Lecture Notes in Abstract Algebra

Lecture Notes in Abstract Algebra

Lectures by Dr. Sheng-Chi Liu

Throughout these notes, signifies end proof, and N signifies end ofexample.

Table of Contents

Table of Contents i

Lecture 1 Review of Groups, Rings, and Fields 11.1 Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.4 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

Lecture 2 More on Rings and Ideals 52.1 Ring Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Review of Zorn’s Lemma . . . . . . . . . . . . . . . . . . . . . . . 8

Lecture 3 Ideals and Radicals 93.1 More on Prime Ideals . . . . . . . . . . . . . . . . . . . . . . . . 93.2 Local Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.3 Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Lecture 4 Ideals 114.1 Operations on Ideals . . . . . . . . . . . . . . . . . . . . . . . . . 11

Lecture 5 Radicals and Modules 135.1 Ideal Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145.2 Radical of an Ideal . . . . . . . . . . . . . . . . . . . . . . . . . . 145.3 Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

Lecture 6 Generating Sets 176.1 Faithful Modules and Generators . . . . . . . . . . . . . . . . . . 176.2 Generators of a Module . . . . . . . . . . . . . . . . . . . . . . . 18

Lecture 7 Finding Generators 197.1 Generalising Cayley-Hamilton’s Theorem . . . . . . . . . . . . . 197.2 Finding Generators . . . . . . . . . . . . . . . . . . . . . . . . . . 217.3 Exact Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

Notes by Jakob Streipel. Last updated August 15, 2020.

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mailto:[email protected]

TABLE OF CONTENTS ii

Lecture 8 Exact Sequences 238.1 More on Exact Sequences . . . . . . . . . . . . . . . . . . . . . . 238.2 Tensor Product of Modules . . . . . . . . . . . . . . . . . . . . . 26

Lecture 9 Tensor Products 269.1 More on Tensor Products . . . . . . . . . . . . . . . . . . . . . . 269.2 Exactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289.3 Localisation of a Ring . . . . . . . . . . . . . . . . . . . . . . . . 30

Lecture 10 Localisation 3110.1 Extension and Contraction . . . . . . . . . . . . . . . . . . . . . 3210.2 Modules of Fractions . . . . . . . . . . . . . . . . . . . . . . . . . 34

Lecture 11 Exactness of Localisation 3411.1 Exactness of Localisation . . . . . . . . . . . . . . . . . . . . . . 3411.2 Local Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

Lecture 12 Primary Decomposition 3712.1 Local Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . 3712.2 Primary Decomposition . . . . . . . . . . . . . . . . . . . . . . . 38

Lecture 13 More on Primary Decomposition 4013.1 First Uniqueness Theorem . . . . . . . . . . . . . . . . . . . . . . 40

Lecture 14 Ring Extensions 4414.1 Second Uniqueness Theorem . . . . . . . . . . . . . . . . . . . . . 4414.2 Integral Ring Extensions . . . . . . . . . . . . . . . . . . . . . . . 45

Lecture 15 Ring Extensions continued 4715.1 Last Lecture, concluded . . . . . . . . . . . . . . . . . . . . . . . 47

Lecture 16 The Going-Up Theorem 4816.1 Integral Dependence . . . . . . . . . . . . . . . . . . . . . . . . . 4816.2 The Going-Up Theorem . . . . . . . . . . . . . . . . . . . . . . . 50

Lecture 17 The Going-Down Theorem 5117.1 The Going-Down Theorem . . . . . . . . . . . . . . . . . . . . . . 51

Lecture 18 Valuation Rings 5418.1 Between Rings and Fields . . . . . . . . . . . . . . . . . . . . . . 54

Lecture 19 Chain Conditions 5719.1 Valuation Rings and Integral Closures . . . . . . . . . . . . . . . 5719.2 Chain Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

Lecture 20 Noetherian Rings 6120.1 When are Submodules Noetherian? . . . . . . . . . . . . . . . . . 61

Lecture 21 More on Noetherian Rings 6321.1 Noetherian Rings Have Decomposable Ideals . . . . . . . . . . . 63

Lecture 22 Artinian Rings 66

TABLE OF CONTENTS iii

22.1 Length of Modules . . . . . . . . . . . . . . . . . . . . . . . . . . 66

Lecture 23 Structure of Artinian Rings 7023.1 Dimension of Ring . . . . . . . . . . . . . . . . . . . . . . . . . . 7023.2 Structure of Artinian Rings . . . . . . . . . . . . . . . . . . . . . 7123.3 Discrete Valuation Rings and Dedekind Domains . . . . . . . . . 73

Lecture 24 Discrete Valuation Rings 7424.1 Connections between Discrete Valuation Rings and Noetherian

Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

Lecture 25 Dedekind Domains 7625.1 Toward the Definition of Dedekind Domains . . . . . . . . . . . . 7625.2 Dedekind Domain . . . . . . . . . . . . . . . . . . . . . . . . . . 7825.3 Completions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

Lecture 26 Graded Rings and Filtrations 7926.1 Graded Rings and Modules . . . . . . . . . . . . . . . . . . . . . 7926.2 Inverse Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

Lecture 27 Closure 8327.1 Closure and Completion . . . . . . . . . . . . . . . . . . . . . . . 8327.2 Consequences of the Krull Intersection Theorem . . . . . . . . . 84

References 86

Index 87

REVIEW OF GROUPS, RINGS, AND FIELDS 1

Lecture 1 Review of Groups, Rings, and Fields

1.1 Groups

Definition 1.1.1 (Group). The set G equipped with a binary operation ·,together denoted (G, ·), is a group if

(i) the operation is associative, i.e. (g1 ·g2) ·g3 = g1 ·(g2 ·g3) for all g1, g2, g2 ∈G;

(ii) there exists an identity element e ∈ G such that e · g = g · e = g for all g;and

(iii) for every g ∈ G there exists an inverse denoted g−1 such that g · g−1 =g−1 · g = e.

Note that these are the properties we require if we wish to be able to solve linearequations of the form ax = b. Note also that · being a binary operation veryimportantly implies that G is closed under ·, i.e., g1 · g2 ∈ G for all g1, g2 ∈ G.

Definition 1.1.2 (Abelian group). A group (G, ·) is called abelian if g1 · g2 =g2 · g1 for all g1, g2 ∈ G.

When a group is abelian we typically use + to denote the group operationand 0 to denote the identity element. Moreover we write −g instead of g−1.

Remark 1.1.3. The groups we consider in this course will all be abelian (thehint’s in the name—commutative algebra).

Examples 1.1.4. Some examples of groups are Z, Q, R, or C with the usual ad-dition, the symmetric group of order n, (Sn, ◦), polynomial rings, e.g. (Z[x],+),(Z/nZ,+), etc. N

1.2 Rings

Definition 1.2.1 (Ring). The set R equipped with two binary operations +and ·, denoted (R,+, ·), is a ring if

(i) (R,+) is an abelian group;

(ii) it is closed under ·, i.e. r1 · r2 ∈ R for all r1, r2 ∈ R;

(iii) multiplication is associative, meaning that (r1 · r2) · r3 = r1 · (r2 · r3) forall r1, r2, r3 ∈ R; and

(iv) multiplication is distributive over addition, i.e. r1 ·(r2+r3) = r1 ·r2+r1 ·r3

and (r2 + r3) · r1 = r2 · r1 + r3 · r1 for all r1, r2, r3 ∈ R.

Remark 1.2.2. We again only consider commutative rings in this course. More-over all rings we consider will be rings with identity, meaning that multiplicationhas a neutral element. Therefore we have the additional properties:

(v) r1 · r2 = r2 · r1 for all r1, r2 ∈ R; and

(vi) there exists 1 ∈ R such that 1 · r = r · 1 = r for all r ∈ R.

Date: August 22, 2017.


Remark 1.2.3. As already indicated in the note, + is usually called additionand · is usually called multiplication, as per usual. In addition we often writer1 · r2 as r1r2 for convenience.

Remark 1.2.4. Finally note that when we use the term ring in this course wewill always mean a commutative ring with identity.

Examples 1.2.5. Some examples of rings are (Z,+, ·) (as well as Q, R, C, etc.)and (Z[x],+, ·). N

1.3 Fields

Definition 1.3.1 (Field). A field F is a commutative division ring , i.e. Fis a commutative ring with identity and for each a 6= 0, a ∈ F there exists anelement a−1 ∈ F such that aa−1 = 1.

Example 1.3.2. Note that the ring (Z,+, ·) referred to previously is not a fieldsince it does not include inverses for all elements but −1 and 1. On the otherhand Q, R, and C with the same operations are fields. N

Example 1.3.3. Let k be a field and let M be a maximal ideal of k[x]. Thenk[x]/M is a field.

This gives us an easy means of generating fields of prime power number ofelements, since k = Z/pZ, p prime is of prime order. Now if we let M = (f)where f is an irreducible polynomial in k[x] of degree n, then |k[x]/M | = pn. N

1.4 Motivation

Commutative algebra is the study of the structures of commutative rings andof modules, with their applications in algebraic geometry and algebraic numbertheory.

Example 1.4.1 (Algebraic number theory). The structure of the ring of in-tegers. First note that Z ⊂ Q has the property that for n ∈ Z we haven = ±pα1

1 pα22 · . . . · p

αk

k , with pi being primes. In particular this factorisa-tion is unique. This means that Z is a unique factorisation domain. Moreoverall ideals in Z, which are of the form nZ, are principal. This means that Z is aprincipal ideal domain.

On the other hand consider Z[√−5], a group of algebraic integers. In this

field we don’t have unique factorisation, since for example 6 = 2 · 3 = (1 +√−5)(1−

√−5), all factors of which are irreducible. Furthermore not all ideals

are principal, consider e.g. I = (2, 1 +√−5). N

Example 1.4.2 (Algebraic geometry). The structure of the set of solutions to{x+ 2y + 3z = 0

5x+ y + 2z = 0

in R3.Let V ⊂ R3 be the set of solution. This has certain structure:

(i) v1 + v2 ∈ V for all v1, v2 ∈ V ;


(ii) (v1 + v2) + v3 = v1 + (v2 + v3) for all v1, v2, v3 ∈ V ;

(iii) 0 ∈ V since the planes go through the origin;

(iv) if v ∈ V , then −v ∈ V ;

(v) for r ∈ R we have r · v ∈ V for all v ∈ V ;

(vi) (r1 + r2) · v = r1 · v + r2 · v for all r1, r2 ∈ R and v ∈ V ;

(vii) r · (v1 + v2) = r · v1 + r · v2 for all r ∈ R and v1, v2 ∈ V ;

(viii) 1 · v = v for all v ∈ V .

Now this seems familiar: observations (i) through (iv) are the axioms of a group,meaning that solutions to these equations form a group, and observations (v)through (viii) are the axioms of a vector field! N

Remark 1.4.3. Of course for an abstract definition of vector space we do notneed V to specifically be a subset of R3.

Remark 1.4.4. We only use the addition and multiplication; we did not use thedivision in the definition. Therefore we may replace the field R with a ring A.If we do this we get the definition of a module instead of a vector space.

Example 1.4.5. Replace the linear equations with polynomial equations ofhigher degree. Let k be a field and let A := k[x1, x2, . . . , xn]. Further letf(x1, x2, . . . , xn) ∈ A.

Now calling V (f) := { (a1, a2, . . . , an) ∈ kn | f(a1, a2, . . . , an) = 0 } is thesolution set to f(x1, x2, . . . , xn) = 0.

The structure of V (f) can be recovered by the quotient ring A/I whereI = (f) ⊂ A. N

Example 1.4.6. Consider f(x, y) = x ∈ R[x, y]. Then V (f) = { (0, y) |y ∈ R }.On the other hand if g(x, y) = x2 we have V (g) = { (0, y) | y ∈ R }. These

look the same!However the structure is subtly different, since for one of them we have a

double root. Indeed R[x, y]/(x) has no nontrivial idempotent elements, whereasR[x, y]/(x2) has some, for instance x. N

Exercise 1.4.7. Let p = (a1, a2, . . . , an) ∈ Rn. Let us now define the mapφ : R[x1, x2, . . . , xn]→ R by φ(f) = f(p). Then

(i) φ is a ring homomorphism, and

(ii) kerφ is the maximal ideal Mp = (x1 − a1, x2 − a2, . . . , xn − an).

This gives a correspondence between maximal ideals in A = R[x1, x2, . . . , xn]and the points of Rn. In fact this correspondence is one-to-one by the Nullstel-lensatz.

Solution. First recall that for two rings R and S, a mapping φ : R→ S is a ringhomomorphism if it preserves the ring structure, i.e.

(i) φ(a+ b) = φ(a) + φ(b) for all a, b ∈ R,

(ii) φ(ab) = φ(a)φ(b) for all a, b ∈ R, and


(iii) φ(1R) = 1S .

To see the first property, recall that the polynomial ring R[x1, x2, . . . , xn] isabelian, so by distributivity and commutativity we have for any two polynomialsf and g that

φ(f + g) = (f + g)(P ) = f(P ) + g(P ) = φ(f) + φ(g)

andφ(fg) = (fg)(P ) = f(P )g(P ) = φ(f)φ(g).

For the final property simply note that φ(1) = 1 since 1 does not depend on xand so evaluates to itself regardless of P .

To show that kerφ is the idealMP = (x1−a1, x2−a2, . . . , xn−an) we considertwo steps: it is clear that MP ⊂ kerφ since MP consists only of polynomialsthat can be written on the form

(x1 − a1)f1 + (x2 − a2)f2 + . . .+ (xn − an)fn (1.4.1)

for polynomials f1, f2, . . . , fn ∈ A, and these of course all evaluate to 0 inP = (a1, a2, . . . , an).

That the other inclusion is also true requires a little more work. Since all ofxi − ai are monic, it means that we have a division algorithm for them. Thismeans that we can write any f ∈ A as

f(x1, x2, . . . , xn) = p1(x1, x2, . . . , xn)(x1 − a1) + r1(x2, x3, . . . , xn),

and in turn

r1(x2, x3, . . . , xn) = p2(x2, x3, . . . , xn)(x2 − a2) + r2(x3, x4, . . . , xn),

and so on untilrn−1(xn) = pn(xn)(xn − an) + rn,

where finally rn is a constant.This means that we may write

f(x1, x2, . . . , xn) = p1(x1, x2, . . . , xn)(x1 − a1) + p2(x2, x3, . . . , xn)(x2 − a2)+

+ . . .+ pn(xn)(xn − an) + rn,

and if f ∈ kerφ then the left-hand side must be 0, which forces the constant rnin the right-hand side to be 0 as well, and so if f ∈ kerφ we can write it on theform (1.4.1), and therefore kerφ ⊂MP . Thus kerφ = MP .

To see that this ideal is indeed maximal, note that since φ is surjective (itevaluates constants to themselves, so certainly it is), by the first isomorphismtheorem we have

A/ kerφ ∼= R.

But for any commutative ring with identity A, an ideal I of A is maximal if andonly if A/I is a field.

Thus since A/ kerφ ∼= R is indeed a field, kerφ = MP is a maximal ideal. �

MORE ON RINGS AND IDEALS 5

Lecture 2 More on Rings and Ideals

2.1 Ring Fundamentals

Definition 2.1.1 (Ring homomorphism). Let A and B be two commutativerings with identity. A ring homomorphism is a map φ : A → B such thatthe ring structure is preserved, i.e.

(i) φ(x+ y) = φ(x) + φ(y) for all x, y ∈ A,

(ii) φ(xy) = φ(x)φ(y) for all x, y ∈ A,

(iii) φ(1A) = 1B .

Definition 2.1.2 (Ideal). An ideal of a ring A is a subset I ⊂ A such that

(i) I is an additive subgroup,

(ii) aI ⊂ I for all a ∈ A.

The reason we might want to construct such a thing is to guarantee that A/Iis a ring. Cf. quotients of groups, where we wish for the thing we divide by tobe normal.

Remark 2.1.3. Let I ⊂ A be an ideal of A. Then A/I inherits a ring structurefrom A, and we call A/I a quotient ring . Secondly if 1 ∈ I, then I = (1) = A,and the other way around.

In fact if I contains any unit, then I is the whole ring.

Definition 2.1.4 (Principal ideal). An ideal I generated by one element x iscalled a principal ideal , denoted I = (x).

Proposition 2.1.5. Let φ : A→ B be a ring homomorphism. Then

(i) kerφ is an ideal of A;

(ii) if I ⊂ A is an ideal, then the map φ : A→ A/I defined by φ(a) = a+ I isa surjective homomorphism, with kerφ = I;

(iii) the map ϕ : { ideals in A/I } → { ideals in A containing I } defined byϕ(J) := φ−1(J) is a one-to-one correspondence.

Proof. For (i), take two elements x and y in the kernel of φ, meaning thatφ(x) = φ(y) = 0. Then x+y is also in the kernel, since φ(x+y) = φ(x)+φ(y) =0 + 0 = 0. In addition φ(ax) = φ(a)φ(x) = φ(a)0 = 0 for all a ∈ A. Ergo kerφis an ideal of A.

Next for (ii) we need to verify both that this φ is a homomorphism and thatit is surjective, and finally that its kernel is I.

Firstly taking x, y ∈ A we have φ(x+ y) = (x+ y) + I = (x+ I) + (y+ I) =φ(x) + φ(y), and likewise for φ(xy). Moreover φ(1A) = 1A + I = 1A/I .

To see that this is surjective, note that all elements of A/I can be writtenon the form a+ I for some a ∈ A, and it is clear.

The kernel is clearly I itself since the 0 element in A/I is I itself, andtherefore the coset representative can be anything from I and nothing else.

Finally for (iii): Later.



Definition 2.1.6 (Zero-divisor). An element x 6= 0 in a ring A is called azero-divisor if there exists some A 3 y 6= 0 such that xy = 0.

Example 2.1.7. Consider the ring Z/6Z. In this ring we have 2 ·3 = 0, despite2 6= 0, 3 6= 0. N

Definition 2.1.8 (Integral domain). A ring with no zero-divisors is called anintegral domain .

From this, the key property of integral domains is that we get cancellation laws.That is, if ax = ay with a 6= 0, then this implies that x = y since a(x− y) = 0,and since there are no zero divisors we must have x− y = 0.

Definition 2.1.9 (Nilpotency). An element x in a ring A is called nilpotentif xn = 0 for some n > 0.

Hence a nilpotent element is a special kind of zero-divisor, since xn = x ·xn−1 =0, but of course not all zero divisors are nilpotent.

Example 2.1.10. Consider the ring Z/n2Z. Then n2 = 0 despite n 6= 0.Note moreover, in view of the previous claim, that the 2 in Z/6Z from the

previous example will never be 0 even if we take greater and greater powers ofit. N

Definition 2.1.11 (Unit). An element x in a ring A is called a unit if thereexists some y ∈ A such that xy = 1. We write y = x−1.

Definition 2.1.12 (Field). A field is a ring in which all nonzero elements areunits.

Proposition 2.1.13. Let A be a ring. The following are equivalent:

(i) A is a field.

(ii) The only ideals in A are { 0 } and A.

(iii) Every homomorphism from A onto (meaning it’s surjective) a nonzeroring B is one-to-one.

Proof. For (i) implying (ii), clearly { 0 } is an ideal regardless, and any idealdifferent from the trivial ideal would have to contain a nonzero element, and allnonzero elements of a field are units, so the ideal is the whole field.

For (ii) implying (iii), consider something. . .Finally for (iii) implying (i), let x ∈ A be any element. Define ϕ : A→ A/(x)

by a 7→ a + (x). This is surjective, as we’ve seen before, and so by (iii) ϕ isinjective.

Therefore kerϕ = (0) since only one element can map to 0 and that mustthen be 0, and therefore (x) = (0) and thus x = 0.

Definition 2.1.14 (Prime ideal). An ideal P ⊂ A is a prime ideal if p 6= (1)and if xy ∈ P then x ∈ P or y ∈ P .

Cf. how if p | xy, with p a prime, then p | x or p | y.

Definition 2.1.15 (Multiplicative set). A subset S of a ring A is called amultiplicative set if 1 ∈ S and for all x, y ∈ S we have xy ∈ S.


Exercise 2.1.16. An ideal P is prime if and only if S = A\P is a multiplicativeset.

Solution. First let P be a prime ideal. We then need to show that 1 ∈ S, whichis clear by definition: since a prime ideal can’t be (1) it mustn’t contain 1, andso 1 is in the complement.

Moreover let x, y ∈ S. Now suppose xy 6∈ S. This means that xy ∈ Pinstead, and since P is a prime ideal, this implies that x ∈ P or y ∈ P . Butif this is the case one of x or y (or maybe both) belong to both P and A \ P ,which is clearly impossible, since they are disjoint.

Therefore we have a contradiction and so xy ∈ S, and therefore S is amultiplicative set.

For the converse, let S = A \ P be a multiplicative set. Suppose xy ∈ P ,meaning that xy 6∈ S. Now if both x and y are in S, then xy ∈ S = A \P sinceS is multiplicative, so that cannot be. Consequently at least one of x and y is inP , and so P is prime by definition. (We also know P 6= (1) since by definition1 ∈ S.) �

Definition 2.1.17 (Maximal ideal). An ideal M ⊂ A is a maximal ideal ifM 6= A and there exists no ideal I such that M ( I ( A.

Proposition 2.1.18. (i) P is a prime ideal if and only if A/P is an integraldomain.

(ii) M is a maximal ideal if and only if A/M is a field.

Remark 2.1.19. Since all fields are integral domains, but not the converse, wesee from this that all maximal ideals are prime ideals, but not the other wayaround.

Proof. For (i), suppose P is a prime ideal, and consider the quotient ring A/P ,consisting of a+ P for a ∈ A. Now suppose (a+ P )(b+ P ) = ab+ P = 0 + P ,meaning that ab ∈ P . We need to show that either a+P = 0+P or b+P = 0+Pso that there are no zero-divisors, or in other words a ∈ P or b ∈ P .

But this is true directly by the definition of P being a prime ideal!For the opposite directions we suppose that A/P is an integral domain,

meaning that if (a+ P )(b+ P ) = ab+ P = 0 + P (meaning that ab ∈ P ), theneither a ∈ P or b ∈ P .

This is the definition of P being prime again, with the caveat that we haven’tshowed that P 6= (1). But if P = (1), then A/P is (isomorphic to) the zeroring, and the zero ring isn’t an integral domain.

Next for (ii), we know that there exists a bijection between ideals of A/Mand ideals of A containing M . But reading that last part carefully, if M ismaximal there is no ideal of A containing M apart from A and M themselves,meaning that there are exactly two ideals of A/M . Now since a ring is alwaysits own ideal and all rings also have the trivial ideal, if A/M has exactly twoideals it can only be A/M and { 0 }, and therefore it is a field.

The same argument works in reverse: if A/M is a field, then it has only{ 0 } and A/M as ideals, and so by the correspondence there is no ideal strictlybetween M and A, and so M is maximal.

Definition 2.1.20 (Spectra). (i) The prime spectrum or Spec of a ring Ais the set of all prime ideals of A.


(ii) The maximal spectrum of m-spec of a ring A is the set of all maximalideals of A.

Example 2.1.21. Let k be a field. Then Spec(k) = { 0 } = m-spec(k). N

Example 2.1.22. We have Spec(Z) = { 0 } ∪ { (p) | p prime }. N

Example 2.1.23. Let k be a field and let A = k[x]. Then Spec(A) = { 0 } ∪{ (f) | f irreducble polynomial in A }, since k is a field meaning that A = k[x]is a Euclidean domain. N

2.2 Review of Zorn’s Lemma

Let Σ be a partially ordered set and let S ⊂ Σ be a totally ordered subset.An upper bound of S is an element u ∈ Σ such that s < u for all s ∈ S.A maximal element of Σ is an element m ∈ Σ such that m < s does not

hold for any s ∈ Σ.

Example 2.2.1. Let Σ = {A | A ⊂ R }, ordered by inclusion. Then S ={ (−m,m) |m ∈ Z+ } is a totally ordered subset. N

Lemma 2.2.2 (Zorn’s lemma). Let Σ 6= ∅ be a partially ordered set. Supposeany totally ordered subset S ⊂ Σ has an upper bound in Σ. Then Σ has amaximal element.

Remark 2.2.3. Zorn’s lemma, the axiom of choice, and the well-ordering prin-ciple are all equivalent.

Theorem 2.2.4 (Existence of maximal ideal). Let A be a ring and I ( A bean ideal. Then there exists a maximal ideal M in A containing I.

Proof. Let Σ be the set of ideals J ( A containing I, ordered by inclusion.Then Σ 6= ∅ since at least I ∈ Σ.

For any totally ordered subset S = { Jλ | λ ∈ Λ } ⊂ Σ define J∗ =⋃λ∈Λ

Jλ.

Two points of order: J∗ 6= A since none of Jλ contain 1 (otherwise they’d beall of A, which is not true by assumption) and secondly J∗ is an ideal since Jλare totally ordered, ergo the union is one of the elements itself!

Now J∗ is clearly an upper bound of S since it contains all of S.Therefore by Zorn’s lemma, Σ has a maximal element M , and thus M is a

maximal ideal containing I.

Corollary 2.2.5. Every non-unit element of A is contained in a maximal ideal.

Proof. If a is the non-unit element, consider the maximal ideal containing theideal (a). By the previous theorem this maximal ideal exists.

Corollary 2.2.6. Let A be a ring and let A× be the set of units in A. ThenA = A× t

⋃λ

Mλ, where Mλ are the maximal ideals and t denotes a disjoint

union.

IDEALS AND RADICALS 9

Lecture 3 Ideals and Radicals

3.1 More on Prime Ideals

Theorem 3.1.1. Let A be a ring and S a multiplicative subset of A. Let I bean ideal of A with I ∩ S = ∅ (i.e. they are disjoint). Then there exists a primeideal P of A containing I, and P ∩ S = ∅.

Remark 3.1.2. So with I = (0), any multiplicative set yields a prime ideal.

Proof. Let Σ = { J ideal of A | J ⊃ I, J ⊃ S = ∅ }, which is partially orderedby inclusion. Since each totally ordered subset is a chain, they have a maximalelement which is also an ideal, and so we may apply Zorn’s lemma, which tellsus that Σ has a maximal element P .

We now claim that P is prime. To see this, take a, b 6∈ P , in which case weneed to show that ab 6∈ P .

Consider the ideals J1 = P + (a) and J2 = P + (b), where J1, J2 ( P sincea, b 6∈ P . This implies that Ji ∩ S 6= ∅ for i = 1, 2. Now suppose p + ac1 ∈ Sand q + bc2 ∈ S for p, q ∈ P and c1, c2 ∈ A. Then

S 3 (p+ ac1)(q + bc2) = pq + pbc2 + ac1q︸︷︷︸∈P

+abc1c2.

Thus if ab ∈ P this product is in P ∩ S, but this is a contradiction sinceP ∩ S = ∅. Therefore ab 6∈ P and so P is prime.

3.2 Local Rings

Recall that we can decompose any ring A as A = A× t (⋃λmλ) where mλ are

all the maximal ideals of A. We are interested in the case when A = A× tM ,i.e. A has exactly one maximal ideal.

Definition 3.2.1 (Local ring). A ring A with exactly one maximal ideal M iscalled a local ring .

The field k = A/M is called the residue field of A.

Proposition 3.2.2. The following are equivalent:

(i) A is a local ring (i.e. has exactly one maximal ideal).

(ii) The set of all non-units of A form an ideal.

(iii) A has a maximal ideal M such that 1 +M ⊂ A×, i.e. 1 + x ∈ A× for allx ∈M .

Proof. For (i) implying (ii), we remark that the set of all non-units is M sincewe know by assumption that A \ A× = M , which is a maximal ideal, so it iscertainly an ideal.

For (ii) implying (iii), note that 1 + x 6∈ A× implies that 1 + x ∈M , wherewe let M be the set of all non-units, and x ∈ M implies that 1 ∈ M , which is


IDEALS AND RADICALS 10

a unit so 1 + x ∈ A×. This M is an ideal by assumption, and it is maximal bythe decomposition.

Finally for (iii) implying (i), we need to show that for any x ∈ A\M , x is aunit. Since M is maximal, the ideal generated by M and x is A (since the onlyideal containing M and being bigger than M must be A, by the maximality ofM). This implies that 1 = xy + t for some y ∈ A and y ∈M , which rearrangedgives xy = 1− t ∈ 1 +M ⊂ A×, whereby x is a unit.

Definition 3.2.3 (Semi-local ring). A ring with only a finite number of maximalideals is called semi-local .

Example 3.2.4. Consider

Z(2) = { q ∈ Q | q = a/b, a, b ∈ Z, gcd(a, b) = 1, 2 - b }.

In there q = a/b is a unit if and only if 2 - a, meaning that q−1 = b/a ∈ Z(2).Hence the set of all non-units of Z(2) is the set { a/b ∈ Z(2) | 2 | a }, and

1 +a

b=b+ a

b

so 2 - b + a implies that 1 + a/b is a unit. Therefore Z(2) is a local ring withmaximal ideal 2Z(2), so

Z(2)

2Z(2)

∼=Z2Z

= F2.

N

The above is a special case of what is called localisation.

3.3 Radicals

Definition 3.3.1 (Nilradical). Let A be a ring. The set of all nilpotent elementsof A is called the nilradical , denoted R or nilrad(A).

We will show in a moment that, interesting, this is the intersection of all primeideals of A.

Proposition 3.3.2. Let A be a ring. Then nilrad(A) is an ideal and the quotientA/ nilrad(A) has no nonzero nilpotent elements.

Proof. If x, y ∈ nilrad(A), meaning that xn = 0 and ym = 0 for some n,m > 0,then (x + y)n+m = 0, whereby x + y ∈ nilrad(A). Similarly if a ∈ A andx ∈ nilrad(A), then (ax)n = anxn = an · 0 = 0, so ax ∈ nilrad(A). Hencenilrad(A) is an ideal of A.

Now let x ∈ A/ nilrad(A), and suppose xn, implying that xn = 0. Thismeans that xn ∈ nilrad(A), whence (xn)m = 0, or xnm = 0, so x ∈ nilrad(A),meaning that x = 0 ∈ A/nilrad(A).

Proposition 3.3.3. Let A be a ring. Then

nilrad(A) =⋂

P∈Spec(A)

P.

IDEALS 11

Proof. Let I =⋂

P∈Spec(A)

P . We claim that nilrad(A) ⊂ I. If x ∈ nilrad(A),

then xn = 0 ∈ P for every P ∈ Spec(A), by P being prime. Therefore x ∈ Pfor all P ∈ Spec(A), and so nilrad(A) ⊂ I.

For the other inclusion, note that x ∈ I implying x ∈ nilrad(A) is equivalentwith x 6∈ nilrad(A) implying x 6∈ I. Therefore it suffices to show that forx ∈ nilrad(A), there exists a prime ideal P such that x 6∈ P .

Consider {xn | n ≥ 0 }, a multiplicative set. By the first theorem of thelecture, since (0) ∩ S = ∅ there exists a prime ideal P such that P ∩ S = ∅.Now 0 6∈ S since x assumed not nilpotent, and therefore x 6∈ P , since x ∈ S.

Example 3.3.4. By this intersection, nilrad(Z) = { 0 }. N

Definition 3.3.5 (Jacobson radical). Let A be a ring. Then

J(A) =⋂

m∈m-spec(A)

m

is called the Jacobson radical of A.

We can classify the Jacobson radical quite interestingly:

Proposition 3.3.6. Let A be a ring. Then x ∈ J(A) if and only if 1−xy ∈ A×for every y ∈ A.

Proof. For the left implication, suppose 1− xy is not a unit. Then 1− xy ∈Mfor some maximal ideal M , by partition. Since x ∈ J(A) ⊂M , this implies that1 ∈M , which is impossible for a maximal ideal.

For the right implication, suppose x 6∈ M for some maximal ideal M . Notethat the ideal generated by M and x is A since M is maximal. Thereforexy+u = 1 for some y ∈ A and u ∈M , which rearranged yields u = 1−xy ∈ A×by assumption. Therefore M = A, which is impossible!

Remark 3.3.7. In a local ring, J(A) is not very useful, since J(A) = M . Notealso that 1− xy ∈ A× in general rings, versus 1 + x ∈ A× in local rings.

Lecture 4 Ideals

4.1 Operations on Ideals

Let I, J be two ideals in a ring A. We define

I + J := {x+ y | x ∈ I, y ∈ J }

to be the smallest ideal containing I and J .Similarly we define IJ to be the ideal generated by xy for all x ∈ I and

y ∈ J .

Example 4.1.1. Let A = Z. Since this is a principal ideal domain, we haveI = (m) and J = (n) for some m,n ∈ Z. Then I+J = (k) where k = gcd(m,n),IJ = (mn), and I ∩ J = (h), where h = lcm(m,n). N


IDEALS 12

Note that in the above example, I ∩J = IJ if and only if m and n are relativelyprime. This gives us a general definition of the notion of coprime ideals:

Definition 4.1.2 (Coprime). Two ideals I and J of a ring A are said to becoprime if I + J = A, in which case I ∩ J = IJ .

Let A be a ring and I1, I2, . . . , In be ideals of A. Define the map

φ : A→n∏i=1

A

Ii

by φ(x) = (x+ I1, x+ I2, . . . , x+ In).This is a ring homomorphism, and clearly it’s injective precisely if kerφ =

n⋂i=1

Ii = (0).

This gives us another generalisation of a familiar concept from elementarynumber theory:

Proposition 4.1.3 (Chinese remainder theorem). Let A be a ring and Ii andφ be defined as above. Then

(i) φ is surjective if and only if Ii and Ij are coprime for all i 6= j,

(ii) φ is injective if and only ifn⋂i=1

Ii = (0).

Proof. For the forward direction of (i) we need to show that I1 and I2 arecoprime, given that φ is surjective. Now surjectivity implies that there existssome x ∈ A such that φ(x) = (1, 0, 0, . . . , 0), which means that x ≡ 1 (mod I1)and x ≡ 0 (mod I2), so

1 = (1− x)︸︷︷︸∈I1

+ x︸︷︷︸∈I2

∈ I1 + I2,

hence I1 + I2 = A since it contains a unit (1 in particular), and therefore I1 andI2 are coprime. Clearly the same general argument holds for any pair Ii, Ij , fori 6= j.

For the converse it suffices to show that there exists some x ∈ A such thatφ(x) = (1, 0, . . . , 0). Since I1 + Ii = (1) = A for each i = 2, 3, . . . , n, there existssome ui + vi = 1 with ui ∈ I1 and vi ∈ Ii.

Now take

x =

n∏i=1

(1− ui) =

n∏i=2

vi

where the latter is in Ik for all k = 2, 3, . . . , n. Therefore x ≡ 1 (mod I1) andx ≡ 0 (mod Ik), for k ≥ 2, whereby φ(x) = (1, 0, . . . , 0).

Since the same thing can be constructed for a 1 in any position, and sincethey generate the whole space, we have surjectivity.

For (ii) simply note that kerφ =k⋂i=1

Ii, and by a previous proposition φ is

injective if and only if kerφ = (0).

RADICALS AND MODULES 13

Proposition 4.1.4. (i) Let P1, P2, . . . , Pn be prime ideals. Let I ⊂n⋃i=1

Pi be

an ideal. Then I ⊂ Pi for some i.

(ii) Let I1, I2, . . . , In be ideals. Suppose there exists a P ⊃n⋂i=1

Ii be a prime

ideal. Then P ⊃ Ii for some i. Hence if P =n⋂i=1

Ii, then P = Ii for some

i.

Proof. Next time!

Lecture 5 Radicals and Modules

First we present the proof of the proposition stated last lecture.

Remark 5.0.1. Note first of all that if I ⊂ (J1∪J2), all being ideals, then I ⊂ J1

or I ⊂ J2, even if J1 and J2 aren’t prime.

Proof. (i) is equivalent to saying that if I 6⊂ Pi for every i, then I 6⊂n⋃i=1

Pi.

We prove this by induction on n. The case of n = 1 is trivially true, and soassume the statement true for n− 1.

For n, consider choose an element x1 that isn’t in the union of P2∪P2∪ . . .∪Pn, with x1 ∈ I. Similarly choose an element x2 in the union of all Pi exceptP2, with x2 ∈ I, and so on, until xn in the union of all Pi but the last one, withxn ∈ I as well.

Now if xi 6∈ Pi for some i, then xi 6∈ P1 ∪ P2 ∪ . . . ∪ Pn. Hence assume thatxi ∈ Pi for all i = 1, 2, . . . , n.

Consider

y =

n∑i=1

x1x2 . . . xi−1xi+1 . . . xn.

Then y ∈ I since by construction all xi ∈ I. Now consider y modulo Pi,

y ≡ x1x2 . . . xi−1xi+1 . . . xn 6≡ 0 (mod Pi)

since that is the only summand not containing xi, and all factors in that sum-mand don’t belong to Pi by construction. Moreover Pi is a prime ideal, so if aproduct is in it, one factor must be, but none of the factors are so the productcan’t be.

Ergo, y 6∈ Pi for every i = 1, 2, . . . , n. This therefore implies that I 6⊂n⋃i=1

Pi.

(ii) Assume P 6⊃ I for all i = 1, 2, . . . , n. Then there exists some xi ∈ Iiwith xi 6∈ P for every i = 1, 2, . . . , n.

Now considern∏i=1

xi ∈n⋂i=1

Ii ⊂ P

since Ii are ideals contained in P , and since P is prime, one of the factors ximust be in P , which is a contradiction.

Date: September 5, 2017.


Finally, if

P =

n⋂i=1

Ii,

then

Ii ⊂ P =

n⋂i=1

Ii ⊂ Ii

for some i, meaning that P = Ii.

5.1 Ideal Quotients

Definition 5.1.1 (Ideal quotient). Let I and J be ideals in a ring A. Theirideal quotient is

(I : J) = {x ∈ A | xJ ⊂ I }.

Proposition 5.1.2. The ideal quotient (I : J) is an ideal of A.

Proof. Take an element x ∈ (I : J), meaning that xJ ⊂ I. Then xJ are allelements of an ideal, and ideals are closed under multiplication from any elementa ∈ A, so axJ ⊂ I as well.

Moreover let y ∈ (I : J). Then x + y ∈ (I : J) since (x + y)J ⊂ I since xJand yJ are.

Definition 5.1.3 (Annihilator). Given an ideal I in a ring A, the annihilatorof I is

Ann(I) = (0 : I) = {x ∈ A | xI = 0 }.

Note for the record that if J = (x) is a principal ideal, then we will write(I : J) = (I : x).

Remark 5.1.4. The set of all zero divisors in A is

F =⋃x 6=0

Ann(x).

Example 5.1.5. Let A = Z, and take, say, I = (144) and J = (12). Then(I : J) = (144 : 12) = (12) = 12Z.

In general, if I = (m) and J = (n), then (I : J) = (k) with k = m/ gcd(m,n).N

5.2 Radical of an Ideal

Definition 5.2.1 (Radical). Let I be an ideal of A. The radical of I is

rad(I) = {x ∈ A | xn ∈ I for some n > 0 }.

Some authors prefer the notation r(I) of√I.

Example 5.2.2. We have nilrad(A) = rad(0).Moreover by the natural homomorphism φ : A → A/I. Then since this is

surjective,rad(I) = φ−1(nilrad(A/I))


is well-defined, and we are taking pre-images of elements that are all nilradical inthe quotient, they are in I for the same power that made them 0 in the quotient.By a proposition of homomorphisms from the beginning of this course, this isan ideal. N

Definition 5.2.3 (Radical ideal). An ideal I of a ring A is called a radicalideal if I = rad(I).

Proposition 5.2.4. Given a ring A with an ideal I, we have

rad(I) =⋂

P∈Spec(A)P⊃I

P.

Proof. Let φ : A → A/I be the natural homomorphism. Note that the primeideals of A containing I are exactly the ideals φ−1(Q) with Q prime ideal inA/I (this since ideals are one-to-one, and the homomorphism preserves multi-plication, and so preserved primality of ideals).

Thereforenilrad(A/I) =

⋂Q∈Spec(A/I)

Q,

and by the above argument taking pre-images yields

rad(I) = φ−1(nilrad(A/I)) =⋂

P∈Spec(A)P⊃I

P.

5.3 Modules

Definition 5.3.1 (Module). Let A be a ring. An A-module is an abeliangroup M with a multiplication map

· : A×M →M

(a · x) 7→ ax

satisfying

(i) a(x+ y) = ax+ ay for all a ∈ A and x, y ∈M ,

(ii) (a+ b)x = ax+ bx for all a, b ∈ A and x ∈M ,

(iii) (ab)x = a(bx) for all a, b ∈ A and x ∈M ,

(iv) 1Ax = x for 1A ∈ A and all x ∈M .

Remark 5.3.2. If A = k is a field, then a module M over A is a vector spaceover k.

Remark 5.3.3. In a module, we might not have a basis. We do, however, havea basis if the module is over a principal ideal domain.

Example 5.3.4. An ideal I of a ring A is an A-module. N

Example 5.3.5. Any module over Z is an abelian group. N


Definition 5.3.6 (Homomorphism of module). Let M,N be A-modules. Amapping f : M → N is an A-module homomorphism or an A-linear mapif

(i) f(x+ y) = f(x) + f(y) for every x, y ∈M ,

(ii) f(ax) = af(x) for every a ∈ A and x ∈M .

Remark 5.3.7. If A = k is a field, then an A-module homomorphism is a lineartransformation of vector spaces.

Proposition 5.3.8. Let M,N be A-modules, and f : M → N be a homomor-phism between them. Then

(i) ker f ⊂M is a submodule of M and im f ⊂ N is a submodule of N ,

(ii) M/ ker f ∼= im f .

The proofs of these are exactly the same as the proofs for linear maps in vectorspaces.

Definition 5.3.9 (Quotient module). Let M be an A-module and let N ⊂ Mbe a submodule. Then M/N is an A-submodule defined by

a(x+N) = (ax) +N

for a ∈ A and x ∈M .

Theorem 5.3.10 (Isomorphism theorems). Let A be a ring.

(i) If L ⊂M ⊂ N are all A-modules, then

L/N

M/N∼=

L

M.

(ii) If M,N ⊂ L are submodules, then

M +N

N∼=

M

M ∩N.

Remark 5.3.11. If N 6⊂ M , then M/N makes no sense. There are two naturalways to resolve this: either enlarge M to contain N (giving M + N as thesmallest alternative), or shrink N until it is contained in M (yielding M ∩N).(ii) above says that these are equivalent.

Proof. (i) Define ϕ : L/N → L/M by ϕ(x + N) = x + M . This is well-defined(i.e. doesn’t depend on the choice of coset representative) since if x+N = y+Nwe have x− y ∈ N ⊂M and therefore

ϕ(x+N) = x+M = y +M = ϕ(y +N).

By definition of coset operations, ϕ is A-linear and surjective.If we can now show that kerϕ = M/N , then we are done by the Isomorphism

theorem. To see this, consider x+N ∈ kerϕ, if and only if ϕ(x+N) = x+M =M if and only if x ∈M .

GENERATING SETS 17

(ii) The strategy is to construct a homomorphism from M to (M + N)/Nthat is surjective and has M ∩N as its kernel.

Consider

ψ : M ↪→M +N → M +N

N.

Now kerψ = M ∩ N since the elements in the kernel must be elements of N ,and since everything is from M only the intersection suffices. Moreover it issurjective since elements in (M + N)/N are of the form m + n + N = m + Nsince n ∈ N , and so by the Isomorphism theorem we’re done.

Lecture 6 Generating Sets

6.1 Faithful Modules and Generators

Recall that if V and W are vector spaces over a field k, then the set of all lineartransformations f : V → W is itself a k-vector space. The same thing is truefor modules:

Proposition 6.1.1. Let M,N be A-modules. Let

Hom(M,N) := { f : M → N | f is an A-modle homomorphism }.

Then Hom(M,N) is also an A-module with

(f + g)(x) := f(x) + g(x)

and(af)(x) := af(x)

for all x ∈M and a ∈ A.

The proof of this is straightforward computation.

Exercise 6.1.2. Show that Hom(A,M) ∼= M .

We have several objects in modules that are completely analogous to that ofrings.

Definition 6.1.3 (Module quotient). Let N,P be submodules of a module Mover a ring A. We call

(N : P ) := { a ∈ A | aP ⊂ N }

a module quotient , which is an ideal of A.

Definition 6.1.4 (Annihilator). The annihilator of a module M over a ringA is

Ann(M) := (0 : M) = { a ∈ A | aM = 0 }.

Note that if I ⊂ Ann(M) with I an ideal, then M is an A/I-module: if a ∈ A/I,then ax := ax for x ∈M . This is well-defined because (a+ I)M = aM + IM =aM , since I is inside the annihilator of M .

Definition 6.1.5 (Faithful module). Let A be a ring. An A-module M is calledfaithful is Ann(M) = 0.

Exercise 6.1.6. If Ann(M) = I, then M is a faithful A/I-module.


GENERATING SETS 18

6.2 Generators of a Module

Given an A-module M and x1, x2, . . . , xk ∈M , consider

(x1, x2, . . . , xk) :=

k∑i=1

Axi =

{k∑i=1

aixi

∣∣∣∣∣ ai ∈ A}

is a submodule of M .This becomes harder to think about if we’re indexing over a set I that isn’t

finite, or even countable.If

M =∑i∈I

Axi =

{∑i∈I

aixi

∣∣∣∣∣ ai ∈ A, ai 6= 0 only for finitely many i

}

then we call {xi }i∈I a set of generators of M .A module M is called a finitely generated A-module if it has a finite set

of generators.

Definition 6.2.1 (Direct sum, Direct product). Let M,N be A-modules. Thedirect sum

M ⊕N := { (x, y) | x ∈M,y ∈ N }

is an A-module with addition and scalar multiplication defined by

(x1, y1) + (x2, y2) := (x1 + x2, y1 + y2)

anda(x, y) := (ax, ay).

If (Mi)i∈I is a family of A-modules, their direct sum⊕i∈I

Mi = { (xi)i∈I | xi ∈Mi, xi 6= 0 only for finitely many i }.

The direct product is ∏i∈I

Mi = { (xi)i∈I | xi ∈Mi }.

Note that if |I| <∞, then these coincide.Let I be an index set. Then we write

A|I| := { (ai)i∈I | ai ∈ A, ai 6= 0 only for finitely many i }.

Consider (xi)i∈I ⊂M . Define ϕ : A|I| →M by

(ai)i∈I 7→∑i∈I

aixi

(cf. how (a1, a2, . . . , an) corresponds to a1v1 +a2v2 + . . . anvn in vector spaces).Note that if ϕ is surjective, then {xi }i∈I is a set of generators of M .

FINDING GENERATORS 19

Definition 6.2.2 (Free module, Basis). An A-module M is said to be free isϕ is an isomorphism and {xi }i∈I is called a basis of M . In this case M ∼= A|I|.

In particular, a finitely generated free A-module is isomorphic to An.

Example 6.2.3. Let A be a ring. Take I 6= (0) ⊂ A, an ideal. Then A/I is anA-module, finitely generated by 1 = 1 + I. However A/I 6∼= A and so A/I is notfree. N

Example 6.2.4. Let A = k[x, y], with k a field. Let M = (x, y), which is amaximal ideal of A. This maximal ideal M is an A-module, but M is not free(x and y are linearly independent, but they do not produce a basis: considerfor example xy ∈ M). The map ϕ : A2 → M by (a, b) 7→ ax + by has thekernel kerϕ = { (cy,−cx) | c ∈ A } 6= 0, and by the first isomorphism theoremA2/ kerϕ ∼= M . N

Astonishingly, this is the only type of isomorphisms we have:

Proposition 6.2.5. Let A be a ring. M being a finitely generated A-module isequivalent with M being isomorphic to a quotient of An for some n > 0.

Proof. Let {x1, x2, . . . , xn } be generators of M . The map φ : An →M definedby

φ(a1, a2, . . . , an) = a1x1 + a2x2 + . . .+ anxn

is surjective since {x1, x2, . . . , xn } is a generating set. Hence An/ kerφ ∼= M .For the converse, if M ∼= An/I, consider φ : An → An/I ∼= M . Then

{φ(ei) }ni=1, with ei = (0, . . . , 0, 1, 0, . . . , 0) with the 1 in the ith position is a setof generators of M .

Recall the Cayley-Hamilton theorem from linear algebra. It says that the char-acteristic polynomial of a matrix evaluates to 0 in that matrix itself.

We have something similar for modules.

Proposition 6.2.6. Let M be a finitely generated A-module (generated by nelements). Let I ⊂ A be an ideal, and let φ : M → M be an A-module homo-morphism such that φ(M) ⊂ IM . Then φ satisfies an equation of the form

φn + a1φn−1 + . . .+ an = 0

where ai ∈ Ii for i = 1, 2, . . . , n.

Lecture 7 Finding Generators

7.1 Generalising Cayley-Hamilton’s Theorem

We start by proving the proposition stated at the end of the last lecture.

Proof. Let x1, x2, . . . , xn be a set of generators of M . Then φ(xi) ∈ IM and so

φ(xi) =

n∑j=1

aijxj



with aij ∈ I since x1, x2, . . . , xn generate M . Rearranging we therefore have

n∑j=1

(δijφ− aij)xj = 0

which we can write in matrix form1 asφ− a11 −a12 · · · −a1n

−a21 φ− a22 · · · −a2n

......

. . ....

−an1 an2 · · · φ− ann

x1

x2

...xn

=

00...0

. (7.1.1)

We treat the matrix on the left-hand side as a matrix ∆ with elements fromEnd(M), the set of endomorphisms of M , i.e. homomorphisms from M toitself.

Recall how in linear algebra, i.e. in a vector space, we have that A−1 =1/ det(A)A∗, where A∗ is the adjoint of A. Rearranging this we have A∗A =det(A)In, where notably we only require multiplication, addition, and subtrac-tion to compute the adjoint, meaning that this latter statement is true in a ringas well.

Therefore let ∆∗ denote the adjoint matrix of ∆. Then ∆∗∆ = det(∆)In.Now multiply both sides of (7.1.1) by ∆∗, yielding

det(∆)In

x1

x2

...xn

=

00...0

which in turn means that det(∆)x = 0 for all x ∈M since x1, x2, . . . , xn generateM .

Expanding the determinant of ∆ we get the polynomial required in theproposition, and noting that the coefficient in front of φn−k has k elementsfrom I, we clearly have ai ∈ Ii as requested.

Using this we can in fact prove a generalisation of Cayley-Hamilton’s theoremon modules.

Theorem 7.1.1 (Cayley-Hamilton). Let A be a commutative ring with unity.Let N = (aij) be an n × n matrix with entries aij ∈ A. Moreover let PN (x) =det(xIn −N) be the characteristic polynomial of N .

Then PN (N) = 0.

Proof. Let M = An, an A-module. Define φ : M → M by φ(x) = Nx. Thenφ(M) ⊂ AM , and so by the proposition PN (N) = det ∆ = 0.

Corollary 7.1.2. Let M be a finitely generated A-module. Let I ⊂ A be anideal such that IM = M . Then there exists some x ≡ 1 (mod I) such thatxM = 0.

1Note that matrix form isn’t unique in a module, but still something we can work with.


Proof. Take φ = IdM , the identity function on M , i.e. φ(m) = m for all m ∈M .By the proposition above

φn + a1φn−1 + . . .+ an = 0

and so if we apply m we get

m+ a1m+ . . .+ anm = 0

and therefore(1 + a1 + a2 + . . .+ an)m = 0.

Taking the element in the parentheses to be x we clearly have xM = 0, and alsox ≡ 1 (mod I).

Corollary 7.1.3 (Nakayama’s lemma). Let M be a finitely generated A-module.Let I ⊂ A be an ideal such that I ⊂ J(A). Suppose IM = M . Then M = 0.

Proof. By the last corollary there exists an x such that x ≡ 1 (mod I) andxM = 0. This then implies that 1 − x ∈ I ⊂ J(A), and since 1 − x is in theJacobson radical, 1− (1−x) = x is a unit, so x−1 exists. Therefore x−1xM = 0meaning that M = 0.

Corollary 7.1.4. Let M be a finitely generated A-module and N ⊂ M a sub-module. Let I ⊂ J(A) be an ideal. Suppose M = IM +N . Then M = N .

Proof. Consider the quotient module M/N . Then

IM

N=IM

N=IM +N

N=M

N

and therefore this satisfies the assumptions of Nakayama’s lemma since I ⊂J(A), and therefore M/N = 0 and so M = N .

7.2 Finding Generators

We begin with a special case. Let A be a local ring (i.e. a ring with exactly onemaximal ideal m). Let k = A/m be the residue field.

Now let M be a finitely generated A-module. Then mM ⊂M is a submod-ule. Consider M/(mM). It is an A-module, and

m ⊂ Ann( M

mM

)meaning that M/(mM) is an A/m-module, and since A/m is a field, it is in facta vector space. This lets us lift a basis from the vector space to a generatingset in the module.

Proposition 7.2.1. Let x1, x2, . . . , xn ∈M be a set whose image in M/(mM)forms a basis in the k-vector space. Then {x1, x2, . . . , xn } generates M .

Proof. Let N = (x1, x2, . . . , xn) ⊂M be a submodule. Consider the map

φ : N ↪→M → M

mM.

This is surjective since {x1, x2, . . . , xn } is a basis in M/(mM). Therefore N +mM = M , and since A is a local ring J(A) = m and therefore N = M by theprevious result.


If A isn’t local, then we would have to first localise, but more on that in thefuture.

7.3 Exact Sequences

Definition 7.3.1 (Exact sequence). Suppose L, M , and N are A-modules andthat

L M Nf g

is a sequence of homomorphisms. The sequence is called exact at M if im f =ker g.

A longer sequence

· · · M1 M2 M3 · · ·

is called exact if it is exact at each Mi.

Example 7.3.2. The sequence 0→ Lf−→M is exact if and only if f is injective,

since the kernel of f must be trivial.

Similarly Lf−→M → 0 is exact if and only if f is surjective, since everything

in M is mapped to 0, making the kernel all of M , and this must then be theimage of f . N

Example 7.3.3. If

0 L M N 0f g

is a short exact sequence (meaning three modules between the zeros), thenthe cokernel is

coker(f) :=M

im f=

M

ker g= N

where the first equality is by exactness and the second is from the first isomor-phism theorem. N

This means that if we know that L ⊂M and we understand M/L, then wecan learn something about M . The best case scenario is M = L ⊕M/L, sincethen we know everything.

One wonders, then, when this happens.

Proposition 7.3.4. Let

0 L M N 0f g

be a short exact sequence of A-modules. Then the following are equivalent:

(i) There exists an isomorphism M ∼= L ⊕N under which f : m 7→ (m, 0) isthe natural embedding and g : (m,n) 7→ n is the natural projection.

(ii) There exists a map h : N →M such that g ◦ h = IdN .

(iii) There exists a map k : M → L such that k ◦ f = IdL.

Remark 7.3.5. If these conditions hold, then the sequence is called a split exactsequence .

EXACT SEQUENCES 23

Lecture 8 Exact Sequences

8.1 More on Exact Sequences

We ended last lecture by stating a result but didn’t prove it—we do so now.

Proof. To see that (i) implies (ii) and (iii), just take h : N → M = L ⊕ N tobe h(n) = (0,m) and k : M → L by (m,n) 7→ m.

For (ii) implying (i), we have

0 L M N 0f g

h

where f is injective by exactness, h is injective since g ◦ h = IdL. We then wishto show that M = f(L)⊕ h(N) ∼= L⊕N .

For m ∈M we have m 7→ g(m) 7→ h(g(m)) ∈ h(N), whereby

m = (m− h(g(m)))︸︷︷︸∈f(L)=ker g

+h(g(m))︸︷︷︸∈h(N)

where the second membership is clear, but the first one needs some work:

g(m− h(g(m))) = g(m)− g ◦ h ◦ g(m) = g(m)− g(m) = 0

since g ◦ h = IdL. Therefore we have the requisite decomposition. For it to bethe desired direct sum we also need to verify that the two parts have a trivialintersection. Let m ∈ f(L)∩h(N), meaning that m = h(n) for some n ∈ N . Wealso have m ∈ f(L) = ker g and so g(m) = 0, meaning in turn that g(h(n)) = n,so n = 0 and m = 0.

(iii) implying (i) is quite similar.

Let ϕ : M → N be an A-module homomorphism. Let L be an A-module. Thenϕ induces two homomorphisms:

(i) First ϕ : Hom(L,M) → Hom(L,N) by f 7→ ϕ ◦ f , i.e. ϕ(f) = ϕ ◦ f . Tosee this we draw the commutative diagram corresponding to f : L → Mand ϕ : L→M , and follow the arrows:

M N

L

ϕ

fφ◦f

(ii) Similarly ϕ : Hom(N,L)→ Hom(M,L) defined by ϕ(f) = f ◦ϕ, again byjust filling in the diagram:

M N

L

ϕ

f◦ϕf


EXACT SEQUENCES 24

Proposition 8.1.1. (i) The sequence

0 L M N

is exact if and only if the sequence

0 Hom(P,L) Hom(P,M) Hom(P,N)

is exact for every A-module P .

(ii) The sequence

L M N 0

is exact if and only if the sequence

0 Hom(N,P ) Hom(M,P ) Hom(L,P )

is exact for every A-module P .

Remark 8.1.2. This says that Hom is left exact , i.e. it only preserved theinjectivity, but not the surjectivity.

We can preserve surjectivity with an additional requirement:

0 L M N 0

is split exact if and only if

0 Hom(P,L) Hom(P,M) Hom(P,N) 0

is split exact for every A-module P .

Proof. For the forward direction of (i), we have

0 L M Nf g

and

0 Hom(P,L) Hom(P,M) Hom(P,N)f g

Now the first step is to show that ker f = 0. Suppose h ∈ ker f . Then

0 L M

P

f

hf◦h

and f(h) = 0 since h is in the kernel, and so since f is injective h(x) = 0 for allx ∈ P .

Secondly let us show that im f = ker g by considering the two inclusions. Sofirst im f ⊂ ker g, for which we have the diagram

0 L M N

P

f g

kh

g(h)=g◦h

EXACT SEQUENCES 25

where h ∈ im f , and by following the diagram we have g(h) = g◦h = g◦f ◦k = 0since g ◦ f = 0 since im f = ker g. Therefore h ∈ ker g.

Secondly for ker g ⊂ im f consider the same diagram, with h ∈ ker g. Letx ∈ P , so that x 7→ h(x) 7→ g(h(x)) = 0 since g ◦ h = 0. Moreover there existssome y ∈ L such that k(x) = y since ker g = im f by exactness. Thereforef(k) = h, so h ∈ im f .

For the opposite direction, first verify that ker f = 0. We take P = ker f ⊂ L,with

0 L M

P = ker f

f

inclusion `f◦`=0

But f is injective, meaning that ` = 0, whereby ker f = 0.Secondly we verify exactness. First check im f ⊂ ker g by taking P = L.

0 L M N

P = L

f g

inclusion ` fg◦f=0

with f = f(`) ∈ im f . Since im f = ker g, we have g ◦ f = 0, and so g ◦ f(`) =g(f(`)) = 0 implying that im f ⊂ ker g.

For the opposite inclusion, take P = ker g:

0 L M N

P = ker g

f g

hinclusion `

g◦`=0

Following the diagram along we have ` ∈ ker g = im f , and so x ∈ im f ifx ∈ ker g.

(ii) is similar.

Proposition 8.1.3 (Snake lemma). Let

0 L M N 0

0 L′ M ′ N ′ 0

f

α

g

β γ

f ′ g′

be a commutative diagram of A-modules with rows being exact. Then there existsan exact sequence

0 kerα kerβ ker γ

cokerα cokerβ coker γ 0

f g

δ

f ′ g′

with δ being the connecting homomorphism and cokerα = L′/ imα, andsimilarly for the other cokernels.

TENSOR PRODUCTS 26

The connecting homomorphism stems from the fact that g is surjective, meaningthat there exists a y such that g(y) = x ∈ ker γ, and we map that y over to M ′

using β and use surjectivity to fetch w ∈ cokerα = L′/ imα.

8.2 Tensor Product of Modules

Definition 8.2.1 (Tensor product of modules). Let M and N be A-modules.Let

C =

{k∑i=1

ai(xi, yi)

∣∣∣∣∣ xi ∈M,yi ∈ N

}be a free A-module in M×N . Moreover let D be the submodule of C generatedby all elements of the following times: (x+x1, y)− (x, y)− (x1, y), (x, y+ y1)−(x, y)− (x, y1), (ax, y), and (x, ay) for a ∈ A, x, x1 ∈M , and y, y1 ∈ N .

We define M ⊗N = C/D, called the tensor product of M and N .Then M ⊗ N is generated by x ⊗ y for x ∈ M and y ∈ N , and we have

linearity in both coordinates.

Remark 8.2.2. Note that the linearity in the coordinates follows directly fromconstruction, since we quotient away the expressions we want to be 0 for thatto hold.

Remark 8.2.3. Roughly why one wants to do these sort of things is becausetensor products are, in some sense, more well behaved than direct products.Take for instance the continuous functions on X and the continuous functionson Y , C(X) and C(Y ) respectively. Then the continuous functions on X × Y ,C(X×Y ) are not C(X)×C(Y ), but—at least with some good assumptions—areC(X)⊗ C(Y ).

Lecture 9 Tensor Products

9.1 More on Tensor Products

Recall from last time that for two A-modules M and N , M ⊗ N is generatedby x⊗ y for x ∈M and y ∈ N , with bilinearity, i.e. linearity in both variables,meaning that (x+ x1)⊗ y = x⊗ y + x1 ⊗ y, x⊗ (y + y1) = x⊗ y + x⊗ y1, andfinally a(x⊗ y) = (ax)⊗ y = x⊗ (ay) for all a ∈ A.

This is equivalent with the following universal property: Let ` : M × N →M ⊗N defined by `(x, y) = x⊗ y. Then for any bilinear map f : M ×N → Pthere exists a unique homomorphism f : M ⊗ N → P called the tensor suchthat f = f ◦ `. This comes from the following commutative diagram:

M ×N M ⊗N

P

`

ff

Proposition 9.1.1. Let M , N , and P be A-modules. Then

(i) M ⊗N ∼= N ⊗M ,


TENSOR PRODUCTS 27

(ii) (M ⊗N)⊗ P ∼= M ⊗ (N ⊗ P ) ∼= M ⊗N ⊗ P ,

(iii) (M ⊕N)⊗ P = (M ⊗N)⊕ (N ⊗ P ), and

(iv) A⊗M ∼= M .

All of these follow by the natural isomorphisms.

Definition 9.1.2 (Extension of Scalars). Let M be an A-module. SupposeB ⊃ A is a ring (meaning that we can view B as an A-module). Then MB :=B ⊗M is an A-module. Moreover it is also a B-module—we take

b(b1 ⊗ x) = (bb1)⊗ x

for all b, b1 ∈ B and x ∈M . In other words we extend scalars in A to scalars inB.

Proposition 9.1.3. If M is a finitely generated A-module, then MB is a finitelygenerated B-module.

Proof. Let x1, x2, . . . , xn generate M over A. Then 1 ⊗ x1, 1 ⊗ x2, . . . , 1 ⊗ xngenerate MB over B since M 3 b ⊗ x = b(1 ⊗ x) where x is a linear sum ofx1, x2, . . . , xn.

Theorem 9.1.4 (Adjoint associativity). Let M , N , and P be A-modules. Then

Hom(M ⊗N,P ) ∼= Hom(M,Hom(N,P ))

by the isomorphism φ, defined in the following way. Let f : M ⊗ N → P , i.e.f ∈ Hom(M ⊗N,P ). Then for x ∈M and y ∈ N we have(

φ(f)(x))(y) := f(x⊗ y).

Sketch of proof. Let S be the set of bilinear maps M × N → P . We havean injective correspondence between S and Hom(M ⊗ N,P ) by the universalproperty. Therefore we

M ×N P

M ⊗N

g

f

Fixing x ∈M we have g(x, ·) : N → P where g(x, ·) ∈ Hom(N,P ) by linearity.Given ψ : M → Hom(N,P ), ψ lifts to the bilinear map ψ : M × N → P byψ(x, y) =

(ψ(x)

)(y).

We then have that Hom(M,Hom(N,P ))↔ S, and combining this with theuniversal property yields our bijection.

Remark 9.1.5. Note that the important takeaway from this theorem is this:tensor product ⊗ is the adjoint functor to Hom.

TENSOR PRODUCTS 28

9.2 Exactness


L M N 0f g

be exact. Then

L⊗ P M ⊗ P N ⊗ P 0f⊗1 g⊗1

is exact for any A-module P .

Remark 9.2.2. This says that the tensor product is right-exact .

Proof. We use the left-exactness of Hom, i.e. Proposition 8.1.1 from last lecture.In other words

L M N 0

is exact if and only if

0 Hom(N,R) Hom(M,R) Hom(L,R)

is exact for any R-module, which again is true if and only if

0 Hom(P,Hom(N,R)) Hom(P,Hom(M,R)) Hom(P,Hom(L,R))

is exact for every A-module P . Finally by the adjoint associativity we have thatthis is true if and only if

P ⊗ L P ⊗M P ⊗N 0

is exact for any A-module P .

Remark 9.2.3. It is not true in general that

L M N

being exact implies that

L⊗ P M ⊗ P N ⊗ P

is exact. In other words, the surjectivity of the last map is important.

Example 9.2.4. Consider A = Z. The following sequence

0 Z Zf

with f(x) = 2x is exact, but

0 Z⊗ Z2Z

Z2Z

f⊗1

is not exact since f ⊗ 1 is not injective. To see this, consider

(f ⊗ 1)(x⊗ y) = (2x)⊗ y = x⊗ (2y),

but then 2y ∈ Z/(2Z), meaning that it is 0, and so the above is x ⊗ 0 = 0,meaning that f ⊗ 1 is identically 0 and certainly not injective. N

TENSOR PRODUCTS 29

Definition 9.2.5 (Flat module). An A-module P is called flat if for any exactsequence

L M N

the sequence

L⊗ P M ⊗ P N ⊗ Pis also exact.

Example 9.2.6. (i) Vector spaces are flat.

(ii) Free modules are flat. Note however that this is very restrictive; therearen’t all that many free modules that aren’t also vector spaces.

(iii) Projective modules are flat. We won’t discuss projective modules much(if at all) in this course, for reference a module is projective if we havethe situation

N

P M

surjective

where we can lift to N .N

Proposition 9.2.7. The following are equivalent:

(i) P is flat.

(ii) If

0 L M N 0

is exact then

0 L⊗ P M ⊗ P N ⊗ P 0

is exact.

(iii) If f : M → N is injective, then (f ⊗ 1) : M ⊗ P → N⊗ P is injective.

(iv) If f : M → N is injective and M and N are finitely generated, then(f ⊗ 1) : M ⊗ P → N ⊗ P is injective.

Proof. (i) being equivalent to (ii) is immediate by definition since we specificallyask for injectivity in 0 → L ⊗ P . Likewise (ii) is equivalent to (iii) again.Moreover (iii) implies (iv) trivially. That leaves (iv) implying (iii), which isthe interesting result.

We claim that ker f = 0. We have

ker f 3 u =

n∑i=1

xi ⊗ pi

since we can view u ∈M ⊗ P , and M is finitely generated by assumption.Now take M0 ⊗ P ⊂M ⊗ P to be the subset generated by xi ⊗ pi. Then

(f ⊗ 1)(u) =

n∑i=1

f(xi)⊗ pi

TENSOR PRODUCTS 30

and we take N0 ⊂ N to be generated by f(xi).Therefore

f ⊗ 1

∣∣∣∣M0⊗P

: M0 ⊗ P → N0 ⊗ P

being f ⊗ 1 restricted to M0 ⊗ P is a map between two finitely generated A-modules, which by assumption make them injective. Therefore u ∈ M0 ⊗ Pimplies that u = 0.

9.3 Localisation of a Ring

The goal of this part is to generalise the idea of fractions—in other words wewish to generalise the construction of quotient fields to not require an integraldomain.

That is to say, let A be an integral domain. Then as we know A has a fieldof fractions or quotient field , made up of elements a/b, with a, b ∈ A andb 6= 0. But the representation for elements here is not unique: just like how inQ we have, say, 1/2 = 2/4, we have

a

b=au

bu

for u ∈ A, u 6= 0. The way we reconcile this is to say that

a

b=s

t⇐⇒ at− bs

bt= 0.

In other words we write them with common denominator and check whetherthe numerator is 0. We say that

k = { (a, b) =a

b| a ∈ A, b ∈ A, b 6= 0 }

under the equivalence relation (a, b) ∼ (s, t) if and only if at − bs = 0. Whenproving transitivity of this relation we require the absence of zero divisors inthe integral domain. Meaning that (a, s) ∼ (b, t) and (b, t) ∼ (c, u) implies(a, s) ∼ (c, u) we have that at− bs = 0 and bu− tc = 0, and if we multiply thefirst one by u and the second one by s and add the two we have

0 = atu− bsu+ bus− tcs = atu− tcs = t(au− cs)

which, since we have no zero-divisors, means that t = 0 (which can’t be sinceit’s the denominator of a fraction) or au− cs = 0, i.e. (a, s) ∼ (c, u).

Our goal, therefore, is to generalise this in such a way that we have ameaningful equivalence relation of this variety despite potentially having zero-divisors.

Definition 9.3.1. Let A be a ring and let S ⊂ A be a multiplicative set (recallthat this means 1 ∈ S and ab ∈ S if a, b ∈ S). Define ∼ on A× S by

(a, s) ∼ (b, t) ⇐⇒ (at− bs)u = 0

for some u ∈ S.

LOCALISATION 31

The relation ∼ so defined is an equivalence relation. It is clearly reflexive ,since (a, s) ∼ (a, s) since by commutativity as− sa = 0. Similarly it is triviallysymmetric since (a, s) ∼ (b, t) implies (b, t) ∼ (a, s) also by commutativity.

The interesting one—and the one where previously we required the absenceof zero-divisors—is transitivity . Suppose (a, s) ∼ (b, t) and (b, t) ∼ (c, u).Show that (a, s) ∼ (c, u).

The proof of this is similar to the previous one, we just need to keep track of afew extra terms, namely the elements of the multiplicative set. So (at−bs)v = 0for some v ∈ S, and (bu− tc)w = 0 for some w ∈ S. Then multiplying the firstone by uw and the second one by sv we have

0 = atvuw − bsvuw + buwsv − tcwsv = tvw(au− cs)

where tvw is in S since t, v, w ∈ S. Therefore (a, t) ∼ (c, s).This, therefore, is how we define our generalisation of the quotient ring:

Definition 9.3.2. Given a ring A and a multiplicative set S ⊂ A, then S−1Ais the commutative ring of fractions a/s with a ∈ A and s ∈ S, with the usualarithmetic operations:

a

s+b

t=at+ bs

stand

a

s· bt

=ab

st.

Remark 9.3.3. If A is an integral domain and S = A \ { 0 }, then S−1A = k isexactly the quotient field of A.

Example 9.3.4. Let A be a ring and P ⊂ P be a prime ideal. Then, as provenbefore, S = A \ P is a multiplicative set, so we can use it to define the above.We then get that S−1A is a local ring—it has exactly one maximal ideal. N

Remark 9.3.5. Define f : A → S−1A by a 7→ a/1. Then f is a ring homomor-phism, but f is not injective in general, since if (a − b)u = 0 and a − b is azero-divisor, then a/1 = b/1, so two distinct a, b ∈ B can map to the sameelement in S−1A.

Lecture 10 Localisation

Recall that A is a ring, S ⊂ A is a multiplicative subset, and that

S−1A ={ as

∣∣∣ a ∈ A, s ∈ S }is a ring of fractions in which

a

s=b

t⇐⇒ (at− bs)u

stu= 0

for some u ∈ S if and only if (at− bs)u = 0 for some u ∈ S.There is a natural map f from A to S−1A, namely a 7→ a/1. Then

ker f = { a ∈ A | as = 0 for some s ∈ S }.


LOCALISATION 32

Moreover if s ∈ S, then f(s) = s/1 is a unit in S−1A.Consider A = k[x, y]/(xy), with k a field, so that xy = 0 in A. Let S =

{ 1, x, x2, . . . }. ThenS−1A ∼= k[x, x−1].

Also Im f = f(A) = k[x], since f(y) = y/1 = yx/x = 0.

10.1 Extension and Contraction

Definition 10.1.1 (Extension and Contraction). Let f : A → B be a ringhomomorphism. If I is an ideal in A, then f(I) is not necessarily an ideal inB, but we can make it one by extending to f(I)B. In other words there is acorrespondence

e : { ideals in A } → { ideals in B }

defined by e(I) = f(I)B which we call extension . It is also sometimes writtenIe.

Similarly we can contract an ideal in B to an ideal in A,

c : { ideals in B } → { ideals in A }

by c(J) = f−1(J) called the contraction or restriction . This is sometimeswritten Jc.

Proposition 10.1.2. Let f : A→ S−1A be defined by f : a 7→ a/1.

(i) For any ideal J in S−1A, we have e(c(J)) = J .

(ii) For any ideal I in A, we have

c(e(I)) = { a ∈ A | sa ∈ I for some s ∈ S }.

(iii) If P is a prime ideal in A, and P ∩ S = ∅, then e(P ) is a prime ideal inS−1A.

For an ideal I ⊂ A, we will denote e(I) by S−1I.

Proof. (i) That e(c(J)) ⊂ J is trivial. For the other direction, suppose a/s ∈ J .Then s/1 · a/s = a/1 ∈ J , and so a ∈ f−1(J) = c(J), meaning that a/1 ∈e(c(J)).

(ii) Call { a ∈ A | sa ∈ I for some s ∈ S } = B. First let us show thatc(e(I)) ⊂ B. If a ∈ c(e(I)), then f(a) ∈ e(I), and in turn f(a) = a/1 = b/s forsome b ∈ I and s ∈ S since f(a) ∈ S−1I. This means that (as − b)u = 0 forsome u ∈ S, and so asu = bu ∈ I. Since b ∈ I and s, u ∈ I, we have that a ∈ B.

For the opposite direction, take a ∈ B, meaning that as ∈ I for some s ∈ S.Then

f(a) =a

1=as

s∈ e(I)

since as ∈ I, and so a ∈ c(e(I)).(iii) Suppose x/s · y/t ∈ e(P ), where x, y ∈ A and s, t ∈ S. Then

xy

st=p

u

LOCALISATION 33

with p ∈ P and u ∈ S, so (xyu− stp)v = 0 for some v ∈ S, and so

(xy)(uv) = stpv ∈ P

since p ∈ P , and since u, v ∈ S we must have xy ∈ P since P ∩ S = ∅, andbecause P is prime, x ∈ P or y ∈ P . Therefore x/s ∈ e(P ) or y/t ∈ e(P ).

Corollary 10.1.3. Localisation commutes with nilradical, i.e. nilrad(S−1A) =S−1(nilrad(A)).

Proof. We write the nilradical as the intersection of prime radicals:

nilrad(A) =⋂

P∈Spec(A)

P.

ThereforeS−1(nilrad(A)) = S−1

( ⋂P∈Spec(A)

P).

Now if P ∩ S 6= ∅, then S−1P = S−1A since the former will contain some s/1,which is a unit. Therefore for the intersection it suffices to consider prime idealsP with P ∩ S = ∅. Then

S−1(nilrad(A)) = S−1( ⋂P∈Spec(A)

P)

=⋂

P∈Spec(A)

S−1P

=⋂

P∩S=∅S−1P =

⋂Q∈Spec(S−1A)

Q = nilrad(S−1A).

Note that we have not yet proved that intersection and localisation commute.We’ll do this next time.

Speaking of prime ideals, they allow us to construct a special kind of localisation.Let P be a prime ideal of A. Then S = A \ P is a multiplicative set, andtherefore we can localise using it. We let AP = S−1A, called the localisationat P localisation .

Proposition 10.1.4. An element a/s ∈ AP is a unit if and only if a 6∈ P .Hence AP is a local ring with maximal ideal e(P ) = S−1P = PAP .

Proof. Start by assuming a 6∈ P . By construction this means that a ∈ S,whereby s/a ∈ AP since it’s a valid denominator, and so a/s · s/a = 1, and a/sis a unit.

In the other direction, let a/s be a unit, meaning that a/s · b/t = 1 for somea ∈ A and t ∈ S. This means that (ab − st)u = 0 for some u ∈ S, and soabu = stu, which is in S since s, t, u ∈ S and S is multiplicative. Thereforeabu 6∈ P and so we must have a 6∈ P for otherwise abu ∈ P since P is an ideal.

In other words AP = A× t S−1P , so it is a local ring.

Corollary 10.1.5. Let P be a prime ideal in A. The prime ideals of AP arein one-to-one correspondence with the prime ideals of A contained in P .

Example 10.1.6. Let A = Z and P = (p), p a prime. Then

Z(p) ={ ab∈ Q

∣∣∣ p - b, a, b ∈ Z}

is a local ring with maximal ideal

M ={ ab∈ Z(p)

∣∣∣ p | a} = PZ(p). N

EXACTNESS OF LOCALISATION 34

10.2 Modules of Fractions

Definition 10.2.1 (Module of fractions). Let M be an A-module. Let S ⊂ Abe a multiplicative set. Define S−1M to be the S−1A-module as follows:

The equivalence relation ∼ on M×S is defined by (m, s) ∼ (n, t) if and onlyif (tm − sn)u = 0 for some u ∈ S. Then S−1M = (M × S)/ ∼. As before wehave

m

s± n

t=mt± sn

st

anda

t·ms =

am

ts

with a ∈ A, s, t ∈ S, and m ∈M .

As before, if S = A\P with P being a prime ideal of A, then we write S−1M =MP . In fact, it turns out that MP = S−1A⊗AM .

Let f : M×N be an A-module homomorphism. This homomorphism inducesan S−1A-module homomorphism

S−1f : S−1M → S−1N

by

S−1f(ms

)=f(m)

s,

i.e. it preserves the denominator. Clearly S−1(f ◦ g) = S−1f ◦ S−1g.

Lecture 11 Exactness of Localisation

11.1 Exactness of Localisation

Proposition 11.1.1. Let L, M , and N be A-modules. If

L M Nf g

is an exact sequence, then

S−1L S−1M S−1NS−1f S−1g

is an exact sequence.

Proof. We want to show that ImS−1f = kerS−1g, so start by assuming theformer is a subset of the latter. Now S−1f ◦ S−1g = S−1(f ◦ g), but since thefirst sequence is exact, Im f = ker g, so f ◦ g = 0, and therefore S−1(f ◦ g) = 0.

Next we show that kerS−1g ⊂ ImS−1f . Let m/s ∈ kerS−1g, meaning that

S−1g(ms

)=g(m)

s= 0.



This means that there exists some t ∈ S ⊂ A such that tg(m) = 0. But g isA-linear, being an A-module homomorphism, so tg(m) = g(tm) = 0, meaningthat tm ∈ ker g = Im f . Therefore tm = f(x) for some x ∈ L, so

m

s=tm

ts=f(x)

ts= S−1f

( xts

)∈ ImS−1f,

and we are done.

Corollary 11.1.2. Let Ni, N , and L be submodules of M . Then

(i) S−1(N + L) = S−1N + S−1L;

(ii) S−1(⋂

i

Ni

)=⋂i

S−1Ni; and

(iii) As S−1A-modules,

S−1(MN

)∼=S−1M

S−1N.

Proof. (i) is trivial by the very definition of fractions: just split a given fractioninto two parts.

For (ii), let m/s ∈ S−1(⋂iNi). This means that m ∈ bigcapiNi, and in turn

m ∈ Ni for all i. Therefore m/s ∈ S−1Ni for all i, and so m/s ∈⋂i S−1Ni.

For the opposite direction, if m/s ∈⋂i S−1Ni, then m/s ∈ S−1Ni for all

i. Now this does not immediately imply that m ∈ Ni for all i, since we areworking with equivalence classes, but it does mean that m/s = ni/si for someni ∈ Ni, and (sim− sni)ui = 0 for some ui ∈ S. Therefore siuim = suini ∈ Nisince ni ∈ Ni and Ni is a module. Therefore

m

s=siuim

siuis∈ S−1Ni

for all i.(iii) Consider

0 N M MN 0.

This is exact since N ⊂M , and so by the previous proposition

0 S−1N S−1M S−1(MN

)0

is exact. The first isomorphism theorem now tells us that

S−1M

S−1N∼= S−1

(MN

).

Proposition 11.1.3. Let M be an A-module. Then S−1M ∼= S−1A ⊗AM asan S−1A-module.

The isomorphism is given by f : S−1A⊗M → S−1M ,

f(as⊗m

)=am

s

with a ∈ A, s ∈ S, and m ∈M .


Proof. That f is a homomorphism is obvious, since we just distribute and sim-plify fractions. Moreover that it is surjective is clear too, since taking a = 1suffices to yield all of S−1M .

That it is injective is less obvious. Consider an element a/s⊗m = 1/s⊗(am).Then all elements in S−1A ⊗M can be written as 1/a ⊗m. Now let such anelement be in the kernel of f , i.e.

f(1

s⊗m

)=m

s= 0.

This implies that tm = 0 for some t ∈ S, and therefore

1

s⊗m =

t

st⊗m =

1

st⊗ (tm) = 0

since tm = 0.

Corollary 11.1.4. The A-module S−1A is flat.

Recall that this means that tensor products preserve exactness.

Proof. Suppose

0 L M N 0

is exact. Then we’d like to show that

0 S−1A⊗ L S−1A⊗M S−1A⊗N 0

is exact. But we just showed that S−1A ⊗ L ∼= S−1L, and so on for M andN , and we know from before that localisation is an exact functor, so the resultfollows.

Proposition 11.1.5. If M and N are A-modules, then S−1M ⊗S−1A S−1N ∼=

S−1(M ⊗A N) as S−1A-modules.The isomorphism is given by f : S−1M ⊗ S−1N → S−1(M ⊗N),

f(ms⊗ n

t

)=m⊗ nst

.

In particular, for any prime ideal P ⊂ A, MP ⊗APNP = (M ⊗A N)P .

Proof. That f is a homomorphism is obvious. Surjectivity is clear by definition,and so is injectivity.

11.2 Local Property

A property K of a ring A (or an A-module M) is called a local property if A(or M) has K if and only if AP (or MP ) has K for all prime ideals P ⊂ A.

Example 11.2.1. A homomorphism f : M → N is injective (or surjective) ifand only if f : MP → NP is injective (or surjective) for every prime ideal Pof A. This means that injectivity and surjectivity, and thereby bijectivity, is alocal property. We’ll prove this later. N

PRIMARY DECOMPOSITION 37

Lecture 12 Primary Decomposition

An aside: We spent a significant amount of time toying with commutativediagrams and so on and so forth. In the opening scene of the movie It’s My Turnfrom 1980, a mathematics professor by the name of Kate Gunzinger (portrayedby Jill Clayburgh) is proving the Snake lemma on the board, to which DanielStern’s character Cooperman responds by saying that ‘This stuff is just garbage,’asking when they’ll move on to interesting stuff.

So here we are, moving on to interesting stuff!

12.1 Local Properties

As established last time, a property K of a ring A or A-module M is called localif the ring or module has the property if and only if AP (or MP ) has the sameproperty for all prime ideals P ⊂ A.

Proposition 12.1.1. Let M be an A-module. Then the following are equivalent:

(i) M = 0.

(ii) MP = 0 for all prime ideals P of A.

(iii) Mm = 0 for all prime ideals m of A.

Proof. It is clear that (i) implies (ii) implies (iii), in the second case sincemaximal ideals are prime. It remains to show that (iii) implies (i), therefore.

Suppose x 6= 0 is some element in M . Let I = Ann(x) = { a ∈ A |ax = 0 } ⊂A, which is an ideal. Then I 6= A = (1), since otherwise 1x = 0 which meansthat x = 0. Hence I ⊂ m ⊂ A, and m is maximal. So we localise! Thus

x

1∈Mm = 0,

by assumption, and so sx = 0 for some s ∈ S = A \ m, implying that s ∈Ann(x) = I ⊂ m. In other words s ∈ m and s 6∈ m, which is impossible, and soM contains no nonzero elements.

In other words, the 0-module is a local property. This might not seem terriblyinteresting or useful, but it is:

Proposition 12.1.2. Let φ : M → N be an A-module homomorphism. Thenthe following are equivalent:

(i) φ is injective.

(ii) φP : MP → NP is injective for all prime ideals P of A.

(iii) φm : Mm → Nm is injective for all maximal ideals m of A.

The same is true if we replace injectivity by surjectivity.

Date: October 3, 2017.


Proof. For (i) implying (ii), we have that

0 M Nφ

being exact (i.e. φ is injective) implies that

0 MP NPφP

is exact since we’ve proved previously that localisation is an exact functor.Next (ii) implying (iii) is trivial since maximal ideals are prime ideals.Finally assume (iii) and consider L = kerφ. Then

0 L M Nφ

is exact by embedding L in M . But then

0 Lm Mm Nmφm

is exact since localisation is an exact functor, and Lm = kerφm = 0 since φm isinjective by assumption. But this means that L = 0 by last proposition, since0-module is a local property, and so φ itself is injective.

For surjectivity, we do almost exact the same thing, but consider the diagram

M N 0φ

instead.

12.2 Primary Decomposition

In Z, prime ideals are (0) and P = (p), for p primes. These then model the ideathat if p | xy, then either p | x or p | y.

Suppose instead that pn | xy. Then possibly pn | x, but that doesn’t haveto be the case—some factors of p could be in x and some in y—and we couldinstead have pn - x, in which case we must have p | y. That does not accountfor all n factors of p, however, so we have pn | yn.

Generalising this to ideals in the same way we did with primes to primeideals we get the primary ideals, which in Z are Pn = (p)n = (pn).

Definition 12.2.1 (Primary ideal). An ideal Q of A is primary if Q 6= A andfor x, y ∈ A, if xy ∈ Q, then x ∈ Q or yn ∈ Q for some n > 0. Note that yn ∈ Qfor some positive power simply means that y ∈ rad(Q).

Another way to look at it therefore is that Q is primary if and only if all zero-divisors of A/Q, which is nonzero since Q 6= A, are nilpotent. This is because ifwe consider a product xy ∈ Q, this in A/Q will be xy = 0. Then either x ∈ Q,meaning x = 0, or yn ∈ Q, or yn = 0, so y is nilpotent.

Example 12.2.2. Every prime ideal is primary. A quick way to see this: if Pis a prime ideal of A, then A/P is an integral domain, which contain no zerodivisors, and so all zero-divisors are nilpotent is vacuously true. N


Example 12.2.3. The contraction of a primary ideal is primary, i.e. if f : A→B and Q ⊂ B is primary, then f−1(Q) ⊂ A is primary. To see why, consider

A B B/Qf

φ

where φ is the composite map. Then A/f−1(Q) ∼= Imφ ⊂ B/Q by the firstisomorphism theorem, and since every zero-divisor is nilpotent in B/Q since Qis primary in B, then by isomorphism the same is true in A/f−1(Q), so f−1(Q)is primary in A. N

Proposition 12.2.4. Let Q be a primary ideal of A. THen P = rad(Q) is thesmallest prime ideal contained in Q.

Proof. It suffices to show rad(Q) is prime, since

rad(S) =⋂

P⊂S prime

P

and so it is automatically the smallest. Now let xy ∈ rad(Q). Then (xy)n =xnyn ∈ Q for some n > 0. Since Q is primary, either xn ∈ Q or ynm ∈ Q forsome positive power m. Therefore x ∈ rad(Q) or y ∈ rad(Q), so rad(Q) is aprime ideal.

Definition 12.2.5 (P -primary ideal). An ideal Q of a ring A is called P -primary if Q is primary and rad(Q) = P .

Example 12.2.6. The primary ideals in Z are (0) or (pn) for n > 0 andp prime. Now consider ideals generated by composites instead. That is, letn = pn1

1 pn22 · · · p

nk

k . Then

(n) = (pn11 ) ∩ (pn2

2 ) ∩ . . . ∩ (pnk

k ). N

There are two natural questions that follow. Firstly, in what rings can we de-compose ideals as intersections of primary ideals (this being the titular primarydecomposition)? Secondly, given a primary ideal Q in a ring A, what sort ofring must A be in order for Q = Pn for some prime ideal P and power n?

The answers—and we’ll explore at least the first one more later on in thiscourse—is that in Notherian rings we have primary decomposition, and inDedekind domains we can write all primary ideals as powers of prime ideals.

Example 12.2.7. Let A = k[x, y], with k a field. Let Q = (x, y2), and considerA/Q = k[y]/(y2), since x is killed. Here every zero-divisor is nilpotent since itmust contain y, and y2 = 0 in the quotient. Therefore Q is primary in A.

We also have rad(Q) = (x, y) = P , which is prime, but clearly P ) Q ) P 2

since the first one contains a lone y, which the middle one does not, and themiddle one contains a lone x, which the last one does not, since P 2 = (x2, xy, y2).Therefore Q is not a prime power. N

Next, let us demonstrate that a prime power isn’t necessarily primary.

MORE ON PRIMARY DECOMPOSITION 40

Example 12.2.8. Let A = k[x, y, z]/(xy − z2), with k a field. Let x, y, and zdenote images of x, y, and z in A, and let P = (x, z) ⊂ A.

We have A/P = k[y], which since k is a field is an integral domain, and soP is a prime ideal. Now consider P 2 = (x2, xz, z2), and let us work in A/P 2.

In here xy = z2 ∈ P 2, since xy − z2 = 0 in A itself. Therefore xy = 0in A/P 2, which makes x and y zero-divisors. They are not both nilpotent,however: x is, since x2 = 0 in here, but y is not zero for any power in A/P 2,and so P 2 is not primary.

We still have rad(P 2) = P , however, which is why we specifically require Qto be primary in the definition of P -primary ideals above. N

Proposition 12.2.9. Suppose Q is P -primary and that P is finitely generated.THen P ⊃ Q ⊃ Pn for some n > 0.

Proof. Let P = (x1, x2, . . . , xk) = rad(Q). Then for every i = 1, 2, . . . , k wehjave xni

i ∈ Q for some ni > 0. Taking n = n1 +n2 + . . .+nk, we have thereforethat Pn ⊂ Q.

Note, by the way, that this is not true for every ideal I.However, if instead we’re considering a maximal ideal m, all of these good

properties hold:

Proposition 12.2.10. If m = rad(Q) is maximal, then Q is m-primary. Inparticular the powers of a maximal ideal m are m-primary.

Proof. Consider A → A/Q. Then m is mapped to m, which is still maximal,and m = rad(Q) corresponds to nilrad(A/Q) = m. But then we have

A

Q=(AQ

)×t m

and so any zero-divisor in here must belong to the maximal ideal m, since itcan’t be a unit. Therefore Q is primary.

The second part, about mk being primary, is trivial given the above sincerad(mk) = m.

Lecture 13 More on Primary Decomposition

13.1 First Uniqueness Theorem

Lemma 13.1.1. If Q1 and Q2 are P -primary ideals, then Q = Q1 ∩ Q2 isP -primary as well.

Proof. First establish that Q has the correct radical:

rad(Q) = rad(Q1 ∩Q2) = rad(Q1) ∩ rad(Q2) = P ∩ P = P.

Next we show that Q is primary. Let xy ∈ Q = Q1 ∩ Q2. Then xy ∈ Q1 andxy ∈ Q2. If x ∈ Q, then we’re done, so assume it isn’t. Then x doesn’t belongto one Q1 or Q2, let’s say x 6∈ Q1. But this, by Q1 being primary, implies thaty ∈ rad(Q1) = P = rad(Q), so Q is primary.



Remark 13.1.2. We can of course iterate on this, so that if Q1, Q2, . . . , Qk areP -primary, then Q1 ∩Q2 ∩ . . . ∩Qk is P -primary.

Definition 13.1.3 (Primary decomposition). Let A be a ring and I ⊂ A anideal. A primary decomposition of I is

I = Q1 ∩Q2 ∩ . . . ∩Qk

with each Q1, Q2, . . . , Qk being primary. This primary decomposition is calledminimal (or shortest , irredundant , reduced , normal , et cetera) if

(i) rad(Qi) are all distinct for i = 1, 2, . . . , k, and

(ii) Qi 6⊂⋂j 6=i

Qj , i.e. each Qi is necessary.

Note that if two primary ideals Qi and Qj have the same radical P , then we cancombine them as per the previous lemma. Therefore a minimal decompositionalways exists if a decomposition exists.

Definition 13.1.4 (Decomposable ideal). An ideal I ⊂ A is said to be decom-posable if it has a primary decomposition.

These primary decompositions are not quite unique, but the radicals of theprimary ideals are.

Theorem 13.1.5 (First uniqueness theorem). Let I be a decomposable idealof A. Let I = Q1 ∩ Q2 ∩ . . . ∩ Qk be a minimal primary decomposition. LetPi = rad(Qi) for i = 1, 2, . . . , k. Then

{P1, P2, . . . , Pk } = { rad(I : x) prime | x ∈ A }

is determined only by the ideal I, whereby the Pi are uniquely determined by theideal I and do not depend on the decomposition.

Proof. Recall (I : x) = { y ∈ A | xy ∈ I }. Since fractional ideal commutes withintersection, we have for x ∈ A that

(I : x) =(⋂

i

Qi : x)

=⋂i

(Qi : x).

Therefore if x ∈ Qi, then Qi : x) = A since Qi is an ideal. For this reason itsuffices to take the intersection over Qi not containing x, i.e.

(I : x) =⋂x 6∈Qi

(Qi : x).

We now wish to take radicals of both sides, but for that we first note that ifx 6∈ Qi, then rad(Qi : x) = Pi. To see this, note that

Qi ⊂ (Qi : x) ⊂ Pi.

The first inclusion is always true. For the second one, note that if y ∈ (Qi : x),then xy ∈ Qi ⊂ Pi, and since x 6∈ Qi, it implies that y ∈ rad(Qi) = Pi.

Taking radicals of all three we squeeze rad(Qi : x) between Pi and Pi.


All this to say:

rad(I : x) =⋂x 6∈Qi

rad(Qi : x) =⋂x6∈Qi

Pi.

Now suppose rad(I : x) = P is prime. Then

P =⋂x 6∈Qi

Pi

which by a theorem long in the past implies that P = Pj for some j. Ergo

{ rad(I : x) prime } ⊂ {P1, P2, . . . , Pk }.

For the opposite inclusion, note that by minimality of the decomposition thereexists some xi 6∈ Qi, yet

xi ∈⋂j 6=i

Qj .

Then rad(I : xi) = Pi since all other Pj contain xi by the above intersection.

Remark 13.1.6. (i) For each I, there exists some xi ∈ A such that (I : xi) isPi primary.

(ii) Consider the A-moduile A/I. The prime ideals Pi are precisely the primeideals that occur as radicals of annihilators of elements of A/I since (I : x)corresponds to (0 : x) = Ann(x).

Definition 13.1.7 (Associated prime). The prime ideals Pi in the above the-orem are said to belong to I or associated with I.

Remark 13.1.8. The ideal I is primary if and only if it has only one associatedprime ideal.

Example 13.1.9. Let A = k[x, y], with k a field. The ideal I = (x2, xy) isnot primary since y is a zero divisor but not nilpotent. We can decompose itas I = P1 ∩ P 2

2 with P1 = (x) and P2 = (x, y). The first one is a prime ideal,and the second one is maximal, making P 2

2 primary. Therefore P1 and P2 areunique, but the decomposition itself is not:

(x2, xy) = (x) ∩ (x2, xy, y2) = (x) ∩ (x2, y)

are two valid primary decompositions, but the radicals are of course the same.N

There is another aspect of the decomposition which is unique: the componentcorresponding to the smallest prime.

Definition 13.1.10 (Minimal prime, isolated prime). The minimum elementsof {P1, P2, . . . , Pk } is called the minimal or isolated prime ideals belongingto I.

The other prime ideals associated with I are called embedded prime ideals.

Proposition 13.1.11. Let I be a decomposable ideal in A. Then any primeideal P ⊃ I contains a minimal prime ideal associated with I.

Hence the minimal prime ideals associated with I are precisely the minimalelements in the set {P prime | P ⊃ I }.


Proof. We have P ⊃ I =⋂iQi. Taking radicals, P ⊃

⋂i rad(Qi) =

⋂i Pi,

whereby P ⊃ Pj for some j.

Proposition 13.1.12. Let I be a decomposable ideal and

I =

n⋂i=1

Qi

be a minimal primary decomposition with rad(Qi) = Pi. Then

n⋃i=1

Pi = {x ∈ A | (I : x) ) I }.

In particular, if (0) is decomposable, then the set D of zero-divisors is the unionof prime ideals associated with (0).

Proof. Consider A → A/I. If I is decomposable, then (0) is decomposable inA/I. Indeed

I =n⋂i=1

Qi

is minimal if and only if

(0) =

n⋂i=1

Qi

is minimal. Therefore it suffices to consider I = (0), which means we’re consid-ering D = {x ∈ A | (0 : x) ) (0) }, the set of all zero-divisors, which is the sameas ⋃

x 6=0

rad(0 : x).

So let us check if this is equal to the union of prime ideals:

rad(0 : x) = rad( ⋂P∈SpecA

P : x) =⋂x6∈Pi

Pi ⊂ Pj

and Pi = rad(0 : x) for some x by the previous theorem.

Proposition 13.1.13. Let S be a multiplicative set, and let Q be a P -primaryideal. Then

(i) If S ∩ P 6= ∅, then S−1Q = S−1A;

(ii) If S ∩P = ∅, then S−1Q is S−1P -primary and its contraction in A is Q.

That is to say, localisation preserves the structure of primary ideals.

Proof. For (i), if the intersection is nonempty, then s ∈ S ∩ P , meaning thatsn ∈ S ∩Q for some n, whereby S−1Q is S−1A since it contains a unit.

For (ii), recall that for A→ S−1A, we have the correspondence

{ I | a ∈ I if as ∈ I for some s ∈ S } ←→ {S−1I | ideals in S−1A }.

If as ∈ Q, s ∈ S, and Q ⊃ P with S∩P = ∅, then S∩Q = ∅, and so a ∈ Q, sinceotherwise if a 6∈ Q we have s ∈ rad(Q) = P , a contradiction. Therefore Q is inone-to-one correspondence with S−1Q. Moreover rad(S−1Q) = S−1(rad(Q)) =S−1P .

RING EXTENSIONS 44

Lecture 14 Ring Extensions

14.1 Second Uniqueness Theorem

Recall how when we localise a ring A into S−1A, we carry over the ideals Isuch that as ∈ I for all s ∈ S implies a ∈ I, and such ideals are in one-to-onecorrespondence with the ideals S−1I of S−1A.

We move between these ideals by extension and contraction, and naturallythe correspondence is true for prime ideals as well and, as we proved at the endof the last lecture, also for primary ideals.

For any ideal I in A, the contraction in A of the ideal e(I) is denoted byS(I), i.e. S(I) = c(e(I)).

Proposition 14.1.1. Let A be a ring and S ⊂ A a multiplicative subset. Let Ibe a decomposable ideal and let

I =

n⋂n=1

Qi

be a minimal primary decomposition. Moreover let Pi = rad(Qi) for each i =1, 2, . . . , n.

Assume Pi ∩S = ∅ for all i = 1, 2, . . . ,m and Pi ∩S 6= ∅ for the remainingi = m+ 1,m+ 2, . . . , n.

Then

S−1I =

m⋂i=1

S−1Qi

and

S(I) =

m⋂i=1

Qi.

Before we prove this, note that it is always possible to arrange the conditionsabove: just reorder the primary decomposition so that the ideals whose radicalsdon’t meet S are at the front.

Proof. Both of these are fairly straight forward:

S−1I = S−1( n⋂i=1

Qi

)=

n⋂i=1

S−1Qi =

m⋂i=1

S−1Qi

since if Qi ∩ S 6= ∅, then Qi contains a unit in S−1A, so S−1Qi = S−1A, andso doesn’t influence the intersection.

The second one is similar. Since we have correspondence between primaryideals,

S(I) = c(S−1I) = c( m⋂i=1

S−1Qi

)=

m⋂i=1

c(S−1Qi) =

m⋂i=1

Qi.

With this in hand we are able to prove the second uniqueness theorem forprimary decompositions:


RING EXTENSIONS 45

Theorem 14.1.2 (Second uniqueness theorem). Let

I =

n⋂i=1

Qi

be a minimal primary decomposition with Pi = rad(Qi). Suppose Pj is a min-imal element of {P1, P2, . . . , Pn } under inclusion. Let S = A \ Pj. ThenQj = S(I) = c(S−1I) is uniquely determined by I and Pj, i.e. the primarycomponent belonging to a minimal prime is uniquely determined and does notdepend on the primary decomposition.

Proof. First note that since Pj is a minimal element of {P1, P2, . . . , Pn }, wehave that Pj 6⊃ Pi for i 6= j, implying that Pi ∩ S 6= ∅ for all i 6= j.

Therefore by the previous proposition, S(I) = Qj .

Example 14.1.3. Let A = k[x, y]. Then I = (x2, y2) = (x) ∩ (x, y)2 = (x) ∩(x2, y), with rad(x) = (x) = P1 and rad(x, y)2 = (x, y) = rad(x2, y) = P2. Wehave P1 ⊂ P2, so P1 is minimal and Q1 = (x) must therefore be in any primarydecomposition of I. N

Note that since ⊂ is a partial order, we can have multiple minimal elements.

14.2 Integral Ring Extensions

The goal of this next part is to try to generalise the idea of algebraic fieldextensions to the setting of rings. Recall how if we have a field extension Fof a field k, then u ∈ F is called algebraic over k if there exists a nonzeropolynomial

f(x) = anxn + an−1x

n−1 + . . .+ a1x+ a0 ∈ k[x]

with an 6= 0 such that f(u) = 0. Since k is a field, all nonzero elements are units,meaning that they have multiplicative inverses, so we can always multiply bya−1n and get a monic polynomial. This is the prototype for our generalisation:

Definition 14.2.1 (Integral element). Let B ⊃ A be a ring extension. Anelement b ∈ B is integral over A if there exists a monic polynomial

f(x) = xn + an−1xn−1 + . . .+ a1x+ a0 ∈ A[x]

such that f(b) = 0.

Example 14.2.2. Let A = Z ⊂ B = Q. Suppose

b =r

s∈ Q

is integral over Z, with gcd(r, s) = 1. Thus(rs

)n+ an−1

(rs

)n−1+ . . .+ a1

(rs

)+ a0 = 0

with ai ∈ Z. If we multiply by sn, then

rn + an−1rn−1s+ . . .+ a1rs

n−1 + a0sn = 0,

which means that if we move all the terms with s over to the other side, wesee that s | rn. Since s and r are coprime, this means s = ±1, and so b ∈ Z.Therefore all rational numbers integral over Z are Z, motivating the name. N

RING EXTENSIONS 46

We will eventually study the same structure for algebraic number fields, i.e.finite extensions of Q, so

Q ⊂ K

Z ⊂ O

with O being the ring of integers of the extensions.

Proposition 14.2.3. Let B ⊃ A be a ring extension. Let x ∈ B. The followingare equivalent:

(i) x ∈ B is integral over A.

(ii) A[x] is a finitely generated A-module.

(iii) A[x] is contained in a subring C ⊂ B such that C is a finitely generatedA-module.

(iv) There exists a faithful A[x]-module M which is finitely generated as anA-module.

Proof. That (i) implies (ii) is seen as follows: Since x ∈ B is integral over A,we have

xn + an−1xn−1 + . . .+ a1x+ a0 = 0

with ai ∈ A. Therefore

xn = −(an−1xn−1 + . . .+ a1x+ a0),

whereby all xm with m ≥ n can be expressed in terms of 1, x, . . . , xn−1. There-fore A[x] is finitely generated by these as an A-module.

To see that (ii) implies (iii), just take C = A[x].For (iii) implying (iv), take M = C. Suppose b ∈ Ann(M), meaning that

bM = 0. We have 1 ∈ M = C ⊂ B, so b · 1 = 0, implying b = 0, meaning thatAnn(M) = 0, so M is faithful.

Finally (iv) implies (i): Consider φ : M → M by φ(m) = mx. This is anA-module homomorphism. By Cayley-Hamilton’s theorem, we have

φn + an−1φn−1 + . . .+ a1φ+ a0 = 0

with ai ∈ A. Therefore

(xn + an−1xn−1 + . . .+ a1x+ a0)M = 0,

so xn + an−1xn−1 + . . .+ a1x+ a0 ∈ Ann(M) over A[x]. But Ann(M) = 0 since

M is faithful, whereby

xn + an−1xn−1 + . . .+ a1x+ a0 = 0

so x is integral over A.

Corollary 14.2.4. Let x1, x2, . . . , xn ∈ B be integral over A.Then A[x1, x2, . . . , xn] is a finitely generated A-module.

RING EXTENSIONS CONTINUED 47

Proof. This follows immediately by induction—we add one xi at a time and usethe previous proposition.

Corollary 14.2.5. Let B ⊃ A be a ring extension. Let

C = {x ∈ B | x integral over A }.

Then X is a subring of B containing A.

Proof. For x, y ∈ C, we wish to show that x ± y ∈ C and that xy ∈ C. NowA[x, y] is a finitely generated A-module, and A[x ± y] ⊂ A[x, y], so by theprevious proposition x± y are integral over A, and similarly for xy.

Note that it is quite cumbersome to prove this directly, i.e. given two polyno-mials with x and y as roots, constructing the polynomials with x ± y or xy asroots.

Definition 14.2.6 (Integral closure). The ring C above is called the integralclosure of A in B.

If C = A, then A is called integrally closed in B.If C = B, then C is called integral over A.

Example 14.2.7. Consider Q ⊃ Z again. We know Z[1/3] is not integral overZ, so by the above Z[1/3] is not finitely generated as a Z-module. (Indeed weknow this already, from an exercise!) N

Example 14.2.8. Let A be an integral domain. Let f ∈ A be a nonzero non-unit in A. Then A[1/f ] is not integral over A since it is not finitely generatedas an A-module. N

Example 14.2.9. The golden ratio

x =1 +√

5

2

is integral over Z, since it is a root to the polynomial equation x2−x−1 = 0. N

Example 14.2.10. The element

x =1 +√

3

2

is not integral over Z, since it is a root of 2x2 − 2x− 1, which is not monic. N

Example 14.2.11. Let k be a field. Then k[x] ⊃ k[x2] = A is an integralextension, since x is a root of f(t) = t2 − x2 ∈ A[t], so x is integral over A. N

Lecture 15 Ring Extensions continued

15.1 Last Lecture, concluded

Corollary 15.1.1. Let A ⊂ B ⊂ C be ring extensions. Suppose B is integralover A and C is integral over B. Then C is integral A.


THE GOING-UP THEOREM 48

In other words, the property of being integral is transitive.

Proof. Let x ∈ C. Then we have

xn + bn−1xn−1 + . . .+ b1x+ b0 = 0

with bi ∈ B. Now consider B′ = A[b0, b1, . . . , bn−1] ⊂ B. Then x is inte-gral over B′, and B′[x] is a finitely generated B′-module. Moreover B′[x] =A[b0, b1, . . . , bn−1, x], so B′ is a finitely generated A-module by the propositionlast lecture.

Therefore B′[x] is a finitely generated A-module, and so A[x] is integral overA.

Corollary 15.1.2. Let A ⊂ B be a ring extension. Let C be the integral closureof A in B. Then C is integrally closed in B.

Lecture 16 The Going-Up Theorem

16.1 Integral Dependence

Proposition 16.1.1. Let A ⊂ B be a ring extension. Assume B is integralover A.

(i) Let J ⊂ B be an ideal and let I = J ∩A. Then B/J is integral over A/I.

(ii) Let S be a multiplicative subset of A. Then S−1B is integral over S−1A.

Proof. In the proof of both we will let x ∈ B which, being integral over A,satisfies an equation

xn + an−1xn−1 + . . .+ a1x+ a0 = 0

with ai ∈ A.For (i), we reduce this equation modulo J , whereby x + J is integral over

A/I since it satisfies such an equation.For (ii), let x/s ∈ S−1B, with x ∈ B as above and s ∈ S. We multiply the

integral equation by 1/sn, so(xs

)n+an−1

s

(xs

)n−1+ . . .+

a1

sn−1

(xs

)+a0

s= 0.

Now ai/sn−i ∈ S−1A, so x/s is integral over S−1A.

Proposition 16.1.2. Let A ⊂ B be integral domains, and assume B is integralover A. Then B is a field if and only if A is a field.

Proof. First assume B is a field, and let take any nonzero x ∈ A ⊂ B. Thenx−1 = y ∈ B since B is a field. We want to show that y ∈ A as well. Now y isintegral over A since it is in B, so

yn + an−1yn−1 + . . .+ a1x+ a0 = 0



with ai ∈ A. We solve for y:

y = −y−(n−1)(an−1yn−1 + . . .+ a1x+ a0)

= −(an−1 + an−2y−1 + . . .+ a1y

−(n−2) + a0y−(n−1))

= −(an−1 + an−2x+ . . .+ a1xn−2 + a0x

n−1)

which is in A since ai and x are all in A. Therefore x−1 ∈ A, and so A is a field.For the converse, take y 6= 0 in B. Since B is integral over A, we again have

yn + an−1yn−1 + . . .+ a1x+ a0 = 0

with ai ∈ A. Assume a0 6= 0, otherwise factor out and cancel a y—we’re inan integral domain, so we have cancellation laws. Thus a0 is a unit in A, soa−1

0 ∈ A. We then multiply by a−10 rearrange to solve for the resulting constant

1:1 = −ya−1

0 (yn−1 + an−1yn−2 + . . .+ a2y + a1),

and all terms are in B, so y−1 ∈ B, and so B is a field.

Corollary 16.1.3. Let A ⊂ B be a ring extension. Assume B is integral overA. Let Q be a prime ideal of B, and let P = Q ∩ A. Then Q is maximal in Bif and only if P is maximal in A.

Proof. First note that P is prime in A since Q is prime in B, since Q is thepreimage of P under the inclusion.

By the first proposition of the lecture, B/Q is integral over A/P , and sinceQ and P are prime ideals, these quotients are integral domains.

So by the previous result, B/Q is a field if and only if A/P is a field, and soQ is maximal in B if and only if P is maximal in A.

Example 16.1.4. Consider, as in the past, a number field K with its ring ofintegers:

K ⊃ O ⊃ P 6= 0 prime ideal

Q ⊃ Z ⊃ (p).

Now since (p) is a maximal ideal in Z, by the last result P must also be maximalin O.

In particular, consider K = Q(√d), with d being square free. Then

K = Q(√d) ⊃ O = Z[α]

Q ⊃ Z.

where α is (1 +√d)/2 is d ≡ 1 (mod 4) and α =

√d is d 6≡ 1 (mod 4). N

Corollary 16.1.5. Let A ⊂ B be a ring extension. Assume B is integral overA, and let Q1 and Q2 be prime ideals of B such that Q1 ⊂ Q2. Moreoversuppose Q1 ∩A = Q2 ∩A = P . Then Q1 = Q2.


In other words, if we have two prime ideals above P , with one contained in theother, then the prime ideals coincide. If one is not contained in the other, thisneed not be the case.

Proof. Localising at P , we have that S−1P is the maximal ideal in the localring AP . Hence by the previous result S−1Q1 and S−1Q2 are maximal in BP ,and since one is contained in the other they must coincide, since they are bothmaximal.

Finally by the one-to-one correspondence of ideals, Q1 = Q2.

Theorem 16.1.6. Let A ⊂ B be a ring extension. Suppose B is integral overA. Let P ⊂ A be a prime ideal. Then there exists a prime ideal Q of B suchthat Q ∩A = P .

Proof. The strategy is to localise both A and B at P , yielding

B BP

A AP .

ϕ

The reason for this is that if we take any maximal ideal m in BP , its intersectionwith AP is a maximal ideal in AP , but AP is a local ring and has only themaximal ideal S−1P .

But then ϕ−1(m) = Q, and so it is prime in B, and since the diagramcommutes we must also have Q ∩A = P .

16.2 The Going-Up Theorem

Theorem 16.2.1 (Going-up theorem). Let A ⊂ B be a ring extension. AssumeB is integral over A. Let P1 ⊂ P2 ⊂ P3 ⊂ . . . ⊂ Pn be a chain of prime idealsin A, and let Q1 ⊂ Q2 ⊂ . . . ⊂ Qm be a shorter chain of prime ideals in B, i.e.m < n, and moreover Qi∩A = Pi for i = 1, 2, . . . ,m. Then the chain of Qi canbe extended to Q1 ⊂ Q2 ⊂ . . . ⊂ Qn such that Qi ∩A = Pi for i = 1, 2, . . . , n.

Before proving this we’ll demonstrate the objective in a diagram. We want tofind Q2 in the diagram

B ⊂ Q1 Q2

A ⊂ P1 ⊂ P2

in such a way that Q1 ⊂ Q2. By the previous proposition we already know thatwe can find something above P2, so the clincher is the inclusion in Q1.

Proof. It suffices to consider n = 2 and m = 1, since we could then repeat.The way to accomplish the inclusion we desire is to work in A = A/P1 and

B = B/Q1, since all ideals that remain after taking quotients are ones thatcontain the ideal we divided by. Now we know that B is integral over A, andso by the previous proposition there exists a prime ideal Q2 ⊂ B such thatQ2 ∩ A = P2.

Let ϕ : B → B/Q1, and like before let Q2 = ϕ−1(Q2). Then Q2 ⊂ B is aprime ideal such that Q2 ∩A = P2, and we are done.

THE GOING-DOWN THEOREM 51

Proposition 16.2.2. Let A ⊂ B be a ring extension, and let C be the integralclosure of A in B. Let S ⊂ A be a multiplicative subset. Then S−1C is theintegral closure of S−1A in S−1B.

Proof. Suppose b/s ∈ S−1B, which we know is integral over S−1A. Then( bs

)n+an−1

sn−1

( bs

)n−1+ . . .+

a1

s1

( bs

)+a0

s0= 0,

with ai/si ∈ S−1A, i.e. ai ∈ A and si ∈ S. Let t = s0s1 · . . . · sn−1 ∈ S, andmultiply the integral equation above by (st)n, whereby

(bt)n + dn−1(bt)n−1 + . . .+ d1(bt) + d0 = 0

with di ∈ A, meaning that bt is integral over A, and so bt ∈ C since C is theintegral closure of A. Therefore

b

s=bt

st∈ S−1C

hence S−1C is the integral closure of S−1 over S−1B.

Definition 16.2.3 (Integrally closed). An integral domain is called integrallyclosed or normal if it is integrally closed in its field of fractions.

Example 16.2.4. We have seen already that Z is integrally closed over Q, andsince Q is its field of fractions, it is integrally closed.

In fact, any unique factorisation domain is integrally closed—the proof isalmost identical to that of Z over Q, modulo some small details. In particulark[x1, x2, . . . , xn] with k a field is integrally closed. N

Proposition 16.2.5. Let A be an integral domain. Then the following areequivalent:

(i) A is integrally closed.

(ii) AP is integrally closed for every prime ideal P of A.

(iii) Am is integrally closed for every maximal ideal m of A.

Proof. Let k be the field of fractions of A, and let C be the integral closure ofA in k. Consider the embedding f : A ↪→ C, which induces fp : AP ↪→ CP andfm : Am ↪→ Cm.

Now A is integrally closed if and only if A = C, and likewise AP = CP ifand only if AP is integrally closed, and the same for Am. In other words Ais integrally closed if and only if f is surjective, and we know surjectivity is alocal property, so this is true if and only if fP is surjective, if and only if fm issurjective, and we are done.

Lecture 17 The Going-Down Theorem

17.1 The Going-Down Theorem

Lemma 17.1.1. Let A ⊂ B be a ring extension. Suppose B is integral over A.Let I ⊂ A be an ideal and b ∈ B. Then the following are equivalent:



(i) b is integral over I;

(ii) bn is integral over I for some n ≥ 1;

(iii) bn ∈ IB for some n ≥ 1 (i.e. b ∈ rad(IB)).

In particular, if I is integrally closed in B, then I = rad(I).

Proof. That (i) and (ii) are equivalent is trivial: if b is integral certainly bn is,and if bn is integral, then its integral equation is an integral equation for b, justwith other powers.

Let us then prove that (ii) implies (iii). If bn is integral over I, then wehave

(bn)m + am−1(bn)m−1 + . . .+ a1(bn) + a0 = 0

for ai ∈ I. If we rearrange this to solve for bnm we get

bnm = −(am−1(bn)m−1 + . . .+ a0),

which is in IB since the coefficients are in I and the powers of b are in B.Therefore bnm ∈ IB, and we have (iii).

Next for (iii) implying (i) and (ii), we have bn ∈ IB, meaning that

bn =

k∑i=1

aibi

for ai ∈ I and bi ∈ B. Let M = A[b1, b2, . . . , bk], which is a finitely generatedA-module since b1, b2, . . . , bk are integral over A.

If we now define φ : M → M by φ(x) = bnx, then φ(M) ⊂ IM . By thegeneralised Cayley-Hamilton theorem, we have

φm + a1φm−1 + . . .+ am = 0

for ai ∈ I. Now evaluate this at 1 ∈M , giving us

(bn)m + a1(bn)m−1 + . . .+ a0 = 0,

whereby b and bn are integral over I.Finally assume I is integrally closed in B. The inclusion I ⊂ rad(I) is trivial,

since this is true for all ideals. The opposite inclusion, rad(I) ⊂ I, we get bynoting that if a ∈ rad(I), then an ∈ I ⊂ IB. By (iii) implying (i), we havethat a is integral over I, whereby a ∈ I, since I is integrally closed.

Lemma 17.1.2. Let A ⊂ B be a ring extension, and I ⊂ A an ideal. Let Cbe the integral closure of A in B, and let e(I) = CI, the extension of I in C.Then the integral closure of I in B is rad(e(I)) = rad(IC)

Proof. By the above, the integral closure of I in C is rad(IC). So if b ∈ B isintegral over I ⊂ A, then b ∈ C as well, so then integral closure of I in B israd(IC) as well.


Definition 17.1.3 (Algebraic element). Let F be a field extension of a field k.An element a ∈ F is algebraic over k if a is a root of a nonzero polynomial

f(t) = antn + an−1t

n−1 + . . .+ a1t+ a0 ∈ k[t],

with an 6= 0. Since k is a field we may assume an = 1, otherwise multiplythrough by a−1

n . If f(t) is the polynomial in k[t] of smallest degree such thatf(a) = 0 and f is monic, then we call f the minimal polynomial of a. Thispolynomial is unique.

Proposition 17.1.4. Let A ⊂ B be integral domains, and assume A is inte-grally closed. Let I ⊂ A be an ideal, and suppose b ∈ B is integral over I. Thenb is algebraic over the field of fractions k of A, and its minimal polynomial overk is

f(t) = tn + an−1tn−1 + . . .+ a1t+ a0

with ai ∈ rad(I).

Proof. We want to show that ai ∈ rad(I). Let L be the splitting field of f(t)over k, wherein we have, say,

f(t) = (t− b1)(t− b2) . . . (t− bn).

Since b is integral over I, there exists some g(t) ∈ I[t] such that g(b) = 0. There-fore f(t) | g(t) in k[t] by the division algorithm since f is minimal. Moreoverthis implies g(bi) = 0 for all i = 1, 2, . . . , n, and therefore bi are integral over I.

Expanding the polynomial f we have

f(t) = (t− b1)(t− b2) . . . (t− bn) = tn + an−1tn−1 + . . .+ a1t+ a0

where ak are polynomials in bi. Now since bi are integral over I, so are the ak,and so by the previous lemma we have ak ∈ rad(I), since I is its own extensionsince A is integrally closed.

Theorem 17.1.5 (Going-down). Let A ⊂ B be integral domains. Assume B isintegral over A and that A is integrally closed. Let P1 ⊃ P2 be prime ideals ofA and Q1 a prime ideal in B such that Q1 ∩A = P1. Then there exists a primeideal Q2 of B such that Q1 ⊃ Q2 and Q2 ∩A = P2.

To prove this we use the following lemma:

Lemma 17.1.6. Let φ : A → B be a ring homomorphism. Let P ⊂ A be aprime ideal. Then P is the contraction of a prime ideal of B if and only ifc(e(P )) = P .

Proof.

Proof of theorem. By the lemma, it suffices to show that BQ1P2 ∩ A = P2, so

by the lemma there exists a prime Q2 ⊂ BQ1such that Q2 ∩ A = P2. Take

Q2 ∩B := Q2.Suppose BQ1

P2 ∩ A 6= P2. Then there exists x ∈ BQ1P2 ∩ A with x 6∈ P2.

Note that since x ∈ BQ1P2, we have x = y/s for y ∈ BP2 and s = S = B \Q1.Then by the earlier lemma, specifically (iii) implying (i), we know that y is

VALUATION RINGS 54

integral over P2, and so by the proposition the minimal polynomial of y overthe quotient field k of A is of the form

f(t) = tn + an−1tn−1 + . . .+ a1t+ a0

with ai ∈ rad(P2) = P2 since P2 is prime. Therefore

yn + an−1yn−1 + . . .+ a1y + a0 = 0,

which if we multiply by 1/xn becomes(yx

)n+an−1

x

(yx

)n−1+ . . .+

a1

xn−1

(yx

)+a0

xn= 0.

Now y/x = s, so

sn +an−1

xsn−1 + . . .+

a1

xn−1s+

a0

xn= 0

and sog(t) = tn +

an−1

xtn−1 + . . .+

a1

xn−1t+

a0

xn∈ k[t]

is the minimal polynomial of s in k[t] (if it isn’t minimal, we could reduce thedegree by f , but f is minimal, so that can’t be the case).

Now s ∈ S = B \Q1. Since s ∈ B is integral over A, we have

an−1

xi∈ rad(A) = A

and soan−ixi

= di ∈ A.

Moreover an−i = dixi, where the left-hand side if in P2 but xi is not, so by P2

being prime we have di ∈ P2. Therefore g(t) ∈ P2[t], so s is integral over P2,implying that sn ∈ P2B ⊂ P1B ⊂ Q1, but s 6∈ Q1, which is a contradiction.Therefore, recalling that our assumption was BQ1

P2 ∩A 6= P2, we must insteadhave BQ1

P2 ∩A = P2.

Counterexample 17.1.7. To see that this does not work if A isn’t integrallyclosed, consult page 32 of ‘Commutative Algebra’ by Matsumaia. N

Lecture 18 Valuation Rings

18.1 Between Rings and Fields

Theorem 18.1.1. Let A be a subring of the field K and h : A → C a ringhomomorphism where C is an algebraically closed field. If α 6= 0 is in K \ A,then either h can be extended to a ring homomorphism h : A[α] → C or toh : A[α−1]→ C.


VALUATION RINGS 55

Proof. We reduce to the case when A is a local ring and F = h(A) is a subfieldof C. Let P be the kernel of h. Then P ⊂ A is a prime ideal (since 0 is aprime ideal in C, and kerh is the pullback of it). Extend h to g : AP → C byg(a/s) = h(a)/h(s) with s ∈ A \P . Therefore h(s) 6= 0, so the inverse of g(a/s)exists in C. Moreover AP ⊂ K as well, and ker g = PAP is the maximal idealin AP , so by the First isomorphism theorem

g(AP ) = Im g ∼=AP

ker g=

APPAP

,

a field. So we replace (h,A) with (g,AP ).Thus we assume A is local and that F = h(A) a subfield of C. First extend

h to a surjective homomorphism of polynomial rings h : A[x] → F [x] in thefollowing way: the elements in A[x] are

f(x) =

n∑i=1

aixi ∈ A[x]

and so

h(f) =

n∑i=1

h(ai)xi.

We can’t just replace x by α since it might not make h well-defined (we couldadd any polynomial which vanishes in α).

LetI = { f ∈ A[x] | f(α) = 0 }.

Then J = h(I) is an ideal of F [x] (this is easy to see—if two polynomials vanishin α, so do their sum, and if one polynomial vanishes in α, so does the productof it and anything else). Since F is a field, F [x] is a principle ideal domain bythe division algorithm, meaning that J = (p(x)) for some p(x) ∈ F [x].

We have three possibilities. First, p(x) is nonconstant. Then p(x) has a rootβ ∈ C since C is algebraically closed. Define h : A[α] → C by h(α) = β. Thenh is well-defined, since if g(α) = g(α) + f(α) for f ∈ I, we have

h(g(α)) = h(g(α) + f(α)) = h(g(α)) + h(f(α)).

Since h(f) ∈ J = (p(x)), we have h(f) = p(x)q(x) for some q(x) ∈ F [x] whichmeans that

h(f(α)) = p(β)q(β) = 0,

so h is well-defined.Secondly, if p(x) is identically 0, i.e. J = (0), then α is a free variable—

everything goes to 0. So defining h(α) = β with β arbitrary is sufficient.Third and finally, suppose p(x) = c 6= 0. This means that J = (1), the whole

space, and so there exists some f ∈ I such that h(f) = 1. If

f(x) = arxr + ar−1x

r−1 + . . .+ a1x+ a0 ∈ A[x]

thenr∑i=0

aiαi = 0 (18.1.1)

VALUATION RINGS 56

since f ∈ I. Moreover h(f) = 1, so h(a0) = 1 and h(ai) = 0 for i = 1, 2, . . . , r.Let r be the smallest degree amongst such f , and consider α−1.

If α−1 is in the first or second case, then we extend h : A[α−1]→ C. Hencewe can assume α−1 is also in the third case, so we have

s∑i=0

bi(α−1)i = 0 (18.1.2)

with bi ∈ A, h(b0) = 1, and h(bi) = 0 for i = 1, 2, . . . , s, and s is the smallestdegree amongst these.

Without loss of generality we assume r ≥ s. Note that h(b0) = 1 = h(1),meaning that 1 − b0 ∈ kerh ⊂ M , the unique maximal ideal of A. Therefore1− (1− b0) = b0 is a unit in A, so b−1

0 ∈ A.Multiplying (18.1.2) by b−1

0 αs we get

αs + b−10 b1α

s−1 + . . .+ b−10 bs = 0 (18.1.3)

where the coefficients are in A. Now if r > s, then (18.1.1) minus αr−s times(18.1.3) produces a smaller degree r satisfying (18.1.1), which is a contradictionsince we assumed minimality.

Next if r = s, then (18.1.1) minus ar times (18.1.3) is 0, so a0 = arb−10 bs,

buth(a0) = h(arb

−10 bs) = 0

where the left-hand side if 1 but the right-hand side is 0 because of bs, which isa contradiction.

Definition 18.1.2. A subring B of a field K is called a valuation ring of Kif for every non-zero α ∈ K either α ∈ B or α−1 ∈ B (or both).

Example 18.1.3. The field K is a valuation ring of itself. N

Example 18.1.4. Take K = Q and p ∈ Z a prime. Then

Zp =

{prm

n∈ Q

∣∣∣∣ r ≥ 0, gcd(m, p) = gcd(n, p) = 1

}is a valuation ring of Q. N

Let K be a field and C an algebraically closed field. Let

Σ = { (A, f) |A a subring of K, f : A→ C a ring homomorphism }.

We define a partial order on Σ as follows: (A, f) ≤ (B, g) if and only if A ⊂ Band g

∣∣A

= f . That is to say, (B, g) is an extension of (A, f).It is easy to verify that Σ satisfies the conditions of Zorn’s lemma, meaning

that Σ has at least one maximal element.

Theorem 18.1.5. Let (B, g) be a maximal element of Σ. Then B is a valuationring of the field K.

Proof. Let α 6= 0, α ∈ K. Then g has an extension to B[α] or B[α−1]. Bymaximality of (B, g) we must therefore have B[α] = B or B[α−1] = B, meaningthat α ∈ B or α−1 ∈ B.

CHAIN CONDITIONS 57

Proposition 18.1.6. Let B be a valuation ring of K. Then

(i) The fractional field of B is K.

(ii) If A is a subring such that B ⊂ A ⊂ K, then A is a valuation ring of K.

(iii) B is a local ring.

(iv) B is integrally closed in K.

Proof. For (i), simply note that α ∈ K implies α ∈ B or α−1 ∈ B. In the firstcase we are done, and in the second we note that 1/α−1 = α, and again we aredone.

For (ii), if α ∈ K we have α ∈ B or α−1 ∈ B, but since A ⊃ B, the elementsin B are also in A.

Taking on (iii), let M be the set of non-units in B. Hence x ∈ M if andonly if x = 0 or x−1 6∈ B. We claim that M is an ideal of B, in which case B isa local ring.

First we check that BM ⊂ M , which is true since ax ∈ M for nonzeroa ∈ B and x ∈ M , for otherwise ax 6∈ M implies (ax)−1 ∈ B, but then(ax)−1a = x−1 ∈ B, a contradiction.

Secondly, if nonzero x, y ∈ M , then we need to show that x + y ∈ M . Butwe have either x/y ∈ B or y/x ∈ B since B is a valuation ring. Without loss ofgenerality assume the former. Then

x+ y = y(xy

+ 1)∈ BM ⊂M,

hence M is an ideal in B.Finally for (iv), suppose x ∈ K is integral over K. Then

xn + bn−1xn−1 + . . .+ b1x+ b0 = 0

with bi ∈ B. If x ∈ B, we are done, and if x 6∈ B then we must have x−1 ∈ B,in which case the above integral equation multiplied by (x−1)n−1 yields

x = −(bn−1 + bn−2x−1 + . . .+ b0(x−1)n−1)

where the left-hand side is in B, so x ∈ B, which is a contradiction.

Lecture 19 Chain Conditions

19.1 Valuation Rings and Integral Closures

Proposition 19.1.1. Let B be a valuation ring of the field K. For any idealsI and J in B, we have either I ⊂ J or J ⊂ I. In other words, the ideals of avaluation ring are ordered by inclusion.

Proof. Suppose I 6⊂ J , i.e. there exists some a ∈ I such that a 6∈ J . Since allideals contain 0, this automatically means a 6= 0.

We claim that J ⊂ I.


CHAIN CONDITIONS 58

For any nonzero b ∈ J , we have a/b ∈ K, so either a/b ∈ B or b/a ∈ B sinceB is a valuation ring. In the case of the first one, b · a/b = a ∈ J since b ∈ Jand J is an ideal, which is a contradiction since a 6∈ J . Therefore we must haveb/a ∈ B, meaning that a · b/a = b ∈ I, so J ⊂ I, and we are done.

The converse is true as well:

Proposition 19.1.2. Let B be an integral domain and K be its field of fractions.Suppose the ideals of B are totally ordered by inclusion. Then B is a valuationring.

Proof. Suppose α ∈ K, α 6= 0. Then we can write α = a/b with a, b ∈ B sinceK is its quotient field. Let I = (a) and J = (b), which by assumption meansthat either I ⊂ J or J ⊂ I.

If (a) ⊂ (b), then a = bc for some c ∈ B, and therefore α = a/b = c ∈ B.If on the other hand (b) ⊂ (a), then b = ac for some c ∈ B, meaning that

α−1 = b/a = c ∈ B.

Corollary 19.1.3. Let B be a valuation ring of the field K. If P ⊂ B is aprime ideal, then BP and B/P are valuation rings of their respective quotientfields.

Proof. This follows directly from the past two propositions: B being a valuationring means that its ideals are totally ordered by inclusion, and that order remainsafter localisation or quotient.

Proposition 19.1.4. Let A be a subring of a field K. Then the integral closureA of A in K is the intersection of all valuation rings B of K such that B ⊃ A,i.e.

A =⋂B⊃A

B valuation ring

B.

Proof. We start by showing A ⊂⋂B⊃A

B. For a ∈ A, a is integral overA, meaning

that a is integral over B, but B is integrally closed since it is a valuation ring,so a ∈ B.

For⋂B⊃A

B ⊂ A, we note that this is true if and only if a 6∈ A implies a 6∈ B,

for some valuation ring B ⊃ A.Suppose a 6∈ A. Then a 6∈ A′ = A[a−1], otherwise

a = bn(a−1)n + bn−1(a−1)n−1 + . . .+ b1a−1 + b0

with bi ∈ A. If we multiply this by an we get

an+1 = bn + bn−1a+ . . .+ b1an−1 + b0a

n.

Therefore a is integral over A, so a ∈ A, which is a contradiction.Thus a−1 is not a unit in A′, meaning that a−1 ∈ M ′ ⊂ A′, with M ′ being

a maximal ideal of A′.

CHAIN CONDITIONS 59

Now let C be an algebraic closure of the field k′ = A′/M ′, and considerh : A′ → A′/M ′ ↪→ C. Then by last lecture h has a maximal extension h : B →C for some evaluation ring B of K. When B ⊃ A′ ⊃ A, we have

A′ A′/M ′ C

B

h

h

Now since a−1 ∈M ′, we have h(a−1) = h(a−1) = 0. If a ∈ B, then

1 = h(1) = h(a · a−1) = h(a)h(a−1) = 0,

so 1 = 0, a contradiction, whereby a 6∈ B.

Corollary 19.1.5. Let A be an integral domain with fraction field K. Then Ais integrally closed if and only if

A =⋂B⊃A

B valuation ring

B.

19.2 Chain Conditions

If we have a k-vector space V with finite dimension, and V has a subspace W ,then we know that W automatically has finite dimension as well.

If, however, M is an A-module, then even if M is finitely generated, asubmodule N of M might not be. We will concern ourselves in the near futurewith the special case where the submodules do inherit the finite generation.

First a statement purely about set theory:

Proposition 19.2.1. Let Σ be a set that is partially ordered by ≤. Then thefollowing conditions on Σ are equivalent:

(i) Σ satisfies the ascending chain condition, meaning that every chainx1 ≤ x2 ≤ x3 ≤ . . . in Σ is stationary, i.e. there exists some n such thatxn = xn+1 = . . ..

(ii) Every nonempty subset of Σ has a maximal element.

Proof. To see that (i) implies (ii), suppose S ⊂ Σ is nonempty with no maximalelement. So for s1 ∈ S there must exist some s2 ∈ S such that s1 < s2, and for s2

there must exist an s3 with s2 < s3, and so forth, and since there is no maximalelement in S this cannot stop, which is a contradiction since s1 < s2 < s2 < . . .is a chain, which by (i) must be stationary.

For the converse, take a chain x1 ≤ x2 ≤ x3 ≤ . . . and let S = {xi }∞i=1.Then by (ii) S has a maximal element, say xn, and so xn = xn+1 = . . ., and weare finished.

This can also be cone with ≥, in which case we consider the descending chaincondition , which is then equivalent with S 6= ∅ having a minimal element.

CHAIN CONDITIONS 60

Definition 19.2.2 (Noetherian and Artinian rings). Let Σ be the set of sub-modules of a module M , ordered by ⊂ and ⊃.

Then M is called Noetherian if Σ satisfies the ascending chain condition,and M is called Artinian if Σ satisfies the descending chain condition.

Example 19.2.3. A finite abelian group (read: Z-module) satisfies both theascending chain condition and the descending chain condition.

The ring Z, seen as a Z-module, satisfies the ascending chain conditionbut not the descending chain condition. This follows from (n) ⊃ (m) beingequivalent with m | n, and we couldn’t keep on factoring distinct terms forever.However for a 6= 0,±1, we have

(a) ) (a2) ) (a3) ) . . .

which never stops.Let k be a field. The ring k[x] satisfies the ascending chain condition but

not the descending chain condition, for exactly the same reason as with Z—it’sa principal ideal domain, and

(x) ) (x2) ) (x3) ) . . . . N

We think of a ring A as an A-module, wherein the submodules are the ideals ofA. Therefore

Definition 19.2.4 (Notherian and Artinian rings). A ring A is called Noethe-rian if it is Noetherian as an A-module. Similarly A is called Artinian if it isArtinian as an A-module.

Remark 19.2.5. We shall prove later on that if a ring A is Artinian (i.e. hasthe descending chain condition) then it is also Noetherian (meaning it has theascending chain condition).

Proposition 19.2.6. An A-module M is Noetherian if and only if every sub-module of M is finitely generated.

Proof. For the forward direction, let N be a submodule of M . Let

Σ = {N ′ ⊂ N |N ′ is a finitely generated submodule }.

Then Σ 6= ∅ since at least 0 ∈ Σ. Hence Σ has a maximal element, say N0. Weclaim that N0 = N .

Suppose N0 ) N . Then there exists x ∈ N such that x 6∈ N0. Consider thesubmodule N ′ = N0 + Ax. This is finitely generated, and N0 ( N ′, which is acontradiction, since N0 is maximal.

For the converse, letM1 ⊃M2 ⊂M3 ⊂ . . .

be a chain of submodules of M . Then

N =

∞⋃n=1

Mn

is a submodule of M—note that unions of submodules aren’t generally sub-modules, but due to the inclusions this is the case here. Hence N is finitelygenerated since every submodule is, say by x1, x2, . . . , xk, with xi ∈Mni

. Taken = max{ni }, whereby x1, x2, . . . , xk ∈ Mn. This means that Mn = Mn+1 =. . ., since Mn contains all of the generators, so the chain is stationary.

NOETHERIAN RINGS 61


0 L M N 0α β

be a short exact sequence of A-modules. Then

(i) M is Noetherian if and only if L and N are Noetherian.

(ii) M is Artinian if and only if L and N are Artinian.

Proof. For the forward direction of (i), we have that every ascending chain inL is ascending in M , and since α is injective we can think of L as submodulesof M , whereby the chain is stationary, so L is Noetherian.

Next let N1 ⊂ N2 ⊂ N3 ⊂ . . . be an ascending chain in N . Taking thepullback, we have

β−1(N1) ⊂ β−1(N2) ⊂ β−1(N3) ⊂ . . .

as an ascending chain in M . Since this chain is in M , which is Noetherian,it is stationary, stopping at, say, β−1(Nk), i.e. β−1(Nk) = β−1(Nk+1) = . . ..Mapping back through β we have β(β−1(Nk)) = Nk, and our chain is stationaryin N as well.

For the converse, let M1 ⊂ M2 ⊂ . . . be an ascending chain in M . ThenL ∩M1 ⊂ L ∩M2 ⊂ . . . is an ascending chain in L, whereby it’s stationary.

Next consider β(M1) ⊂ β(M2) ⊂ . . ., an ascending chain in N , whereby itis stationary.

Now take the largest index n from the chain in L and N , and show thatMn = Mn+1.

Lecture 20 Noetherian Rings

We start by finishing off the proof from last time.

Proof continued. We left off with trying to show that Mn = Mn+1 = . . ., i.e.that our chain is stationary.

Since our chain is ascending by assumption, it suffices to show Mn+1 ⊂Mn.For x ∈ Mn+1, we have β(x) ∈ β(Mn+1) = β(Mn). Therefore there existssome y ∈ Mn such that β(x) = β(y), meaning that β(x − y) = 0 since β is ahomomorphism, and so x− y ∈ kerβ = Imα = L by exactness. In other wordsx ∈Mn+1, and y ∈Mn ⊂Mn+1, so x− y ∈ L∩Mn+1 = L∩Mn, meaning thatx = (x− y) + y ∈Mn.

20.1 When are Submodules Noetherian?

Corollary 20.1.1. If M1,M2, . . . ,Mn are Noetherian (respectively Artinian)A-modules, then

n⊕i=1

Mi

is Noetherian (respectively Artinian).


NOETHERIAN RINGS 62

Proof. Since the sequence

0 M1 M1 ⊕M2 M2 0

is exact, the previous proposition gives us the corollary for n = 2. For n > 2,we use induction on

0 M1

⊕nn=1Mi

⊕ni=2Mi 0.

Example 20.1.2. Any principal ideal domain is Noetherian, since all ideals—which we view as the submodules—are generated by a single element, hencefinitely generated. N

Example 20.1.3. Let k be a field and consider A = k[x1, x2, . . .]. This is nota Noetherian ring, since

(x1) ⊂ (x1, x2) ⊂ (x1, x2, x3) ⊂ . . .

isn’t stationary.But A is an integral domain, meaning that it has a quotient field K, which is

Noetherian—it has only two ideals, (0) and (1) = K, both of which are clearlyfinitely generated. N

This serves to indicate that a subring of a Noetherian ring doesn’t have to beNoetherian, essentially because they aren’t necessarily ideals.

Proposition 20.1.4. Let A be a Noetherian (respectively Artinian) ring. Let Mbe a finitely generated A-module. Then M is Noetherian (respectively Artinian).

Proof. We think about what it means for a module to be finitely generated. IfM is finitely generated, then M ∼= An/N for some n and submodule N . Thenthe sequence

0 N An An

N∼= M 0

is exact. By assumption A is Noetherian, meaning that An is Noetherian, andby our result from last time, then, An/N ∼= M is Noetherian.

Proposition 20.1.5. Let A be a Noetherian ring and I ⊂ A an ideal. ThenA/I is a Noetherian ring.

Proof. The sequence

0 I A AI 0

is exact, meaning that A/I is Noetherian as an A-module. Since I annihilatesit, we can carry this process over and view A/I as a Noetherian A/I-module,and thus as a ring.

MORE ON NOETHERIAN RINGS 63

Lecture 21 More on Noetherian Rings

21.1 Noetherian Rings Have Decomposable Ideals

Recall that we know by now that the ring A being Noetherian means that itsatisfies the following equivalent conditions:

(i) Every nonempty set of ideals in A has a maximal element.

(ii) Every ascending chain of ideals in A is stationary.

(iii) Every ideal in A is finitely generated.

Proposition 21.1.1. Let φ : A→ B be a surjective ring homomorphism. If Ais Noetherian, then B is Noetherian.

Proof. Since φ is surjective, we have by the first isomorphism theorem thatB ∼= A/ kerφ, and this quotient is Noetherian since all of its ideals come fromideals in A, which are finitely generated.

Proposition 21.1.2. Let A ⊂ B be rings. Suppose A is Noetherian and B isfinitely generated as an A-module. Then B is Noetherian.

Proof. Since A is Noetherian, B is Noetherian as an A-module. Therefore everysubmodule of B is finitely generated over A, meaning that it is also fintelygenerated over A since A ⊂ B. Thus B is a Noetherian B-module as well, andso it’s a Noetherian ring.

Proposition 21.1.3. Let A be a Noetherian ring and S ⊂ A a multiplicativeset. Then S−1A is Noetherian.

Proof. Let ϕ : A→ S−1A, and recall that we have a one-to-one correspondencebetween ideals I in A and S−1I = e(I) in S−1A. Now ϕ(I) is a generator ofe(I), and I itself is finitely generated, so ϕ(I) is too.

Corollary 21.1.4. If A is Noetherian and P ⊂ A a prime ideal, then AP isNoetherian.

Remark 21.1.5. The converse is false—though counterexamples are very muchnontrivial—meaning that Noetherian is not a local property.

Recall how if k is a field, then k[x] is a principle ideal domain, and thereforeNoetherian. Surprisingly, the same is true for polynomial rings over Noetherianrings as well.

Theorem 21.1.6 (Hilbert basis theorem). If A is Noetherian, then A[x] isNoetherian.

Proof. Let I ⊂ A[x] be an ideal and define

Jn = { a ∈ A | there exists f(x) ∈ I such that f(x) = axn + . . .. }.

In other words Jn is the set of coefficients a ∈ A such that there are degree npolynomials in I with a as their leading coefficients. It is not hard to see that

Date: November 2, 2017.


Jn is an ideal in A, effectively since I itself is an ideal in A[x]. Moreover wehave an ascending chain, since Jn ⊂ Jn+1 since I is an ideal, meaning that apolynomial in Jn will be in Jn+1 by multiplying it by x.

ThusJ1 ⊂ J2 ⊂ J3 ⊂ . . .

in A, which is Noetherian, so it is stationary. Thus Jn = Jn+1 = . . . for some n,and all Jm are finitely generated, say by am,1, am,2, . . . , am,km . So there existsfm,j(x) = am,jx

m + . . . in I.We claim that the set

{ fm,j(x) } 1≤m≤n1≤j≤km

generates I.To prove this, let F (x) = axs + . . . ∈ I. If deg f = s ≥ n, then a ∈ Js =

Jn = (an,1, an,2, . . . , an,kn). Therefore

a =

kn∑i=1

bian,i,

and therefore

f(x)−kn∑i=1

bifn,i(x)xs−n

is a polynomial in I with smaller degree—specifically less than or equal to s−1.Repeat this until s = n.

Then, for deg f = s ≤ n, we again have a ∈ Js = (as,1, as,2, . . . , as,ks), with

f(x)−ks∑i=1

bifs,i(x)

in I of degree less than or equal to s− 1, so we can again continue inductivelyuntil s = 0, and so f(x) is a linear combination of fm,j(x).

We can extend this inductively:

Corollary 21.1.7. If A is Noetherian, then A[x1, x2, . . . , xn] is Noetherian.

Definition 21.1.8 (Algebra). Let A and B be rings. Then B is an A-algebraif

(i) B is an A-module, and

(ii) a(b1b2) = (ab1)b2 for a ∈ A and b1, b2 ∈ B.

Basically, we want the operations in B to be compatible with those of A.

Corollary 21.1.9. Let B be a finitely generated A-algebra. If A is Noetherian,then B is a Noetherian ring. In particular every finitely generated algebra overZ or a field k is Noetherian.

Proof. Much like how a finitely generated module of A can be identified asAn/N for some submodule N , we can identify B as B ∼= A[x1, x2, . . . , xn]/Nfor some subalgebra, and this is Noetherian by the Hilbert basis theorem.


A while ago we went through great pains to prove several interesting resultsabout primary decompositions, but each theorem began with assuming decom-posability. It turns out that if we work over Noetherian rings, this is not aproblem:

Theorem 21.1.10. Let A be a Noetherian ring. Then every ideal of A has aprimary decomposition.

We’ll prove this by means of two lemmas, but first:

Definition 21.1.11 (Irreducible ideal). An ideal I ⊂ A is irreducible if I =J1∩J2, J1 and J2 ideals, then I = J1 or I = J2. An ideal which is not irreducibleis called reducible .

Lemma 21.1.12. Let A be a Noetherian ring. Then every ideal in A is a finiteintersection of irreducible ideals.

Proof. Suppose this is not the case, and let Σ be the set of ideals in A that cannotbe written as finite intersections of irreducible ideals. Then Σ is nonempty byour supposition, and since A is Noetherian, Σ has a maximal element, call it I.Then I is reducible, so I = J1 ∩ J2 with J1 ) I and J2 ) I.

By maximality of I, we have J1, J2 6∈ Σ, meaning that J1 and J2 are finiteintersections of irreducible ideals, and so I = J1 ∩ J2 is as well, which is acontradiction.

Lemma 21.1.13. Let A be a Noetherian ring. Then every irreducible ideal inA is primary.

Proof. First note that I ⊂ A is irreducible if and only if (0) is irreducible inA/I. The forward direction of this is quite clear. The converse direction followsfrom noting that if (0) = J1 ∩ J2, then J1 ∩J2 = I since I = (0) in the quotient.

Next, I ⊂ A is primary if and only if (0) is primary. It suffices to show thatif (0) in A is irreducible, then (0) is primary. Hence suppose xy = 0, and y 6= 0.We therefore need to show that xn = 0 for some n ≥ 1. Consider the ideals

Ann(x) ⊂ Ann(x2) ⊂ Ann(x3) ⊂ . . .

which, since A is Noetherian, is stationary, so Ann(xn) = Ann(xn+1) = . . . forsome n. We claim (xn) ∩ (y) = (0). Suppose a ∈ (xn) ∩ (y). Then a ∈ (y)implies a = by for some b, so ax = byx = 0 since yx = 0 Since a ∈ (xn), we alsohave a = cxn for some c, which means that if we multiply both sides by x, weget ax = cxn+1 = 0 by the above.

Thus c ∈ Ann(xn+1) = Ann(xn), meaning that cxn = 0 = a, so a = 0, andtherefore (0) = (xn) ∩ (y), (0) is irreducible, and (y) 6= (0), so we must have(xn) = 0, meaning that xn = 0, meaning that (0) is primary.

Proposition 21.1.14. Let A be Noetherian and I ⊂ A an ideal. Then I ⊃rad(I)n for some n ≥ 0.

Since I ⊂ rad(I) for all I, thus means that in the particular case of Noetherianrings, we have

rad(I)n ⊂ I ⊂ rad(I).

ARTINIAN RINGS 66

Proof. Since rad(I) is an ideal of A, which is Noetherian, rad(I) must be finitelygenerated, say by x1, x2, . . . , xm. Moreover since they are in the radical, we havexa11 , xa22 , . . . , xamm ∈ I for some powers ai. Letting n = a1 + a2 + . . . + am, wehave that rad(I) is generated by xrii with r1 + r2 + . . . + rm = n, and xrii ∈ I,so indeed rad(I)n ⊂ I.

Corollary 21.1.15. Let A be Noetherian. Then nilrad(A) is nilpotent.

Proof. The proof is straight forward: nilrad(A) = rad(0).

Corollary 21.1.16. Let A be Noetherian and m ⊂ A be a maximal ideal. LetQ ⊂ A be any ideal. Then the following are equivalent:

(i) Q is m-primary.

(ii) rad(Q) = m.

(iii) mn ⊂ Q ⊂ m for some n > 0.

Proof. We have proved in the past that (i) and (ii) are equivalent. That (ii)implies (iii) is the previous proposition, and if we assume (iii) then (ii) followsby taking radicals.

Lecture 22 Artinian Rings

22.1 Length of Modules

The goal of this and the next lecture is to prove that a ring is Artinian if andonly if it is a Noetherian ring of dimension 0.

To accomplish this we first, of course, need to explain what the dimensionmeans, and to that end we need to establish a fair amount of results on Artinianrings:

Proposition 22.1.1. Let A be an Artinian ring. Then every prime ideal of Ais maximal.

Proof. Let P ⊂ A be a prime ideal. Then A/P is an Artinian integral domain—the former because quotient maintains Artinian, and the latter because P isprime. For x 6= 0 in A/P , we have the descending chain

(x) ⊃ (x2) ⊃ (x3) ⊃ . . .

which must therefore have a minimal element, i.e. (xn) = (xn+1) = . . . for somen. Thus xn = yxn+1, and we are in an integral domain, meaning that we havecancellation laws, so 1 = yx, making x a unit, thus finally A/P is a field and Pis a maximal ideal.

An immediate corollary of this is

Corollary 22.1.2. Let A be Artinian. Then nilrad(A) = J(A).


ARTINIAN RINGS 67

Proof. This follows directly since the former is the intersection of all primeideals, and the latter is the intersection of all maximal ideals, but these are thesame in an Artinian ring.

Proposition 22.1.3. Let A be Artinian. Then A has only finitely many max-imal ideals.

Proof. Let Σ be the set of all finite intersections

m1 ∩m2 ∩m3 ∩ . . . ∩mr

where mi are maximal ideals. Then since this is nonempty, Σ has a minimalelement since A is Artinian, say

M1 ∩M2 ∩M3 ∩ . . . ∩Mn.

We claim that M1,M2, . . . ,Mn are all the maximal ideals in A.To prove this, take any maximal ideal m and consider

m ∩M1 ∩M2 ∩M3 ∩ . . . ∩Mn ⊂M1 ∩M2 ∩M3 ∩ . . . ∩Mn,

which by minimality of the latter means that

m ∩M1 ∩M2 ∩M3 ∩ . . . ∩Mn = M1 ∩M2 ∩M3 ∩ . . . ∩Mn.

Now since all ideals here are maximal, they are in particular prime, so

m ⊃M1 ∩M2 ∩M3 ∩ . . . ∩Mn

and thus m ⊃ Mi for some i, but since Mi is maximal we must then havem = Mi.

Thus there are only finitely many maximal ideals.

Example 22.1.4. We have discussed this before in view of a descending chain,but we now have two new ways of demonstrating that Z is not Artinian: first,(0) is a prime ideal that is not maximal. Secondly, (p) is a prime ideal for allprime numbers p, and there are infinitely many of them. N

Proposition 22.1.5. Let A be Artinian. Then nilrad(A) is nilpotent.

Proof. By the descending chain condition we have

nilrad(A) ⊃ nilrad(A)2 ⊃ . . .

being stationary, meaning that

I = nilrad(A)n = nilrad(A)n+1 = . . .

for some n.Now if I = (0) we are done, since we have then demonstrated that nilrad(A)

to some power is 0, i.e. it is nilpotent.Assume, therefore, that I 6= (0), and consider

Σ = { J ⊂ A ideal | IJ 6= (0) }.

ARTINIAN RINGS 68

Now at the very least I ∈ Σ since I2 = I 6= (0), so Σ is nonempty, so it has aminimal element, say J . Then there exists x ∈ J such that xI 6= 0, whereby(x)I 6= (0). But (x) ⊂ J , meaning that (x) = J by the minimality of J , and so(xI)I = xI2 = xI 6= 0, ergo

xI ⊂ (x) = xA = J.

Therefore x = xy for y ∈ I = nilrad(A)n, meaning that y is nilpotent, sayym = 0. Now consider

x = xy = (xy)y = xy2 = . . . = xym = 0,

so x = 0, which is a contradiction since we assumed xI 6= 0, and we are done.

Definition 22.1.6 (Chain, length, and composition series). (i) A chain ofsubmodules of a module M is a sequence

M = M0 )M1 )M2 ) . . . )Mn = 0.

The chain!length of this chain is n.

(ii) A composition series of M is a maximal chain, i.e. Mi−1/Mi is simple ,i.e. has no submodules except 0 and itself.

Example 22.1.7. The prototypical example to keep in mind is a finite dimen-sional vector space V , which thus have a basis, say v1, v2, . . . , vn. Then

V = V0 ⊃ V1 ⊃ V2 ⊃ . . . ⊃ Vn−1 ⊃ Vn = 0

where Vi is the subspace generated but the last i basis vectors is a compositionseries of length n. If we skip some of the intermediate subspaces, we have achain that is not a composition series. N

Proposition 22.1.8. Suppose M has a composition series of length n. Thenevery composition series of M has length n.

Moreover every chain in M can be extended to a composition series.

Proof. Let `(M) denote the least length of a composition series (we let `(M) =+∞ if M has no composition series).

We will prove the proposition by means of two claims.First: If N (M is a proper submodule, then `(N) < `(M) = n.To see this, let {Mi } be a composition series of M of minimal length, and

consider the submodules Ni = N ∩ Mi of N . Then {Ni } is a chain for N .Moreover

Ni−1

Ni=N ∩Mi−1

N ∩Mi=

N ∩Mi−1

N ∩Mi ∩Mi−1

since Mi−1 is contained in Mi. Now this is isomorphic to

(N ∩Mi−1) +Mi

Mi⊂ Mi−1

Mi,

which is simple. Therefore Ni−1/Ni is a submodule of a simple module, so itmust either be 0 or isomorphic to Mi−1/Mi itself. In the former case, Ni−1 = Ni,so we remove one of them from the chain. In the latter case, Ni−1/Ni is simple

ARTINIAN RINGS 69

Thereby we obtain a composition series with `(N) ≤ `(M). Suppose now`(N) = `(M). Then Ni−1/Ni ∼= Mi−1/Mi for every i. Consider the last term,Nn−1/Nn ∼= Mn−1/Mn, with Mn = 0 = Nn, implying that Nn−1 = Mn−1.Then look at the previous term Nn−2/Nn−1

∼= Mn−2/Mn−1 implying thatNn−2 = Mn−2, and so on, finally implying that N = M , which is a contra-diction since N (M , so `(N) < `(M) as claimed.

For the second claim which finishes the proof, we claim that any chain in Mhas length less than or equal to `(M). Hence by minimality of `(M) we havethat all composition series of M have the same length.

To prove the claim, let

M = M0 )M1 ) . . . )Mk = 0

be a chain of length k. By claim 1, `(M) > `(M1) > `(M2) > . . . > `(Mk) = 0.Since each step must increase the length by at least 1, we must have `(M) ≥k.

Proposition 22.1.9. A module M has a composition series if and only if Msatisfies both the ascending chain condition and the descending chain condition.

Proof. The forward direction is trivial since `(M) <∞, meaning that any chainis finite, and so must stop.

The reverse direction requires a little more work. We construct a compositionseries of M as follows: Let M0 = M , and consider Σ = {N (M submodules }.By the ascending chain condition Σ has a maximal element, say M1, and simi-larly M1 has a maximal submodule M2, and so on, with

M = M0 )M1 )M2 ) . . .

which by the descending chain condition is stationary—hence we have a com-position series of M .

Definition 22.1.10 (Module of finite length). A module satisfying both theascending chain condition and the descending chain condition is called a moduleof finite length . The length `(M) is the length of any composition series ofM .

Proposition 22.1.11. Let k be a field. For a k-vector space, the followingconditions are equivalent:

(i) Having finite dimension.

(ii) Having finite length.

(iii) Satisfying the ascending chain condition.

(iv) Satisfying the descending chain condition.

Corollary 22.1.12. Let A be a ring. Suppose M1M2 . . .Mn = 0 for somemaximal ideals Mi ⊂ A, not necessarily distinct. Then A is Noetherian if andonly if A is Artinian.

STRUCTURE OF ARTINIAN RINGS 70

Proof. Consider

A ⊃M1 ⊃M1M2 ⊃ . . . ⊃M1M2 . . .Mn = 0.

ThenM1M2 . . .Mi−1

M1M2 . . .Mn

is an A/Mi module, so a vector space, since Mi contains the annihilator ofthe module. Hence the ascending chain condition and the descending chaincondition are equivalent in this quotient, as an A-module. Next consider theexact sequence

0 M1M2 . . .Mn−1 M1M2 . . .Mn−2M1M2...Mn−2

M1M2...Mn−10.

Now we can quotient the left-side of this by M1M2 . . .Mn = 0, since thatchanges nothing, but in doing so we see that we have another vector space,so the ascending and descending chain conditions are equivalent on both sidesof M1M2 . . .Mn−2, and so they are equivalent in it as well.

Repeat this with smaller indices, and inductively we get that the ascendingand descending chain conditions are equivalent in A itself.

Lecture 23 Structure of Artinian Rings

23.1 Dimension of Ring

Recall that our goal is to prove that a ring A is Artinian if and only if it isNoetherian and of dimension 0.

We are now ready to define what we mean by the dimension of a ring.

Definition 23.1.1 (Dimension of a ring). Let A be a ring. A chain of primeideals of a ring A is a sequence

P0 ⊂ P1 ⊂ P2 ⊂ . . . ⊂ Pn

where Pi are prime ideals of A. The length of this chain is n—note that thereare n+ 1 prime ideals; we count one less.

The dimension of A is is the maximal length of all chains of prime idealsin A.

Example 23.1.2. Let k be a field. Then dim k = 0 since there are only twoideals, of which 0 is prime.

We have dimZ = 1, since the longest chain of prime ideals we can make is(0) ⊂ (p) for p a prime.

We have dim k[x1, x2, . . . , xn] = n, since

(0) ⊂ (x1) ⊂ (x1, x2) ⊂ . . . ⊂ (x1, x2, . . . , xn)

is the longest chain of primes we can muster.On the other hand dimZ[x1, x2, . . . , xn] = n+ 1, because we can do almost

as above, but also include I = (p);

(0) ⊂ I ⊂ I[x1] ⊂ I[x1, x2] ⊂ . . . ⊂ I[x1, x2, . . . , xn]. N



Theorem 23.1.3. The ring A is Artinian if and only if A is Noetherian anddimA = 0.

Proof. For the forward direction we assume that A is Artinian. Therefore everyprime ideal in A is maximal, so dimA = 0 since we can’t have any nontrivialchain of prime ideals. Let M1,M2, . . . ,Mn be all distinct maximal ideals in A.Then

n∏i=1

Mki ⊂

( n⋂i=1

Mi

)k= J(A)k = (nilradA)k = 0

for some k, since the nilradical of an Artinian ring is nilpotent. Hence we havea product of maximal ideals that equals the zero ideal, which means that theascending chain condition and the descending chain condition are equivalent,which means that A is Noetherian.

For the reverse direction, we make the following claim: since A is Noetherian,it has only finitely many minimal prime ideals.

To prove this, note that (0) has a primary decomposition—every ideal isdecomposable in a Noetherian ring. Say 0 = Q1 ∩ Q2 ∩ . . . ∩ Qn is a minimalprimary decomposition, with rad(Qi) = Pi all distinct by the first uniquenesstheorem. Taking radicals we therefore have rad(0) = P1 ∩ P2 ∩ . . . ∩ Pn. Nowlet P be a minimal prime ideal of A, meaning that 0 ⊂ P implies rad(0) ⊂ P ,further implying

P1 ∩ P2 ∩ . . . ∩ Pn ⊂ P.

Hence Pi ⊂ P for some i, and so by minimality of P we have Pi = P . That isto say, all minimal prime ideals of A appear in the primary decomposition of 0.

Now let P1, P2, . . . , Pn be all of the minimal prime ideals in A. Since dimA =0, we must have that all of these primes are also maximal, since we couldn’tform a chain of them any longer than just the one prime. Therefore

nilrad(A) = ∩ni=1Pi,

where it suffices to consider only the minimal primes since others wouldn’tcontribute to the intersection, since they contain the above. Thus

n∏i=1

P ki ⊂( n⋂i=1

Pi

)k= (nilradA)k = 0

for some k, since A is Noetherian, meaning that the nilradical is nilpotent.(Note for the record that strictly speaking the inclusion above is an equalitysince the prime powers are pairwise coprime, but we don’t require equality forthis argument.)

Hence some product of maximal ideals is 0, making the ascending and de-scending chain conditions equivalent, and we are finished.

23.2 Structure of Artinian Rings

Let A be an Artinian local ring with maximal ideal M . Then nilrad(A) =J(A) = M . We know that in a local ring

A = A× tM.


Thus in the case of an Artinian local ring, everything in the maximal ideal Mis nilpotent, i.e. every non-unit is nilpotent. One might then ask oneself whatthe relation is between M and dimensionality of 0.

Proposition 23.2.1. Let A be a Noetherian local ring with maximal ideal M .Then exactly one of the following statements is true:

(i) Mn 6= Mn+1 for all n = 1, 2, . . ., in which case A is not Artinian (we havean infinite descending chain).

(ii) Mn = 0 for some n, in which case A is Artinian.

Proof. Suppose Mn = Mn+1 for some n—if not we clearly have (i). Then sinceJ(A) = M we have

Mn = Mn+1 = J(A)Mn,

which by Nakayama’s lemma means that Mn = 0.Hence also a finite product of maximal ideals is 0, so A is Artinian. Another

way to see this is to consider any prime ideal P of A. Then Mn = 0 ⊂ P ⊂M ,which if we take radicals yields rad(Mn) ⊂ P ⊂ rad(M) which is the same asM ⊂ P ⊂M , so P = M which makes dimA = 0, and again A is Artinian.

The reason we want to understand Artinian local rings is this:

Theorem 23.2.2 (Structure theorem for Artinian rings). An Artinian ring Ais isomorphic to a finite direct product of Artinian local rings. In other words,

A ∼=n∏i=1

Ai,

where Ai are Artinian local rings.

Proof. Let M1,M2, . . . ,Mn be the distinct maximal ideals of A. Then

n∏i=1

Mki = 0

for some k, and Mki are pairwise coprime. Consider the map

ϕ : A→n∏i=1

A/Mki .

This is clearly surjective, so

kerϕ =

n⋂i=1

Mki =

n∏i=1

Mki = 0

by coprimality. But then ϕ is also injective, having a trivial kernel, so ϕ is anisomorphism.

If we can show that Ai := A/Mki is an Artinian local ring, we are done.

Certainly it is Artinian, since quotient preserves the property of being Ar-tinian (and Noetherian, for that matter). Let ψ : A → A/Mk

i . There is aone-to-one correspondence between ideals I ⊃ Mk

i in A and ideals in A/Mki .


Thus if M is a maximal ideal in A/Mki , then M is a maximal ideal in A such

that M ⊃ Mki . Now since M and Mi are prime ideals, we have Mi ⊂ M , and

therefore Mi = M since both are maximal.Therefore the only maximal ideal comes from Mi/M

ki , so A/Mk

i has a uniquemaximal ideal Mi/M

ki , making it local.

Example 23.2.3. The trivial example of a local Artinian ring is Z/pnZ, withmaximal ideal pZ/pnZ. N

Consider the following situation. LetA be a local ring with maximal idealM andresidue field k = A/M . Consider the A-module M/M2. Since M(M/M2) = 0,M contains its annihilator, meaning that we can view it as an A/M -module.

But A/M is a field, so then M/M2 is a vector space over k.Now if M is a finitely generated A-module, then dimk(M/M2) <∞, and in

fact { x1, x2, . . . , xk } spans M/M2 if and only if {x1, x2, . . . , xk } generate M .Let us now relate these considerations with the idea of local Artinian rings:

Theorem 23.2.4. Let A be an Artinian local ring with maximal ideal M . Thenthe following are equivalent:

(i) Every ideal in A is principal.

(ii) The maximal ideal M is principal.

(iii) dimk(M/M2) ≤ 1.

Proof. That (i) implies (ii) is trivial, since the maximal ideal is an ideal. Like-wise (ii) implies (iii), since M is generated by a single element, and so certainlythe dimension of M/M2 is at most 1.

The only nontrivial part is (iii) implying (i), thus. First let dimk(M/M2) =0. Then M = M2 = J(A)M , which by Nakayama’s lemma means that M = 0.Hence the only maximal ideal of A is trivial, making A a field. In a field (0)and (1) = A are the only ideals, which are both clearly principal.

If dimk(M/M2) = 1 instead, thenM = (x). For any ideal I ⊂ A with I 6= (0)we have 0 = Mk ⊂ I ⊂M for some k. Let r ∈ N such that I ⊂Mr = (xr) butI 6⊂Mr+1.

Then there exists some y ∈ I with y 6∈ Mr+1, meaning that y = axr witha ∈ A and a 6∈ M . Therefore since A = A× tM , a must be a unit. Thusxr = a−1y ∈ I, so I = (xr) = Mr.

So not only are all ideals principal in Artinian local rings, but in fact every idealis a power of the maximal ideal.

23.3 Discrete Valuation Rings and Dedekind Domains

When classifying rings, in some sense fields are the easiest case—everythingworks well, and we understand them quite well.

The next easiest situation is Artinian rings, which we know is the same asNoetherian rings of dimension 0. In here we have shown that every prime idealis maximal.

The next simplest case is a Noetherian domain with dimension 1. In otherwords we have (0) = P0 ⊂ P1 as the prototypical chain of prime ideals, and inhere therefore every nonzero prime ideal is maximal.

DISCRETE VALUATION RINGS 74

In such a ring we have the remarkable property that we can factor everyideal uniquely.

Proposition 23.3.1. Let A be a Noetherian domain of dimension 1. Then forany ideal I 6= 0 in A,

I =

n∏i−1

Qi

where Qi are primary ideals and rad(Qi) = Pi are all distinct. Moreover suchan expression is unique.

Proof. The ring A being Noetherian implies that any ideal I has a primarydecomposition, say

I =

n⋂i=1

Qi

is a minimal primary decomposition with rad(Qi) = Pi. Since dimA = 1, Piare maximal, and so Pi are distinct and pairwise coprime. This in turn makeQi pairwise coprime, and so

I =

n⋂i=1

Qi =

n∏i=1

Qi.

Now for the uniqueness: since Pi are maximal ideals, all Pi in {P1, P2, . . . , Pn }are minimal (since one couldn’t be contained in the other due to maximality).This means that all Pi are isolated primes, and so all Qi are uniquely determinedby I according to the second uniqueness theorem of primary decompositions.

Example 23.3.2. In Z we have Qi = P ki . This is not true in general.However if we assume Q = P k, then A → AP then every ideal in in the

latter is a power of PAP = M , and by the one-to-one correspondence I ⊂ Pcorresponding to IP must be I = P k. N

Lecture 24 Discrete Valuation Rings

24.1 Connections between Discrete Valuation Rings andNoetherian Rings

We previously discussed Noetherian domains of dimension 1. We establishedthat in such a domain A we have that every nonzero prime ideal is maximal,and given I 6= 0 an ideal in A we have

I =∏i

Qi

where Qi are primary ideals, and this factorisation is unique. Moreover callingPi = rad(Qi), in general we do not have Qi = P ki , unlike in the integers oralgebraic extensions of them.

However if we assume that Qi = P ki , then the localisation AP of A has onlyideals of the form Mk, where M = PAP . We want to classify this kind of localring:


DISCRETE VALUATION RINGS 75

Definition 24.1.1 (Discrete valuation). Let k be a field. A discrete valua-tion on k is a surjective map v : k∗ → Z such that

(i) v(xy) = v(x) + v(y), i.e. v is a group homomorphism,

(ii) v(x+ y) ≥ min{ v(x), v(y) }.

Note that this doesn’t give us a valuation of 0—the convention is to let v(0) =+∞.

Example 24.1.2. Let k = Q, and let p be a prime. For x ∈ Q, write x = pa ·y,where a ∈ Z and y ∈ Q, with both numerator and denominator coprime to p. Forexample, if we let p = 2, we have x = 3/4 = 2−2 ·3, and, say, x = 5/6 = 2−1 ·5/3.Then we define the valuation vp(x) = a.

Hence

Z(p) ={ ab∈ Q

∣∣∣ gcd(a, b) = 1, p - b}

= {x ∈ Q | vp(x) ≥ 0 }.

In this localisation, the unique maximal ideal is

M ={ ab∈ Z(p)

∣∣∣ p | a} = {x ∈ Q | vp(x) ≥ 1 }. N

This is true in general:

Proposition 24.1.3. Let v be a discrete valuation on the field k. Then A ={ a ∈ k | v(a) ≥ 0 } is a valuation ring with the unique maximal ideal M = { a ∈k | v(a) ≥ 1 }.

Proof. The two properties of v, along with the convention v(0) = +∞ ≥ 0,guarantee that A is a ring. To see that it’s a valuation ring, note that v is agroup homomorphism, so v(a−1) = −v(a), so if v(a) ≥ 0 we’re happy, and ifnot v(a−1) > 0 instead, so either way each nonzero element or its inverse is inA.

Next, if u ∈ A is a unit, i.e. u−1 ∈ A, then v(u) ≥ 0 and v(u−1) ≥ 0,meaning that v(u) = 0. Indeed the converse is true: if v(u) = 0, then u is aunit—this too follows from v(a−1) = −v(a) as discussed above.

Hence M is the set of all non-units in A, so it’s the unique maximal ideal.

Definition 24.1.4 (Discrete valuation ring). An integral domain A is a dis-crete valuation ring if there is a discrete valuation v of its field of fractions ksuch taht A is the valuation ring of v. In other words, A = {x ∈ k | v(x) ≥ 0 }.

An element π ∈ A with v(π) = 1 is called a uniformiser or prime ele-ment .

Proposition 24.1.5. Let A be a discrete valuation ring and let π be a uni-formiser. Then the maximal ideal is M = (π), meaning in particular that M isprincipal. Conversely, if M = (π′), then π′ is a uniformiser.

Proof. Since M is the unique maximal ideal we clearly have (π) ⊂M . Now forany x ∈M we have v(x) ≥ 1, meaning that v(xπ−1) = v(x)− v(π) ≥ 1− 1 = 0,so xπ−1 = a ∈ A. Hence x = π ·a ∈ (π), so M ⊂ (π) as well, and thus M = (π).

Now suppose M = (π′). We wish to show that v(π′) = 1. We know thatM = (π), so v(π) = 1, and since the ideals (π) and (π′) are equal by assumption,

DEDEKIND DOMAINS 76

we have π = a · π′ for some a ∈ A. Hence v(π) = 1 = v(a) + v(π′), and sincev(a) ≥ 0 and v(π′) ≥ 0, we have only two options: either v(a) = 0 and v(π′) = 1,or v(a) = 1 and v(π′) = 0.

But in the latter case, π′ is a unit since it has valuation 0, but this isa contradiction since π′ ∈ M but M is the set of all non-units and nothingelse.

Proposition 24.1.6. Let A be a discrete valuation ring and k its field of frac-tions. Let π be a uniformiser of A. Then every nonzero element x ∈ k can beexpressed uniquely as x = u · πn, with u a unit and n ∈ Z. Hence k = S−1Awhere S−1 = { 1, π, π2, . . . }.

Proof. Let n = v(x). Then v(xπ−n) = v(x) − nv(π) = 0 since v(π) = 1 bydefinition. Therefore xπ−n = u ∈ A is a unit, and multiplying by πn we havex = u · πn.

For uniqueness, if x = u1πm = u2π

n, then v(x) = m = n, and moreoveru1π

n = u2πn implues u1 = u2 since we are in an integral domain, which means

we have cancellation laws.

Proposition 24.1.7. Let A be a discrete valuation ring with maximal ideal M .Then every ideal I 6= 0 in A is of the form I = Mn, with n ≥ 0 being unique.We write v(I) = n.

Proof. Take a ∈ I such that n = v(a) ≤ v(x) for all x ∈ I, i.e. the smallestpossible element in the ideal I, as per the valuation. Then a = u · πn, takingπ to be a uniformiser of A, and u being a unit in A. Hence πn = u−1a ∈ I,meaning that (πn) = Mn ⊂ I.

On the other hand, for x ∈ I we have v(x) = k ≥ n and so x = u1πk with

u1 a unit in A. Now write x = u1πk−nπn ⊂ (πn) = Mn, so I = Mn.

To establish uniqueness, suppose I = Mn = Mn+1. In that case, since weare in a local ring, we have J(A) = M , and Mn = J(A)Mn, so by Nakayama’slemma Mn = 0, which is a contradiction since we assumed I 6= 0.

Remark 24.1.8. Note that if A is a discrete valuation ring we therefore have

A ⊃M ⊃M2 ⊃M3 ⊃ . . .

which is a complete list of all the ideals. Thus in particular A is Noetherian,since any ascending sequence of ideals must end in what we have above.

Lecture 25 Dedekind Domains

25.1 Toward the Definition of Dedekind Domains

We want to classify rings in which every ideal is the power of maximal ideals.

Proposition 25.1.1. Let A be a Noetherian local ring of dimension 1 withmaximal ideal M and residue field k = A/M . The following are equivalent:

(i) A is a discrete valuation ring.


DEDEKIND DOMAINS 77

(ii) A is integrally closed.

(iii) M is principal.

(iv) dimk(M/M2) = 1.

(v) Every nonzero ideal is of the form Mk, k ≥ 0.

(vi) There exists π ∈ A such that every nonzero ideal is of the form (πk),k ≥ 0.

Proof. Recall two properties of A. First, if I 6= 0 is an ideal in A, then I isM -primary since M is the only nonzero prime ideal, and therefore the radicalof I, which is prime, must be M . Moreover I ⊃Mn for some n.

Second, Mn 6= Mn+1 for all n ≥ 0, for otherwise by Nakayama’s lemma we’dhave Mn = J(A)Mn, implying Mn = 0, and in turn M = 0.

We start with (i) implying (ii): Since A is a discrete valuation ring, A is avaluation ring, and so A is integrally closed.

Next (ii) implies (iii): Let a 6= 0, a ∈ M . If M = (a), we’re done. IfM 6= (a), then by the first property above we have Mn ⊂ (a) for some n, andMn−1 6⊂ (a), by choosing the smallest n. Now take b ∈ Mn−1 and b 6∈ (a),and consider π = a/b ∈ K, the field of fractions of A. Then π−1 ∈ A, sinceotherwise π−1 = b/a ∈ A, implying that b ∈ (a), a contradiction.

Hence since A is integrally closed, π−1 is not integral over A, meaning thatπ−1M 6⊂ M . Otherwise M is a faithful A[π−1]-module, and M is finitely gen-erated since A is Noetherian, and so π−1 is integral over A, which is a contra-diction.

But π−1M = b/aM ⊂ 1/aMn ⊂ A, meaning that π−1M = A since M isthe unique maximal ideal, and thus M = πA = (π).

For (iii) implying (iv), take M = (π) and so dimk(M/M2) is either 0 or 1.The former happens only if M = M2, but by the second property this cannotbe, and so the dimension is 1.

Now in somewhat of a twist we prove that (iv) implies (iii). Since by (iv),dimk(M/M2) = 1, we have a generator π + M2 of the k-vector space M/M2,and therefore M = (π).

We have proved (i) implying (v) before.To prove (v) implying (iii), take π ∈ M/M2, which exists by the second

property, and so by (v) we have (π) = Mk for some k, but by choice of π ∈Mwe don’t have π ∈M2, so k = 1.

Next (iii) implies (vi). Assume M = (π), and let I 6= 0 be an ideal in A.Then there exists some n such that I ⊂Mn and I 6⊂Mn+1 by the first property,and so take x ∈ I \Mn+1. This means that x ∈ Mn, meaning that x = uπn

with u 6∈ M and u ∈ A, so u is a unit. Hence πn = u−1x ∈ I since x ∈ I.Therefore (πn) ⊂ I ⊂Mn = (πn), so I = (πn).

Finally (vi) implies (i): Take M = (π) and (πk) 6= (πk+1). For a 6= 0, a ∈ A,we have (a) = (πk). Define v : K∗ → Z by v(a) = k if a ∈ A, and moreoverdefine v(a/b) = v(a)− v(b). Then v is a discrete valuation on K∗ and A is thevaluation ring of v since a ∈ A if and only if v(a) ≥ 0.

DEDEKIND DOMAINS 78

25.2 Dedekind Domain

We now consider the same result as above, except with the condition of localitydropped.

Theorem 25.2.1. Let A be a Noetherian domain of dimension 1. The followingare equivalent:

(i) A is integrally closed.

(ii) Every primary ideal is a prime power.

(iii) AP is a discrete valuation ring for every prime ideal P 6= 0 in A.

Definition 25.2.2 (Dedekind domain). A ring satisfying these equivalent con-ditions is called a Dedekind domain .

Proof. First (i) is equivalent to (iii), since A is integrally closed if and only ifAP is integrally closed for all prime ideals P (since integrally closed is a localproperty), which in turn is equivalent to AP being a discrete valuation ring forevery P by the previous classification.

Next (ii) is equivalent to (iii) since every idael in A is a prime power if andonly if every primary ideal in AP is a prime power for all P by the one-to-onecorrespondence between primary ideals contained in P in A and primary idealsin AP , which in turn is equivalent to AP being a discrete valuation ring by theprevious classification.

Corollary 25.2.3. Let A be a Dedekind domain. Then every nonzero ideal inA has a unique factorisation as a product of prime ideals.

Proof. In a Noetherian domain of dimension 1 we have

I =∏i

Qi

where Qi are primary ideals, and this factorisation is unique by dimension 1.Moreover since A is Dedekind, each of these primary ideals is a prime power, so

I =∏i

P kii .

Example 25.2.4. Any principal ideal domain is Dedekind. N

Example 25.2.5. Let A be a Dedekind domain. Then A is a principal idealdomain if and only if A is a unique factorisation domain.

Of particular interest for number theorists, therefore, is this: If K is a finiteextension of Q, i.e. a number field, then O is the integral closure of Z over K,called the ring of integers of K.

In Z we have n = pa11 · · · pakk uniquely, but a more useful way to write the

same property is(n) = (p1)a1 · · · (pk)ak

because this property generalises to O. N

GRADED RINGS AND FILTRATIONS 79

25.3 Completions

Definition 25.3.1 (Topological abelian group). We say that G is a topologicalabelian group if G is both a topological space and an abelian group such thatG × G → G defined by (a, b) 7→ a + b and G → G defined by a 7→ −a arecontinuous, i.e. the group structure preserves the topological structure.

Definition 25.3.2. A topological space X is called complete if every Cauchysequence converges to a point in X.

Note that whilst we don’t necessarily have a metric, we do have a group structurecompatible with our topology, so by {xn } being Cauchy we mean that xn−xmis in a neighbourhood of 0 for all m and n.

Example 25.3.3. Consider Q with the usual absolute value |·|, inducing themetric d(x, y) = |x − y|. Then (Q,+) is a topological abelian group, but Qis not complete, since there are Cauchy sequences that converge to irrationalnumbers. Its completion is well-known, namely R. N

A very useful property when trying to establish a topology in this way is this:if G is a topological abelian group, fix a ∈ G and define Ta : G→ G by Ta(g) =g+a. Then Ta is a homeomorphism of G onto G, i.e. Ta and its inverse are bothcontinuous. So if U is a neighbourhood of 0 in G, then a+U is a neighbourhoodof a in G, and conversely every neighbourhood in G is of this form.

Hence the topology of G is uniquely determined by the neighbourhoods of 0in G.

Lecture 26 Graded Rings and Filtrations

26.1 Graded Rings and Modules

Definition 26.1.1 (Graded ring). A graded ring is a ring A such that

(i) A =∞⊕n=0

An, where An are subgroups of A and

(ii) AmAn ⊂ Am+n for m,n ≥ 0.

Hence A0 is a subring of A and An is an A0-module.

Example 26.1.2. Perhaps the simplest example is A = k[x1, x2, . . . , xn]. LetAn be the set of all homogeneous polynomials of degree n. Then A =

⊕∞n=0An

and AnAm ⊂ An+m, so A is a graded ring. For instance,

f(x) = anxn + an−1x

n−1 + . . .+ a0,

where anxn ∈ An, an−1x

n−1 ∈ An−1, et cetera. Similarly,

g(x1, x2) = a1x1x2 + a2x21 + a3x1 + a4x2,

wherein a1x1x2 + a2x21 ∈ A2, and a3x1 + a4x2 ∈ A1. N

Date: December 5, 2017.


Definition 26.1.3 (Graded module). Let A be a graded ring. A graded A-module M is expressible as

M =

∞⊕n=0

Mn

with AmMn ⊂Mn+m.An element x ∈ M is called homogeneous if x ∈ Mn for some n. We call

n the degree of x.

Note that any element y ∈ M can be written uniquely as y =∑n≥0 xn, with

xn ∈Mn, where xn = 0 for all but finitely many n.

Definition 26.1.4 (Filtration of a ring). Let A be a ring. Then {An } is afiltration of A if

A = A0 ⊃ A1 ⊃ A2 ⊃ A3 ⊃ . . .

are subgroups with AmAn ⊂ Am+n.

Definition 26.1.5 (Filtration of a module). LetM be a module over the filteredring A. Then {Mn } is a filtration of M if

M = M0 ⊃M1 ⊃M2 ⊃M3 ⊃ . . .

are submodules with AmMn ⊂Mm+n.

Example 26.1.6. Let I be an ideal of a ring A, and let M be an A-module.The I-adic filtration of A and M are given by A0 = A = I0, An = In forn ≥ 1, and M0 = M with Mn = InM for n ≥ 1. Hence AmNn = ImInM =Im+nM = Mm+n. N

Proposition 26.1.7. Let A =⊕

n≥0An be a graded ring. The following areequivalent:

(i) A is Noetherian.

(ii) A0 is Noetherian and A is finitely generated as an A0-algebra.

Proof. First show that (i) implies (ii). Let I =⊕

n≥1An, which is an idealof A, and A0

∼= A/I. Hence A being Noetherian implies A0 is Noetherian,since the quotient of a Noetherian ring remains Noetherian. Now I is finitelygenerated since it is an ideal of a Noetherian ring A, say by x1, x2, . . . , xk, whichwe may take to be homogeneous of degree m1,m2, . . . ,mk (if not, take linearcombinations to make them). Let A′ = A0[x1, x2, . . . , xk] ⊂ A, an A0-algebra—think of it as a subring.

We claim that A′ = A. By construction it suffices to show An ⊂ A′ for alln. First, therefore, note that for n = 0 we have A0 ⊂ A′, which is fine. Nextassume Am ⊂ A′ for all m ≤ n − 1, and consider n. Then if y ∈ An ⊂ I, wehave

y = a1x1 + a2x2 + . . .+ akxk,

with ai ∈ A. Note that y is homogeneous, so since xi ∈ Ani , we must haveai ∈ An−ni

. Now since the degree of ai is less than n, we have by the inductionhypothesis that

ai = bi1x1 + bi2x2 + . . .+ bikxk


with bij ∈ A0. Hence y is a polynomial in x1, x2, . . . , xn with coefficients in A0,meaning that y ∈ A′, so A = A′ as claimed.

Now for (ii) implying (i), note that A is a finitely generated A0-algebraimplies that A is some quotient of A0[x1, x2, . . . , xn]. Now A0 is Noetherian, soby Hilbert basis theorem A0[x1, x2, . . . , xn] is Noetherian as well, and since Ais some quotient of this, A in turn must also be Noetherian.

Definition 26.1.8 (I-filtration). Let M be a filtered A-module with filtration{Mn }. Let I ⊂ A be an ideal. Then {Mn } is called an I-filtration if IMn ⊂Mn+1.

Moreover an I-filtration {Mn } with IMn = Mn+1 for n ≥ N for some N iscalled I-stable .

Remark 26.1.9. The I-adic filtration is I-stable.

Proposition 26.1.10. Let A be a Noetherian ring and let M be a finitelygenerated A-module. Suppose {Mn } is an I-filtration of M . The following areequivalent:

(i) {Mn } is I-stable.

(ii) Define a graded ring A∗ and a graded A∗-module M∗ by A∗ =⊕

n≥0 In

and M∗ =⊕

n≥0Mn. Then M∗ is a finitely generated A∗-module.

Proof. Let Qn =⊕n

i=0Mi ⊂ M∗. Since Mi are finitely generated A-modulesfrom the hypotheses of the proposition, we have that Qn are finitely generatedA-modules. In general Qn is not an A∗-submodule of M∗. But Qn does generate

Q∗n = M0 ⊕M1 ⊕M2 ⊕ . . .⊕Mn ⊕ IMn ⊕ I2Mn ⊕ . . . .

Having I-stability implies that this stops, so this is a finitely generated A∗-submodule of M∗ with the same generators as Qn over A.

Note moreover thatM∗ =

⋃n≥0

Q∗n

where Q∗0 ⊂ Q∗1 ⊂ Q∗2 ⊂ . . ..Therefore (ii) is equivalent with M∗ being a finitely generated A∗-module;

since A is Noetherian, A∗ is Noetherian, so M∗ is a Noetherian module. Thisis true if and only if the chain of Q∗i stops, if and only if M∗ = Q∗m for some m,if and only if Mm+k = IkMn for all k ≥ 1. This finally is the same as {Mn }being I-stable, i.e. (i).

The example to keep in mind with discussing these things is usually the I-adicfiltration.

Proposition 26.1.11 (Artin-Rees lemma). Let A be a Noetherian ring andI ⊂ A an ideal. Let M be a finitely generated A-module and {Mn } be an I-stable filtration. Let N be a submodule of M . Then {Nn = N ∩Mn } is anI-stable filtration of N .

Proof. Let A∗ =⊕

n≥0 In and M∗ =

⊕n≥0Mn, as above. Finally let N∗ =⊕

n≥0Nn.


The filtration {Mn } being I-stable, by the above proposition, implies thatM∗ is a finitely generated A∗-module, with A∗ being Noetherian. This in turnimplies that M∗ is a Noetherian A∗-module, so all of its submodules are finitelygenerated.

Now in particular N∗ is a submodule, so it is finitely generated over A∗. Theconverse direction of the proposition therefore implies that {Nn } is I-stable.

Corollary 26.1.12. There exists m ∈ N such that

(Im+kM) ∩N = Ik((ImM) ∩N)

for all k ≥ 0.

Proof. Take the filtration Mm = ImM . Then

(Im+kM) = Mm+k ∩N = Nm+k = IkNm = Ik((ImM) ∩N).

Remark 26.1.13. There is a topological interpretation of this. Let M be anA-module, and I ⊂ A an ideal. Consider the I-adic filtration { InM }, where

M ⊃ IM ⊃ I2M ⊃ I3M ⊃ . . . ⊃ (0),

so we use these to define open neighbourhoods of 0—we let them be a basis forthe neighbourhoods of 0. By translation we moreover let {x+ InM } be a basisfor the neighbourhoods of x.

These induce a topology on M where module operations are continuous. Thetopology is called the I-adic topology on M .

The corollary says that the I-adic topology on N coindices with the topologyon N induced by the I-adic topology on M .

This can be generalised beyond just the I-adig filtration:

Remark 26.1.14. Let M be a filtered A-module with filtration {Mn }. The fil-tration determines a topology on M with {Mn } forming a basis for the neigh-bourhoods of 0.

26.2 Inverse Systems

Definition 26.2.1 (Inverse system). Suppose we have {Mn }n≥0, a collectionof A-modules, with A-module homomorphisms ϑn : Mn →Mn−1 for n ≥ 1. Thecollection of modules and homomorphisms is called an inverse system .

A sequence (xn) ∈∏nMn is called coherent if ϑn+1(xn+1) = xn for all

n ≥ 0. In other words, we have the diagrammatic picture

M0 M1 M2 M3 M4 · · ·

x0 x1 x2 x3 x4 · · ·

ϑ1 ϑ2 ϑ3 ϑ4 ϑ5

The collection M of all coeherent sequences is called the inverse limit of theinverse system, denoted

M = lim←−Mn.

The inverse limit M is an A-module with coordinatewise addition, (xn)+(yn) =(xn+yn), which is also coherent since ϑn is a homomorphism, and likewise scalarmultiplication is coordinatewise, a(xn) = (axn), which again is coherent.

CLOSURE 83

Lecture 27 Closure

27.1 Closure and Completion

Proposition 27.1.1. Let M be a filtered A-module with filtration {Mn }. If Nis a submodule of M , the closure of N (in M) is

N =

∞⋂n=0

N +Mn.

Proof. Let x ∈ M . Then x 6∈ N if and only of (x + Mn) ∩N 6= ∅ for some n,since N is closed, and so its complement is open.

We claim that this is equivalent with x 6∈ N +Mn for some N . This in turnis equivalent with N =

⋂∞n=0(N +Mn), so it’s all down to the claim.

For the forward direction of the claim, let us assume (x+Mn) ∩N = ∅. Ifx ∈ N +Mn, then x = y + z with y ∈ N and z ∈Mn, hence x+ (−z) = y is inx+Mn and also in N , so x+ (−z) ∈ N ∩ (x+Mn), but this is empty, so thisis a contradiction. Hence x 6∈ N +Mn.

For the reverse direction, assume x 6∈ N + Mn. If y ∈ (x + Mn) ∩ N 6= ∅,then y = x + z, with z ∈ Mn, meaning that x = y − z ∈ N + Mn, anothercontradiction.

Corollary 27.1.2. The topology induced from this filtration is Hausdorff ifand only if

∞⋂n=0

Mn = { 0 }.

Recall that a topology is called Hausdorff if for any two points x and y, we canfind a neighbourhood U of x and a neighbourhood V of y such that U ∩V = ∅.

Proof. A topological space is Hausdorff if and only if every singleton is closed,but our topology is induced by the neighbourhoods of 0, so this is true if andonly if { 0 } is closed, which is true if and only if

{ 0 } = { 0 } =

∞⋂n=0

Mn.

Definition 27.1.3 (Completion). Let {Mn } be a filtration of the A-moduleM , meaning that we have an induced topology on M .

A sequence {xn } in M is called Cauchy if for every k ∈ N there existssome N ∈ N such that xn − xm ∈Mk for every n,m ≥ N .

Moreover if {xn } and { yn } are two Cauchy sequences in M , we call themequivalent if for every k ∈ N there exists some N ∈ N such that xn− yn ∈Mk

for every n ≥ N .Finally the completion of M is the equivalence classes of Cauchy sequences

in M , denoted by M .

Date: December 7, 2017.

CLOSURE 84

Note that if {xm } is a Cauchy sequence in M , then {xm } in M/Mn is ulti-mately constant, say ξn, since xm−xk ∈Mn is equivalent to xm+Mn = xk+Mn.Now define ϑn+1 : M/Mn+1 →M/Mn by ϑn+1(ξn+1) = ξn. We therefore have

M/M0 M/M1 M/M2 M/M3 M/M4 · · ·

ξ0 ξ1 ξ2 ξ3 ξ4 · · ·

ϑ1 ϑ2 ϑ3 ϑ4 ϑ5

Then { ξn } is a coherent sequence in the inverse system (M/Mn, ϑn).

Remark 27.1.4. If {xn } and { yn } are equivalent Cauchy sequences in M , thenthey define the same coherent sequence { ξn }.

On the other hand, given any coherent sequence { ξn } in the inverse system(M/Mn, ϑn), take xn to be any element in ξn + Mn. Then {xn } is Cauchy inM .

Hence

M = lim←−M

Mn.

Example 27.1.5. Let A = M = Z, and I = (p), with p being prime. Let Mn =In = pnZ, and ϑn+1 : M/Mn+1 →M/Mn defined by ϑn+1(a+ In+1) = a+ In.

Then the inverse limit

lim←−M

Mn= lim←−

ZpnZ

= Zp

is the p-adic integers. N

Proposition 27.1.6. Suppose

0 L M N 0f g

is an exact sequence of A-modules. Let {Mn } be a filtration of M such that{Mn } induces a filtration {L ∩ f−1(Mn) } on L and { g(Mn) } on N . Then

0 L M N 0,

where of course the completions are under the respective induced topologies, isexact.

27.2 Consequences of the Krull Intersection Theorem

We close off this course by mentioning, without proof, some consequences of theKrull intersection theorem.

Corollary 27.2.1. Let A be a Noetherian domain and I ⊂ A an ideal. Then

∞⋂n=0

In = { 0 }

meaning that the I-adic topology induced by I is Hausdorff.

CLOSURE 85

Corollary 27.2.2. Let A be a Noetherian ring and I ⊂ J(A) an idael. Let Mbe a finitely generated A-module. Then

∞⋂n=0

InM = { 0 },

meaning that the I-adic topology on M is Hausdorff.

Theorem 27.2.3. Let A be a Noetherian ring and I an ideal. Then the I-adiccompletion A of A is Noetherian.

Corollary 27.2.4. Let A be a Noetherian ring. Then A[[x1, x2, . . . , xn]] isNoetherian.

We have proven this directly before, but in the interest of fun we provide a veryquick proof as a consequence of the above:

Proof. By Hilbert basis theorem A[x1, x2, . . . , xn] is Noetherian. Consider I =(x1, x2, . . . , xn). The I-adic completion of A[x1, x2, . . . , xn] is A[[x1, x2, . . . , xn]],and by the above it is Noetherian.

REFERENCES 86

References

[AM94] M. F. Atiyah, I. G. MacDonald. Introduction To Commutative Algebra.Westview Press, 138 pages, 1994.

Index

algebra, 64algebraic, 53annihilator, 14, 17Artinian

module, 60ring, 60

basis, 19

Cauchy sequence, 83equivalent, 83

chain, 68of prime ideals, 70

chain conditionascending, 59descending, 59

closure, 83cokernel, 22completion, 83composition series, 68contraction, 32

Dedekind domain, 78direct product, 18direct sum, 18discrete valuation, 75discrete valuation ring, 75domain

dedekind, 39

elementalgebraic, 45integral, 45

endomorphism, 20exact

left, 24right, 28

exact sequence, 22short, 22split, 22

extension, 32

field, 2, 6residue, 9

field of fractions, 30filtration

I, 81I-adic, 80

I-stable, 81module, 80ring, 80

generator, 18graded

module, 80ring, 79

group, 1abelian, 1

Hausdorff, 83homogeneous element, 80

degree, 80homomorphism

connecting, 25

ideal, 5coprime, 12decomposable, 41irreducible, 65maximal, 7P-primary, 39primary, 38prime, 6principal, 5quotient, 14radical, 15reducible, 65

integral closure, 47integral domain, 6

integrally closed, 51normal, 51

inverse limit, 82inverse system, 82

Jacobson radical, 11

local property, 36

maximal element, 8minimal polynomial, 53module, 3, 15

faithful, 17finite length, 69finitely generated, 18flat, 29free, 19homomorphism, 16

87

INDEX 88

projective, 29quotient, 17simple, 68

multiplicative set, 6

nilpotent, 6nilradical, 10Noetherian

module, 60ring, 60

number field, 46

orderpartial, 8total, 8

primary decomposition, 39, 41irredundant, 41minimal, 41normal, 41reduced, 41shortest, 41

prime element, 75prime ideal

associated, 42embedded, 42isolated, 42minimal, 42

quotient field, 30

radical, 14reflexive, 31restriction, 32ring, 1

dimension, 70division, 2homomorphism, 5integral, 47integrally closed, 47local, 9Noetherian, 39quotient, 5semi-local, 10valuation, 56

sequencecoherent, 82

spectrummaximal, 8prime, 7

symmetric, 31

tensor, 26tensor product, 26topological abelian group, 79topological space

complete, 79topology

I-adic, 82transitivity, 31

uniformiser, 75unit, 6upper bound, 8

zero-divisor, 6

Lecture Notes in Abstract Algebra

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