universal if each Tn with n > 0is effaceable, meaning that every object of C is the source of a monomorphism u in Cwith Tn(u) = 0.

PROOF. We only define for every objectA a morphism F i(A) : T i(A)→ T ′i(A),leaving to you the proof that they are functors. Suppose n > 0 and that the naturaltransformations F i : T i ⇒ T ′i have been constructed for i < n. Given A ∈ C, thenchoose a monomorphism u : A → B such that Tn(u) = 0. Denote by B → C thecokernel of u so that we have a short exact sequence 0 → A → B → C → 0. Nowconsider the commutative diagram with exact rows

Tn−1(B) −−−−→ Tn−1(C) −−−−→ Tn(A)0−−−−→ Tn(B)yFn−1(B)

yFn−1(C)

T ′n−1(B) −−−−→ T ′n−1(C) −−−−→ T ′n(A)


So Tn−1(C) → Tn(A) is an epimorphism whose kernel is contained in the kernelof the composite Tn−1(C) → T ′n−1(C) → T ′n(A). It follows that there is uniquemorphism Tn(A) → T ′n(A) which makes this diagram commute. To prove inde-pendence of u, let u′ : A → B′ be a second monomorphism with Tn(u′) = 0. Ifwe have a factorization u′ : A

u−→ B → B′, then it easily checked that we get thesame morphism Tn(A) → T ′n(A). But we can always reduce to this case by fac-toring u and u′ through the monomorphism (u, u′) : A → B ⊕ B′ (observe thatTn(U, u′) = 0). So we have a well-defined morphism Tn(A)→ T ′n(A).

We get nothing new when C has enough injectives:

COROLLARY 2.18. Assume C has enough injectives. Then for every left exact func-tor F : C → C′ the system of its right derived functors (RiF, δi)i is a universal δ-package and conversely, any universal δ-package is of this form (so that T i is the ithderived functor of T 0).

PROOF. Every object A admits a monomorphism u : A → I with injective.Then for n > 0, RnF (u) : RnF (A) → RnF (I) = 0 and so by Proposition 2.17,(RiF, δi)i is universal as a δ-package.

Conversely, if (T i, δi)i is a universal δ-package, then T 0 is left exact, and sogives rise to a δ-package. Since this package is universal and has the same functorin degree zero, it must be isomorphic to (T i, δi)i.

Here are two applications.

PROPOSITION 2.19. Let f : X → Y be a continuous map of topological spaces sothat we have defined the left exact functor f∗ : AbX → AbY . Then the right derivedfunctor Rkf∗ : AbX → AbY applied to an abelian sheaf F on X is the sheafificationof the presheaf on Y which assigns to an open subset V ⊂ Y , Hk(f−1V,F).

PROOF. The sheafification process defines for any k ≥ 0 a functor T k : AbX →AbY . For k = 0 this functor is simply f∗. The Tk are part of a delta package: fora fixed open subset V ⊂ Y , the functor F 7→ Hk(f−1V,F) is clearly part of one andthis then remains true if we let V vary by passing to Tk. For an injective sheaf I onX, T k(I) is zero for k > 0 and hence is universal. Now apply Corollary 2.18.

Let R be ring and L a free R-module. Then the functor HomR(L,−) is exactand so ExtiR(L,−) = 0 for all i > 0.

PROPOSITION 2.20. For every R-module M , we have natural R-homomorphisms

ExtiR(M ⊗ L,N)∼=−→ ExtiR(M,HomR(L,N)).

and these are isomorphisms when L is of finite rank.

PROOF. We have a natural transformation of functors

ModR ⇒ModR : HomR(M ⊗ L,−)⇒ HomR(M,HomR(L,−))

and this is an isomorphism in case L is of finite rank. Both functors are part of δ-packages with T i being represented by ExtiR(M⊗L,−) resp. ExtiR(M,HomR(L,−)).According to Corollary 2.18 the former is universal, and so the natural transforma-tion extends to one of δ-packages ExtiR(M ⊗ L,−) ⇒ ExtiR(M,HomR(L,−)) andthis extension is unique. To see that the latter is also universal we check that the

168 4. APPENDIX

ExtiR(M,HomR(L,−)) are effaceable (and apply Proposition 2.17). If N is an R-module, and N → I embeds N in an injective module, then HomR(L, I), beinga direct sum of copies of I, is also injective and so ExtiR(M,HomR(L, I)) = 0 fori > 0. So when L is of finite rank, the transformations above are isomorphisms.

The following proposition is usually derived with the help of a spectral se-quence.

PROPOSITION 2.21. Let F : C → C′ and G : C′ → C′′ be left exact functorsbetween abelian categories of which the source categories C and C′ have enough in-jectives. Then for every object A of C and every integer k ≥ 0 there are naturalhomomorphisms

R(GF )k(A)→ G(RkF (A)).

Assume that every object of C admits a monomorphism to an injective object I withthe property that F (I) is G-acyclic. This is an isomorphism when RjF (A) is G-acyclicfor j < k. If RjF (A) = 0 for j 6= k, then we have even for every integer l a naturalisomorphism

R(GF )l(A) ∼= RGk+l(RcF (A)).

PROOF. Choose an injective right resolution A → I• and put C• := F (I•), sothat RjF (A) = hj(C•) and Rj(GF )(A) = hj(G(C•)). Let Zj := ker(Cj → Cj+1)and Bj = Im(Cj−1 → Cj). Since hk(C•) is the cokernel of the monomorphismBk → Zk we have a long exact sequence

0→ G(Bk)→ G(Zk)→ G(hk(C•))→ R1G(Bk)→ R1G(Zk)→ · · ·

As hk(G(C•)) is the cokernel of G(Ck−1) → G(Zk) (which factors as G(Ck−1) →G(Bk)→ G(Zk)) is follows that we have a natural morphism

Rk(GF )(A) = hk(G(C•))→ G(hk(C•)) = G(RkF (A)).

We also see that this is an isomorphism if R1G(Bk) = 0 and R1G(Zk−1) = 0(for G(Ck−1) → G(Bk) is then an epimorphism). We verify that this is so un-der the given hypotheses. We then may assume that each Cj is G-acyclic. Ifwe feed this in the long exact sequence associated with the short exact sequence0 → Zj−1 → Cj−1 → Bj → 0 we find that we have for i ≥ 1 isomorphismsRiF (Bj)

∼=−→ Ri+1F (Zj−1). Consider the strings

R1G(Bk))∼=−→ R2G(Zk−1)←R2G(Bk−1)

∼=−→ · · · ← Rk+1G(B0) = 0,

R1G(Zk−1)← R1G(Bk−1))∼=−→R2G(Zk−2)← R2G(Bk−2)

∼=−→ · · · ← RkG(B0) = 0.

Since R2G(hk−1(C)) = · · · = Rk+1G(h0(C)) = 0, the left headed arrows in the firstline are all isomorphisms so that R1G(Bk)) = 0. Similarly, since R1G(hk−2(C)) =· · · = RkG(h0(C)) = 0, the left headed arrows in the second line are ismorphisms,except that R1G(Zk−1) ← R1G(Bk−1)) may only be just onto. The conclusion isstill that R1G(Zk−1) = 0. The first assertion follows.

Assume now that RjF (A) = 0 for j 6= k. Then the above argument showsthat for l > 0, RlG(Bk) = 0. This implies that RlG(Zk) → RlG(hk(C•)) =RlG(RkF (A)) is an isomorphism for l > 0. On the other hand, the sequence 0 →Zk → Ck → Ck+1 → · · · is exact and so we may regard this as a G-acyclic resolu-tion of Zk. It follows that for l > 0, RlG(Zk) = hk+l(G(C•)) = Rk+l(GF )(A).

3. SHEAF COHOMOLOGY 169

3. Sheaf cohomology

The category of modules over a ringed space. In this subsection we fix aringed space (X,O).

DEFINITION 3.1. Let (X,O) be ringed space. AnO-module is an abelian sheafAon X such that for every open U ⊂ X, A(U) comes with the structure of an O(U)-module and for U ′ ⊂ U the restriction map A(U) → F(U ′) is a homomorphism ofO(U)-modules (where O(U) acts on A(U ′) via the ring homomorphism O(U) →O(U ′)).

We say that a sheaf-homomorphism between abelian sheaves φ : A → B thatare O-modules is a O-homomorphism if for ever open U ⊂ X, φ(U) : A(U)→ B(U)is a homomorphism of O(U)-modules.

It is clear that we have thus defined a category of O-modules ModO. Noticethat when X is a singleton, then O is given a by a single ring R := O(X) and ModOis then simply the category of R-modules. The goal of this subsection is to showthat in general the category ModO has many properties in common with a categoryof modules over a ring.

PROPOSITION 3.2. The category ModO is abelian. For its structure as an abeliancategory, a sequence of O-homomorphisms · · · → Ak → Ak+1 → · · · is exact if andonly for every x ∈ X, the stalk sequence · · · → Akx → Ak+1

x → · · · is so.

PROOF. The category has an evident zero object: given an open U ⊂ X, thenthis assigns to U the zero module. The biproduct is also obvious: given O-modulesA and B, then A ⊕ B assigns to U the direct sum A(U) ⊕ B(U). The kernel ofφ : A → B assigns to U ker

(φ(U) : A(U) → B(U)

). But if we assign to U the

cokernel of φ(U) we get in general only a presheaf and we must sheafify to getCoker(φ). The last assertion is left as an exercise.

We will usually write HomO(A,B) (or just Hom(A,B) for the set of sheaf ho-momorphisms A → B; the structure of an abelian group that is intrinsic to factthat ModO is an abelian category is of course the obvious one. But there is alsoan object HomO(A,B) of ModO which assigns to U , the O(U)-module of sheafhomomorphisms HomO|U (A|U,B|U) (this is a sheaf indeed). In particular, an O-homomorphism φ ∈ HomO(A,B) is a section of HomO(A,B) over all of X.

We further have a tensor product of two objects A and B of ModO (whichshould not be confused with the biproduct: it is not a product or a sum in ModO):this is the sheaf associated to the presheaf U 7→ A(U)⊗O(U) B(U).

The category ModR of modules over a ring R has enough injectives: for anyR-module A there exists a monomorphism A→ I with I injective. The same is truefor ModO:

PROPOSITION 3.3. The category ModO has enough injectives: for everyO-moduleA there exists a monomorphism of A to an injective object I of O. This I has theproperty that every local section is the restriction of a global section.

PROOF. Given x ∈ X, then the category ModOx is injective and so there exists amonomorphism of Ox-modules αx : Ax → Ix to an injective Ox-module Ix. We thedefine a sheaf I on X as construction which ignores the topology on X: it assignsto an open U ⊂ X the product

∏x∈U Ix, so that a section s of I over U is given

170 4. APPENDIX

by a collection (sx ∈ Ix)x∈U . There is an evident O-homomorphism α : A → I: ifa ∈ A(U), then α(a) = (αx(ax))x∈U . This is clearly a monomorphism.

It remains to see that I is injective, that is, we must show that HomO(−, I) isexact. If F is an arbitraryO-module, then to give a homomorphism φ : HomO(F , I)amounts to giving a collection of stalk homomorphisms (φx ∈ HomOx(Fx → Ix)x∈X .If 0→ F ′ → F → F ′′ → 0 is an exact sequence of O-modules, then so is every stalk0 → F ′x → Fx → F ′′x → 0 of this sequence. Since Ix is injective, this remains so ifwe apply HomOx(−, Ix). But this tells us that HomO(−, I) is exact when applied toour exact sequence.

If we take forO the constant sheaf ZX , then we find that the functor Γ : AbX →Ab which assigns to an abelian sheaf A the abelian group Γ(X,O) is left exact. Wewill denote its right derived functors by Hi(X,−) and call it sheaf cohomology:Hi(X,F) = RiΓ(F).

REMARK 3.4. One can prove that for every paracompact Hausdorff space X,the singular cohomology of X with values in M is naturally isomorphic with thesheaf cohomology of X of the constant sheaf ZX . But for X a scheme endowedwith its structure sheaf OX , the use of these coholomology modules will be of adifferent kind.

Flasque sheaves. An abelian sheaf F on X is said be flasque if for every opensubset U ⊂ X, the restriction map F(X) → F(U) is onto. In other words, everysection of F over U must extend to one over X. The injective objects of ModOconstructed in Proposition 3.3 have evidently this property. (Indeed, one can showthat any injective object of ModO is flasque.)

It is clear that if F is flasque, then for any open subset U ⊂ X, F|U is alsoflasque and that so is the sheaf which assigns to an open V ⊂ X, F(U ∩ V ) (if theinclusion is denoted j : U ⊂ X, then these are j−1F and j∗j−1F respectively).

LEMMA 3.5. Let F be a sheaf on a space X and let F ′ ⊂ F a flasque subsheaf.Then for every open subset U ⊂ X, the map F(U)→ (F/F ′)(U) is onto.

PROOF. Put F := F/F ′ and let s ∈ F(U) for some open U ⊂ X. Considerthe collection of pairs (U ′, s ∈ F(U ′)) with the property that U ′ is an open subsetof U and s maps to s|U ′. This collection is partially ordered in an evident manner.By Zorn’s lemma there exists an maximal element (U ′, s ∈ F(U ′)). We show thatU ′ = U (which evidently suffices). If not, then choose x ∈ U − U ′ and let sx ∈ Fxmap to sx ∈ Fx. We then have a neighborhood Ux of x in U on which sx is definedand maps to s|Ux. Then s|U ′∩Ux−sx|U ′∩Ux is an element of F ′(U ′∩Ux) and henceextends to all of X. If we denote this extension s′ and put s′x := sx+s′|Ux, then wesee that now s|U ′∩Ux = s′x|U ′∩Ux. So this defines an element of F(U ′∪Ux) whichmaps to s|U ′ ∪ Ux. This contradicts the maximality property of (U ′, s ∈ I(U ′)) andwe conclude that U ′ = U .

PROPOSITION 3.6. Let F be a flasque abelian sheaf F on a space X. Then F isacyclic for the global section functor: Hi(X,F) = 0 for all i > 0. More generally, iff : X → Y is a continuous map between topological spaces, then F is acyclic for thedirect image functor: Rif∗F = 0 for all i > 0.


PROOF. We only prove the first assertion; the second then follows from Propo-sition 2.19. Let F → I be a monomorphism to an injective sheaf that we con-structed in Proposition 3.3: first choose for every x ∈ X, an injective monomor-phism Fx → Ix and then let I(U) consist of the collections (sx ∈ Ix)x∈U . We putG := I/F . In view of Lemma 3.5, the sequence

0→ Γ(X,F)→ Γ(X, I)→ Γ(X,G)→ 0,

is then exact. Since I is acyclic for Γ, the long exact sequence yields: H1(X,F) = 0and Hi(X,G) → Hi+1(X,F) for all i ≥ 1. We now prove with induction on n ≥ 1that for any flasque sheaf H, Hi(X,H) = 0 for i ≤ n (we just checked the casen = 1). We do this by proving that G is flasque as well: if Hi(X,G) = 0 for i ≤ n, itthen follows that Hi(X,F) = 0 for i ≤ n+ 1.

Any s ∈ G(U) is by Lemma 3.5 the image of some s = (sx)x∈U ∈ I(U). But itis clear that we can extend s to an element s ∈ I(X) by putting sx = 0 when x /∈ U(I is flasque). Then the image of s in F(X) restricts to s and so G is flasque.

The advantage of the flasqueness property over injectivity is that it is one thatit ignores any additional structure the abelian sheaf might have. We exploit this inthe following corollary.

COROLLARY 3.7. For a ringed space (X,O), denote by ΓO : ModO → ModO(X)

the global section functor for O-modules. Then for any O-module M, the abeliangroup underlying RiΓO(M) is Hi(X,M).

PROOF. We noted that the injective O-resolutionM→ I• produced in Propo-sition 3.3 is also one by flasque sheaves. By definition RkΓ(M) = Hk(Γ(X, I•)).According to Proposition 3.6, each Ik is acyclic for the global section functorAbX → Ab for abelian sheaves and so by the Abstract De Rham theorem 2.14,this also computes Hi(X,M).

This justifies our writing Hi(X,M) forRkΓO(M); it then should be understoodthat Hi(X,M) comes with the structure of a O(X)-module. In particular, if R is aring and we take for O the constant sheaf RX , then Hi(X,M) is a natural mannera RX(X)-module, and hence a R-module.

Comparison with Cech cohomology. In this subsection X is a space and Fis an abelian sheaf on X. Let A be a set. A k-simplex of A is simply an elementof Ak+1, that is a sequence of (k + 1)-elements σ = (α0, . . . , αk). We have fori = 0, . . . , k a map ∂i : Ak+1 → Ak which forgets the element in slot αi. If Ck(A)denotes the free Z-module spanned by Ak+1, then we have a boundary operatord : Ck(A)→ Ck−1(A) defined by

d(σ) =

k∑i=0

(−1)i∂i(σ).

This defines a complex C•(A) that can be identified with the simplicial (homolog-ical) complex of the abstract simplex spanned by A. If A 6= ∅, then this simplexis contractible, so has only homology in degree zero, namely Z, but it is also fairlystraightforward to prove this directly (forA = ∅we get of course the zero complex).

172 4. APPENDIX

Suppose now that A indexes an open covering U = Uαα∈A of X. We thendefine find for every k-simplex σ = (α0, . . . , αk) of A an open subset Uσ = ∩ki=0Uαi ,which may or may not be empty. We put

Ck(U,F) :=∏

σ∈Ck(A)

F(Uσ)

(note that when Uσ = ∅, then F(Uσ) = 0 and so such a factor can be omitted).In other words, an element of Ck(U,F) is given by a map s : σ ∈ Ak+1 7→ sσ ∈F(Uσ). We make it a (cohomological) complex C•(U,F) by letting dk : Ck(U,F)→Ck+1(U,F) be defined by

dks : σ ∈ Ak+2 7→k+1∑i=0

(−1)is∂i(σ)|Uσ ∈ F(Uσ).

This is called the Cech complex of U with values in F . Its cohomology is called theCech cohomology of F with respect to the covering U and is denoted by H

•(U,F). It

is immediate from the definitions that H0(U,F) = Γ(X,F) = H0(X,F).

EXERCISE 98. Let Ckalt(U,F) consist of the elements of Ck(U,F) that transformunder permutations according to the sign character, i.e., the set of (sσ ∈ F(Uσ))σsuch that sτ(σ) = sign(τ)sσ for all τ ∈ Sk+1. Show that this defines a subcomplexC•alt(U,F) ⊂ C•(U,F) and that this inclusion is a chain homotopy equivalence,hence yields the same cohomology. (This subcomplex is often more practical fordoing computations.)

REMARK 3.8. This construction makes sense in much greater generality. Note that Adefines a category: its objects are the finite sequences in A and its morphisms are the maps∂i. Any contravariant functor from A to an abelian category C gives rise to such a complex.Such a functor is called a semisimplical object in C.

We also have a sheaf analogue. For any open jU : U ⊂ X consider the sheafjU∗j

−1U F on X. So this sheaf assigns to the open subset V ⊂ X the group F(U∩V ).

We putCk(U,F) := ⊕σ∈Ck(A)jUσ∗j

−1UσF

and define the sheaf homomorphism dk : Ck(U,F) → Ck+1(U,F) in the obviousway. Notice that then Γ(X, Ck(U,F)) = Ck(U,F).

PROPOSITION 3.9. This defines a resolution of F . In other words, the complex

0→ F → C0(U,F)d0−→ C1(U,F)

d1−→ C2(U,F)d2−→ · · ·

is exact. In particular, we have natural homomorphisms Hk(U,F)→ Hk(X,F).

PROOF. Exactness is tested on stalks. For x ∈ X, let Ax ⊂ A be the set of α ∈ Afor which x ∈ Uα. Then the stalk at x of the complex in question is Hom(C•(A),Fx).This is acyclic, because C•(A) is. The last assertion is an application of Proposition2.14.

COROLLARY 3.10. If F is flasque, then Hk(U,F) = 0 for k 6= 0.

PROOF. In that case jU∗j−1U F is flasque for any open subset j : U ⊂ X. Hence

each Ck(U,F), being a direct sum of flasque sheaves, is also flasque. Now apply theabstract De Rham theorem 2.14 and conclude that Hk(U,F) = Hk(X,F). Accord-ing to Proposition 3.6 the latter is zero when k 6= 0.


COROLLARY 3.11. Suppose that Hk(Uσ,F) = 0 for all k > 0 and all σ. Then thenatural maps Hk(U,F)→ Hk(X,F) are isomorphisms.

PROOF. The complex of sheaves C•(U,F) is a resolution of F . It is acyclic forthe global section functor, for if l > 0, then for every k

Hl(X, Ck(U,F)) = ⊕σ∈Ak+1 Hl(Uσ,F)

and this is 0 by assumption. Hence the natural map from hk(Γ(X, C•(U,F)) =

Hk(U,F) to Hk(X,F) is an isomorphism.

Absolute Cech cohomology. Let v : A→ A′ be a map of sets. Then v extends toa map between the simplices spanned by A and A′ and thus gives rise to map onchain complexes v• : C•(A) → C•(A

′). Suppose now that A′ indexes the membersof another open covering U′ = (U ′α′)α′∈A′ of X and that v is a ‘refiner map’ in thesense that Uα ⊂ U ′v(α). Then for any k-simplex σ of A, we have Uσ ⊂ U ′v(σ) so thatwe have a well-defined restriction homomorphism F(U ′v(σ)) → F(Uσ). This givesrise to a homomorphism

Ck(U′,F) :=∏

σ′∈Ck(A′)

F(U ′σ′)→∏

σ∈Ck(A)

F(Uσ) = Ck(U,F)

which sends (sσ′ ∈ F(U ′σ′))σ′ to (sv(σ)|Uσ ∈ F(Uσ))σ. This is in fact a chainmap C•(U′,F) → C•(U,F) so that v induces a maps on Cech cohomology groupsHk(U′,F)→ Hk(U,F). We claim that this last map is independent of v: if w : A→A′ is another refiner function, then we use the homotopies defined as follows: forσ = (α0, . . . , αk) ∈ Ak+1, and j = 0, . . . , k, put

(v ∗j w)(σ) := (v(α0), . . . , v(αj), w(αj), . . . , w(αk)).

Observe that then Uσ ⊂ U(v∗jw)(σ). If we define hk : Ck(A)→ Ck+1(A′) by

σ ∈ Ak+1 7→k∑j=0

(−1)j(v ∗j w)(σ) ∈ Ck+2(A)k+2,

then hk−1dk + dk+1hk = w − v. In addition hk induces a map

hk+1(U,F) : (sσ′)σ′ ∈ Ck+2(U′,F) 7→( k∑j=0

(−1)js(v∗jw)|Uσ)σ∈ Ck+1(U,F)

We have hk+1dk + dk−1h

k = wk(U,F) − vk(U,F) and hence h•(U,F) defines ahomotopy between the chain maps defined by v and w. In particular, they inducethe same map Hk(U′,F)→ Hk(U,F).

Since every two open coverings U and U′ admit a common refinement (namelythe collection U ∩ U ′ with U a member of U and U ′ a member of U′), we can formthe direct limit:

Hk(X,F) := lim−→U

Hk(U,F).

This is the (absolute) Cech cohomology of X with values in F . So an element ofHk(X,F) is given by some c ∈ Hk(U,F) with the understanding that c′ ∈ Hk(U′,F)represents the same element if c and c′ have the same image in Hk(U′′,F) for somecommon refinement U′′ of U and U′.

174 4. APPENDIX

The natural maps Hk(U,F)→ Hk(X,F) are compatible under refinement: if U′

is refined by U, then the natural map Hk(U′,F)→ Hk(X,F) is just the compositionHk(U′,F)→ Hk(U,F)→ Hk(X,F). Hence we have also natural maps

Hk(X,F)→ Hk(U,F).

We know that this is an isomorphism for k = 0. This is also the case for k = 1:

PROPOSITION 3.12. The natural map H1(X,F)→ H1(X,F) is an isomorphism.

PROOF. Choose a monomorphism F → I with I flasque and denote by G itscokernel. Then the long exact sequence for cohomology produces an isomorphismH0(X,G) ∼= H1(X,F). On the other hand, for every open covering U = (Uα)α ofX, C•(U, I) is an acyclic complex and so the short exact sequence of complexes

0→ C•(U,F)→ C•(U, I)→ C•(U, I)/C•(U, I)→ 0

yields an isomorphism h0(C•(U, I)/C•(U, I)) ∼= H1(U,F). Notice that the grouph0(C•(U, I)/C•(U, I)) is simpy the set of (sα ∈ I(Uα)/F(Uα))α with the propertythat sα and sβ have the same image in I(Uα ∩ Uβ)/F(Uα ∩ Uβ) for all α, β. Inparticular, such a system defines a section of G. Indeed, the resulting map

h0(C•(U, I)/C•(U,F))→ H0(X,G).

corresponds to the natural map H1(U,F) → H1(X,F). The image consists of thesections s of G with the property that for every α, s|Uα lifts to a section of I|Uα.But for any given s there always exists an open covering of X for which this is thecase. On the other hand, if (sα ∈ I(Uα)/F(Uα))α lies in the kernel, then thereis a refinement U′ of U such that (sα ∈ I(Uα)/F(Uα))α maps already to zero inh0(C•(U′, I)/C•(U′,F)). So if we pass to the direct limit over all the coverings wefind that H1(X,F)→ H1(X,F) is an isomorphism.

Morphisms between ringed spaces. We first recall an elementary fact fromcommutative algebra. If we are given a ring homomorphism φ : R′ → R, thenany R-module M can be regarded as R′-module. This defines a functor ModR →ModR′ , called restriction. But we can also go in the opposite direction: from anR′-module M ′ we get an R-module: R ⊗R′ M ′. Moreover, any R′-homomorphismα : M ′ → M extends in an obvious manner to R-homomorphism R ⊗R′ M ′ → Mby (r ⊗R′ m′) 7→ φ(r)α(m′) and this in fact establishes an bijection

HomR′(M′,M)

∼=−→ HomR(R⊗R′ M ′,M)

(the inverse is obtained by composing an R-homomorphism R ⊗R′ M ′ → M withthe obvious R′-homomorphism m′ ∈ M ′ → 1 ⊗R m′ ∈ R ⊗R′ M ′). In categoricallanguage we may express this by saying that the functor R⊗R′− : ModR′ →ModR(which is called induction) is a left adjoint of the functor ModR →ModR′ .

This generalizes in a straightforward manner to the sheaf setting: let f :(X,O) → (X ′,O′) be a morphism between ringed spaces. We recall that thisamounts to giving a continuous map X → X ′ (also denoted f) and a sheaf ho-momorphism of rings f−1O′ → O on X, or equivalently, a sheaf homomorphismO′ → f∗O of rings on X ′. If M is an O-module, then f∗M is in a natural man-ner a f∗O-module and hence via O′ → f∗O a O′-module. This defines a functorModO → ModO′ . It has a left adjoint ModO′ → ModO, namely the functor which


assigns to an O′-moduleM′ the O-module O⊗f−1O′ f−1M′. We shall write f∗M′

for this O-module so that we obtain a bijection

HomO′(M′, f∗M)∼=−→ HomO(f∗M′,M).

Notice that on stalks this is just the construction recalled above: we have

(f∗M′)x = Ox ⊗O′f(x)Mf(x).

Corollary 3.7 and its proof (including its use of Proposition 3.6) extend in astraightforward manner to this setting:

PROPOSITION 3.13. Let f : (X,O) → (X ′,O′) be a morphism of ringed spacesand denote by f : X → X ′ the underlying (continuous) map. Then for any O-moduleM, the abelian sheaf underlying the O′-module Rkf(M) is Rkf∗(M).

Sheaf cohomology modules and their relative versions behave contravariantlyin the following sense:

PROPOSITION 3.14. Let be given a commutative square of ringed spaces:

(X ′,OX′)g−−−−→ (X,OX)

f ′y f

y(Y ′,OY ′)

g−−−−→ (Y,OY )

Then any OX -moduleM gives rise to natural OY ′ -homomorphisms

g∗Rjf∗(M)→ Rjf ′∗(g∗M), j = 0, 1, 2, . . .

PROOF. We first do the case for sheaf cohomology. LetM→ I• be an injectiveresolution. Then g−1M → g−1I• is a right resolution by g−1OX -modules (thatneed not be injective). By Proposition-definition 2.12, we therefore have a natu-ral homomorphism hj(Γ(X ′, g−1I•)) → Hj(X ′, g−1M). The obvious chain mapΓ(X, I•)→ Γ(X ′, g−1I•) gives rise to homomorphisms of O(X)-modules

Hj(X,M) = hj(Γ(X, I•))→ hj(Γ(X ′, g−1I•))→ Hj(X ′, g−1M).

This is easily checked to be independent of the choice of the injective resolution.Now compose this with the homomorphism Hj(X ′, g−1M) → Hj(X ′, g∗M) in-duced by the sheaf homomorphism g−1M → g∗M. This yields natural homo-morphism of O(X)-modules Hj(X,M) → Hj(X ′, g∗M) and hence one of O(X ′)-modules O(X ′)⊗O(X) Hj(X,M)→ Hj(X ′, g∗M).

The general case then follows from Proposition 2.19, for we have shown thatfor every open V ′ ⊂ Y ′ and every open V ⊂ Y with V ′ ⊂ g−1V , we have naturalO(V ′)-homomorphisms O(V ′)⊗O(V ) Hj(f−1V,M)→ Hj(f ′−1V ′, g∗M).

Vector bundles and invertible sheaves over a ringed space. We fix a ringedspace (X,O).

DEFINITION 3.15. An O-module F on X is said to be free of rank n if F ∼= Onand is said to be locally free of rank n if we can cover X by open subsets U suchthat F|U ∼= On|U . An invertible sheaf on X is locally free O-module of rank one.

176 4. APPENDIX

Let F be a locally free on X of rank n. We may think of F as a sheaf of sectionsof an object that may be thought as a vector bundle: by definition X admits anopen covering U = (Uα)α so that there exists an isomorphism of OUα -moduleshα : F|Uα ∼= On|Uα (in vector bundle parlance, a local trivialization). We thenobtain on an intersection Uα ∩ Uβ the isomorphism

gαβ : On|Uα ∩ Uβhα←−−−−∼= F|Uα ∩ Uβ

hβ−−−−→∼= On|Uα ∩ Uβ .

Note that gαβ is given by an n× n-matrix with values in Rαβ := O(Uα ∩ Uβ). Sincegαβ is invertible, we may regard it as an element of GLn(Rαβ) (for the zero ring thisis by definition the trivial group). The collection (gαβ ∈ GLn(Rαβ))α,β satisfies thefollowing two properties: besides the obvious fact that

(i) gαα = 1n ∈ GLn(Rα) for all α, it satisfies(ii) the cocycle condition: gαβ g

βγ g

γα = 1n ∈ GLn(Rαβγ) for all α, β, γ.

where Rα := O(Uα) and Rαβγ = O(Uα ∩ Uβ ∩ Uγ).The collection (gαβ ) ∈ GLn(Rαβ))α,β can be understood as the system of glueing

data needed to reconstruct F (in the context of vector bundles called the transitionfunctions). The sheaf F can be reconstructed from this system: a section s of Fover an open subset U ⊂ X corresponds to a collection (sα ∈ O(U ∩ Uα)n)α withthe property that gαβ |U ∩ Uα ∩ Uβ takes sα|U ∩ Uα ∩ Uβ ∈ Γ(U ∩ Uα ∩ Uβ ,On) tosβ |U ∩ Uα ∩ Uβ . In fact, given U, then any system (gαβ ∈ GLn(Rαβ))α,β satisfyingthe two properties above thus defines a locally free O-module of rank n over X thatis trivial on each member of U.

We can express this in the language of sheaf theory: we have a sheaf GLn(O)with values in a group (nonabelian when n > 1) and we then refer to such asystem as a Cech 1-cocycle with values in this sheaf. We denote the collection ofsuch systems by Z1(U,GLn(O)). Note that for n = 1, GLn(O) = O× is a sheafof abelian groups, and then this is an ordinary Cech 1-cocycle. If we choose fora given U a different set of local trivializations h′α : F|Uα → On|Uα, then we canwrite h′α = gαhα for some gα ∈ GLn(Rα) and so the associated cocycle is given byg′αβ := (gβ |Uα∩Uβ)gαβ (g−1

α |Uα∩Uβ). We then say that the two Cech 1-cocycles (gαβ )

and (g′αβ) are cohomologous. In particular, when F is trivial, then we can choose aglobal trivialization h : F ∼= On and use h′α to be the restriction of h to Uα. Theng′αβ is the identity matrix and so it follows that there exist (gα ∈ GLn(Rα))α suchthat gαβ = (gα|Uα ∩Uβ)(g−1

β |Uα ∩Uβ) for all α, β. So the set of cohomology classes,denoted H1(U,GLn(O)), can be understood as the set of isomorphism classes oflocally free sheaves on X of rank n that are trivial on each member of U.

For n > 1 this is just a set with a distinguished element (represented by the triv-ial sheaf), but when n = 1, being cohomologous indeed means that g′αβ(gαβ )−1 =

(gβ |Uα∩Uβ)(g−1α |Uα∩Uβ) is a coboundary and so H1(U,GLn(O)) is then the usual

Cech cohomology group H1(U,O×). The group structure has then a simple inter-pretation: if L and L′ are invertible sheaves on X that are both trivial on everymember of U, then L⊗O L′ is an invertible sheaves with the same property. In fact,local trivializations for both invertible sheaves relative to U determine local trivi-alizations for their tensor product and the Cech 1-cocycle for the latter is just theproduct of the Cech 1-cocycles of L and L′. Thus the group structure on H1(U,O×)


comes from the tensor product. In particular, if we take the direct limit over allcoverings, we find

PROPOSITION-DEFINITION 3.16. The isomorphism classes of the invertible sheaveson X make up a group under tensor product and this group may be identified withH1(X,O×). This group is called the Picard group of X and usually denoted Pic(X)(although Pic(X,O) would be more correct).

PROOF. The preceding argument shows that the group if question can be iden-tified with lim−→U

Hk(U,O×) = H1(X,O×) and according to Proposition 3.12 thelatter maps isomorphically onto H1(X,O×).

Let f : (X,O)→ (X ′,O′) be a morphism of ringed spaces. If L′ is an invertibleO′-module, then f∗L′ = O ⊗f−1O′ L′ is an invertible O-module. Its isomorphismclass only depends on the isomorphism class of L′ and indeed, we thus have defineda group homomorphism Pic(X ′)→ Pic(X).

Cohomological dimension of a noetherian space. Let be given a X is aspace. Suppose Y ⊂ X is a closed subset with complement U . We denote byi : Y ⊂ X and j : U ⊂ X the inclusions. For any abelian sheaf F on X, we have anevident adjunction homomorphism F → i∗i

−1F (we recall that i∗ : AbY → AbXis a left adjoint of i−1 : AbX → AbY ). On the level of stalks, this is the identityFx = Fx when x ∈ Y and the zero map 0x → Fx if x ∈ U .

There is a complementary construction involving U as follows. For any abeliansheaf G on U , we denote by j!G the sheaf on X obtained by extension by zero: itis the sheaf associated to the presheaf that assigns to an open subset V ⊂ U , G(V )in case V ⊂ U and zero otherwise. In particular, the stalk of j!G at x is Gx whenx ∈ U and is 0 if x ∈ Y . This sheaf is more directly characterized by saying thatj!G(V ) is a subgroup of G(V ∩ U), namely the group of s ∈ G(V ∩ U) for whichthere exists neighborhood N of V ∩ Y in V such s|N ∩ U = 0. This is in fact a leftadjoint j! : AbU → AbX for the functor j−1 : AbX → AbU : we have an evidentbijection Hom(j!G,F) ∼= Hom(G, j−1F). In particular we have a natural adjunctionhomomorphism j!j

−1F → F . On the level of stalks, this is identity Fx = Fx whenx ∈ U and 0→ Fx if x ∈ Y .

We thus have obtained an exact sequence

0→ j!j−1F → F → i∗i

−1F → 0.

It is clear from the definition that i∗ sends flasque sheaves to flasque sheaves. Inparticular, it sends a flasque resolution of i−1F to one of i∗i−1F so that Hk(Y, i−1F) =

Hk(X, i∗i−1F) for all k. Hence we have a long exact sequence

· · · → Hk(X, j!j−1F)→ Hk(X,F)→ Hk(Y, i−1F)→ Hk+1(X, j!j

−1F)→ · · ·

We will use this exact sequence in the proof of Grothendieck’s theorem below.Recall that a topological space X is said to be noetherian of dimension ≤ n if

for any strictly descending chain X ⊃ X0 ) X1 ) · · · ) Xm of closed irreduciblesubsets we have m ≤ n.

PROPOSITION 3.17 (Grothendieck). If X is a noetherian space of dimension ≤ n,then for every abelian sheaf F on X we have Hk(X,F) = 0 for k > n.

178 4. APPENDIX

PROOF. We first give the proof in case F is locally finitely generated in the sensethat we can cover X by open subsets U with the property that there exists anepimorphism ZNU F|U for some N . Note that such sheaves make up an abeliansubcategory Modlfg

X of ModX . We then use induction on the dimension on X. Theinduction starts trivially when n = −1, for then X = ∅.

Let F be locally finitely generated sheaf on X. If C1, . . . , Cr are the irreduciblecomponents of X, then each Cs contains an open-dense subset Us such that F|Ui isa constant and hence equal to AsUs , where As := F(Us). We may of course assumethat the Us’s are pairwise disjoint. Let U := ∪sUs and i : Y = X − U ⊂ X. Theexact sequence above becomes

· · · → ⊕s Hk(Cs, js!AsUs)→ Hk(X,F)→ Hk(Y, i−1F)→ · · · ,

where js : Us ⊂ Cs is the inclusion. By induction assumption Hk(Y, i−1F) = 0 fork > n (even for k ≥ n) and so it remains to see that Hk(Cs, js!A

sUs

) = 0 for k > n.For this we use the short exact sequence

0→ js!AsUs → AsCs → is∗A

sYs → 0,

where is : Ys := Cs − Us ⊂ Cs. It gives rise to an exact sequence

· · · → Hk−1(Ys, AsYs)→ Hk(Cs, js!A

sUs)→ Hk(Cs, A

sCs)→ · · ·

A constant sheaf on an irreducible space is clearly flasque and so Hk(Cs, AsCs

) = 0

when k 6= 0. By the induction hypotheses Hk−1(Ys, AsYs

) = 0 for k − 1 > n− 1 andso the desired vanishing follows.

The general result then follows from the observation that an arbitrary abeliansheaf is the union of its locally finitely generated subsheaves and the fact that forabelian sheaves on a noetherian space, taking cohomology commutes with forminga direct limit (see Hartshorne [7], Ch. III Prop. 2.9).

Ext modules. Let (X,O) be a ringed space as before and write R for the ringΓ(X,O). For a O-moduleM, we have covariant functors

HomO(M,−) : ModO →ModO,

HomO(M,−) = Γ(X,HomO(M,−)) : ModO →ModR.

Both are left exact functors, which, since ModO possesses enough injectives, haveright derived functors. We call these the local resp. global Ext functors and in-stead of writing RiHomO(M,−) resp. Ri HomO(M,−) we usually denote themExt iO(M,−) resp. ExtiO(M,−).

REMARKS 3.18. We make some observations (which sometimes boil down tosimple exercises) concerning these modules.

(i) In case X is a singleton, then the datum of O just amounts to giving a ringR and then ModO = ModR. If M corresponds to the R-module M , then bothfunctors coincide with HomR(M,−) whose right derived functors are the familiarmodules ExtiR(M,−).

(ii) If we take M = O, then these two functors are just the identity functorN 7→ N resp. the global section functor N 7→ Γ(X,N ) and hence we find thatExt iO(O,N ) is zero unless i = 0, whereas ExtiO(O,N ) = Hi(X,N ). It also followsthat if M is a free O-module of finite rank, then Ext iO(M,N ) is zero unless i = 0(in which case we get Homi

O(M,N )).


(iii) For a given O-module N , the Exti(−,N ) make up a δ-functor from ModOto ModR. For let N → I• be an injective right resolution. Then for any shortexact sequence of O-modules 0 → M′ → M → M′′ → 0, the short sequence ofcomplexes 0 → HomO(M′′, I•) → HomO(M, I•) → HomO(M′, I•) → 0 is stillexact and so gives rise to a long exact cohomology sequence. This is just

0→ HomO(M′′,N )→ HomO(M,N )→ HomO(M′,N )→ Ext1O(M′,N )→ · · ·

· · · → ExtiO(M′′,N )→ ExtiO(M,N )→ ExtiO(M′,N )→ Exti+1O (M′,N )→ · · ·

It is however not always possible to interpret the Exti(−,N ) as derived functors ofHomO(−,N ).

(iv) Let L be a locally free O-module. Then HomO(L,−) is an exact functorand hence Ext i(L,−) = 0 for all i > 0. This has the following consequence. LetM and N be O-modules and let L• → M be a left resolution by locally free O-modules. Then the Abstract De Rham theorem 2.14 (or rather its dual form) showsthat Ext iO(M,N ) = hi(HomO(L•,N )). Proceeding as in the proof of Proposition2.20 we also find for every O-moduleM natural transformations

Ext iO(M⊗O L,−)⇒ Ext iO(M,HomO(L,−)),

ExtiO(M⊗O L,−)⇒ ExtiO(M,HomO(L,−))

that are isomorphisms when L is locally of finite rank.(v) Let j : U ⊂ X be an open subset and let M,N be O-modules. Then we

have a natural isomorphism j∗HomO(M,N ) ∼= Homj∗O(j∗M, j∗N ), which weregard as an isomorphism of functors

j∗HomO(M,−) ∼= Homj∗O(j∗M, j∗−) : ModO →Modj∗O,

that are parts of δ-packages with T i equal to j∗ Ext iO(M,−) resp. Ext ij∗O(j∗M, j∗−).These δ-packages are universal, for if I is an injective O-module, then for i > 0,Ext iO(M, I) = 0 and, since j∗I is also injective, Ext ij∗O(j∗M, j∗I) = 0. Hence forall i and all O-modules N we have a natural isomorphism

j∗ Ext iO(M,N ) ∼= Ext ij∗O(j∗M, j∗N ).

(vi) Let M and N be pair O-modules. We can regard ExtkO(M,−) as the kthderived functor of the composite functor ΓHom(M,−) and so by Proposition 2.21we have a natural R-homomorphism ExtkO(M,N ) → Γ(X, ExtkO(M,N )). Thatsame proposition tell us that this is an isomorphism if Ext lO(M,N ) is Γ-acyclic forl < k, once we observe that an injective resolution N → I• as constructed inProposition 3.3 has the property that HomO(M, Ik) is flasque for every k.

If we combine this with the observations made in (ii), then it also follows thatwhen L is a locally free O-module of finite rank, then ExtjO(L,N ) = 0 unless j = 0,in which case we getHomj

O(L,N ). So if the O-moduleM admits a right resolutionby such modules L• →M, then (iii) combined with the abstract De Rham theoremimplies that ExtjO(M,N ) = hj(HomO(L•,N )).

(vii) Let L,M,N be O-modules. Then for all i, j ≥ 0 there are natural homo-morphisms

Ext iO(L,M)⊗O ExtjO(M,N )→ Ext i+jO (L,N ),

ExtiO(L,M)⊗R ExtjO(M,N )→ Exti+jO (L,N ).

180 4. APPENDIX

In particular, we find (by taking L = O) natural R-homomorphisms

Hi(X,M)⊗R ExtjO(M,N )→ Hi+j(X,N ).

This is a special case of a much more general result, which is best presented in thesetting of derived categories (which we here avoid). Let us just give the proof forthe global Ext.

LetM→ I•. The point is that replacingM by I• yields the same Ext modules,even when M appears as the first variable: let N → J • be a resolution by injec-tive O-modules and form the complex of O-modules HomO(I•,J •): an element ofHomO(I•,J •) of degree k is a chain map φ : I• → J •(k) and d(φ) : I• → J •(k+1)is the chain map which sends a ∈ Is to dφ(a) − φ(da). Then the chain mapof O-modules M → I• induces a chain map of R-modules HomO(I•,J •) →HomO(M,J •) which induces an isomorphism on cohomology, so that

hj(HomO(I•,J •)) ∼= hj(HomO(M,J •)) = ExtjO(M,N ).

Now observe that we have an evident chain map of R-modules

HomO(L, I•)⊗R HomO(I•,J •)→ HomO(L,J •).(the source is a tensor product of complexes of R-modules) and as is well-known,we then have an induced Kunneth map on cohomology:

hi(HomO(L, I•))⊗R hj(HomO(I•,J •))→ hi+j(HomO(L,J •)).By definition, we may identify hi(HomO(L, I•)) with ExtiO(L, I•) and likewisehi+j(HomO(L,J •)) with Exti+jO (L,N ). The assertion follows.

Bibliography

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