Commutative Rings and Integral Domainswarrenb/Courses/ring.pdf · Commutative Rings and Integral...

Chapter 1

Commutative Rings and IntegralDomains

1. Commutative Rings with 1

Let’s start out with some simple reminders of what was learned in 633-634. We should remark thatthroughout this course the underlying assumption is that rings will be assumed to commutativeand with an identity (and 1 6= 0). We will forego the definition of a ring and ideals and leave thatto the reader. Most of this information may be found in [DF] and [fs]. At other times other booksshall be referenced explicitly. Throughout R and S will be used to denote rings and P , I and Jwill be used to denote ideals.

Definition 1.1. A (proper) ideal P is called prime if for any a, b ∈ R a · b ∈ P implies that eithera ∈ P or b ∈ P . A (proper) ideal P is called maximal if whenever P ≤ I ≤ R then either I = P orI = R.

Definition 1.2. The ring R is called an integral domain if whenever a b ∈ R and a · b = 0 theneither a = 0 or b = 0. This is known as the “zero-divisor property”.

In general a non-zero element a ∈ R satisfying a · b = 0 implies b = 0 is called a non zero-divisoror regular. Thus, a ring is an integral domain precisely when every non zero element is regular.

An element a ∈ R is called a unit if there exists b ∈ R such that a · b = 1. It is known that aunit is regular. Notice that the ideal generated by a unit is all of R. (In fact, this characterizesunits.) If every nonzero element is a unit then we call R a field. Thus, fields are integral domains.Another useful characterization is that a ring is a field if and only it has no nontrivial ideals.

Proposition 1.3. In a commutative ring with identity maximal ideals exist. Moreover, everyproper ideal is contained in a maximal ideal.

Proposition 1.4. The ideal P is prime if and only if R/P is an integral domain.

Proposition 1.5. The ideal P is maximal if and only if R/P is a field.

Corollary 1.6. Every maximal ideal is prime.

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Definition 1.7. A prime ideal P is called minimal if whenever P1 ≤ P where P1 is a prime idealthen P1 = P . A Zorn’s Lemma argument ensures us that every prime ideal contains a minimalprime ideal.

We use Spec(R), Max(R), and Min(R) to denote the set of all prime, maximal, and minimalprime ideals (respectively.) Each of these may equipped with the hull-kernel topology. We shalltalk about this later.

Definition 1.8. Given a collection of ideals {Ij}j∈B the intersection is again ideal. Thus, thecollection of all ideals of a ring forms a complete lattice when partially ordered under inclusion.Thus, given a subset A ⊆ R we may speak of the ideal generated by A. It is simply the interesectionof all ideals containing the set A. We shall use (A) to denote this ideal and is called the idealgenerated by A.

If A = {a} we simply write (a). Ideals of this type are called principal ideals. Given an ideal Iif there exists a finite subset A of I such that I = (A) then I is called a finitely generated ideal.

The subcollection of finitely generated ideals forms an upper semilattice of the lattice of allideals, i.e., the join of two finitely generated ideals is again finitely generated. In general thesubcollection of principal ideals is simply a partially-ordered set where sups and infs may not exist.

Proposition 1.9. Given a set A ⊆ R

(A) = {Σni=1riai|n ∈ N, a1, · · · an ∈ A, r1, · · · , rn ∈ R}.

In other words the ideal generated by a set is simply the R-linear combinations of elements inthe set.

2. Elements and Integral Domains

At this point we urge the reader to reread Chapter 7 of [DF]. In particular the sections on polyno-mial rings. Hopefully, at this point the section on the Isomorphism Theorem is cake or easy as pieif you prefer something fruity.

Definition 2.1. First off, for elements a, b ∈ R We say a divides b if there exists an x ∈ R suchthat ax = b. Divisibility is an equivalence relation. The elements a, b ∈ R are associates if thereexists a unit u such that a = ub. It follows that (a) = (b). We shall say more about this later inthe context of groups of divisibility.

A nonzero, nonunit a ∈ R is called prime if whenever px = ab for some x, a, b ∈ R, then thereis a y ∈ R such that py = a or py = b. This is equivalent to saying that principal ideal generatedby p is a prime ideal.

The non-zero, nonunit p ∈ R is called irreducible if whenever p = ax then either a or x is aunit. The reader should verify that an element is irreducible precisely when the principal ideal itgenerates is a maximal principal ideal. Clearly a prime element is always irreducible.

If R is a domain then the set of nonzero elements is denoted by R∗.

Definition 2.2. Given elements a, b ∈ R an element d is called greatest common divisor forthe pair a, b if d divides both a and b and whenever e divides both a and b, then d divides e. It

Commutative Rings and Integral Domains 3

follows that two elements have a gcd if and only if in the set of principal ideals their respectiveprincipal ideals have a sup. Observe that this sup may not be the same sup in the context of thelattice of all ideals.

If the only common divisors of two elements a and b are units, then we say that a and b areco-prime. Moreover, if R = (a, b), then we call a and b relatively prime.

Least common multiples are defined analogously.

Definition 2.3. A domain is called a GCD-domain if every pair of elements has a greatest commondivisor. Equivalently, R is a GCD domain if and only if the set of principal ideals forms a lattice.Even in this case, it should be noted that it may not be a sublattice of the lattice of all ideals.

Some rumblings: The class of GCD domains is knowm more for its subclasses then the classitself. In my humble opinion this unfortunate as the class GCD is very interesting in its own rightand recently others have figured this out. From a lattice-orederd group perspective we shall showlater exactly why it is interesting. But also from a ring theoretic point of view there are lots ofinteresting question and results out there to be discovered. Ahhh only time will tell.

Definition 2.4. A domain R is a Euclidean Domain if there exists a function N : R → N suchthat for every a, b ∈ R (b 6= 0) there exists unique q, r ∈ R such that

a = qb+ r r = 0 or N(r) < N(b).

In other words it satisifes a generalized Division Algorithm.

Definition 2.5. The domainR is a Discrete Valuation Domain if there exists a function v : R∗ → Nsuch that

(i) v is a surjection;

(ii) v(xy) = v(x) + v(y) for all x, y ∈ R∗;

(ii) v(x+ y) ≥ min{v(x), v(y)} for all x, y ∈ R∗.

Definition 2.6. The domain R is a Principal Ideal Domain if every ideal is principally generated.We shall simply write PID to mean the above.

Definition 2.7. The domain R is a Bezout Domain if every finitely generated ideal is principal.Observe that here this says that that the set of principal ideals is a sublattice of the lattice of allideals.

Definition 2.8. The domain R is a Unique Factorization Domain (or simply a UFD) if eachelement of A may be written uniquely (up to relabeling and associates) as a product of irreducibles.

Definition 2.9. The domain R is called a Valuation Domain if the lattice of ideals forms a chain.We shall give another characterization once we describe the field of quotients. Valuation domainsare example of local ring, that is, a ring with a unique maximal ideal.

Theorem 2.10. Every Euclidean Domain is a PID.

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Theorem 2.11. Every PID is a Bezout Domain.

Theorem 2.12. Every Bezout domain is a GCD domain.

Theorem 2.13. Every Discrete Valuation Domain is a Euclidean Domain. (See pg. 273 [DF].)

Theorem 2.14. A domain is a PID if and only if it a UFD and a Bezout Domain. (See Exercise11, pg. 295 [DF].)

Theorem 2.15. Every UFD is a GCD domain.

Theorem 2.16. Every valuation domain a Bezout domain.

Proposition 2.17. In a PID, an element is prime if and only if the ideal it generates is a maximalideal.

Proposition 2.18. In a UFD, an element is irreducible if and only if it is prime.

Proposition 2.19. The ring R is a domain if and only R[x] is a domain.

Theorem 2.20. For a domain R the following are equivalent:

(i) R is a field.

(ii) R[x] is a Euclidean domain.

(iii) R[x] is a PID.

(iv) R[x] is a Bezout domain.

Proof: The proof of the equivalence of the first three is shown in Corollary 8 of Chapter 8 andand Theorem 3 of Chapter 9 in [DF]. Clearly (iii) implies (iv). Thus, all we need to do is showthat (iv) implies (i).

By means of contradiction let a ∈ R be a nonzero, nonunit. Consider the ideal of R[x] generatedby a and x. By (iv), there exists d ∈ R[x] such that (d) = (a, x). Thus, a = ds for some s ∈ R[x].But due to degree considerations deg(d) = 0 and so d ∈ R. Similarly, x = df . Since deg(d) = 0 itfollows that def(f) = 1 and so f = ux+ v for some u, v ∈ R. Next,

x = df = dux+ dv.

Then dv = 0 whence v = 0. Then du = 1 so that d is a unit which says that a and x arerelatively prime. Thus, 1 may be written as an R[x]-linear combination of a and x, i.e., there existss(x) = a0 + a1x · · ·+ amx

m and t(x) = b0 + b1x · · ·+ bmxn such that

1 = s(x)a+ t(x)x.

A simple calculation shows that m = n + 1 and that aa0 = 1, contradicting the fact that a is anonunit.�


Theorem 2.21. The ring R is a UFD if and only R[x] is a UFD.

Example 2.22. (1) Z is a discrete valuation domain which is not a field.

(2) R[x] is a euclidean domain which is not a discrete valuation domain. (Consider the elementsf(x) = x and g(x) = x+ 1.

(3) Z[x] is a UFD that is neither a PID nor a Bezout domain.

(4) Let R = Z[1+√−19

2 ]. This is given in the last example of section 8.2 of [DF] as an exampleof a PID which is not a euclidean domain.

3. R-Modules

We now move on to modules. Observe that the definition is for left R-modules. A similar definitioncould be given for right R-modules. As we will assume that our rings are commutative it is apparentthat left and right will mean the same thing and thus we shall drop the left and right business. Butit would be wrong of me to not point out that things are not so nice and some extra care would beneeded if we were to treat this subject from a not necessarily commutative point of view.

Definition 3.1. Let R be a ring (not necessarily commutatve nor with 1). A left R-module isan abelian group (M,+) together with an action of R on M (denoted by r ·m, for all r ∈ R andm ∈M such that

(1) (r+s)m=rm+sm for all r, s ∈ R and m ∈M

(2) (rs)m = r(sm) for all r, s ∈ R and m ∈M

(3) r(m+n)=rm+rn for all r ∈ R and m,n ∈M

(4) If 1 ∈ R then 1m = m for all m ∈M .

We shall consider this treatment categorically. We let R-Mod denote the category of all R-modules and R-module homomorphisms between them. (See Chapter 10.2 [DF].) Also, all moduleshall be considered nonzero.

Example 3.2. The simplest example of R-modules is that every ideal of R is an R-module wherethe action is simply multiplication. At this point we shall direct the student to Chapter 10.1 of[DF]. In particular, the first 6 examples given in 10.1. In particular, it is apparent that Z-Mod isprecisely Abel. Similarly, Q-Mod is precisely DivAbel.

When R is a field we shall emply the terminology of vector spacess instead of modules.

Definition 3.3. Let M be an R-module. The set of R-submodules of M forms a complete lattice.In particular given any set A ⊆M , we let RA dente the module generated by A (as the intersectionof all submodules containing A.) As before, RA is simply the set of (finite) R-linear combinationsof elements from A. If A = {m} is a singleton set, then we write this Rm and call any module ofthis form cyclic. We similarly define a finitely generated (sub)module.

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Definition 3.4. A module M is called simple or irreducible if it has no nontrivial submodules.The reader should be able to determine all simple Z-modules. See below.

Definition 3.5. Similarly, the reader should look up the definitions for torsion element, torsionmodule, annihilator of M in R, annihilator of I in M , torsion-free module, quotient module.

Definition 3.6. Let I be an ideal of R and M and R-module with S ⊆M . Define

IS = {Σni=1risi|ri ∈ I, si ∈ S}.

IS is a submodule of M .

Proposition 3.7. M/IM is an R/I-module under the action

(r + I)(m+ IM) = rm+ IM.

Proposition 3.8. Suppose M is a cyclic module. Then there exists an ideal J of R such thatM ∼= R/J (as R-modules).

Proof: Set M = Rm and J = Ann(M) = {r ∈ R : rm = 0}. Define φ : R → M by φ(r) = rm.Then by the axioms of an R-module it follows that φ is an R-module homomorphism. The factthat M is cyclic implies that φ is surjective. By the First Isomorphism Theorem

R/ kerφ ∼= M

But kerφ = J .�

Corollary 3.9. Suppose M is an irreducible R-module. Then Ann(M) is a maximal ideal of R.

Proof: Since M is irreducible it follows that M = Rm for any nonzero element m ∈ M . By theprevious proposition it follws that M ∼= R/Ann(M). This means that R/Ann(M) has no nontrivialsubmodules. Terming this in terms of ideals it means that R/Ann(M) has no nontrivial ideals, i.e.,R/Ann(M) is a field. Thus, Ann(M) is a maximal ideal.�

Definition 3.10. Let {Mi}i∈I be a collection of R-modules. We can make the (external) directproduct Πi∈IMi, and the (external) direct sum ⊕i∈IMi, into R-modules by the doing the obivousthing: r(mi) = (rmi).

The projection of either Πi∈IMi or ⊕i∈IMi onto Mi shall be denoted by πi. The injection ofMi into Πi∈IMi or ⊕i∈IMi shall be denoted by ιi.

If we consider the sum of two modules, say M and N then we shall use πM , πN and ιM and ιNas the projections and inclusions of the respective modules.

Example 3.11. We recall the Hom functors. Let A be a fixed R-module. Hom(A, ·) is a covariantfunctor from R-Mod to itself. Similarly, Hom(·, A) is a contravariant functor from R-Mod to itself.The only thing new here is that for each R-module B, Hom(A,B) and Hom(B,A) are both R-modules.


4. Pushout

We now give the construction of a pushout. In principle the idea is that given the diagram ofR-modules A,B,C and R-module homomorphisms f, g

A - B

?

C

g

f

We would like to pushout to make a commutative square, i.e., find an R-module L and R-modulehomomorphisms f ′, g′ so that the following diagram commutes:

A - B

?

C - L

?

g

f

g′

f ′

Construction!! Consider the submodule W of C ⊕B defined by

W = {(f(a),−g(a) : a ∈ A}.

Let L = C ⊕ B/W and let π : C ⊕ B → L be the natural projection. Define g′ : C → L byg′(c) = (c, 0) + W . Simlarly, define f ′ : B → L by f ′(b) = (0, b) + W . Since g′ = π ◦ ιC andf ′ = π ◦ ιB it follows that f ′ and g′ are module homomorphisms.

Clearly, for every a ∈ A, (g′ ◦ f)(a) = g′(f(a)) = (f(a), 0) + W and (f ′ ◦ g)(a) = f ′(g(a)) =(0, g(a)) + W . Since (f(a), 0) − (0, g(a)) ∈ W it follows that g′ ◦ f = f ′ ◦ g, whence the squarecommutes.

Furthermore, we now demonstrate that L is projectively the largest such module that pushesout the original diagram, i.e., if M is a module together with a pair of homomorphisms such that

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A - B

?

C - M

?

g

f

β

α

is a commutative square then there exists a unique R-module homomorphism p : L→M such thatp ◦ g′ = β and p ◦ f ′ = α.

Suppose M,α, β exist. Define p : L→M by

p((c, b) +W ) = β(c) + α(b).

Notice that we must make sure that p is well-defined. Suppose that (c1, b1)+W = (c2, b2)+W .Then (c1− c2, b1− b2) ∈W , whence there is an a ∈ A such that c1− c2 = f(a) and b1− b2 = −g(a).A simple calculation

(β(c1) + α(b1))− (β(c2) + α(b2)) = β(c1)− β(c2) + α(b1)− α(b2)= β(c1 − c2) + α(b1 − b2)= β(f(a)) + α(−g(a)= (β ◦ f)(a)− (α ◦ g)(a)= 0

showing that p is well-defined. (Observe that in the last equation we used that hypothesis thatα ◦ g = β ◦ f .) We leave it to the reader to verify that p is an R-module homomorphism.

Now, for any c ∈ C and b ∈ B we have (p ◦ g′)(c) = p(g′(c)) = p((c, 0) + W ) = β(c) and(p ◦ f ′)(b) = p(f ′(b)) = p((0, b) + W ) = α(b). Thus, simultaneously we have demonstrated thatp ◦ g′ = β and p ◦ f ′ = α. As for uniqueness of the map p, if p′ were any other such map satisfyingp′ ◦ g′ = β and p′ ◦ f ′ = α:

p′((c, b) +W ) = p′((c, 0) +W ) + p′((0, b) +W= p′(g′(c)) + p′(f ′(b))= β(c) + α(b)= p((c, b) +W )

The pushout has the property that if g is injective (surjective) then so is g′. We prove thispresently.

Injectivity: Suppose g′(c) = 0, i.e., (c, 0) +W = W . Thus, (c, 0) ∈ W hence there is an a ∈ Asuch that (c, 0) = (f(a),−g(a)). Notice that equality of the second coordinate forces a = 0 and soc = f(0) = 0.

Surjectivity: Let (c, b) + W ∈ L. Since g is onto there is an a ∈ A such that g(a) = b. Now,f(a) ∈ C and so c+ f(a) ∈ C. Observe that g′(c+ f(a)) = (c+ f(a), 0) +W and

(c+ f(a), 0)− (c, b) = (f(a),−b) = (f(a),−g(a))

where this last element lies in W so that g′(c+ f(a)) = (c, b) +W , whence g′ is onto.

Chapter 2

More Category Theory

1. Limits

Definition 1.1. Let C be a category and r : A → B a morphism. We call r a C-retract if thereexists a C-morphism g : B → A such that r ◦ g = 1B, i.e., r has a right inverse. A composition ofretractions is a retraction.

Similarly, the morphism s : A → B is a C-section if there is a morphism f : B → A such thatf ◦ s = 1A, i.e., s has a left inverse. A composition of sections is a section.

Finally, a morphism that is both a C-retraction and section is called an isomorphism. Obvioulsya composition of isomorphisms is again an isomorphism.

Definition 1.2. Suppose that C is any category. Let {Ai}i∈I be a set of objects in C. The productof the Ai is an object A, together with a family of morphisms pi : A → Ai such that, for each setof morphisms {gi : B → Ai}i∈I , there is a unique morphism g : B → A, such that pi · g = gi, foreach index i. (There is no reason, a priori, that the product should exist! Consider Field.)

If a product does exist then it is unique up to an isomorphism. To see this consider two products.Then by the universality condition there exist maps between them. Then the composition mustequal the respective identities due to uniqueness.

The dual notion is that of a coproduct. Simply turn the arrows around.

Exercise 1.3. We leave it to the reader to verify that in CRng1 and R-mod the product is simplythe cartesian product together with the projections. As for coproducts we have

Example 1.4. Let {Mj}j∈J be a collection of R-modules. Let M = ⊕j∈JMj and let ιj : Mj →Mbe defined by

ιj(m)(i) ={m, if i = j;0, otherwise.

It is apparent that for each j ∈ J , ιj is an R-module homomorphism.Now, suppose that there is an R-module N together with a collection of maps{gj : Mj → N .

Let m ∈M . Define φ : M → N by φ(m) = Σj∈Jgj(m(j)). Since m is nonzero in only finitely manyplaces it follows that this map is well defined. It is tedious yet straightforward to show that φ isan R-module homomorphism. The difficuly lies in the fact that the sum of two elements in M maybe zero on some coordinate although each element is nonzero at said coordinate.

Next, it follows that for each j ∈ J and m ∈ Mj , φ ◦ ιj(m) = gj(m) and so φ ◦ ιj = gj .Uniqueness is again straighforward.

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Exercise 1.5. (1) Let BanSp be the category of Banach spaces together with bounded lineartransformations. Let A and B be Banach spaces. Let P1 be the cartesian product of A and Bequipped with the sup norm. Let P2 be the cartesian product of A and B equipped with the sumnorm. Show that both P1 and P2 together with the natural projections are products of A and Bin BanSp.

Finally, show that if we change to BanSp< the category of Banach spaces and norm decreasinglinear transformations then P2 is a product but P1 is not.

(2) Show that in the category of torsion groups that the product is not necessairly the cartesionproduct.

A more general notion of products is that of limits and colimits. We need some sources andsinks.

Definition 1.6. Let C be a category, and I be a small category; that is, one so that obj(I) is aset. A diagram in C is a covariant functor D : I → C.

For example, to obtain the diagram that is used to define the product one chooses a set I, andtreats it as a category in which the only morphisms are the identities on each object i ∈ I.

Definition 1.7. Suppose that D : I −→ C is a diagram. A source for D is an object A ∈ obj(C)together with morphisms αi : A −→ D(i), which are compatible with D, meaning, that if τ : i −→ jis a morphism, then

D(τ) · αi = αj .

A sink for D is the dual concept; that is, a C–object B with morphisms βi : D(i) −→ B, such that,for each τ : i −→ j, βi = βj ·D(τ).

We now have all the preliminaries in place to define the general concept of a limit in a category.

Definition 1.8. Suppose that D : I → C is a diagram. We say that D has a limit if there is auniversal source for D; that is to say, there is a source (δi : L −→ D(i))i∈I , with the property that,for any source (fi : A −→ D(i))i∈I there is a unique C–morphism f : A −→ L such that, for eachi ∈ I, δi · f = fi.

D is said to have a colimit if there is a universal sink for D. (We leave the fleshing out of“universal sink” to the reader.)

Example 1.9. In this context the pushout from the previous section is simply the colimit of thediagram D : I → R-Mod where Obj(I) = 1, 2, 0 and Mor(I) = {m : 0 −→ 1, n : 0 −→ 2}. Thediagram is given by D(0) = A,D(1) = B,D(2) = C and D(m) = g,D(n) = f .

Definition 1.10. An upward-directed set is a partially-ordered (I,≤) set such that for every i, j ∈ Ithere is a k ∈ I for which i, j ≤ k. A diagram D from an upward-directed set into a category C iscalled a direct system in C. A colimit (L, li) for the diagram D is called a direct limit.

Analogously we may define downward directed set, inverse system, and inverse limit.Exercises 8-11 of Chapter 7.6 and Ex. 25, 26 of Chapter 10.3 ([DF]) show how to construct the

direct and inverse limit for groups, rings and R-modules.

We should point out that the notion of a direct limit is in fact a colimit and an inverse limit isa limit. I know....


2. Natural Transformations

At this point in one’s mathematical career the reader has probably encountered the use of thephrase naturally isomorphic. We wish to explain what this means in all its glory.

Definition 2.1. Let F,G : A −→ B be two covariant functors between categories. A naturaltransformation from to F to G is a map η : Obj(A) −→Mor(B) satisfying the following condition

1) For each A ∈ Obj(A), η(A) : F (A) −→ G(A)2) For each A-morphism f : A→ A′ the following diagram commutes:

F (A) -G(A)

?

F (A′) -G(A′)?

η(A)

F (f)

η(A′)

G(f)

If for each A-object A, η(A) is a B-isomorphism then we call η a natural isomorphism or naturalequivalence and say F and G are naturally isomorphic or equivalent. We shall often write F ∼= G.

Suppose that C and D are categories. We say that C and D are equivalent categories if thereexist covariant functors F : C −→ D and G : D −→ C such that G · F ∼= 1C, and F ·G ∼= 1D.

If the functors in the previous paragraph are contravariant, then we say that the categories aredual, and the functors in question are called dualities.

Example 2.2. (1) Let R be a fixed field and consider R-Mod. For a vector space V , let V ∗ =HomR(V,R) called the dual of V . Define a covariant functor G from R-Mod to itself by settingG(V ) = V ∗∗. And if f : V −→ W is a linear transformation then G(f) = Hom(Hom(f,R), R).Now, Hom(·, R) is a contravariant functor but when applied twice the second dual functor iscovariante. Let F be the identity functor. Finally, for a vector space V define η(V ) : V −→ V ∗∗ byη(V )(v) = Ev, where Ev is the evaluation map. It is not hard to check that the following diagramcommutes:

V - V ∗∗

?

W - W ∗∗?

η(V )

f

η(W )

G(f)

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If we restrict our attention to the full subcategory of finite dimensional vector spaces then thisnatural transformation is in fact a natural isomorphism. (See Theorem 19, Ch. 11.3 [DF].)

(2) Let F be the functor from Grp to itself which assigns to each group G, its commutatorsubgroup. Then there is a natural transformation between F and the identity functor where for agroup G, η(G) : F (G) −→ G is simply the inclusion homomorphism. We leave it to the reader toverify that the appropriate square commutes.

(3) Let Top312

and KTop2 denote the categories of Tychonoff spaces and compact Hausdorffspaces, respectively. Let β : Top31

2−→ KTop2 be the functor which assigns to a Tychonoff space

its Stone-Cech compacitification. Let T : KTop2 −→ Top312

be the inclusion functor.There exists two natural transformations:

η : 1Top312

−→ T ◦ β

andε : β ◦ T −→ 1KTop2

where for each Tychonoff space X, ηX is the natural embedding of X into its Stone-Cech compac-itification and for each compact hausdorff space Y , εY is the unique homeomorphism induced bythe identity on Y .

Example 2.3. The categories BA of boolean algebras and homomorphisms is dual to KZero, thecategory of compact zero-dimensional spaces and continuous maps between them. The functors inquestion are F which assigns to a compact zero-dimensional spaces its boolean algebra of clopensets, and G the functor which asssigns to a boolean algebra its topological space of maximal idealsendowed with the hull-kernel topology.

3. A Projective Partial Order

We would like to use this section to define what we mean by “projectively the largest” object ina category. Let C be a category and let P be a collection of C-objects. Let be a property P ofmorphisms. Define a “partial order” (observe that P may not be a set) on P by saying that

A ≤ B

for A,B ∈ P if there is a morphism f : B −→ A satisfying P . (To ensure that ≤ is anti-symmetricone would have to mod out by isomorphism classes.) If there exists an object in P (unique upto ismorphism), say P , such that P ≥ A for all other A ∈ P, then we say (any element of theequivlence class containing) P is projectively the largest.

Example 3.1. Consider the category Top312

consisiting of all Tychonoff spaces and continuousmaps. Fix a Tychonoff space X and let P be the class of all compactifications of X, that is,K ∈ P is K contains a dense subspace which is homeomorphic to X. To each K ∈ P call thishomeomorphism iK : X −→ K. Let P be the property that if K1,K2 ∈ P and f : K1 −→ K2 thenf ◦ iK1 = iK2 . Then βX is projectively the largest element of P.


Example 3.2. We return to the pushout. Fix R-modules A,B,C together with map g : A −→ Band f : A −→ C as in Chapter 1, § 4. Let P be the class of all R-modules M such that there existR-module homomorphism g′M : C −→ M and f ′M : B −→ M making the square commute. Let Pbe the property that if p : M −→ N then p ◦ g′M = g′N and p ◦ f ′M = f ′N . Then the pushout isprojectively the largest such R-module.

We could have defined the above using diagrams instead of propertys. For example, βX isuniversal for a triangle in its appropriate category and the pushout is universal for a square.

In the future to understand a certain pair of functors HomR(B, ·) and B⊗R (·) it shall be usefulto consider adjoint situations.

Definition 3.3. Suppose that C and C’ are categories. Consider functors F : C −→ C′ andG : C′ −→ C. The pair (F,G) is called an adjunction – or an adjoint situation, or an adjointpair – if there exist natural transformations σ : 1C −→ G · F and τ : F ·G −→ 1C′ , satisfying thefollowing conditions:

(adj1) For each A ∈ obj(C′), the composite

G(A)σG(A)−→ G(F (G(A)))

G(τA)−→ G(A),

is the identity morphism on G(A); that is,

G(τA) · σG(A) = 1G(A).

(adj2) The composite

F (X)F (σX)−→ F (G(F (X)))

τF (X)−→ F (X),

is the identity morphism on F (X), for each C–object X; that is to say,

τF (X) · F (σX) = 1F (X).

The above adjoint situation is encapsuled as a quadruple (F,G, σ, τ), and it is also common to write:F a G. In the adjoint situation (F,G, σ, τ), the functor F is frequently called the front (or left)adjoint, while G is referred to as the back (or right) adjoint. Likewise, the natural transformationsσ and τ are called the front and back adjunctions, respectively.

Definition & Remarks 3.4. Suppose that (F,G, σ, τ) is an adjunction, with F : C −→ C′ andG : C′ −→ C. Let X ∈ obj(C). For each C–morphism g : X −→ G(B), consider g′ = τB · F (g) :F (X) −→ B. Note that, by the natural property of σ and then (adj1),

G(g′) · σX = G(τB) · (G · F (g)) · σX = G(τB) · σG(B) · g = g.

All of which shows that there exists a morphism g′ : F (X) −→ B such that the triangle belowcommutes.

X -σX

@@

@@

@@

@R

g

G(F (X)) F (X)

?

G(B) B

?

G(g′) g′(†)

14 Math 734

g′ is unique with respect to this property, for if G(h) · σX = g, then, owing first to the naturalfeature of τ , and then (adj2),

g′ = τB · F (g) = τB · (F ·G(h)) · F (σX) = h · τF (X) · F (σX) = h.

We have proved half of the following theorem:

The reference (†) in the formulation is to the diagram above.

Theorem 3.5. Suppose that F : C −→ C′ and G : C′ −→ C are two functors, and that σ :1C −→ G ·F and τ : F ·G −→ 1C′ are natural transformations so that (F,G, σ, τ) is an adjunction.Then for each C–morphism g : X −→ G(B) there exists a unique C’–morphism g′ : F (X) −→ Bsuch that the triangle (†) commutes.

Chapter 3

Introduction to Homological Algebra

1. Free Modules

In this section we reintroduce free modules and give a nice characterization of them.

Definition 1.1. Let X be a nonempty set. A module F is said to be free over X if there existsan injective map i : X → F such that for any map f : X → M where M is an R-module, thereexists a unique R-module homomorphism f : F →M making the following diagram commute:

F

?

f

X -i

M

��

��

��

f

Free modules over any set X always exists. We shall recall the object but leave the readerto verify the details or check Theorem 6 of Ch. 10.3 of [DF]. Recall that the phrase “extend bylinearity” arises from consideration of free objects.

If X is a nonempty set let F = ⊕x∈XRx where Rx = R for each x ∈ X. Let ex ∈ F be definedby

ex(y) ={

1R, if x 6= y;0, otherwise.

Then define i : X → F by i(x) = ex. It turns out that F is free over X.

We are aiming at an internal characterization of when a module is free over some set X. Thefollowing will be useful.

Definition 1.2. Let M be an R-module and X ⊆ M . We say X is linearly independent if fordistinct x1, · · · , xn ∈ X and r1, · · · , rn ∈ R if

r1x1 + · · ·+ rnxn = 0

15

16 Math 734

then r1 = · · · = rn = 0. A linearly independent set which generates M is called a basis for M .Observe that a subset of a linearly independent set is again linearly independent. Also, if x is amember of a lin. ind. set then x is not torsion since rx = 0 implies r = 0.

A set which is not linearly independent is called linearly dependent.

Lemma 1.3. Let M be a module and suppose X is a basis for M . Then every element of M maybe written uniquely as a finite sum of elements from X.

Proof: Part of the definition of basis says that X generates M . Thus, every element may bewritten as a sum of elements from X. Now suppose r1x1 + · · · rnxn = s1y1 + · · · smym, where0 6= ri, si ∈ R and xi, yi ∈ X. Subtract one side to the other. Now there might be a yi = xj .Relabel and without loss of generality the first k-many have this property. Thus,

(r1 − s1)x1 + · · · (rk − sk)xk + rk+1xk+1 + · · · rnxn − sk+1yk+1 − · · · − smym = 0

Linear independence implies that for all j > k, rj = 0 = sj . But we assumed that these were nonzeroand so n = k = m. Furthermore, linear independence also implies that r1 = s1, · · · , rk = sk. Butthen this says that the expression was unique to begin with.�

Theorem 1.4. Let F be an R-module. The following are equivalent:

(1) F has a nonempty basis.

(2) F is an internal direct sum of cyclic R-modules, each isomorphic to R.

(2’) F is an internal direct sum of cyclic R-modules, each having zero annihilator.

(3) F ∼= ⊕x∈XRx where Rx∼= R for each x ∈ X.

(4) There exists a set X for which F is free over X.

Proof: (1) implies (2). Let X be a nonempty basis for F . For each x ∈ X, the map φx : R→ Rxdefined by φx(r) = rx is an R-module homomorphism. φx is always onto and since x is not torsionit follows that φx is an isomorphism. Consider M = Σx∈XRx. M is a submodule of F . Also, sinceX is linearly indepenedent this sum is direct. Thus, all we need show is that M = F But thatfollows from the fact that X is a basis. Thus, F = ⊕x∈XRx where each of said cyclic modules isisomorphic to R.

(2) and (2’) are clearly equivalent. (2) implies (3) is patent.(3) implies (1). It follows from (3) that in fact F is isomorphic to the free module de-

scribed/constructed in Definition 1.1. Thus, all we need to show is that the set X = {ex : x ∈ X}is a basis for ⊕x∈XR. This we leave to the interested reader.

(1) implies (4). Let X be a basis for F and let i : X ↪→ F be the inclusion map. Supposef : X →M is any map.

Suppose F ∼= ⊕x∈XRx where Rx∼= R for each x ∈ X. Fix each of these isomorphisms. For

each x we let 1x denote the element of Rx which corresponds to 1R. By Lemma 1.3 each element ofF may be written uniquely as a finite R-linear combination of elements from X. Thus, for u ∈ F ,u = Σn

i=1rixi. Define a map f : F → M by f(u) = Σni=1rif(xi). The fact that it is an R-module

homomorphism and that it makes the diagram commute is a similar argument used in showingthat the direct sum of copies of R has said property. Uniqueness follows easily from the R-modulehomomorphism property.

Homological Algebra 17

(4) implies (3) Given i : X ↪→ F , let yx = i(x) and set Y = {yx : x ∈ X}. Now let M =oplusx∈XR and define f : X → M by f(x) = ex. By (4) there exists a unique homomorphismf : F → M . But we know that M is free over X so that there exists a unique homomorphismi : M → F . Recapping f ◦ i = f and i ◦ f = i.

Consider i ◦ f ◦ i. For any x ∈ X we have

i ◦ f ◦ i(x) = i(f(x)) = i(x)

so that (i ◦ f) ◦ i = i. But considering the following diagram

F

?

i

X -i

F

��

��

��

1F

and uniqueness of 1F it follows that 1F = i ◦ f . But this means that f is an isomorphism which iswhat we wanted to conclude (3).�

Corollary 1.5. Free R-modules exist and are unique up to isomorphism.

Corollary 1.6. Every R-module M is the homomorphic image of a free module F . Moreover, ifM is finitely generated then we may take F to be finitely generated.

Proof: Let S be a generating set for M and F be a free module over X where |S| = |X|. Let fbe any bijection from X onto S. We consider f as a map from X into M . Thus, there exists aunique R-module homomorphism f : F → M such that f ◦ i = f . The image of f is a submodulecontaining S and hence is all of M , i.e., f is a surjection.�

2. Vector Spaces

Throughout this section R is a field and vector spaces are assumed to be R-vector spaces.

Lemma 2.1 [AC]. Every linearly independent set of V is contained in a maximal linearly inde-pendent set.

Proof Let X be a lin. ind. set and set S to be the set of lin. ind. sets containing X. SurelyS 6= ∅. If C is a chain in S, then let Y = ∪C. Let x1, · · · , xn be distinct elements from Y . Thenfor each i = 1, · · · , n there is a Yi ∈ C such that xi ∈ Yi. Since it is a chain there is a Z ∈ C suchthat xi ∈ Z for each i = 1, · · ·n. Thus, the fact that Z is linearly independent ensures us that if aR-linear combination of the xi is zero then R elements each must be 0.

Enforcing Zorn’s Lemma we obtain our desired result.�

Lemma 2.2. Every maximal linearly independent set of V is a basis for V .

18 Math 734

Proof: Let X be a maximal linearly independent set of V and set W to be the span of X, asubspace of V . Our aim is to show that W = V .

Suppose not and let a ∈ V \W . Suppose

ra+ r1x1 + · · · rnxn = 0

for ri ∈ R and xi ∈ X (xi distinct). If r 6= 0, then a = r−1(−r1x1 − · · · − rnxn) ∈W . Thus, r = 0.By the linear independence of X it follows that r1 = · · · = rn = 0 and so X ∪ {a} is a linearlyindependet set contradicting the fact that X is maximal.�

Theorem 2.3. Let V be a vector space. V has a basis.

Proof: Let v ∈ V be nonzero. Recall that Ann(v) is an ideal of R and since R is a field it followsthat v is not torsion and so {v} is a linearly independent set. By Lemma 2.1 we may extend thisto a maximal linearly independent set which by Lemma 2.2 is a basis.�

Corollary 2.4. If R is a field, then every R-module is free.

Example 2.5. Consider Z/2Z. Since this is not isomorphic to a direct sum of copies of Z it followsthat it is not a free Z-module.

Proposition 2.6 [Replacement Theorem]. Let V be a finite dimensional vector space, i.e., it hasa finite basis. Then any other basis has the same cardinality.

Proof: Let X = {x1, · · · , xn} and Y = {y1, · · · ym} be bases for V . Suppose for the sake ofargument that n < m.

Now, we may write ym as a linear combination of the xi, say

ym = r1x1 + · · ·+ rnxn.

Let k be the smallest natural such that rk 6= 0. Then

xk = r−1k ym − r−1

k rk+1xk+1 − · · · − r−1k rnxn.

It follows then that X ′ = {ym, x1, · · · , xk−1, xk+1, · · · , xn} spans V .We can continue this process with ym−1, ym−2, · · ·. For any natural p after p many steps in this

process we have that {ym, · · · , ym−(p−1)} together with n− p many of the xi span V . Since n < m,{ym, · · · , ym−(n−1)} spans V (the nth step) which then means that y1 may be written as a linearcombination of {ym, · · · , ym−(n−1)} which then contradicts that Y is a linearly independent set.

If m > n a similar argument yields a contradiction. Thus, n = m.�

Corollary 2.7. Let R be a field. If Rn ∼= Rm, then n = m.

We observe that the more general proof that for any vector space two bases have the samecardinality is also true. But its proof is long and tedious and takes us away from what we wouldlike to do. We suggest [H], Theorem 2.6, Ch. IV. We simply state it as a theorem.

Theorem 2.8. Let F be a free R-module over sets X and Y (not necessarily finite). Then |X| =|Y |.

We end this section by giving the reader something to look forward to. The converse of Corollary2.4 is also true but we need some heavier machinery to handle the task.

Projective Modules 19

3. Projective Modules

Definition 3.1. A composition g ◦ f of R-mod homomorphisms, written

Af−→ B

g−→ C

is said to be exact if Im(f) = ker g. Observe that in this case g ◦ f = 0. A sequence

· · · fi−1−→ Aifi−→ Ai+1

fi+1−→ · · ·

is said to be exact if for each n, Im(fn) = ker fn+1. We often abuse language and say it is asequence of R-modules instead of R-module homomorphisms.

Remark: the notation 0 −→ A and A −→ 0 should be clear even without reference to anyfunctions.

Lemma 3.2. (1) The sequence 0 −→ Af−→ B is exact if and only if f is 1-1.

(2) The sequence Bg−→ C −→ 0 is exact if and only if g is onto.

Definition 3.3. An exact sequence of the form 0 −→ Af−→ B

g−→ C −→ 0 is called short exact.Observe that in this case f is 1-1 and g is onto which means we may consider A as a submoduleof B and C as a quotient module with (by the First Isomorphism Theorem) C ∼= B/ ker(g) =B/Im(f) = B/f(A).

In the above sequence A is called the submodule and C is called the projection.

Example 3.4. Let f : A −→ B be a homomorphism. Then

0 −→ ker f i−→ Aπ−→ B/f(A) −→ 0

is a short exact sequence where i is the inclusion map and π is the natural projection.

Example 3.5. Suppose B = A⊕ C. Then

0 −→ AιA−→ B

πC−→ C −→ 0

is a short exact sequence.

Theorem 3.6. Let 0 −→ Af−→ B

g−→ C −→ 0 be a short exact sequence. The following areequivalent.

(1) There exists an R-module homomorphism h : C −→ B such that gh = 1C .

(2) There exists an R-module homomorphism k : B −→ A such that kf = 1A.

(3) f(A) is a summand of B, i.e., there exists an R-module C ′ ≤ B for which B = f(A)⊕ C ′.

Proof: (2) implies (3). Let C ′ = ker k. We claim B = f(A)⊕ C ′. For any b ∈ B,

b = f(k(b)) + (b− f(k(b)).

Now, f(k(b)) ∈ f(A) and k(b− f(k(b))) = k(b)− kf(k(b)) = k(b)− k(b) = 0. Thus, B = f(A) +C ′

so all we need is to show the sum is direct. Let b ∈ f(A)∩C ′, then there is an a ∈ A with f(a) = b.Then

a = kf(a) = k(b) = 0

20 Math 734

whence b = f(0) = 0.

(3) implies (2). Suppose B = f(A) ⊕ C ′ for some submodule C ′ ≤ B. Define k : B −→ A bythe following. For b ∈ B, then b = x + y for some x ∈ f(A), y ∈ C ′. Let a ∈ A with f(a) = x.Then define k(b) = a which is well defined by uniqueness of the direct sum. It also apparent thatk is an R-module homomorphism. Since for any a ∈ A the representation of f(a) in B is given byf(a) = f(a) + 0 it follows that k(f(a)) = a, whence k ◦ f = 1A.

(1) implies (3). Let C ′ = h(C). We claim B = f(A) ⊕ C ′. Let b ∈ B and notice thatb = b−hg(b) +hg(b), where h(g(b)) ∈ C ′. Since g(b−hg(b)) = g(b)− ghg(b) = g(b)− g(b) = 0 andthe original sequence is exact we obtain that b− hg(b) ∈ ker g = f(A).

Finally, if x ∈ f(A) ∩ C ′, then there is some c ∈ C such that h(c) = x and x ∈ f(A) = ker g soc = gh(c) = g(x) = 0 hence x = 0.

(3) implies (1). Let c ∈ C. Since g is surjective there is a b ∈ B such that g(b) = c. By (3)write b = x+ y where x ∈ f(A), y ∈ C ′. Define h : C −→ B by h(c) = y.

First note if g(b1) = g(b2) = c and b1 = x1 + y1 and b2 = x2 + y2 then we would like to convinceourselves that y1 = y2. Now,

y1 − y2 = (b1 − b2)− (x1 − x2).

Since both y1 and y2 lie in C ′ so does their difference. Similarly, x1 − x2 ∈ f(A). But alsog(b1 − b2) = 0 so b1 − b2 ∈ f(A). Therefore, y1 − y2 ∈ f(A) ∩ C ′ whence y1 = y2. So h iswell-defined. Again it is easy to check that h is an R-mod homomorphism.

Finally, letting c, b, x, y be as above we obtain

g(h(c)) = g(y) = g(b) = c

and so g ◦ h = 1C . (The second equality stems from the fact that g(b) = g(x) + g(y) = g(y) sinceg(x) = 0.)�

Definition 3.7. If either of the conditions of Theorem 3.6 are satisfied then B ∼= A⊕ C. For thisreason such a short exact is said to splt.

Example 3.8. Let 0 −→ Af−→ B

g−→ F −→ 0 be a short exact sequence where F is a free module,say over X ⊆ F . Since g is surjective, for each x ∈ X choose bx ∈ g−1({x}) and let h : F −→ Bbe the unique R-module homomorphism such that h(x) = bx. Then since (g ◦ h(x)) = g(bx) = itfollows by uniquesness that gh = 1F and so the exact sequence splits.

Our main remark is that a free module has a very nice property in that whenever it is placedin the projection spot then the sequence splits. We explore this phenomenon presently,

Recall that for a fixed R-module D, HomR(D, ·) is a covariant functor. For the rest of thissection given an f : A −→ B instead of writing HomR(D, f) we shall employ f ′.

Proposition 3.9. Let D be a fixed R-module. Suppose

0 −→ Af−→ B

g−→ C −→ 0

is a short exact sequence. Then

0 −→ HomR(D,A)f ′−→ HomR(D,B)

g′−→ HomR(D,C)

is exact.


Proof: We need to demonstrate that f ′ is 1-1 and that Im(f ′) = ker g′.Let ψ, φ ∈ HomR(D,A) such that f ′(ψ) = f ′(φ). Then

f ◦ ψ = f ′(ψ) = f ′(φ) = f ◦ φ

and since f is monic it follows that ψ = φ so f ′ is injective.

Two situations:I. Let φ ∈ Imf ′, i.e., there exists ψ ∈ HomR(D,A) such that f ′(ψ) = φ. So

g′(φ) = g′(f ′(ψ)) = g ◦ (f ◦ ψ) = (g ◦ f) ◦ ψ = 0 ◦ ψ = 0.

Therefore, Imf ′ ⊆ ker g′.

II. Let φ ∈ ker g′. So for every d ∈ D, 0 = (g ◦ φ)(d) = g(φ(d)) and thus φ(d) ∈ ker g = Imfsince the original sequence is exact. It follows that there is a unique a ∈ A such that f(a) = φ(d).Define ψ : D → A by ψ(d) = a which is well-defined by uniqueness. Once again it is an R-modulehomomorphism.

(f ◦ ψ)(d) = f(ψ(d)) = f(a) = φ(d)

and sof ′(ψ) = f ◦ ψφ,

whence ker g′ ⊆ Imf ′. Together with I. it follows that Imf ′ = ker g′ and the desired result isobtained.�

Theorem 3.10. For an R-module D the following are equivalent.

(1) HomR(D, ·) preserves short exact sequences.

(2) Given a surjective R-mod homomorphsim g : M −→ N and any R-module homomorphismφ : D −→ N there exists a homomorphism ψ : D −→ M such that the following diagramcommutes.

D

?

φ

M -g

N 0-

��

��

��

�

ψ

(3) If D is a quotient of an R-module M then D ”is” a summand of M .

(4) D is a summand of a free R-module.

22 Math 734

Proof: (1) implies (2). Let g : M → N be surjective. Form the short exact sequence

0 −→ ker g i−→Mg−→ N −→ 0

Let L = ker g. By (1) the following sequence is exact

0 −→ HomR(D,L) i′−→ HomR(D,M)g′−→ HomR(D,N) −→ 0

and so g′ is surjective, i.e., given φ ∈ HomR(D,N) there is some ψ ∈ HomR(D,M) such thatg ◦ ψ = g′(ψ) = φ, which is what we wanted.

(2) implies (3). Suppose D is the quotient of M , i.e., there is a surjective homomorphismg : M −→ D. Consider

0 −→ ker g i−→Mg−→ D −→ 0

(2) implies there exists k : D −→ M such that g ◦ k = 1D. Theorem 3.6 implies D is a summandof M .

(3) implies (4). Duh every module is the quotient of a free module.

(4) implies (1). Let

0 −→ Lf−→M

g−→ N −→ 0

be a short exact sequence. Then by the previous proposition

0 −→ HomR(D,A)f ′−→ HomR(D,B)

g′−→ HomR(D,C) −→ 0

is short exact if and only if g′ is surjective. So let φ : D −→ N be a homomorphism. Let F befree (over the set X) where F = D ⊕Kand π : F −→ D be the natural projection. Consider thediagram


F

?

��

��

��

��

��

��

��

��

π

6

ιD

Dh

?

φ

M -g

N

��

��

��

�0-

ψ

Now, for each x ∈ X, (φ◦π)(x) = g(mx) for some mx ∈M . Let h be the unique homomorphismfrom F to M such that h(x) = mx and note that (φ◦π)(x) = g(mx) = (g◦h)(x) so that φ◦π = g◦h.Let ι : D −→ F be the inclusion map and set ψ = h ◦ ι. Then

(g ◦ ψ)(d) = g(h((ι(d)))= (g ◦ h)(ι(d))= (φ ◦ π)(ι(d))= φ(π(ι(d)))= φ(d)

So g′(ψ) = g ◦ ψ = φ, whence g′ is onto.�

Definition 3.11. Any module D that satisfies any, and hence all of the conditions of the previoustheorem is called projective.

Corollary 3.12. Free modules are projective. If P is a finitely generated, then it is projective ifand only if it is the summand of a finitely generated free module.

Corollary 3.13. Every summand of a projective module is projective.

Proof: Suppose P is projective and P = A⊕B. Then for some module K and free module F wehave F = P ⊕K = A⊕B ⊕K and so A is projective.�

Example 3.14. Suppose P is a projective Z-module. Then P ⊕K is a free Z-module, i.e., a directsum of copies of Z. It follows that P cannot have any elements of finite order.

E.g. Z/2Z is not a projective Z-module. Neither is Q/Z. Later on we will show that if R is aPID then every projective module is free.

Example 3.15. Let R = Z/2Z × Z/2Z. Let P1 = {(a, 0) ∈ R : a ∈ Z/2Z} and similarly P2 ={(0, a) ∈ R : a ∈ Z/2Z}. Then P1 and P2 are projective modules yet they are not free as any freemodule would have to have at least 4 elements.

Theorem 3.16. If R is a PID, then every submodule of a free R–module is free, i.e., every pro-jective module is free.

24 Math 734

Proof: This argument requires the principle of transfinite induction, as well as the set–theoreticaxiom that every set can be well-ordered.

Suppose that F is a free R–module. Then F = ⊕i∈I Ri, where each Ri is an isomorphic copyof RR. Now, we assume that I has been well-ordered. For each j ∈ I, let

Fj =⊕i<j

Ri;

it should be clear that each Fi ≤ Fi+1, and that if j has no predecessor in the ordering, thenFj = ∪i<j Fi. Finally, F = ∪i∈I Fi. (Note that F1 is trivial.)

Now suppose that G is a submodule of F , and let Gi = Fi∩G. Again observe that (i) Gi ≤ Gi+1

and (ii) Gj = ∪i<j Gi, if j has no predecessor. Also, (iii) G = ∪i∈I Gi. By one of the isomorphismtheorems, for each i ∈ I,

Gi+1/Gi = Gi+1/(Fi ∩Gi+1) ∼= (Fi +Gi+1)/Fi,

and the latter is a submodule of Fi+1/Fi, which is isomorphic to R. Because R is a PID, Gi+1/Gi

is isomorphic to an ideal Ji of R, which is either trivial, or else principal, and free in either event.Since free modules are projective, it follows that either Gi = Gi+1, or else the canonical short

exact sequence0 −→ Gi −→ Gi+1 −→ Gi+1/Gi −→ 0

splits, by Theorem 3.6. This means that Gi+1 = Gi⊕Si, where Si is either trivial or an isomorphiccopy of R. Suppose now that, for each i < j, it has been demonstrated that Gi = ⊕i′<i Si′ . (Notethat S1 = G2.) If j is the successor of j′, then Gj = Gj′+1 = Gj′ ⊕ Sj′ , and it follows thatGj = ⊕i<j Si. If j has no predecessor, then it is the union of the Gi (all i < j), each of whichis a direct sum of Si’s, which, in turn, implies that Gj = ⊕i<j Si. By the principle of transfiniteinduction, it follows that

G =⊕i∈I

Si,

and hence free.�

The converse of Theorem 3.16 is not true. We shall point this out in the next section (thoughwithout a proof). We finish this section with the proof of the converse of Corollary 2.4. First auseful lemma regarding cyclic modules (look back at Proposition 3.8).

Lemma 3.17. Suppose I is an ideal of R. If I is a summand then I is a prinicipal ideal I = (e)generated by an idempotent.

Proof: Since I is assumed to be a summand there exists an ideal J of R satisfying R = I + Jwhere the sum is direct. There exists i ∈ I, j ∈ J such that 1 = i+ j. A simple calculation gives usi = i(i+j) = i2 + ij. Now, since ij ∈ I∩J = 0 it follows that i = i2, hence i is idempotent. Clearly,since i ∈ I the ideal generated by i is contained in I. Also if x ∈ I, then x = x(i+ j) = xi+xj = xi(since xj = 0) and so I ⊆ (i) forcing them to be equal.�

Lemma 3.18. Let A be a cyclic module and I = Ann(A). If A is projective, then I is a summandof R.


Proof: Consider the short exact sequence

0 −→ I −→ R −→ A −→ 0.

By the hypothesis A is projective and hence the short exact sequence splits which means that I isa summand of R. Furthermore, by the previous lemma I is generated by a single idempotent.�

Proposition 3.19. Suppose every R-module is free then R is a field.

Proof: Suppose M is a proper maximal ideal. The goal is to show that M = 0. Now as anR-module, the cyclic module R/M is free and hence projective. By the previous lemma M is asummand which means there is an ideal K of R such that R = M ⊕ K. Again by assumptionM and K is both free. Now, R is free of rank 1 so that applying Theorem 2.8 either M must beof rank 1 and K of rank 0, or vice-versa. In the first case, it follows that K = 0 and so R = Mcontradicting that M is proper. The other case implies that M = 0 which is what we had hopedfor.�

4. Nakayama’s Lemma

We feel now is a good time to bring up Nakayama’s Lemma and use it to show that for other kindsof rings f.g. projectives are free. We need a discussion on the Jacobson radical.

Definition 4.1. The Jacobson radical of a ring R is defined to be the intersection of all maximalideals of R. We denote this ideal by J(R).

Proposition 4.2. Let R be a commutative ring with identity. Then

J(R) = {a ∈ R : 1− ra is a unit ∀r ∈ R}.

Proof: Let I be the set on the right. Let a ∈ Rand suppose a /∈ J(R) so a /∈ M for someM ∈ Max(R). Then the ideal generated by a and M is everything and so 1 = ra + m for somem ∈M and r ∈ R. It follows that 1− ra is not a unit. Hence I ⊆ J(R).

Conversely, suppose a ∈ J(R) but for some r ∈ R, 1− ra is not a unit. Since it is not a unit itmust lie in some maximal ideal, say M . Then both ra and 1 − ra are in M and hecne so is 1; acontradiction.�

Proposition 4.3. Suppose R is a commutative ring with identity. Then J(R) = 0 if and only ifR is a subdirect product of fields.

Proof: The student has seen this type of theorem before. Just follow your nose.

Theorem 4.4 [Nakayama’s Lemma]. Suppose that R is a commutative ring with identity. Assumethat M is a finitely generated R-module, J an ideal of R contained in J(R), and that JM = M .Then M = {0}.

Proof: Choose a minimal generating set for M . Call it X = {a1, · · · , an}. If M 6= 0, then a1 6= 0by minimality. Since JM = M it follows that there are j1, · · · , jn ∈ J such that

a1 = j1a1 + · · · jnan.

26 Math 734

Now, in case n = 1 then (1− j1)a1 = 0. But because 1− j1 is a unit it then follows that a = 0; acontradiction. Thus,

(1− j1)a1 = j2a2 + · · ·+ · · · jnan.

And soa1 = (1− j1)−1(j2a2 + · · ·+ · · · jnan).

But this means that a1 is an R-linear combination of {a2, · · · , an} contradicting the minimality ofX.�

Recall that a local ring is a ring with a unique maximal ideal. A ring R is local if and only ifJ(R) is a maximal ideal. For example, valuation domains are local.

Proposition 4.5. If R is a local ring with M the unique maximal ideal, then every finitely gener-ated projective R-module is free.

In fact, Kaplansky proved in 1958 that every projective module over a local ring is free but thatresult would take us away from the scope of the course.

Proof: Let P be a f.g. projective module. There exists a free module F with a finite basis X suchthat P is a quotient of F . We may in fact assume that any other free module satisfying this has abasis no smaller than X. Enumerate X = {x1, · · · , xn}. Let π : F −→ P be the natural projection(as P is projective.) We also know that {π(x1), · · · , π(xn)} is a generating set for P . Let K = kerπ.

If K is not a subset of MF , then there exists a k ∈ K \MF . Now, there are r1, · · · rn such thatk = r1x1 + · · · rnxn. We know that one of and hence without loss of generality r1 /∈ M . But thismeans r1 is a unit in R. So a simple calculation using that k ∈ kerπ we obtain

π(x1) = π(x1 − r−11 k) = π(Σn

i=2 − r−11 rixi) = Σn

i=2 − r−11 riπ(xi)

from which it follows π(x1) is spanned by the set {π(x2), · · · , π(xn)}. But then the free modulewith basis {x2, · · · , xn} would map surjectively onto P contradicting the minimality of X. Thus,K ⊆MF .

Next, we identify P with its isomorphic copy which happens to be a summand of F , i.e., F =P ⊕K. Furthermore F ⊆MF +P . Consider the R-module F/P . If x ∈ F then x = (Σm

i=1rifi)+ pwhere ri ∈M , fi ∈ F and p ∈ P . Then

x+ P = (Σmi=1rifi) + P ∈M(F/P )

which shows that F/P = M(F/P ). Finally, F/P is f.g. so Nakayama’s Lemma implies thatF/P = 0, whence K = 0 so F ∼= P .�

5. Injective Modules

In this section we consider the case where the contravariant functor HomR(·, D) preserves exactsequences. Throughout this section f∗ will denote HomR(f,D), where for any φ ∈ HomR(f,D),f∗(φ) = φ ◦ f . The interesting here to note is that contravariance usually turns arrows around.

Lemma 5.1. If Bg−→ C −→ 0 is exact, then g∗ is injective, i.e., 0 −→ HomR(C,D)

g∗−→HomR(B,D) is exact.

Proof: Suppose g∗(f) = g∗(h), i.e., h ◦ g = f ◦ g. But g is epic so that f = h.�

Injective Modules 27

Theorem 5.2. Suppose 0 −→ Af−→ B

g−→ C −→ 0 is exact. Then

0 −→ HomR(C,D)g∗−→ HomR(B,D)

f∗−→ HomR(A,D)

is exact.

Proof: Let φ ∈ Im g∗. There is some ψ in the domain of g∗ such that φ = g∗(ψ) = ψ ◦ g. Then

f∗(ψ ◦ g) = ψ ◦ g ◦ f = ψ ◦ 0 = 0

so that φ ∈ ker f∗, whence Im g∗ ⊆ ker f∗.Conversely, suppose f∗(φ) = 0. Consider the diagram

B - C

?

D

g

φ

Now, ker g = Imf ⊆ kerφ and so by the Induced Homomorphism Theorem there is a ψ : C → Dfor which φ = ψ ◦ g = g∗(ψ) from which we get the reverse inclusion.�

Definition 5.3. Any module D for which HomR(·, D) preserves exact sequences is called injective.By the previous result D is injective if and only if HomR(·, D) takes injective maps to surjectivemaps.

Theorem 5.4. (1) HomR(·, D) preserves short exact sequences.

(2) Given an injective R-mod homomorphsim f : A −→ B and any R-module homomorphismφ : A −→ D there exists a homomorphism φ : B −→ D such that the following diagramcommutes.

0 - A - B

?

��

��

��

��D

f

φ φ

28 Math 734

(3) Whenever D is a submodule of M , then D is a summand of M , ie., every short exact sequence

0 −→ Df−→M

g−→ N −→ 0 splits.

Proof: That (1) and (2) are equivalent should be patent by now.(2) implies (3). Suppose D is a submodule of M . Consider

D - M

?

��

��

��

��D

i

1D h

where h is given by (2). Thus, there exists an h such that h ◦ i = 1D. Applying Theorem 3.6 to

0 −→ Di−→M

π−→M/D −→ 0

we get that D is a summand of M .

(3) implies (2). Suppose we are given a diagram like that in hypothesis 2 where f is injective.Form the pushout of the diagram § 4 of Chapter 1.:

A - B

?

D - L

?

f

φ

f ′

φ′

Since f is injective it follows that f ′ is also 1-1 so by (3) there is an h : L −→ D such thath ◦ f ′ = 1D. Letting φ = h ◦ φ′ we get that for any a ∈ A

φ(f(a)) = h(φ′(f(a)) = h(f ′(φ(a)) = φ(a)

so φ ◦ f = φ as (2) desires.�

Remark 5.5. The proof of the above may be found in [R].

Lemma 5.6 [Baer’s Criterion]. Suppose that J is an R-module. Then J is injective if and only iffor each ideal K of R, and each R-homomorphism f : K −→ J , there is an extension of f to R.

Injective Modules 29

Proof: The necessity is trivial, so we move on to the sufficiency.Suppose that J satisfies the stated condition, with respect to ideals of the ring. Let A be a

submodule of B, and f : A −→ J be an R-homomorphism. We denote by Ex(f) the set of all pairs(C, h), in which C is a submodule of B containing A, and h : C −→ J is a homomorphism whichextends f . Setting (C, h) ≤ (C ′, h′) by C ≤ C ′, and so that h′, restricted to C is h, we get a partialorder. We shall leave it to the reader to verify that Zorn’s Lemma applies to Ex(f).

Now, let (B∗, g) be a maximal member of Ex(f). We wish to show that B∗ = B.If not so, then pick c ∈ B \ B∗. Let K = { r ∈ R : cr ∈ B∗ }; it is readily verified that

K is an ideal of R. Next, define φ : K −→ J by φ(r) = g(cr), and check that this defines anR-homomorphism. By our assumption, there is an extension φ∗ : R −→ J of φ; let z = φ∗(1) andset

f∗(b+ cr) = g(b) + zr, ∀ b ∈ B∗, r ∈ R.

If this is a well-defined map, then it is easy to verify that it is an R-homomorphism extending g tothe submodule B∗ + cR, which is larger than B∗. This is a violation of the maximality of (B∗, g),and the contradiction them implies that B∗ = B, and finishes the proof.

So what remains is to show that f∗ is well defined: if b + cr = b′ + cr′, with b, b′ ∈ B∗, andr, r′ ∈ R, then c(r − r′) = b′ − b ∈ B∗, and therefore r − r′ ∈ K. Thus,

g(b′ − b) = φ(r − r′) = φ∗(1)(r − r′) = z(r − r′);

that is to say, g(b) + zr = g(b′) + zr′, proving that f∗ is well defined. �

Proposition 5.7. For a commutative ring with identity R, the following are equivalent.

(a) Every R–module is projective.

(b) Every short exact sequence splits.

(c) Every R–module is injective.

Proof: We prove that (a) is equivalent to (b), and leave the equivalence of (b) and (c) to thereader.

If every module is projective, and

0 −→ Af−→ B

g−→ C −→ 0,

is short exact, then (since C is projective) the sequence splits, according to Exercise 3.6. Conversely,if every short exact sequence splits, let P be an R–module. There is a free module F and a surjectivehomomorphism g : F −→ P . The sequence below, with inclusion i,

0 −→ ker(g) i−→ Fg−→ P −→ 0,

then splits. P is a summand of a free module and so again by Theorem 3.6 P is projective. �Previously we determined when every R-module is free. We shall later determine when R

satisfies Proposition 5.7. For now we give a nice characterization of injective modules over PID.

Proposition 5.8. Let R be a PID. Then J is injective if and only if rJ = J for all r ∈ R (r 6= 0).

30 Math 734

Proof: Suppose J is injective. Let r ∈ R be nonzero and q ∈ J . Define φ : (r) −→ J by φ(xr) = xqwhich is well-defined since R is an integral domain. Also, φ ∈ HomR((r), J). Extend φ to all of R.Call the extension φ′. Setting q′ = φ′(1). Then as an extension φ′(r) = q and

φ′(r) = rφ′(1) = rq′.

So q = rq′, whence rJ = J .Conversely, let I = (r) be an arbitrary ideal of R. Without loss of generality, r 6= 0. We shall

apply Baer’s Criterion. Let φ ∈ HomR(I, J). Set q = φ(r). Since rJ = J we may choose a q′ ∈ Jsuch that rq′ = q. Define φ′ : R −→ J by

φ′(s) = sq′.

It is straighforward to show that φ′ ∈ HomR(R,Q).Now if s ∈ I, then s = xr where x is unique. Then

φ′(s) = φ′(xr) = xφ′(r) = xrq′ = xq = xφ(r) = φ(xr) = φ(s)

which shows that φ′ extends φ.�

Corollary 5.9. For any integral domain R every injective module J is divisible, i.e., rJ = J forall nonzero r ∈ R.

Corollary 5.10. A Z-module is injective if and only if it is a divisble abelian group.

Example 5.11. Examples of injective Z-modules include Q and Q/Z. So there exists injectiveswhich are not projectives. Also, Z is projective but not injective.

6. Tensor Product

We’ve come to one of the most important definitions contained in these notes. The tensor product isa difficult construction to interpret in practice; we shall try to do this successfully in this section andin the succeeding chapters as it plays a huge role in algebra. Again we note that most treatments ofthe tensor product are done without the assumption of commutativity but as we shall see it makeseverything a whole lot cleaner.

Definition 6.1. Suppose that A,B and M are R-modules. A map θ : A×B −→M is said to beR-bilinear if both θ(a, ) : B −→M and θ( , b) : A −→M are R-module homomorphisms, for eachfixed a ∈ A, and b ∈ B, such that, in addition, for each r ∈ R,

θ(ar, b) = θ(a, rb),∀ a ∈ A, b ∈ B.

A tensor product of A and B is a pair (T, τ), where T is an R-module, and τ : A×B −→ T anR-bilinear map such that, whenever θ : A×B −→M is an R-bilinear map, then there is a uniqueR-module homomorphism θ∗ : T −→M so that θ∗ · τ = θ.

The first result on tensor products establishes the existence and uniqueness.

Theorem 6.2. For each R-module A and R-module B, the tensor product exists, and it is uniqueup to a R-module isomorphism.

Tensor Product 31

Proof:(Existence) Consider the set X = A×B, and let F be the free R-module over X as a set. We

assume for notational purposed that X ⊆ F . Let K be the subgroup of F , generated by all thefollowing elements:

(a) (ar, b)− (a, rb), ∀ r ∈ R, a ∈ A and b ∈ B;

(b) (a1 + a2, b)− (a1, b)− (a2, b), ∀ a1, a2 ∈ A, b ∈ B;

(c) (a, b1 + b2)− (a, b1)− (a, b2), ∀ a ∈ A, b1, b2 ∈ B.

Let T = F/K, and τ : A×B −→ T be the composition π ◦ i where π is the natural projection andi is the inclusion map of X into F .

τ(a, b) = (a, b) +K, ∀ a ∈ A, b ∈ B.

We leave it to the reader to verify that τ is R–bilinear.Now if θ : A×B −→M is an R-bilinear map into the R-module M , denote by g : F −→M the

unique homomorphism for which g · i = θ whose existence follows by the universal property of thefree module. Next, it follows that K ≤ ker(g). Then apply the Induced Homomorphism Theoremto conclude that there is a unique homomorphism θ∗ : T −→ M , so that θ∗(y + K) = g(y), i.e.,θ∗ ◦ π = g. This means that

θ∗ ◦ τ(a, b) = θ∗((a, b) +K) = g((a, b)) = θ(a, b),

Next, suppose that φ : T −→ M is another R-module homomorphism such that φ ◦ τ = θ. Thenθ = φ ◦ τ = (φ ◦ π) ◦ i so that by the uniqueness of g we have that φ ◦ π = g = θ∗ ◦ π. By theuniqueness of θ∗ it follows that θ∗ = φ.

(Uniqueness) Suppose that (T ′, τ ′) is a pair satisfying the defining conditions of a tensorproduct. On the one hand, there is a homomorphism α : T −→ T ′ so that α · τ = τ ′; likewise, wehave a homomorphism β : T ′ −→ T so that β · τ ′ = τ . Composing, we get:

(β · α) · τ = τ and (α · β) · τ ′ = τ ′,

and by uniqueness it follows that α and β are mutually inverting group isomorphisms.�

Definition 6.3. We shall denote the tensor product of A and B by A⊗ B, (or A ⊗R B) and theimage of the universal map τ on the pair (a, b) by a⊗ b. Equivalently, (a, b) +K = a⊗ b.

The assignment B ⊗ (·) is a covariant functor from R-Mod to R-Mod. For if f : A −→ A′ isa R-homomorphism, then θf (b, a) = b ⊗ f(a), for all a ∈ A and b ∈ B, defines an R-bilinear mapθf on B ×A, into B ⊗A′. This gives rise to a homomorphism

B ⊗ f ≡ θ∗f : B ⊗A −→ B ⊗A′,

such that (B ⊗ f)(b⊗ a) = b⊗ f(a).We leave it to the reader to verify the functorial properties. One should add that the assignment

(·)⊗A defines a covariant functor from R-Mod to R-Mod, in the same way.

Remark 6.4. We point out that not every element of A ⊗ B may be written in the form a ⊗ b.(Such a tensor is called a simple tensor). But every element of A ⊗ B may be written as a finitesum of simple tensors. Of course this representation is not unique. The identity 0⊗ 0 shall simplybe written as 0.

A simple check show that the following are true

32 Math 734

(i) (a1 + a2)⊗ b = a1 ⊗ b+ a2 ⊗ b,∀a1, a2 ∈, b ∈ B;

ii) a⊗ (b1 + b2) = a⊗ b1 + a⊗ b1,∀a ∈ A, b1, b2 ∈ B;

iii) ra⊗ b = a⊗ rb = r(a⊗ b),∀r ∈ R, a ∈ A, b ∈ B;

iv) 0⊗ b = a× 0 = 0 for any a ∈ A, B ∈ B.

Example 6.5. Consider D⊗Z Q where D is a torsion group. let a⊗ mn ∈ D⊗Q. Since a is torsion

there is some natural n such that pa = 0. Then

a⊗ m

n= a⊗ pm

pn= a⊗ p

m

pn= pa⊗ m

pn= 0.

Thus, D ⊗Z Q = 0.

Exercise 6.6. Prove that if A is a torsion-free abelian group then

A⊗Z Q ∼= A.

Proposition 6.7. For any R-modules A and B

A⊗R B ∼= B ⊗R A.

Proposition 6.8.(⊕i∈I

Ai)⊗R B ∼=⊕i∈I

(Ai ⊗R B),

for all R-modules Ai and B.

Proof: First off recall that the maps ιi : Ai −→ ⊕Ai are the coprojections. Consider the collectionof R-modules {Ai⊗B}i∈I and let ηi : Ai⊗B −→ ⊕(Ai⊗B) be the appropriate coprojctions. Thismeans that for n1, · · · , np ∈ Ai, b1, · · · , bp ∈ B

ηi(Σpk=1nk ⊗ bk)(j) =

{Σp

k=1nk ⊗ bk, if i = j;0, otherwise.

For each i ∈ I there is an R-mod homomorphism ιi⊗1B : Ai⊗B −→ (⊕Ai)⊗B. By the universalproperty of the coproduct there exists a unique R-module homomorphism ψ : ⊕(Ai ⊗ B) −→(⊕Ai) ⊗ B such that for each I ∈ I, ιi ⊗ 1B = ψ ◦ ηi. We show that ψ is an isomorphism bydemonstrating its inverse.

Consider the map from α : (⊕Ai)×B −→ ⊕(Ai ⊗B) defined by

α( (m, b) )(i) = ηi(m(i)⊗ b).

First this map is well-defined since for all but finitely many of the i we have m(i) = 0. This mapis clearly bilinear and so there is a unique R-module homomorphsim φ : (⊕Ai)⊗B −→ ⊕(Ai ⊗B)for which φ(m⊗ b)(i) = ηi(m(i)⊗ b)(i) = m(i)⊗ b. We now demonstrate (thoroughly) that φ andψ are mutual inverses.

Take n⊗ b ∈ Ai ⊗B and consider φ ◦ ψ ◦ ηi(n⊗ b).

(φ ◦ ψ ◦ ηi)(n⊗ b)(j) = φ( ιi ⊗ 1B(n⊗ b) )(j)= φ(ιi(n)⊗ b)(j)

Tensor Product 33

= ηj(ιi(n)(j)⊗ b)(j)

where

ηj(ιi(n)(j)⊗ b)(j) ={ιi(n)(j)⊗ b = n⊗ b, if i = j;0, otherwise.

It follows that φ◦ψ◦ηi(n⊗b) = ηi(n⊗b) but the identity on ⊕(Ai⊗B) is the unique homomorphismwhich has this property, thus φ ◦ ψ = 1⊕(Ai⊗B).

Going the other way, let m⊗ b be a simple tensor in (⊕Ai)⊗B. For each i define

mi(j) ={m(i), if i = j;0, otherwise.

so thatm = Σimi. Clearly there are only finitely many i such thatmi 6= 0. Also, m⊗b = Σi(mi⊗n).Consider (ψ ◦ φ)(m⊗ b) = Σi(ψ ◦ φ)(mi ⊗ b).

Now, to make computations a bit easier let hi = φ(mi ⊗ b) ∈ ⊕(Ai ⊗B). So

hi(j) = ηj(mi(j)⊗ b)(j) ={ηi(m(i)⊗ b)(i), if i = j;0, otherwise.

It follows that hi = ηi(m(i) ⊗ b). Therefore, (ψ ◦ φ)(mi ⊗ b) = ψ(hi) = ψ(ηi(m(i) ⊗ b) = (ιi ⊗1B)(m(i)⊗ b) = mi ⊗ b.

(ψ ◦ φ)(mi ⊗ b) = ψ(hi)= ψ(ηi(m(i)⊗ b)= (ιi ⊗ 1B)(m(i)⊗ b)= mi ⊗ b

and since the identity is the only homomophism it follows that ψ ◦ φ = 1(⊕Ai)⊗B.�

Proposition 6.9. For R-modules A,B,C

(A⊗R B)⊗R C ∼= A⊗R (B ⊗R C),

Proposition 6.10. Suppose that R is a commutative ring with identity. Then for each module B,one has a natural isomorphism

R⊗R B ∼= B.

Proof: The map ρB : R × B −→ B given by ρB(r, b) = rb, is easily shown to be R-bilinear, sothat there is a homomorphism ρB : R ⊗R B −→ B for which ρB(r ⊗ b) = rb. One needs to verifythat it preserves scalars, and for this it suffices to show that

r1ρB(r2 ⊗ b) = ρB(r1r2 ⊗ b),

for all r1, r2 ∈ R and b ∈ B. (Why?) But the latter identity should be obvious.On the other hand, the map α : B −→ R⊗RB, defined by α(b) = 1⊗b is an R–homomorphism.

Finally note that ρB(α(b)) = b, for all b ∈ B, and that

α(ρB(r ⊗ b)) = α(rb) = rα(b) = r(1⊗ b) = r ⊗ b,

whence α · ρB = 1R⊗RB.One also needs to verify that ρ is natural, but this we leave to the reader.�

34 Math 734

Proposition 6.11. For any ring R with identity we have:

(a) HomR(R, ·) is naturally equivalent to 1R−Mod.

(b) R⊗R (·) is naturally equivalent to 1R−Mod.

Definition 6.12. Suppose φ : M −→ M ′ and ψ : N −→ N ′ are two R-module homomorphisms.The map (m,n) 7→ φ(m) ⊗ ψ(n) from M × N to M ′ ⊗ N ′ is R-bilinear and so there is a uniqueR-module homomorphism mapping M⊗N into M ′⊗N ′ sending m⊗n 7→ φ(m)⊗ψ(n). We denotethis map by φ⊗ ψ.

7. Flat Modules

Proposition 7.1. (Tensor Products on Short Exact Sequences.) Suppose that

0 −→ Af−→ B

g−→ C −→ 0

is a short exact sequence of R-modules. Then for each R-module M , the following sequence isexact:

M ⊗AM⊗f−→ M ⊗B

M⊗g−→ M ⊗ C −→ 0.

Exactness at M ⊗A fails, in general.

Proof: The proof, for the most part, goes to the heart of the definition of tensor products. Theonly trivial part is the identity (M ⊗ g) · (M ⊗ f) = 0, because M ⊗ ( ) is a functor, and, clearly,M ⊗ 0 = 0.

Exactness at M ⊗ C: each element of M ⊗ C is a sum of tensors of the form m ⊗ c, each ofwhich can be written as m⊗ c = m⊗ g(b), for a suitable b ∈ B, as g is surjective.

Exactness at M ⊗B: we already know that (M ⊗ f)(M ⊗A) ≤ ker(M ⊗ g). Denote

M ⊗B/(M ⊗ f)(M ⊗A) ≡ G

and let µ : M ⊗ B −→ G be the canonical homomorphism. By the Induced HomomorphismTheorem, there is a homomorphism g∗ : G −→ M ⊗ C, such that g∗ · µ = M ⊗ g. On the otherhand, define a map θ : M ×C −→ G by θ(m, c) = µ(m⊗ b), where b is chosen so that g(b) = c. Welet the reader check that this is well defined: µ(m⊗ b) does not depend on the choice of b. It is alsoeasy to verify that θ is R–bilinear. This means that there is a homomorphism θ∗ : M ⊗ C −→ Gwith the property that

θ∗(m⊗ c) = µ(m⊗ b),

whenever g(b) = c. From this we get that

θ∗(g∗(µ(m⊗ b))) = θ∗(m⊗ g(b)) = m⊗ b,

from which we conclude (using the uniqueness provision of the Induced Homomorphism Theorem)that θ∗ · g∗ = 1G. This makes g∗ one–to–one. On the other hand, since M ⊗ g is onto, so is g∗, andtherefore g∗ is an isomorphism. This outcome means that ker(M ⊗ g) = (M ⊗ f)(M ⊗A). �

Flat Modules 35

Definition 7.2. An R-module M is said to be flat if for each short exact sequence

0 −→ Af−→ B

g−→ C −→ 0,

the following sequence is exact:

0 −→M ⊗AM⊗f−→ M ⊗B

M⊗g−→ M ⊗ C −→ 0.

By Proposition 7.1 it follows that M is flat if and only if whenever f : A −→ B is injective thenM ⊗ f : M ⊗A −→M ⊗B is also injective.

Note that it is a consequence of Proposition 6.10 that R is flat. From this observation, and thefact that the functor B⊗R (·) preserves direct sums (that is to say, coproducts), for each R-moduleB, we will, shortly, obtain that every free module is flat.

Let’s consider an example which might help our understanding. Suppose A is a submodule of Band let i : A −→ B be the inclusion map. Consider the map 1M ⊗ i : M ⊗RA −→M ⊗RB and theimage 1⊗ i(M ⊗R A). Call the image L. Flatness is concerned with the relationship of M ⊗A andits image L. In general, this map is onto since L consists of all elements of the form Σn

i=1mi ⊗ ai.But due to the nature of the tensor product when we consider M ⊗ A inside of M ⊗B there maybe some cancellation or other screwy things which may cause the map to not be injective. M isflat precisely when they are isomorphic for every pair A and B.

Example 7.3. For R = Z, consider the short exact sequence

0 −→ Z α−→ Q µ−→ Q/Z −→ 0,

where α is the inclusion map, and µ is the canonical homomorphism. Tensor with M = Z2: byExample 6.5, Z2⊗Q = 0, whereas Z2⊗Z = Z2, according to Proposition 6.10. Thus, Z2⊗α is thezero map, and not one–to–one. Exactness at M ⊗A then, in Proposition 7.1, can indeed fail.

The following shall be useful. It gives us way of creating flat modules.

Proposition 7.4. Let M = ⊕iMi (i ∈ I) be a direct sum of R-modules. Then M is flat if andonly if each Mi is flat.

Proof: (Sufficiency) Recall Proposition 6.8: for any R-module A,

M ⊗A ∼=⊕i∈I

(Mi ⊗A).

Indeed, this isomorphism is natural; let’s be precise about this: let di,A : Mi ⊗ A −→ ⊕i(Mi ⊗ A)denote the i–th coprojection. Now, consider the functor TM defined by setting TM (A) = ⊕iMi⊗A,and, for each R-homomorphism g : A −→ B, TM (g) : TM (A) −→ TM (B) is the unique R-modhomomorphism such that

TM (g) · di,A = di,B · (Mi ⊗ g).

Refer to the diagram which follows for the discussion below.

36 Math 734

Mi ⊗A - TM (A)

?

Mi ⊗B - TM (B)?

di,A

Mi ⊗ g

di,B

TM (g)

- M ⊗A

- M ⊗B

tA

tB

M ⊗ g

?

The full force of Proposition 6.8 is that the functors M ⊗ ( ) and TM are naturally equivalent. Weidentify the natural equivalence, and otherwise leave the details to the reader. The isomorphismfrom ⊕iMi ⊗A −→M ⊗A is simply the unique R-mod homomorphism tA for which

tA · di,A = δi ⊗A,

where δi : Mi −→M is the i–th coprojection. Now t : TM −→M ⊗ ( ) is a natural equivalence.It follows that, to prove that M ⊗ ( ) sends one–to–one maps to one–to–one maps, is equivalent

to proving it for the functor TM . But this uses a property of direct sums: if g : A −→ B is aR–homomorphism which is one–to–one, then, by assumption, each Mi⊗g is one–to–one. From thisit easily follows that TM (g) is one–to–one as well, because in a direct sum of groups Gi an elementg =

∑i gi = 0, if and only if each gi = 0.

(Necessity) What is true is this: if N is a summand of M , and M is flat, then so is N . Theproof involves an argument upon a diagram like the one above. We leave the details to the reader.�

Corollary 7.5. Free and projective R-modules are flat.

Proof: Recall the remark following Definition 7.2 which explains that R is a flat R-module. Henceso is any free module being a direct sum of copies of R. Finally projectives are summands of freeand hence flat modules so they too are flat.�

Definition 7.6. Let A be an R-module and D an abelian group. The group HomZ(A,D) maybe viewed as an R-module when one defines (rf)(a) = f(ra). We leave it to the reader to checkthat this is a well-defined R-module structure. We would like to say more. Now suppose B isalso an R-module and f : A −→ B is an R-homomorphism. We show that f∗ = HomZ(f,D) :HomZ(B,D) −→ HomZ(A,D) is also an R-module homomorphism. We do so by showing thatf∗(rφ) = rf∗(φ) for every φ ∈ HomZ(B,D) and r ∈ R. Note that f∗(φ), f∗(rφ) ∈ HomZ(A,D).

f∗(rφ)(a) = ((rφ) ◦ f)(a)= (rφ)(f(a))= φ(rf(a))= φ(f(ra))= (f∗(φ))(ra)= (r(f∗(φ)))(a)

Thus, rf∗(φ) = f∗(rφ).Continuing with this line of thinking, suppose D happens to be an injective Z-module, i.e., a

divisible abelian group, and0 −→ A

f−→ Bg−→ C −→ 0

Flat Modules 37

is a short exact sequence of R-modules. Then this sequence is in fact a short exact sequence ofZ-modules and so by the injectivity of D

0 −→ HomZ(C,D)g∗−→ HomZ(B,D)

f∗−→ HomZ(A,D) −→ 0

is a short exact sequence of Z-modules. But since all the maps in question are in fact R-modulehomomorphisms we conclude that HomZ(·, D) preserves short eact sequences of R-modules eventhough D may not even be an R-module. We shall return to this later. For now we would like toshow that in certain situations HomR(A,B) may have more structure than simply as an R-module.Whenever an abelain group A is both an R-module and and S-modules for rings R,S we say A isan R,S-bimodule.

Proposition 7.7. Let R and S be commutative rings with identity; A,B, and C are all abeliangroups.

(1) Suppose A is an R,S-bimodule, B is an R-module. Then HomR(A,B) is an R,S-bimoduleby defining (sφ)(a) = φ(sa) for all a ∈ A, φ ∈ HomR(A,B).

(2) Suppose A is an R-module, B is an R,S-bimodule. Then HomR(A,B) is an R,S-bimoduleby defining (sφ)(a) = sφ(a) for all a ∈ A, φ ∈ HomR(A,B).

(3) Suppose A is an R-module, B is an R,S-bimodule. Then A ⊗R B is an R,S-bimodule bydefining s(a ⊗ b) = a ⊗ (sb) for all a ∈ A, b ∈ B and then extending by linearity to all ofA⊗R B.

Theorem 7.8 [Adjoint Isomorphism]. Let R and S be rings, A an R-module, B an R,S-bimodule,and C an S-module. There exists a group isomorphism

τ : HomS(A⊗R B,C) −→ HomR(A,HomS(B,C)).

Moreover, τ is both an R-module and S-module isomorphism.

Proof: First of all by the previous proposition everything makes sense because both groups areR,S-bimodules.

Now, let f ∈ HomS(A ⊗R B,C) and a ∈ A. Define fa : B −→ C by setting fa(b) = f(a ⊗ b).This is a well defined function and is a group homomorphism. Furthermore,

fa(sb) = f(a⊗ (sb))= f(s(a⊗ b)) [Th.7.7, (3)]= (sf)(a⊗ b)= s(f(a⊗ b))= s(fa)(b)

which demonstrates that fa ∈ HomS(B,C). Thus, we may define a map f : A −→ HomS(B,C) byf(a) = fa. Also,

fra+a′(b) = f( (ra+ a′)⊗ b)= f((ra)⊗ b) + f(a′ ⊗ b)= f(a⊗ (rb)) + f(a′ ⊗ b)= fa(rb) + fa′(b)

38 Math 734

= (rfa)(b) + fa′(b) [Th.7.7, (1)]= (rfa + fa′)(b)

which shows that fra+a′ = rfa + fa′ , whence f ∈ HomR(A,HomS(B,C)). Finally, define τ ::HomS(A ⊗R B,C) −→ HomR(A,HomS(B,C)) by τ(f) = f . We leave it to the reader to checkthat this is a group homomorphism and that it is indeed an R-module. That it is an S-modulehomomorphism is even easier to check.

To show that τ is an isomorphism we shall exhibit its inverse. Suppose g : A −→ HomS(B,C) isan R-map. Define θ : A×B −→ C by θ(a, b) = g(a)(b). It is straightforward to check that this mapis S-bilinear and so there exists an S-map g′ : A⊗R B −→ C for which g′(a⊗ b) = g(a)(b). Thenσ(g) = g′. Again straightforward and tedious to check that σ is both an R-module and S-modulemap. Then observe that the composition σ ◦ τ sends f 7→ (f)′. Now,

(f)′(a⊗ b) = f(a)(b) = fa(b) = f(a× b)

so that f = (f)′. Conversely,

g′(a)(b) = g′a(b) = g′(a⊗ b) = g(a)(b)

whence g′ = g.�

Corollary 7.9. For any R-modules A,B and abelian group C we have that

HomZ(A⊗R B,C) ∼= HomR(A,HomZ(B,C)).

We would like to continue with applying the Adjoint Isomorphism and see what else we canget from it. The first is demonstrating that the covariant functor (·) ⊗R A (for any R-module A)preserves direct limits.

Proposition 7.10.(lim−→

Mi)⊗R A ∼= lim−→

(Mi ⊗R A),

where each Mi is an R-module, and A is an R-module.

Proof: Set M = lim−→

Mk. Let {Mj , ρjk} be a directed system and let ρJ : Mj −→ M be the maps

which are provided by the definition of the direct limit. Now, the set {Mj ⊗ A, ρjk ⊗ A} is also a

directed system. Furthermore, the maps ρj ⊗ A : (Mj ⊗R A) −→ M ⊗R A have the property thatρk ⊗A ◦ ρj

k ⊗A = ρj ⊗A. Thus, we need only demonstrate that if X is an R-module together withmaps gj : Mj ⊗R A −→ X for which gk ◦ (ρj

k ⊗A) = gj then there exists a unique map

γ : (lim−→

Mi)⊗R A −→ X

for which γ ◦ (ρj ⊗A) = gj .For each k let sMk

: Mk −→ HomR(A,Mk⊗RA) be the unique map for which sMk(m)(a) = m⊗a.

Observe that for each j ≤ k we have (Hom(A, gk) ◦ sMk) ◦ ρj

k = Hom(A, gj) ◦ sMj so that by thedefinition of the limit M there is a map β : M −→ HomR(A,X) for which the appropriate trianglescommute.

Now, consider the isomorphism τ : HomR(M ⊗R A,X) −→ HomR(M,HomR(A,X)) given byTheorem 7.8. Then γ = τ−1(β) : (lim

−→Mk) ⊗R A −→ X. Since τ is natural it follows that our

desired triangles commute. Uniqueness follows from the fact that τ is a bijection.�

Flat Modules 39

Proposition 7.11. The direct limit of flat modules is flat.

Proof: Let {Mk, φkj }k∈I be a directed system of flat modules over the directed set I and set

M = lim−→

Mk.

Let f : A −→ B be monic. Then there is a commutative diagram

(lim−→

Mk)⊗A -

lim−→

(Mk ⊗A)

?

(lim−→

Mk)⊗B

- lim−→

(Mk ⊗B)

?

M ⊗ f

ρ

where the vertical maps are isomorphisms by the previous proposition. We will leave it as anexercise that if there are maps between directed systems that are all 1-1, then the unique mapbetween the direct limits is also 1-1. It then follows that ρ is 1-1 from which we conclude thatM ⊗ f must be 1-1.�

Corollary 7.12. If every finitely generated submodule of M is flat then M is flat.

Proof: M is in fact the direct limit of its finitely generated submodules.�

Definition 7.13. Suppose that M is an R-module. We let M∗ = HomZ(M,Q/Z), viewed as anR-module, in the usual way. M∗ is sometimes referred to as the character module of M ; we shallsimply refer to it as the dual of M . This is different then the notion of the dual of a vector space.We know that HomZ(·,Q/Z) preserves short exact sequences of R-modules, even though Q/Z isnot necessarily an R-module. We shall show more. First off we restate the Adjoint Isomorphism.

Corollary 7.14. For any R-modules A,B we have

(A⊗R B)∗ ∼= HomR(A,B∗).

Lemma 7.15. If m ∈M is nonzero, then there exists an f ∈M∗ such that f(m) 6= 0.

Proof: The proof hinges on the divisibility of Q/Z. To produce a homomorphism f ∈ M∗ forwhich f(m) 6= 0, it suffices to find one defined on Zm, the cyclic subgroup generated by m. Thenone uses the injectivity of Q/Z (over Z) to extend it to M .

Now there are two cases to consider.

I. The order of m is infinite. Then pick any natural number n > 1, and define f(km) = Z+k/n.It is trivial, to show that f is a homomorphism.

II. The order of m is t > 0. Define f(km) = Z + k/t. One easily proves that f is well defined,and a homomorphism.

40 Math 734

In either case f does the job. �

With this lemma one can show that dualization both preserves and reflects short exactness. Wedo so now.

Proposition 7.16. The following sequence

0 −→ Af−→ B

g−→ C −→ 0

is short exact if and only if0 −→ C∗

g∗−→ B∗ f∗−→ A∗ −→ 0

is short exact.

Proof: The necessity follows from the remarks in 7.6 and 7.7.To prove the sufficiency we shall prove that if ker g∗ = Imf∗ then ker f = Imf without assuming

anything about f and g.Suppose a ∈ A yet f(a) /∈ ker g. Thus, g(f(a)) 6= 0 and so by Lemma 7.15 there is a map

φ : C −→ Q/Z such that φ(g(f(a))) 6= 0. Thus, φ ∈ C∗ and φ ◦ g ◦ f 6= 0, i.e., f∗ ◦ g∗ ◦ φ 6= 0. Thiscontradicts f∗ ◦ g∗ = 0. Whence we obtain that f(A) ⊆ ker g.

Conversely, suppose b ∈ ker g \ f(A). Thus, b + Imf is a nonzero elemnt of B/Imf . Againby Lemma 7.15 we may choose a map φ : B/Imf −→ Q/Z such that φ(b + Imf) 6= 0. Letπ : B −→ B/Imf be the natural projection so that ψ = φ ◦ π ∈ B∗ and has the property thatψ(b) 6= 0 and 0 = ψ ◦ f = f∗(ψ). This last equation means that ψ ∈ Img∗ which further yieldsthat there is some α ∈ C∗ for which ψ = g∗(α) = α ◦ g. In particular, ψ(b) = α(g(b)) = α(0) = 0,a contradiction.�

Using the preceding lemma and proposition, one obtains:

Theorem 7.17. The R-module M is flat if and only if its dual M∗ is an injective module.

Proof: Recall Corollary 7.14 in the following form where the isomorphisms are the vertical maps.

(B ⊗M)∗ -

HomR(B,M∗)

?

(A⊗M)∗

- HomR(A,M∗)

?

(α⊗M)∗

HomR(α,M∗)

If M is flat, and α : A −→ B is an injective homomorphism, then α⊗M : A⊗M −→ B ⊗Mis also one–to–one. Since Q/Z is divisible (injective), it follows that (α ⊗ M)∗ is onto. Usingthe diagram above it follows that HomR(α,M∗) is onto. By Theorem 5.4, it follows that M∗ isinjective.

Conversely, suppose that M∗ is injective, and α : A −→ B is one–to–one. We reverse the stepsin the preceding paragraph: HomR(α,M∗) is onto, and, therefore, so is the naturally equivalent(α⊗M)∗. By the previous proposition, α⊗M is one–to–one, proving that M is flat. �

Flat Modules 41

Theorem 7.18. If M is an R-module such that 0 −→ M ⊗ I −→ M ⊗ R is exact for every f.g.ideal I of R, then M is flat.

Proof: By the remark we made before about 1-1 maps being preserved between direct limits itfollows that the statement is true for all ideals of R (since each ideal is a direct limit of its f.g. subideals). This gives exactness of (M ⊗R)∗ −→ (M ⊗ I)∗ −→ 0, which in turn gives exactness of

HomR(R,M∗) −→ HomR(I,M∗) −→ 0

as in the proof of the previous theorem. Finally, this means that M∗ is injective by Baer’s Criterionwhich in turn says M is flat.�

Remark 7.19. Suppose that M is an R-module, and K is an ideal of R. Recall that

KM ≡ { r1m1 + · · ·+ rkmk : ∀mi ∈M, ri ∈ K }.

KM is an R-submodule of M .As the map (r,m) −→ rm, from K×M into KM is clearly bilinear, there is a unique R-module

homomorphism φK : K ⊗R M −→ KM , for which φK(r ⊗m) = rm. Since the tensors of the formr ⊗m generate K ⊗M , it should be clear that φK is surjective.

Now, look at φK another way. The natural inclusion j : K −→ R induces j ⊗M : K ⊗M −→R⊗M . Then we have the natural isomorphism (guaranteed by the dual of 6.11) τM : R⊗M −→M(which satisfies τM (r ⊗ m) = rm, for each m ∈ M , and each r ∈ R). Observe now that φK =τM ◦ (j ⊗M).

We are now set up to prove the following technical, but rather useful result. For convenience,let us refer to the map φ introduced two paragraphs ago as the internal tensor map associated withthe ideal K and the R-module M .

Next, here is a technical result, which gets to the internal conditions that make a module flat.In some way it is a similar result to Baer’s Criterion; a shorter way of checking flatness.

Proposition 7.20. If M is an R-module, then the following are equivalent.

(a) M is a flat R-module.

(b) For each ideal K of R, the internal tensor map associated with K and M is an isomorphism;

(c) For each finitely generated ideal K of R, the internal tensor map associated with K and Mis an isomorphism.

Proof: It is clear that (b) implies (c).(a) ⇒ (b): If M is flat, then

j ⊗M : K ⊗M −→ R⊗M

is one–to–one, and since φK = τM ◦ (j ⊗M), it follows that φ too is one–to–one. It has alreadybeen observed that it is onto, and therefore an isomorphism.

(c) ⇒ (b): Suppose that K is an ideal. It suffices to show that if r1m1 + · · ·+ rkmk = 0 (withthe scalars ri ∈ K, and mi ∈M) then

(#) r1 ⊗m1 + · · ·+ rk ⊗mk = 0,

42 Math 734

computed in K ⊗ M . However, if K0 is the ideal generated by r1, r2, . . . , rk, then (#) holds,computed in K0 ⊗M , by assumption. But we have the homomorphism θ : K0 ⊗M −→ K ⊗Minduced by the inclusion of K0 in K. Apply θ to (#), calculated in K0 ⊗M , to get (#), computedin K ⊗M .

(b) ⇒ (a): Invoking the identity

φK = τM ◦ (j ⊗M)

yet again, one sees that j ⊗M is one–to–one, if φK is an isomorphism. Since K was an arbitraryideal, Theorem 7.18 implies that M is flat.�

For integral domains we can actually show that flat modules are necessarily torsion-free modules.We then can apply the prior few propositions and conclude some interesting results when R is anice ring (e.g., PID or a Bezout domain).

Theorem 7.21. Let R be a domain and M a flat module. Then M is torsion-free.

Proof: Suppose M is not torsion free and choose a ∈ R and m ∈ M (both nonzero) such thatam = 0. Let K = Ra = (a). Now φK(a⊗m) = am = 0 so by Proposition 7.20 once we show thata⊗m 6= 0 it follows that φK is not 1-1 and hence M is not flat. Thus, the proof is done.

Now the homomorphism r −→ ra, from R −→ K, is an isomorphism, since R is an integraldomain. This means that the homomorphism induced by r⊗m 7→ ra⊗m, from R⊗RM −→ K⊗RMis also an isomorphism. Furthermore, it follows that K ⊗R M,R ⊗R M, and M are all isomorphicand the correspondences are as follows

a⊗m 7→ 1⊗m 7→ m 6= 0.

Thus, a⊗m 6= 0.�

Proposition 7.22. Let K be a principal ideal of R and M a torsion-free R-module. Then φK isan isomorphism.

Proof: Since K is principal choose a ∈ R such that K = (a). Let x ∈ K ⊗R M be such thatφ(x) = 0. Now, there are ri ∈ R and mi ∈M such that

x =n∑

i=1

ri(a⊗mi) =n∑

i=1

a⊗ (rimi) = a⊗ (n∑

i=1

rimi).

Then φK(x) = am = 0 where m =∑n

i=1 rimi. Since M is torsion free it follows that either a = 0or m = 0 in which case x = a⊗m = 0 and whence φ is an isomorphism.�

Proposition 7.23. Suppose R is a Bezout domain. Then a module is flat if and only if it istorsion-free.

Proof: All we need is to show that every torsion-free module M is flat. But since R is Bezoutand each φK is an isomorphism for all f.g. ideal K, Theorem 7.20 forces M to be flat.�

Flat Modules 43

Exercise 7.24. Prove that an R-module M is flat if and only if it satisfies the condition whenever∑ni=1 rimi = 0 then there exist elements nj ∈M and sij ∈ R such that

ai =m∑

j=1

sijnj andn∑

i=1

risij = 0.

(The proof is done in [fs] Chapter VI.9.4.)

Remark 7.25. Let 0 −→ Kj−→ F

β−→ M −→ 0 be a short exact sequence with F flat. For anideal I f R consider the submodule FI ≤ F . We may consider β′ to be the restriction of β to thissubmodule. Then

MI = {∑

mkik : mk ∈M, ik ∈ I}= {

∑β(fk)ik : fk ∈ F, ik ∈ I} since β is epic

= {β(∑

fkik) : fk ∈ F, ik ∈ I}= β′(FI)

and so we get that

0 −→ K ∩ FI j′−→ FIβ′−→MI −→ 0

is short exact. (It is straighforward to show that kerβ′ = K ∩ FI.)Tensoring the original sequence by by the ideal I gives exactness of

K ⊗ Ij⊗I−→ F ⊗ I

β⊗I−→M ⊗ I −→ 0.

Thus, the First Isomorphism Theorem enduces isomorphisms

FI/(K ∩ FI) ∼= MI and F ⊗ I/Im(K ⊗ I) ∼= M ⊗ I.

Since F is flat Proposition 7.20 says that we may identify F ⊗ I with FI via the internal tensorisomorphism f ⊗ i 7→ fi. This then identifies Im(j ⊗ I) with KI. Thus,

FI/KI ∼= M ⊗ I.

Now, it is always true that KI ⊆ K ∩ FI. It is a simple group theoretic fact that FI/KI ∼=FI/(K ∩ FI) via this identification if and only if KI = K ∩ FI.

Moreover, the internal tensor map φI : M ⊗ I −→ MI is an isomorphism precisely whenFI/KI ∼= FI/(K ∩ FI).

Proposition 7.26. Let F be flat and 0 −→ K −→ F −→ M −→ 0 a short exact sequence. Thenthe following are equivalent.

(i) M is flat.

(ii) K ∩ FI = KI for every ideal I.

(iii) K ∩ FI = KI for every f.g ideal I.

44 Math 734

Proof: By the last sentence of the rpevious discussion together with Proposition 7.20 we havethat M is flat if and only if for every ideal I of R, φI : M ⊗ I −→ MI is an isomorphism if andonly if KI = K ∩ FI.�

We finish off this section with a characterization of those rings R having the property that everyR-module is flat. Recall that a ring R is called von-Neumann regular if for every element a ∈ Rthere exists an x ∈ R such that a2x = a. von Neumann rings are Bezout rings.

Theorem 7.27. Every R-module is flat if and only if R is von Neumann regular.

Proof: Assume R is von Neumann regular and let M be an R-module. Let

0 −→ Ki−→ F

g−→M −→ 0

be an exact sequence where F is a free module. By Proposition 7.26 it suffices to prove thatK ∩FI ⊆ KI for every finitely generated ideal I of R. Since R is von Neumann regular and henceBezout it follows that I is principal, say I = (a). Let k ∈ K ∩ FI and set k = fa ∈ Fa. Now,

k = fa = faxa = kxa ∈ Ka = KI.

Therefore M is flat.For the converse, let a ∈ R. The cyclic R-module is flat (by the hypothesis.) Consider the exact

sequence0 −→ (a) −→ R −→ R/(a) −→ 0.

Proposition 7.20 gives us that (a)I = (a)∩RI = (a)∩I for every ideal I of R. in particular, settingI = (a) we obtain that (a)(a) = (a) ∩ (a) and so since a is in the right side and hence the left sideof this equation it follows that a = rasa = (rs)a2, whence R is von Neuamann regular.�

Because of this theorem, von Neumann regular rings are often – especially in the French liter-ature – referred to as absolutely flat rings. Here are some more characterizations of von Neumannregular rings.

Theorem 7.28. Suppose that R is a ring. Then the following are equivalent.

(a) R is von Neumann regular; (meaning that, for each a ∈ R, there is an x ∈ R, such thata2x = a.)

(b) Every principal ideal of A is generated by an idempotent.

(c) Every finitely generated ideal of A is a direct summand.

(d) R is semiprime (that is, no nonzero nilpotent elements), and every prime ideal is maximal.

(e) Every ideal is an intersection of maximal ideals.

(f) Every R-module is flat.

Corollary 7.29. (a) Any direct product of von Neumann regular rings is von Neumann regular.

(b) If R is von Neumann regular then J(R) = n(R) = 0.

Topologizing Spec(R) 45

8. The Prime Spectrum

Proposition 8.1. Suppose that R is a ring. Then the set n(R), consisting of all nilpotent elements,is the intersection of all prime ideals of R.

This proposition prompts a definition.

Definition 8.2. Suppose that r is an ideal of the ring R. Let√

r stand for the set

{x ∈ R : ∃n ∈ N, such that xn ∈ r }.

We shall call√

r the radical of r. Evidently, n(R) =√

0.

We first record the following extension of Proposition 8.1. Its proof depends on Zorn’s Lemma,and is left to the reader.

Proposition 8.3. Suppose that r is an ideal of the ring R. Then√

r is the intersection of all theprime ideals of R that contain r.

On the heels of Proposition 8.3, now is as good a time as any to make the observation whichfollows. We record it as an exercise, with a definition.

Exercise 8.4. Suppose that r is an ideal of the ring R. Then√

r = r if and only if a2 ∈ r impliesthat a ∈ r.

An ideal satisfying this condition is called a semiprime ideal. When {0} is semiprime (that is,when n(R) = 0) the ring R is also said to be semiprime. (Some texts still use the term reducedring in place of semiprime ring.)

The proof is left to the reader.

Proposition 8.5. Suppose that r and s are ideals of the ring R. Then

(a)√

r ≥ r and√

(√

r) =√

r.

(b) r ≤ s implies√

r ≤√

s.

(c)√

rs =√

r ∩ s =√

r ∩√

s.

(d)√

r = R if and only if r = R.

(e) If r is a prime ideal, then√

rn = r, for each n ∈ N.

(f) In general,√

r + s ≥√

r +√

s, but equality need not hold.

Let R be a ring. Denote by Spec(R) the set of all prime ideals of R. In this section we willtopologize Spec(R), as well as a number of its subsets, in order to introduce, in a systematic manner,the interplay between commutative rings and topological spaces.

46 Math 734

Definition & Remarks 8.6. The Hull–kernel Topology. The point here is to define a topologyon Spec(R). What is perhaps unusual about this is the notion that, in this setting, the prime idealsthemselves are to be thought of as points.

Let R be a ring, S be a subset of R. Let V (S) stand for the set of all prime ideals of R containingS. Note that

V (S) = V (RS) = V (√RS),

where RS is the ideal generated by S.In the following proposition we summarize the basic properties of the sets V (S); in view of the

comment just made, we shall assume that S is an ideal.

Proposition 8.7. Suppose that R is a ring, that { si : i ∈ I } is a family of ideals of R, s and t

are ideals of R. Then

(a) s ≤ t implies that V (t) ⊆ V (s).

(b) V (s ∩ t) = V (st) = V (s) ∪ V (t).

(c) V (∑

i si) = ∩i V (si).

(d) V ({0}) = Spec(R) and V (R) = ∅.

Proof: We let the reader worry about (a), (c) and (d). Let’s show that (b) holds.Let p be a prime ideal of R. Then, if a ∈ s\p and st ≤ p, we have that t ≤ p. This immediately

implies that V (st) ⊆ V (s) ∪ V (t). From (a), one has the containments

V (s) ∪ V (t) ⊆ V (s ∩ t) ⊆ V (st),

whence the identities follow. �

Definition & Remarks 8.8. Proposition 8.7 informs us that the collection

{V (s) : s is an ideal of R }

is the collection of closed subsets of a topology, called the Zariski or hull–kernel topology on Spec(R).The open sets are

{U(s) : s is an ideal of R },

where U(s) is the set theoretic complement of V (s). In particular, letting a ∈ R, we have

V (a) = V ({a}) = V (Ra) = { p ∈ Spec(R) : a ∈ p }

andU(a) = { p ∈ Spec(R) : a /∈ p }.

The sets U(a), for all a ∈ R, form a base for the open sets of the topology. With respect to thistopology, one refers to Spec(R) as the prime spectrum of R.

We record an exercise (a sort of companion to Proposition 8.7), the upshot of which is preciselythat the U(a) (a ∈ R) form a base of open sets. Recall that a family of subsets {Ui : i ∈ I } of theset X form a base for the open sets of a given topology τ on X, if

(op–i) for each point p ∈ Ui ∩ Uj there is a Uk so that p ∈ Uk ⊆ Ui ∩ Uj ;

Topologizing Spec(R) 47

(op–ii) X = ∪{Ui : i ∈ I };(op–iii) each open set is a union of some of the Ui.

Exercise 8.9. Suppose that R is a ring. Then

(a) U(a) ∩ U(b) = U(ab), for all a, b ∈ R;

(b) U(a) = ∅ if and only if a is nilpotent;

(c) U(a) = Spec(R) if and only if a is a unit;

(d) U(a) = U(b) if and only if√Ra =

√Rb;

(e) For each ideal s of R,U(s) = ∪{U(a) : a ∈ s }.

Recall that a subset Y of a topological space is compact if every open cover of Y has a finitesubcover. (Warning: compact, in these pages, does not signify “compact & Hausdorff”!!!)

(f) Each U(a) is compact; thus, Spec(R) = U(1) is compact.

(g) An open set in Spec(R) is compact if and only if it is a finite union of members of the baseU(a) (a ∈ R).

Remark 8.10. In general Spec(R) is not a Hausdorff space. Recall that a topological space isHausdorff (or T2) if for each pair of distinct points p and q there exist disjoint open sets U and Vcontaining p and q, respectively.

Dr. Hoffman, during his colloquium talk discussed the Zariski topology on Spec(R). Thistopology plays an important role in algebraic geometry. ONe might consider his course for the fall.

For any ring A, observe that if U is an open set in Spec(R), and p < q are prime ideals, withq ∈ U , then p ∈ U . Now, compute Spec(Z): it consists of the pZ, for all prime numbers p, plus thezero ideal. If p and q are any two distinct positive primes, then if U and V are open sets containingpZ and qZ, respectively, then {0} ∈ U ∩ V .

Proposition 8.11. For a ring R, the following are equivalent:

(a) Spec(R) is Hausdorff.

(b) Every prime ideal is maximal.

(c) R/n(R) is von Neumann regular.

(d) Spec(R) is T1; (which is to say that each singleton set is closed.)

48 Math 734

Chapter 4

Classical Rings

1. Semi-simple Rings

The main goal of this section to characterize those rings which have the property that every R-module is projective. Once that is done the reader should have a thorough understanding of whena ring has the property that all R-modules are (1) free, (2) projective, and (3) flat. In order toachieve our goal we shall need to discuss semi-simple rings and modules. The achievement of thisgoal shall present itself in the form of the Wedderburn-Artin Theorem for commutative rings. Weremind the reader that interesting things occur if we do not consider commutativity and urge theinterested participant to peruse Chapter 18 of [DF]. The Wedderburn-Artin Theorem plays a hugerole in the study of group rings and “character theory”. We shall not have time to discuss suchtopics but the theorem in its own right has withstood the test of time.

Definition 1.1. Recall the definition of a simple module. A module is called semi-simple (orcompltely reducible) if it is a direct sum of (possibly infinite many) simple submodules.

Proposition 1.2. A direct sum of semisimple modules is semisimple.

Lemma 1.3. Suppose M = P ⊕B and P ≤ N ≤M . Then N = P ⊕ (N ∩B).

Proof: Clearly, P ⊕ (N ∩B) ≤ N . Let n ∈ N . We may write n = p+ b where P ∈ P and b ∈ B.Since b = n− p ∈ N ∩B the result follows. �

Theorem 1.4. A module M is semisimple if and only if every submodule is a summand.

Proof: Suppose M is semisimple, say M = ⊕i∈IMi where each Mi is a simple module. LetN ≤M be a submodule. For any subset K ⊆ I define MK = ⊕i∈KMi. A Zorn’s Lemma argumentmay be used to choose a maximal subset K of I such that MK ∩ N = 0. We would like to showthat M = N ⊕MK . It suffices to show that for every i ∈ I we have Mi ≤ N +MK . Clearly this isso for every i ∈ K.

Let i /∈ K. SetK ′ = K∪{i}. Maximality ofK says that there is a nonzero elementm ∈MK′∩N .Since m ∈MK′ this means there is an m1 ∈MK and m2 ∈Mi for which m = m1 +m2. First notethat m2 6= 0. So m2 = m−m1 ∈MK +N . What we have shown is that (MK +N) ∩Mi 6= 0 andhence a nonzero submodule of the simple module Mi. It follows that Mi ≤ MK + N . This holdsfor all i ∈ I whence M = MK ⊕N .

49

50 Math 734

Conversely, suppose every submodule of M is a summand of M . Now, let N be a submodule.We show that N contains a simple submodule. Let n ∈ N (n 6= 0). Choose a submodule P of Nmaximal with respect to not containing n. By the previous lemma it follows that N = P ⊕ S forsome submodule S. We claim that S is simple. Suppose 0 ≤ T < S and again S = T ⊕ T ′ and soN = P ⊕ S = P ⊕ T ⊕ T ′.

Suppose n ∈ P ⊕T, P ⊕T ′. Then n = p1 + t = p2 + t′, thus p1−p2 = t′− t ∈ P ∩S which forcest = t′ ∈ T ∩ T ′, whence t = t′ = 0 and n ∈ P ; a contradiction. Thus, without loss of generalityn /∈ P ⊕ T . Maximality of P implies T = 0 which shows S is simple and the claim is proved.

Another Zorn’s lemma argument (a trickier one) supplies us with a family of simple submodule{Sk}k∈I maximal with respect to the property that the submodule they generate, denoted S, is adirect sum of the Sk’s. By hypothesis M = S ⊕ T . If T 6= 0, then clearly by maximality T is notsimple. Also as above it does contain a simple submodule T1 which also satisfies T = T1 ⊕ T2. Butthen again the set {Sk} ∪ {T1} contradicts maximality. Thus, T = 0 and so M = S, whence issemi-simple.�

Corollary 1.5. Every submodule and every quotient of a semisimple module is again semisimple.

Proof: Suppose N is a fixed submodule of M ; a semisimple module. If L ≤ N ≤ M , then sinceM = L⊕K the previous lemma shows that L is a submodule of N . Hence the theorem implies Nis semisimple.

Similarly, if C is a quotient of M then there is a submodule N of M for which M/N ∼=C. Considering short exact sequences of summands we can conclude that C is isomorphic to asubmodule of M hence is semisimple.�

Definition 1.6. A ring R is called semi-simple if it is semisimple as an R-module. A ring is calledartinian if it satisfied the descending chain condition. More will be said about this class of rings inthe next section.

Theorem 1.7 [Wedderburn-Artin Theorem]. For a commutative ring with identity R, the follow-ing are equivalent.

(a) R is semisimple.

(b) Every R–module is projective.

(c) Every short exact sequence splits.

(d) Every R–module is injective.

(e) Every R-module is semisimple.

(f) R is a direct sum of minimal (simple) ideals, each generaetd by an idempotent, say R =⊕i∈ILi.

(g) R is a (finite) direct product of fields.

(h) R is artinian and J(R) = 0.

Proof: We know that (b), (c), and (d) are equivalent.

(a) implies (e): By Proposition 1.2 every free module is semisimple and hence every R-moduleis semisimple by Corollary 1.5.

Noetherian Rings 51

(e) implies (b): Again every free module is semisimple. Then a similar argument to the oneused in the corollary shows that every module is a summand of a free module and hence projective.

(b) implies (a): Given an ideal I, we know that R/I is projective and hence by Lemma I.3.18we know that I is a summand of R which by Theorem 1.4 forces R to be semisimple.

(f) implies (a): This one is clear.

(a) implies (f): Suppose that R is a direct sum of left ideals R = ⊕i∈ILi, where each Li issimple as as an R-submodule, i.e., each Li is a minimal ideal. We have already seen that if an idealis a summand then it is a principal ideal generated by an idempotent so we may write Li = (ei)and where

1 = e1 + e2 + · · ·+ ek,

with ej ∈ Lij , uniquely, as prescribed by the direct sum. By a similar argument, it also follows that

{Li : i ∈ I } = {Li1 , Li2 , . . . , Lik };

in particular, there are only finitely many such left ideals.

(g) implies (h): We need the following lemma whose proof we leave to the reader.

Lemma 1.8. If {Ri}i∈I is a set of rings then

J(Πi∈IRi) = Πi∈IJ(Ri).

What the lemma gets us is that if R is a direct product of fields then since a field has zeroJacobson radical it follows that a product of fields has zero Jacobson radical. Moreover, since itis a finite direct product of fields it follows that R has only a finite number of ideals and hence isartinian.

(h) implies (g): What we need is that artinian rings have a finite number of maximal ideals.This is Lemma 2.10 which is proved in the next section (without a hint of this theorem). Enumeratethe maximal ideals, say Max(R) = {M1, · · · ,Mm}. Thus, the Chinese Remainder Theorem get usthat

R ∼= R/J(R) ∼= R/M1 ×R/Mm

where the right side is a diret product of fields.

At this juncture we have proved that (a) through (f) are equivalent and that (g) and (h) areequivalent and finally ii is straight forward to show that (g) implies (f). Thus, we are left with(d) implies (g). We leave the proof of this as an exercise to the interested reader. In fact, theseare problems 5-10 of Chapter 18.2 [DF]. The reader must be warned that the result there is fornoncommutative rings and so when doing such a result for commutative rings one must realize thata full n × n matrix ring over a division ring is commutative if and only if n = 1 and the divisionring is a field. �

2. Noetherian Rings

We consider Noetherian rings and modules in this section. The goal is to prove Hilbert’s BasisTheorem. We will then look at in the following sections some specific types of Noetherian do-mains; namely Dedekind domains, which are the natural generalizations of UFD’s, in terms of thearithmetic of ideals.

52 Math 734

Definition 2.1. Suppose that A is a commutative ring with 1. An A-module M is said to beNoetherian (resp. Artinian) if the lattice of submodules of M satisfies the ascending (resp. de-scending) chain condition.

We say that the ring A is Noetherian (resp. Artinian) if it is a Noetherian (resp. Artinian)module over itself. Eventually, in Proposition 2.12, we will show that, for commutative rings, anyArtinian ring is also Noetherian.

The following should seem very familiar and so we leave its proof to the reader.

Proposition 2.2. For an A-module M the following are equivalent.

(a) M is Noetherian.

(b) Every A–submodule is finitely generated.

(c) Every nonempty family of proper A–submodules has a maximal member.

In particular A is a Noetherian ring if and only if every ideal of A is finitely generated.

The “Artinian” counterpart of Proposition 2.2 follows, as an exercise:

Exercise 2.3. The A–module M is Artinian if and only if every nonempty family of nonzeroA–submodules has a minimal member.

Examples 2.4. (A) The group of all complex pn–th roots of of unity (all n ∈ N) is an Artinianmodule but not Noetherian.

(B) All PIDs are Noetherian. Is a UFD necessarily Noetherian?

(C) For any field F , the polynomial ring F [T1, T2, . . .] in an infinite number of indeterminates isneither Noetherian nor Artinian.

(D) An integral domain which is Artinian is a field.

(E) Von Neumann regular rings need not be Noetherian. Example?

The following result is typical of theorems about chain conditions.

Proposition 2.5. Suppose that 0 −→M −→ N −→ P −→ 0 is an exact sequence of A–modules.Then

(a) N is Noetherian if and only if M and P are Noetherian.

(b) N is Artinian if and only if M and P are Artinian.

Proof: We prove (a) only, as the proof of (b) is quite similar.Obviously, an infinite ascending sequence of submodules of M is an ascending sequence of

submodules of N . Likewise, ifP1 < P2 < · · ·

is an ascending sequence of submodules of P , then, by taking inverse images under the map N −→P , we obtain an infinite ascending sequence of submodules of N . Thus, if N is Noetherian, thenso are the other two.

Noetherian Rings 53

Suppose now that N1 < N2 < · · · is an ascending sequence of submodules of N , yet both Mand P are Noetherian. For a suitable index t,

M ∩Nt = M ∩Nt+1 = . . . ,

and a suitable index s,M +Ns = M +Ns+1 = . . .

For u = max (s, t) both sequences become stationary for n ≥ u.By one of the isomorphism theorems, (M +Ni)/M ∼= Ni/(M ∩Ni), for each i. Now, for n ≥ u,

the left side of these identities is constant, while the denominator of the right side is too. ThusNn = Nu, for all n ≥ u, a contradiction. �

Corollary 2.6. Suppose that Mi (i = 1, 2, . . . , k) are A–modules, and that M is their direct sum.Then M is Noetherian (resp. Artinian) if and only if each Mi is Noetherian (resp. Artinian).

Proof: Apply Proposition 2.5 inductively to sequences

0 −→Mi −→M1 ⊕M2 ⊕ · · · ⊕Mi −→M1 ⊕M2 ⊕ · · · ⊕Mi−1 −→ 0.

�Here’s an example of a very natural commutative ring which is, in general, far from being

Noetherian.

Exercise 2.7. Let C(X) be the ring of all continuous real valued functions defined on the compactHausdorff space X. Prove that C(X) is Noetherian if and only if X is a finite space.

Next, we have another basic feature of Noetherian and Artinian modules.

Proposition 2.8. If A is a Noetherian (resp. Artinian) ring, then every finitely generated A–module is Noetherian (resp. Artinian).

Proof: If M is finitely generated, then it is a homomorphic image of a finitely generated freeA–module. Thus, M is a homomorphic image of a finite direct sum of copies of A. Now applyProposition 2.5.�

Definition 2.9. An ideal I of A is called nilpotent if Ik = 0 for some natural k. Clearly, a nilpotentideal is contained in the nilradical.

Lemma 2.10. An Artinian ring A has a finite number of maximal ideals.

Proof: Consider the family of all finite intersections of maximal ideals of A. By the Artinianproperty (Exercise 2.3), this set has a minimal element r = m1 ∩ m2 ∩ · · · ∩ mk. Now, if m is anymaximal ideal of A, then m ∩ r = r, which is to say that r ≤ m. We leave it to the reader to checkthat this implies that m actually contains one of the mi, and and therefore equals it.�

Corollary 2.11. Let A be an artinian ring and M1, · · ·Mn the distinct maximal ideals of A. Then

A/J(A) ∼= F1 × · · · × Fn

where Fi∼= A/Mi is a field.

54 Math 734

Proof: Apply the Chinese Remainder Theorem.�

Proposition 2.12. Let A be a commutative Artinian ring with identity.

(a) Every prime ideal is maximal.

(b) The Jacobson radical equals the nilradical, i.e., n(A) = J(A). Furthermore, J(A) is a nilpotentideal.

(c) A is isomorphic to the direct product of a finite number of Artinian local rings.

(d) A is also Noetherian.

Proof: Let J = J(A). For some natural m we have Jm = Jm+1. By way of contradiction supposethat Jm 6= 0. Let S be the set of all proper ideals of A for which IJm 6= 0. Let I ′ be a minimalelement of S. Thus for some x ∈ I ′ it follows that xJm 6= 0 which by minimality we obtain thatI ′ = (x). Then

((x)J)Jm = (x)Jm+1 = (x)Jm

and so again by minimality (x)J = (x). By Nakayama’s Lemma Theorem 3.4.4 it follows that(x) = 0 and so we have by contradcition that J is a nilpotent ideal. It then follows that J(A) = n(A).

By the previous lemma we have that A/n(A) is isomorphic to a finite direct product of fields.But in a direct product of rings R1×· · ·×Rn an ideal is of the form I1×· · ·×In where each Ii ≤ Ri

is an ideal. Thus, since the only ideals of a field are trivial it follows that the ideals of A/n(A)are finite intersections of the maximal ideals of A and thus the only primes are the maximal onesthemselves (again we are heavily relying on the Chinese Remainder Theorem).

Next, let M1, · · · ,Mn be the distinct maximal ideals of A and suppose m satisfies J(A)m = 0.Then first off since the Mm

1 , · · · ,Mmn are pairwise comaximal it follows that

Mm1 ∩ · · · ,∩Mm

n = Πni=1M

mi

so thatΠn

i=1Mmi = (Πn

i=1Mi)m = J(A)m = 0.

Again CRT gives us thatA ∼= (A/Mm

i )× · · · × (A/Mmn )

where each A/Mmi is an artinian local ring. Thus, once we demonstrate that an artinian local ring

is noetherian it follows that since a direct product of noetherian rings is noetherian that an artinianring is noetherian. �

Noetherian Rings 55

Proposition 2.13. Suppose A is a local artinian ring. Then M is noetherian.

Proof: Let M be the unique maximal ideal of A. Let k ∈ N for which Mk = 0. Then considerthe following sequence of short exact sequences.

0 −→M −→ A −→ A/M −→ 0

0 −→M2 −→M −→M/M2 −→ 0

0 −→M3 −→M2 −→M2/M3 −→ 0

· · ·

0 −→M i −→M i−1 −→M i−1/M i −→ 0

0 −→M i+1 −→M i −→M i/M i+1 −→ 0

· · ·

0 −→Mk−1 −→Mk−2 −→Mk−2/Mk−1 −→ 0

0 −→Mk −→Mk−1 −→Mk−1/Mk −→ 0

Now each of A/M,M/M2, · · · ,M i/M i+1, · · · are all A/M vector spaces defined in the obviousmanner. Since A is artinian Proposition 2.5 says (working your way down) that each elementin the above chain of sequences is an artinian A-module. Now, a subgroup of any of the A/M ,M/M2, · · ·, M i/M i+1, · · · is an A-module if and only if it is an A/M vector space. Thus eachof the A/M,M/M2, · · · ,M i/M i+1 is an artinian vector space. By Problem #2 of HW2 it followsthat each of A/M,M/M2, · · · ,M i/M i+1 is a noetherian A/M vector space whence a noetherianA-module. Since Mk = 0 it is trivially a noetherian A-module. Thus again using Proposition 2.5we may work our way back up and get that A is noetherian.�

Remark 2.14. Proposition 2.12 tells us that an Artinian ring is Noetherian and has Krull dimen-sion 0. We prove the converse of this in §3: a Noetherian ring which has 0 Krull dimension isArtinian.

Theorem 2.15. Hilbert’s Basis Theorem. If A is a Noetherian ring then so is the polynomialring A[T ].

Proof: We prove that every ideal of A[T ] is finitely generated. To that end, let r be an ideal ofA[T ]. The set of leading coefficients of elements in r, say r0, forms an ideal of A. Suppose that{a1, . . . , an} is a generating set for r0. Let fi(T ) be a polynomial in r of degree ri, with leadingcoefficient ai. Then with

r′ ≡ A[T ]f1 + · · ·+A[T ]fn,

we have r′ ≤ r. Let r be the maximum of the ri.Now pick f ∈ r, with leading coefficient a ∈ r0 and degree m. If m ≥ r, write a = u1a1 + · · ·+

unan. Then

f − (n∑

i=1

uifiTm−ri) ∈ r,

and has degree less than m. Continuing in this manner, we may eventually write f = g + h, withg ∈ r, deg(g) < r, and h ∈ r′.

56 Math 734

Next, if M is the A–submodule generated by {1, T, . . . , T r−1}, then the above paragraph showsthat r = (M ∩ r) + r′. Since M is finitely generated and A is a Noetherian ring, it follows thatM ∩ r is finitely generated (as an A–module). Thus, r is finitely generated as an ideal of A[T ], bythe A–generators of M ∩ r and the fi. �

We have two corollaries of Hilbert’s Basis Theorem; one iterates to get the first, and very littleelse to get the second.

Corollary 2.16. If A is Noetherian, then so is A[T1, T2, . . . , Tn], for any finite number of indeter-minates.

One more item should be mentioned in connection with Hilbert’s Basis Theorem. It’s the versionfor power series rings.

Exercise 2.17. If A is a Noetherian ring, prove that A[[T ]] is too. (Hint: Imitate the proof of theBasis Theorem, but instead of the leading coefficients – which may not be there – consider an idealof lowest–term coefficients.)

Corollary 2.18. Suppose that A is a ring and that m1m2 · · ·mk = 0, for a suitable set of maximalideals of A (which aren’t necessarily distinct). Then A is Noetherian if and only if it is Artinian.

Proof: Consider the chain of ideals (A–submodules)

A > m1 ≥ m1m2 ≥ · · · ≥ m1m2 · · ·mk = {0}.

Each factor (m1 · · ·mi−1)/(m1 · · ·mi) can be regarded as an A/mi–vector space. Now, by the exercisein HW 2, each factor satisfies the a.c.c if and only if it does the d.c.c (on A/mi-subspaces, andhence on A-submodules.) By repeated applications of Proposition 2.5, it follows that A has thea.c.c. on ideals if and only if it has the d.c.c. �

3. Primary Ideals and Decomposition

The goal of this section is to introduce primary ideals, and to show that, in a Noetherian ring, everyideal can be expressed as an intersection of primary ideals. This is, in a sense, one ideal theoreticgeneralization of prime–power factorization of natural numbers.

Definition 3.1. Let A be any commutative ring with identity, and q be a proper ideal of A. q isprimary if xy ∈ q implies that x ∈ q or else yk ∈ q, for a suitable k. It is readily seen that q isprimary if and only if, in A/q, every zero-divisor is nilpotent. Thus, if q is primary,

√q is a prime

ideal.

Before getting to some examples, let’s make a basic observation about primary ideals. Theproof is already done, more or less.

Proposition 3.2. If q is a primary ideal of A then√

q is a prime ideal, and (obviously) the smallestprime ideal containing q.

Definition 3.3. If q is a primary ideal of A and p =√

q, then q is said to be p–primary.

Now some examples to disabuse the reader of possible misconceptions.

Noetherian Rings 57

Examples 3.4. (a) In the ring of integers the nonzero primary ideals are precisely those generatedby the power of a prime.

(b) Let K be any field, A = K[T1, T2], with q the ideal generated by T1 and T 22 . Now, A/q =

K[T ]/(T 2); it is easy to verify that in A/q every zero–divisor is nilpotent. Thus, q is primary; itsradical is p = (T1, T2). Notice that p2 < q < p, so that q itself is not the power of a prime ideal.

Therefore, a primary ideal need not be the power of a prime ideal.

(c) Conversely, the power of a prime ideal need not be primary.Let K be a field, and A = K[T1, T2, T3]/(T1T2 − T 2

3 ). We shall denote the cosets of T1, T2

and T3 by x, y and z, respectively. Thus, A = K[x, y, z]. Let p = (x, z); this is a prime ideal, asA/p = K[T ], which is an integral domain. Next, xy = z2 ∈ p2, but x /∈ p2, and y /∈

√p2 = p, which

says that p2 is not primary.

By contrast, we do have the following result.

Exercise 3.5. If√

r is a maximal ideal, then r is a primary ideal. In particular, all the powers ofa maximal ideal are primary.

To establish the primary decomposition of ideals of a Noetherian ring, it’s useful to introducethe following concept.

Definition 3.6. An ideal r of the ring A is said to be irreducible if r = s1 ∩ s2 (for any ideals si ofA) implies that r = s1 or r = s2.

Lemma 3.7. In a Noetherian ring A, every ideal is the intersection of a finite number of irreducibleideals.

Proof: By contradiction: if there is an ideal of A which is not a finite intersection of irreducibleones, then by the maximality principle, there is a maximal one r with this property. Now, r mustbe reducible, so write r = s1 ∩ s2, so that both si’s properly contain r. But then each si can bewritten as a finite intersection of irreducible ideals, and then so can r, a contradiction. (Does thisproof seam familiar?)�

Now we have the existence of primary decompositions. There are several uniqueness conditionsthat can be attached to this kind of decomposition. See as well the concluding exercise in thissection.

Theorem 3.8. In a Noetherian ring A every ideal can be expressed as the intersection of a finitenumber of primary ideals.

Proof: By Lemma 3.7, all we need to do is show that every irreducible ideal in a Noetherian ringis primary.

Next, by passing to factor rings, it suffices to show that if {0} is an irreducible ideal it is alsoprimary. So suppose that xy = 0, but x 6= 0; consider the chain

Ann(y) ≤ Ann(y2) ≤ · · · .

From the a.c.c. we conclude that, for some positive integer n, Ann(yn) = Ann(yn+1) = · · ·. Thismeans that Ax∩Ayn = {0}; for if a = bx = cyn, then ay = 0 and so cyn+1 = 0, whence a = cyn = 0.Since {0} is irreducible, it follows that Ayn = 0 (because Ax 6= 0), and yn = 0. �

58 Math 734

Remark 3.9. If q is a primary ideal and p =√

q, then p is the least prime ideal containing q.Thus in the decomposition of Theorem 3.8, one can reduce the decomposition by dropping thoseprimary ideals whose radicals are not minimal with respect to containing the ideal in question.

This is an informal account, and the reader ought to think this through.

With a little more work we get some nice corollaries. We leave the proof of the following as anexercise.

Proposition 3.10. In a Noetherian ring A, every ideal r contains a power of its radical.

Sketch of the Proof. Pick a (finite) generating set for√

r. Then select a natural number m,large enough so that xm ∈ r, for each x ∈

√r. The reader should then note that (

√r)m ≤ r. �

Corollary 3.11. In a Noetherian ring A, n(A)r = {0}, for some natural number r. (That is, n(A)is a nilpotent ideal.)

Proof: Let r = {0} in 3.10;√

0 = n(A). �

Corollary 3.12. Suppose that m is a maximal ideal of the Noetherian ring A, and q is any idealof A. The following are equivalent:

(a) q is m–primary.

(b)√

q = m.

(c) mk ≤ q ≤ m, for a suitable k.

Proof: By definition (a) implies (b), and the reverse follows from Exercise 3.5. Proposition 3.10gives that (b) implies (c), and by computing radicals, (c) implies (b). �

At long last we have the converse promised in 2.14.

Proposition 3.13. Suppose that A is a ring. Then A is Artinian if and only if it is Noetherianand has zero Krull dimension.

Proof: The necessity has been proved by application of Proposition 2.12. If the Krull dimensionof A is zero and A is Noetherian, then any primary decomposition of the zero ideal of A (whichexists by Theorem 3.8 can be rewritten as an intersection of a finite number of powers of maximalideals, and therefore {0} is a product of maximal ideals. Now invoke Corollary 2.18, and concludethat A is Artinian. �

Chapter 5

Localization

1. Multiplicative Subsets of a ring

Definition 1.1. A subset S of the ring A is called multiplicative if it does not contain 0 and it isclosed under multiplication. The set of units of A, dentoed U(A) is always a multiplicative set; aswell as the set of regular elements.

In the upcoming discussion A is a ring, and S is a multiplicative system of A. Consider the setof all pairs (a, s), where a ∈ A, s ∈ S, and define (a, s) ∼ (b, t), provided there is a u ∈ S so that(at− bs)u = 0. We leave it to the reader to verify that ∼ is an equivalence relation.

Definition 1.2. S−1A denotes the set of all equivalence classes modulo ∼. We denote the equiv-alence class of [a, s] by a/s. Define addition and multiplication on S−1A in the expected manner:

(a/s) + (b/t) = (at+ bs)/st and (a/s)(b/t) = ab/st.

It is left as an exercise to show that these definitions are independent of the representatives used,and that they make of S−1A a commutative ring with identity s/s (for any s ∈ S). Unlike somesources, we do not require the identity of the ring to belong to S.

There is a canonical homomorphism of A into S−1A, denoted by φ(a) = as/s (again, indepen-dent of the choice of s ∈ S).

The following proposition gives the essential properties of the ring S−1A, called the ring offractions of A with respect to S.

Proposition 1.3. Suppose that S is a multiplicative system of A. Let φ : A −→ S−1A be givenby φ(a) = as/s. Then

(a) if s ∈ S, then φ(s) is a unit of S−1A;

(b) ker(φ) = { a ∈ A : ∃ s ∈ S, as = 0 }.

(c) If f : A −→ B is a homomorphism such that f(s) is a unit of B, for each s ∈ S, then thereis a unique homomorphism f∗ : S−1A −→ B, such that f = f ·φ.

Proof: (a) (ts/s)(s/ts) = u/u, for all s, t, u ∈ S.(b) If as = 0, for some s ∈ S, then φ(a) = as/s = 0/s = 0. Conversely, at/t = 0 implies that

at/t = 0/s, for any s ∈ S, which means that there is a u ∈ S, such that (ats− 0)u = 0.

59

60 Math 734

(c) Define f∗(a/s) = f(a)/f(s); this makes sense, since f(s) is invertible. And if a/s = b/t, then(at− bs)u = 0, for some u ∈ S, whence (f(a)f(t)− f(b)f(s))f(u) = 0. Since f(u) is invertible, wemay cancel it; this gives that f(a)/f(s) = f(b)/f(t), proving that the function f∗ is well defined.We let the readers convince themselves that f∗ is a homomorphism of rings, preserving the identity.�

With an additional observation we obtain the following characterization of S−1A.

Corollary 1.4. Suppose that B is a ring, with a homomorphism g : A −→ B so that

(i) g(s) is a unit, for each s ∈ S;

(ii) ker(g) = { a ∈ A : ∃ s ∈ S, as = 0 }; and

(iii) each b ∈ B can be expressed as b = g(a)/g(s), for some a ∈ A and s ∈ S.

Then there is a unique isomorphism h : S−1A −→ B, such that h · φ = g.

Examples 1.5. Here are some important types of rings of fractions by multiplicative systems.

(a) If S = A \ p, where p is a prime ideal, then we denote S−1A = Ap, and call it the localizationof A at p. We shall have plenty of occasion to consider properties of localizations in the nextsection, but for now shall say no more.

(b) If s ∈ A, then S = { sn : n ∈ N } is a multiplicative system, provided s is regular; that is tosay, a non–zero–divisor. We use the notation S−1A = As in this event.

(c) If S = U(A), then S−1A ∼= A.

(d) Letting S be the set of regular elements of A, S−1A is called the classical ring of quotientsof A and is denoted by qA. If A happens to be a domain then qA is its field of fractions.

Definition 1.6. Let A and S be as above. For an ideal I of A, we let Ie

be the ideal generated by I in S−1A. It is not hard to show that

Ie = {as

: a ∈ I, s ∈ S}

and so we often write S−1I instead of Ie. This ideal is called the extension of I to S−1A.Going the other way if J ≤ S−1A is an ideal then the contraction of J to A is Jc = J ∩ A.

Again it is not hard to show Jc is an ideal of A.

Lemma 1.7. Let A and S be as above. For ideals I,K of A and J of S−1A we have

(a) (I +K)e = Ie +Ke.

(b) (IK)e = IeKe

(c) (I ∩K)e = Ie ∩Ke.

(d) Ie = S−1A if and only if I ∩ S 6= ∅.

(e) I ≤ Iec

(f) J = Jce so that every ideal of S−1A is the extension of an ideal from A.

61

(g) If P is a prime ideal of A and P ∩ S = ∅, then P = P ec.

Proposition 1.8. There is an order preserving 1-1 correspondence between the prime ideals ofS−1A and the primes of A which meet S trivially.

Proof: Simply show that if Q ∈ Spec(S−1A), then Qc ∈ Spec(A). Conversely, if P is a prime ofA and P ∩ S = ∅, then P e ∈ Spec(S−1A).�

The notion of ring of fractions can be extended to the modules over that ring.

Definition 1.9. Let M be an A–module, and S be a multiplicative system of A. We denote, asin the case of rings of fractions, by S−1M , the set of all equivalence classes m/s, with m ∈M ands ∈ S, subject to the identity: m/s = n/t if and only if u(tm− sn) = 0, for a suitable u ∈ S.

With addition again defined as

m/s+ n/t = (tm+ sn)/st,

and scalar multiplication defined by

(a/s)(m/t) = am/st,

one makes S−1M into an S−1A–module. S−1M is called the module of fractions with respect to S.Note that S−1M = 0 precisely when there exists an s ∈ S for which sM = 0; i.e., if and only if

Ann(M) ∩ S 6= ∅.S−1( ) becomes a functor, from AMod to S−1AMod. We leave the verification of this, as

well as the remaining details concerning the definition of modules of fractions, to the reader. (Iff : M −→ N is an AMod–morphism, then define S−1f(m/s) = f(m)/s.)

The next proposition tells us that S−1 preserves exactness.

Proposition 1.10. If the sequence of A–modules

Lf−→M

g−→ N

is exact at M , then the sequence

S−1LS−1f−→ S−1M

S−1g−→ S−1N

is exact at S−1M ′.

Proof: Functoriality implies that the image of S−1f is contained in ker(S−1g). As to the reverse,if S−1g(m/s) = 0, then g(m)/s = 0, which implies that g(tm) = tg(m) = 0, for some t ∈ S; that is,tm ∈ ker(g). This means that tm = f(x), for some x ∈ M ; that is, m/s = f(x)/st = S−1f(x/st).�

As a corollary of Proposition 1.10 we get the next item. One needs Proposition 1.10, so thatthe corollary makes sense to begin with.

Corollary 1.11. Suppose that N and P are A–submodules of the A–module M . Then

(a) S−1(N ∩ P ) = S−1N ∩ S−1P ;

62 Math 734

(b) S−1(N + P ) = S−1N + S−1P ;

(c) as S−1A–modules S−1(M/N) ∼= S−1M/S−1N .

And now the piece that cements rings and modules of fractions together:

Theorem 1.12. Let S be a multiplicative system of A, and M be an A–module. Then the functorsS−1 and S−1A ⊗A ( ) are naturally equivalent. More specifically, the map αM : S−1A ⊗A M −→S−1M defined by

αM (a/s⊗m) = am/s,

for all a ∈ A, s ∈ S and m ∈ M , defines a natural equivalence, which, in particular, is an S−1A–isomorphism.

Proof: First of all, S−1A ⊗A M is an S−1A–module, as explained in earlier chapters, becauseS−1A is an S−1A–module over itself. Observe that the map θ(a/s,m) = am/s is A–bilinear. Thisright away implies the existence of αM , unique with respect to the stated condition. It should beclear that αM is surjective.

Next let’s observe the following. A typical element of S−1A⊗AM is of the form∑

i ai/si⊗mi;let s be the the product of all the si and ti be the product of all sj (j 6= i). Then∑

i

ai/si ⊗mi =∑

i

(aiti)/s⊗mi = 1/s⊗ (∑

i

aitimi);

the point of which is that each element in this tensor product may be written as 1/s⊗m. Now, ifαM (1/s⊗m) = 0, then m/s = 0, so that tm = 0, for some t ∈ S. Thus,

1/s⊗m = t/st⊗m = 1/st⊗ tm = 0,

proving that αM is one–to–one.We leave the verification that α is natural to the reader. �

As an immediate corollary of Proposition 1.10 and Theorem 1.12 we have:

Corollary 1.13. For each multiplicative system S, S−1A is a flat A–module.

Proposition 1.14. Let A be a commutative ring with 1 and S ⊆ A a multiplicative systemcontaining 1. Then the following are true.

a) If A is a Bezout ring, then so is S−1A.

b) If A is a valuation domain (or more generally a chaained ring), then so is S−1A.

c) If A is Noetherian, then so is S−1A.

d) If A is a GCD-domain, then so is S−1A.

Chapter 6

Some Nice Rings

1. Invertible Ideals

We are aiming at describing Dedekind domains. Throughout R will denote an integral domain. qRdenotes its classical field of quotients. Superscripted numbers are footnotes which will be discussedlater in the section.

Definition 1.1. Let F be an R-submodule of qR. We say F is a fractional ideal of R if there isan r ∈ R(1) for which rF ⊆ R. Clearly, the ideals of R are fractional ideals of R. These we shallsimply call integral ideals (or even simpler an ideal). Th name comes from the fact that rF beinga submodule of R and hence an ideal means that F = 1

r I for some ideal of R. We let F(R) denotethe set of all fractional ideals. Observe that any cyclic (principal) submodule of qR is fractional.

We are interested in certain kinds of fractional ideals. Observe that if F,G ∈ F(R) then thesubmodule F +G is again fractional and if we define

FG = {n∑

i=1

xiyi : xi ∈ F, yi ∈ G}

then it is not hard to show that FG is an R-submodule of qR and that it is fractional (Note thatF = d−I,G = e−1J and so FG = (de)−1IJ .) Also, since finitely generated modules are sums ofcyclic ones it follows that every f.g. submodule is fractional.

If F and G are fractional then the set

[F : G] = {x ∈ qR : xG ⊆ F}

is a submodule. Moreover, if G(2) is nonzero then it is a fractional ideal.

We summarize

Proposition 1.2. F(R) is closed under addition, intersection, and multiplication. Moreover, it isan abelian monoid with identity R.

Exercise 1.3. Show that qR is fractional if and only if qR = R.

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64 Math 734

Definition 1.4. An element of (F(R), ·) which is a unit is called an invertible ideal of R. Thus, thefractional ideal F is invertible if and only if there is a fractional ideal G such that FG = R. Sinceinverses are unique in a monoid, it follows that we may speak of the inverse of an invertible ideal,which we shall write F−1. Also, recall that if the product of elements in a monoid is invertible theneach piece separately is invertible. Also, if F is invertible then whenever G,H are fractional idealsand FG = FH then it follows that G = H. (Fractional ideals with this property are often calledcancellation ideals. We will not be discussing this class of fractional ideals much.

Lemma 1.5. A fractional ideal F is invertible if and only if there exits x1, · · · , xn ∈ qR andf1, · · · , fn ∈ F such that

(1) xiF ⊆ R

and

(2) 1 =n∑

i=1

xifi.

Proof: Suppose F is invertible, then since FF−1 = R it follows that 1 =∑n

i=1 xifi for somefi ∈ F and xi ∈ qR. Also, since for any f ∈ F xif ∈ FF−1 = R we obtain (1).

Conversely, let G be the f.g. submodule and hence fractional ideal generated by {x1, · · · , xn}.Clearly, FG ≤ R because of condition (1). Condition (2) says that 1 ∈ FG and hence so is everyelement of R.�

Proposition 1.6. (i) If F is invertible then F−1 = [R : F ].

(ii) If F is an invertible ideal then it is finitely generated(3).

Proof: (i) Observe first that F−1 ≤ [R : F ]. Then

R = FF−1 ≤ F [R : F ] ≤ R.

Since F is invertible and hence a cancellation ideal item (i) follows.As for (ii), this follows from (2) in the previous lemma.�

Definition 1.7. The abelian group of invertible ideals is denoted by I(R). The set of principalfractional ideals forms a subgroup P(R) of I(R). The factor group I(R)/P(R) is called the classdivisor group of R. We shall not discuss class groups at any length. We do point out that thestatement in [DF] preceeding Proposition 10 of Chapter 16 is incorrect.

In domain theory fractional ideals play a role in studying the projective modules of R. We stateexactly how presently.

Theorem 1.8. Let I be a nonzero fractional ideal of R. I is invertible if and only if it is a projectiveR-module.

Proof: Assume that I is invertible and setn∑

i=1

xiyi = 1 for some xi ∈ I and yi ∈ [R : I]. Let F be

the free module over a set of n many elements (let {f1, · · · , fn} be a basis for F . Define φ : F −→ Iby

φ(n∑

i=1

rifi) =n∑

i=1

rixi.

Some Nice Rings 65

This map is onto since the xi generate I. Thus, we obtain a short exact sequence

0 −→ kerφ −→ F −→ I −→ 0.

Define ψ : I −→ F by ψ(a) =∑n

i=1(ayi)fi. It is not hard to check that both of these maps areR-module homomorphisms. Furthermore,

φ ◦ ψ(a) = φ(n∑

i=1

(ayi)fi) =n∑

i=1

(ayi)xi = a

n∑i=1

xiyi = a.

It follows that the above sequence splits and so I is a summand of a free module, i.e., a projctiveR-module.

For the converse

Definition 1.9. (1) We call a domain R a Prufer domain if every f.g ideal of R is invertible.Clearly, the domain R is Prufer if and only if every f.g. ideal is projective. If every ideal of R isinvertible then we say R is a Dedekind domain. Again R is a Dedekind domain if and only if everyideal is projective.

Remark 1.10. It should be noted that everything that took place above also is well defined in thecontext of commutative rings and not just integral domains. What one needs though is the is thefollowing

(1) r has to be regular.

(2) G must contain a regular element.

(3) F contains a regular element.

Also, Theorem 1.8 in general is not true for non domains. Thus, we define a ring to be Pruferif every f.g. regular ideal is invertible. We say it is a semi-hereditary ring if f.g. every ideal isprojective. Similarly, we call a ring hereditary if every ideal is projective.

66 Math 734

Chapter 7

Groups of Divisibility

1. Preliminaries

Throughout this chapter we make the tacit assumption that R is an integral domain which is nota field. It also always contains 1. We shall use q(R) to denote its classical ring of quotients, R∗ todenote the set of nonzero elements of R, and U(R) to denote the set of units.

Definition 1.1. Observe that q(R)∗ is a group, in fact one may consider it as the group constructedfrom the semigroup R∗ where 1 is the identity of this semi-group. Then U(R) is a subgroup ofq(R)∗ and one may form the quotient group q(R)∗/U(R) (as U(R) is obviously normal). We useG(R) to denote this group and call G(R) the group of divisibility of R.

Next, we define a partial order which will hopefully illustrate why we have given G(R) sucha name. Let xU(R), yU(R) ∈ G(R). Define xU(R) ≤ yU(R) if y

x ∈ R. Observe that 1) this iswell-defined and 2) it is indeed a partial order. Observe that U(R) ≤ xU(R) precisely when x ∈ R.In other words the positive cone of G(R) is induced by the elements of R. Also, observe that ifx, y ∈ R then xU(R) ≤ yU(R) if and only y

x ∈ R if and only x divides y in R.

Proposition 1.2. The relation defined above on G(R) is in fact a partial order which makes G(R)into a partially-ordered group. Moreover, G(R) is directed.

Proof: We leave the fleshing out of the proof for the interested reader.�

Lemma 1.3. The positive cone of the group of divisibility is the set of cosets whose representativesmay be a chosen element of R, i.e., G(R)+ = {rU(R) : r ∈ R}.

Proof: U(R) ≤ xU(R) if and only if x ∈ R.�

Proposition 1.4. R is a GCD-domain if and only if G(R) is a lattice-ordered group.

Proof: Suppose R is a GCD-domain and let xU(R), yU(R) ∈ G(R). Without loss of generality,we assume that x = a

s , y = bs . Let c = gcd(a, b). Then observe that c

sU(R) ≤ asU(R), b

sU(R). Also,if

d

tU(R) ≤ a

sU(R),

b

sU(R)

then we have that sdU(R) ≤ atU(R), btU(R), i.e., sd|at, bt. Since R is a GCD-domain we knowthat gcd(at, bt) = ct and so sd|ct. Thus, it follows that d

tU(R) ≤ csU(R) which demonstrates that

csU(R) = x ∧ y.

67

Conversely, if G(R) is an `-group then for any a, b ∈ R, we want to show that if cU(R) =aU(R) ∧ bU(R) then c = gcd(a, b). This is straightforward.�

Proposition 1.5. The domain R is a valuation domain if and only if G(R) is totally-ordered.

References

[DF] David S. Dummit and R. M. Foote Abstract Algebra, 2nd ed., Prentice Hall 1999.

[fs] Laszlo Fuchs and Luigi Salce Modules over Non-noetherian Domains, MathematicalMonographs and Surveys, 84, AMS 2001.

[H] Thomas W. Hungerford Algebra, Graduate texts in Mathematics, 73, Springer-Verlag1974.

[R] Joseph J. Rotman An Introduction to Homological Algebra, Pure and Applied Mathe-matics, 85, Academic Press 1979.

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