MAT 535 Problem Set 1 Solutions

MAT 535 Algebra II Jason Starr Midterm 1, Thur. 2/11 Spring 2010 Problem Set 1 Due Thurs. 2/4

Selected Problems. (1) Exercise 14, p. 376. Prove that tensor product commutes with arbitrary direct sums: For every indexing set I, and for every indexed family (Ni)i∈I of R-modules, there is an isomorphism ! M ∼ M M ⊗R Ni = (M ⊗R Ni). i∈I i∈I

(2) Exercise 21, p. 377. For ideal I and J in a commutative ring R, deﬁne the natural R-module homomorphism I ⊗R J → IJ, prove it is surjective, and give an example proving it is not always injective. (3) Exercise 2, p. 403. Prove some special cases of the (long) ﬁve lemma. Let

A −−−→u B −−−→v C −−−→w D     α β γ  y y y δy 0 0 0 A0 −−−→u B0 −−−→v C0 −−−→w D0 be a commutative diagram with exact rows. (a) If α is surjective, β is injective and δ is injective, prove that γ is injective.

(b) If δ is injective, α is surjective and γ is surjective, prove that β is surjective.

(4) Exercise 5, p. 403. Prove that an arbitrary direct sum of ﬂat modules is ﬂat. (5) Exercise 16, p. 405. Prove that every left R-module M is contained in an injective left R- module. Solutions to Selected Problems. Solution to (1) Recall the universal property of a direct sum. There exists an family of (left) R-module homomorphisms indexed by elements i ∈ I, M qi : Ni → Ni, i∈I

1 MAT 535 Algebra II Jason Starr Midterm 1, Thur. 2/11 Spring 2010 Problem Set 1 Due Thurs. 2/4 such that for every R-module L and for every indexed family

ri : Ni → L there exists a unique R-module homomorphism M r : Ni → L i∈I such that for every i in I, ri equals r ◦ qi. Moreover, if we forget the R-module structure and L remember only the underlying structure of Abelian groups, the family (qi : Ni → i∈I Ni)i∈I is still a direct sum of (Ni)i∈I , i.e., it satisﬁes the same universal property.

Similarly, for the indexed family (M ⊗R Ni)i∈I of Abelian groups, there is a direct sum which is an indexed family (ei)i∈I of homomorphisms of Abelian groups M ei : M ⊗R Ni → (M ⊗R Ni) , i∈I satisfying the analogous universal property. By functoriality of tensor product, for every i in I there exists a unique homomorphism of Abelian groups ! M IdM ⊗ qi : M ⊗R Ni → M ⊗R Ni , m ⊗ ni 7→ m ⊗ qi(ni). i∈I So by the universal property of the direct sum there exists a unique homomorphism of Abelian groups ! M M IdM ⊗ q : (M ⊗R Ni) → M ⊗R Ni i∈I i∈I such that for every index i in I, (IdM ⊗ q) ◦ ei equals IdM ⊗ qi. The exercise is precisely to prove that IdM ⊗ q is an isomorphism. The main step is just to deﬁne the map which will turn out to be an inverse isomorphism. For every index i ∈ I, denote by

βi : M × Ni → M ⊗R Ni, (m, ni) → βi(m, ni) = m ⊗ ni the universal R-bilinear map. In particular, for every element m, the induced map

βi,m : Ni → M ⊗R Ni, ni → βi(m, ni) is a homomorphism of Abelian groups. Thus the composition M ei ◦ βi,m : Ni → (M ⊗R Ni) i∈I

2 MAT 535 Algebra II Jason Starr Midterm 1, Thur. 2/11 Spring 2010 Problem Set 1 Due Thurs. 2/4

is also a homomorphism of Abelian groups. By the universal property of the direct sum of (Ni)i∈I , there exists a unique homomorphism of Abelian groups M M βm : Ni → (M ⊗R Ni) i∈I i∈I such that for every i ∈ I, βm ◦ qi equals ei ◦ βi,m. Thus there is a function ! M M β : M × Ni → (M ⊗R Ni) , (m, n) 7→ βm(n). i∈I i∈I Of course there is also the universal R-bilinear map ! ! M M γ : M × Ni → M ⊗R Ni . i∈I i∈I The claim is that β is R-bilinear, and hence determines a unique group homomorphism ! M M b : M ⊗R Ni → (M ⊗R Ni) i∈I i∈I such that b ◦ γ equals β. And of course then there is a further claim that Id ⊗ q and b are inverse maps. First we prove that β is bilinear, and then we prove that β is R-bilinear. By construction, for each element m of M, the map β(m, •) holding the first variable m fixed is linear in the second variable. Thus to prove that β is bilinear, it suffices to prove linearity in the first variable, i.e., for every pair 0 of elements m, m of M, it suffices to prove that βm+m0 equals βm + βm0 . By the universal property of the direct sum of (Ni)i∈I , to prove that βm+m0 equals βm + βm0 , it suffices to prove for every i ∈ I that βm+m0 ◦ qi equals (βm + βm0 ) ◦ qi. Of course (βm + βm0 ) ◦ qi equals βm ◦ qi + βm0 ◦ qi. And by construction, βm+m0 ◦ qi equals ei ◦ βi,m+m0 , resp. βm ◦ qi equals ei ◦ βi,m, βm0 ◦ qi equals ei ◦ βi,m0 . So we are reduced to proving that

ei ◦ βi,m+m0 = ei ◦ βi,m + ei ◦ βi,m0 .

And of course, ei ◦ βi,m + ei ◦ βi,m0 equals ei ◦ (βi,m + βi,m0 ). Thus, it suﬃces to prove that

βi,m+m0 = βi,m + βi,m0 .

But this follows from the fact that βi is bilinear. Therefore β is bilinear. Since β is R-bilinear, to prove that β is R-bilinear, it suﬃces to prove that β(mr, n) equals β(m, rn) for every element r in R. To be precise, scaling by r on the left deﬁnes a group homomorphism M M Lr : Ni → Ni. i∈I i∈I

3 MAT 535 Algebra II Jason Starr Midterm 1, Thur. 2/11 Spring 2010 Problem Set 1 Due Thurs. 2/4

This is the unique group homomorphism such that for every i ∈ I, Lr ◦ qi equals qi ◦ Li,r, where Li,r is the obvious homomorphism

Li,r : Ni → Ni, ni 7→ rni.

And this deﬁnes a map ! ! M M (Id,Lr): M × Ni → M × Ni , (m, n) 7→ (m, Lr(n)). i∈I i∈I Similarly, scaling by r on the right determines a group homomorphism

Rr : M → M, m 7→ mr, which in turn determines a map ! ! M M (Rr, Id) : M × Ni → M × Ni , (m, n) 7→ (Rr(m), n). i∈I i∈I

To prove that β is R-bilinear, we need to prove for every r in R that β ◦(IdM ,Lr) equals β ◦(Rr, Id). This is equivalent to proving for every m in M that βm◦Lr equals βmr. By the same kind of argument as in the previous paragraph, this reduces to proving for every i ∈ I that βi,m ◦ Lr equals βi,mr. And this is a consequence of the fact that βi is R-bilinear. Thus also β is R-bilinear. So, by the universal property of γ, there exists a unique homomorphism b such that β equals b ◦ γ. The next claim is that b and Id ⊗ q are inverse maps, i.e., (Id ⊗ q) ◦ b is the identity map and b ◦ (Id ⊗ q) is the identity map. By the universal property of γ, (Id ⊗ q) ◦ b is the identity map if and only if ((Id ⊗ q) ◦ b) ◦ γ equals Id ◦ γ, i.e., γ. By associativity of composition, it is equivalent to prove that (Id ⊗ q) ◦ (b ◦ γ) equals γ. By the deﬁnition of b, b ◦ γ equals β. So it is equivalent to prove that (Id ⊗ q) ◦ β equals γ. For every element m of M, denote by ! ! M M γm : Ni → M ⊗R Ni , n 7→ γ(m, n). i∈I i∈I

Then it is equivalent to prove for every m in M that (Id⊗q)◦βm equals γm as group homomorphisms ! M M Ni → M ⊗R Ni . i∈I i∈I

By the universal property of (qi)i∈I , this is equivalent to prove that for every index i ∈ I, (Id ⊗ q) ◦ βm ◦ qi equals γm ◦ qi. By the definition of βm, βm ◦ qi equals ei ◦ βi,m. So it is equivalent to prove that (Id ⊗ q) ◦ (ei ◦ βi,m) equals γm ◦ qi. By associativity of composition, (Id ⊗ q) ◦ (ei ◦ βi,m) equals ((Id ⊗ q) ◦ ei) ◦ βi,m. And by definition of Id ⊗ q, (Id ⊗ q) ◦ ei equals Id ⊗ qi. So it suffices to prove

4 MAT 535 Algebra II Jason Starr Midterm 1, Thur. 2/11 Spring 2010 Problem Set 1 Due Thurs. 2/4

that (Id ⊗ qi) ◦ βi,m equals γm ◦ qi. But this is exactly how Id ⊗ qi was deﬁned: via the universal property of βi,m, Id ⊗ qi is the unique group homomorphism such that (Id ⊗ qi) ◦ βi,m equals γm ◦ qi. Therefore (Id ⊗ q) ◦ b equals the identity.

Finally we prove that b ◦ (Id ⊗ q) equals the identity. By the universal property of (ei)i∈I , it suffices to prove that for every index i ∈ I,(b ◦ (Id ⊗ q)) ◦ ei equals ei. By associativity of composition, (b ◦ (Id ⊗ q)) ◦ ei equals b ◦ ((Id ⊗ q) ◦ ei). By definition of Id ⊗ q, (Id ⊗ q) ◦ ei equals Id ⊗ qi. So it suffices to prove that b ◦ (Id ⊗ qi) equals ei. And by the universal property of βi, it is equivalent to prove that (b ◦ (Id ⊗ qi)) ◦ βi equals ei ◦ βi. By the definition of β, ei ◦ βi equals β ◦ (Id, qi). By associativity, (b◦(Id⊗qi))◦βi equals b◦((Id⊗qi)◦βi). By definition, (Id⊗qi)◦βi equals γ ◦(Id, qi). Thus it suffices to prove that b ◦ (γ ◦ (Id, qi)) equals β ◦ (Id, qi). And by associativity once more, it is equivalent to prove that (b ◦ γ) ◦ (Id, qi) equals β ◦ (Id, qi). Thus it suffices to prove that b ◦ γ equals β. But b was defined to be the unique homomorphism such that b ◦ γ equals β. Thus also b ◦ (Id ⊗ q) equals the identity. Therefore, finally, b and Id ⊗ q are inverse isomorphisms, i.e., they give an isomorphism ! M ∼ M M ⊗R Ni = (M ⊗R Ni). i∈I i∈I Remark. As I hope is clear from this argument, checking this kind of compatibility between objects which are defined by a universal property is a straightforward, if tedious, exercise in comparing universal properties. It usually involves a fair amount of diagram chasing, as in the previous 2 paragraphs. But this type of verification quickly becomes routine. One can also justify this isomorphism by working with the “generators and relations” definition of tensor product, although that is also tedious. From that perspective, the key idea is that tensor product of elements commutes L with finite direct sums, and every element in i∈I Ni is a sum of finitely many elements contained in finitely many summands Ni, or more precisely, qi(Ni). Solution to (2) Denote the inclusions of the ideals by

i : I → R, j : J → R.

Denote the multiplication map on R by

m : R × R → R, m(r1, r2) = r1r2.

Then there is an induced map

m ◦ (i, j): I × J → R, (f, g) 7→ fg.

By the deﬁnition of the ideal IJ, the image of m ◦ (i, j) is contained in IJ. Thus we can think of m ◦ (i, j) as a map to IJ, m ◦ (i, j): I × J → IJ. The claim is that m◦(i, j) is an R-bilinear to IJ. This claim is independent of whether we consider m◦(i, j) as a map to R or to its R-submodule IJ. Thus it suﬃces to prove that m◦(i, j) is R-bilinear

5 MAT 535 Algebra II Jason Starr Midterm 1, Thur. 2/11 Spring 2010 Problem Set 1 Due Thurs. 2/4 as a map to R. Since i and j are each R-linear, m ◦ (i, j) is R-bilinear if m is R-bilinear. And m is R-bilinear by distributivity of addition and multiplication and by associativity of multiplication. Thus m ◦ (i, j) is R-bilinear and hence deﬁnes an R-module homomorphism

mI,J : I ⊗R J → IJ.

To prove that mI,J is surjective, it suffices to prove the image generates IJ as an R-module. By the definition of mI,J , it suffices to prove that m ◦ (i, j) generates IJ as an R-module, i.e., IJ is generated as an ideal by all elements of the form fg for f ∈ I and g ∈ J. But this is precisely the definition of IJ. Thereforem mI,J is surjective.

Typically mI,J is not injective. Exercise 17 on p. 376 gives one example. Here is a slightly diﬀerent example. Let R be the ring Z/4Z. Let I and J both be the ideal

I = J = 2Z/4Z = {0, 2}.

Then clearly IJ equals {0}. So to prove that mI,J is not injective, it suﬃces to prove that I ⊗R J is nonzero. But there are isomorphisms,

R/I → I, n 7→ 2n, and similarly for J (well, J equals I so this is the same). So for this particular example, I ⊗R J is isomorphic to R/I ⊗R R/J (of course for most ideals in most rings, this will certainly fail). And as discussed in lecture, R/I ⊗R R/J is canonically isomorphic to R/(I + J). In this case, since I equals J, I + J equals I. So R/I ⊗R R/J is isomorphic to R/I. Since I is not all of R, R/I is nonzero. Solution to (3) There are formal proofs which adapt more easily to more general Abelian categories (where the same result holds, but with the words “injective” and “surjective” replaced throughout by “monomorphism” and “epimorphism”). The following proof is an “element chasing” argument, which is more intuitive than the formal proofs. (a) Let c be an element in the kernel of γ, i.e., an element of C such that γ(c) equals 0. The goal is to prove that c equals 0. Since δ(w(c)) equals w0(γ(c)), also w(c) is in the kernel of δ. By hypothesis, δ is injective. Thus w(c) equals 0. Since the top row is exact, there exists b in B such that c equals v(b). Since γ(v(b)) equals v0(β(b)), and since γ(v(b)) equals γ(c), which is 0 by hypothesis, β(b) is in the kernel of v0. Because the bottom row is exact, there exists an element a0 in A0 such that β(b) equals u0(a0). By hypothesis, α is surjective. Thus there exists an element a of A such that a0 equals α(a). Since u0(α(a)) equals β(u(a)), since α(a) equals a0, and since u0(a0) equals β(b), β(u(a)) equals β(b). By hypothesis, β is injective. Therefore b equals u(a). Thus v(b) equals v(u(a)). Since the top row is a complex, v(u(a)) is 0. Thus v(b) is 0. But c equals v(b). Therefore c equals 0. So γ is injective. (b) Let b0 be an element in B0. The goal is to prove that b0 is in the image of β. Consider v0(b0) as an element in C0. By hypothesis, γ is surjective. Thus there exists an element c in C such that

6 MAT 535 Algebra II Jason Starr Midterm 1, Thur. 2/11 Spring 2010 Problem Set 1 Due Thurs. 2/4

γ(c) equals v0(b0). Since the bottom row is a complex, w0(v0(b0)) equals 0. Thus w0(γ(c)) equals 0. Since w0(γ(c)) equals δ(w(c)), w(c) is in the kernel of δ. By hypothesis, δ is injective. Thus w(c) equals 0, i.e., c is in the kernel of w. Since the top row is exact, there exists an element b in B such that c equals v(b). Thus γ(v(b)) equals γ(c), which equals v0(b0) by hypothesis. Since γ(v(b)) equals v0(β(b)), also v0(β(b)) equals v0(b0). Thus v(b0 − β(b)) equals 0, i.e., b0 − β(b) is in the kernel of v0. Since the bottom row is exact, there exists an element a0 in A0 such that b0 − β(b) equals u0(a0). By hypothesis, α is surjective. Thus there exists an element a in A such that α(a) equals a0. Thus u0(α(a)) equals u0(a0), which equals b0 − β(b). Since u0(α(a)) equals β(u(a)), b0 − β(b) equals β(u(a)). Thus b0 equals β(b + u(a)). So b0 is in the image of β. So β is surjective.

Solution to (4) Let I be an indexing set and let (Ni)i∈I be a family of (left) R-modules. And, for ease of notation, denote the direct sum of this family by N, i.e., there is a family (qi : Ni → N)i∈I of homomorphisms of R-modules which is universal for all such families. Use the same notation as in the Solution to Problem 1. What that exercise proves is that there is a natural isomorphism η between the functors F, G : Modules − R → Abelian Groups given by M F(M) = (M ⊗R Ni) i∈I and G(M) = M ⊗R N.

For every index i of I, denote by Fi the functor,

Fi : Modules − R → Abelian Groups, Fi(M) = M ⊗R Ni.

Now assume that every R-module Ni is flat, i.e., each functor Fi preserves exact sequences. The goal is to prove that also N is flat, i.e., the functor G preserves exact sequences. Because of the natural isomorphism η, it suffices to prove that the functor F preserves exact sequences. But each Fi preserves exact sequences, and F is the direct sum of the functors Fi. Thus since a direct sum of exact sequences is an exact sequence, also F preserves exact sequences. Solution to (5) The following argument is an “adjoint functors” argument in disguise, but at this moment it probably doesn’t help much to use that terminology. The argument depends on some parts of Exercise 15. Let Q be any Abelian group and form the group

H(Q) = HomZ(R,Q). For every element r of R and for every group homomorphism f : R → Q, deﬁne r ∗ f : R → Q by (r ∗ f)(s) = f(sr). Because multiplication sr in R distributes with respect to addition in the variable s, and since f commutes with addition, r ∗ f is a group homomorphism. This deﬁnes a function

µ : R × HomZ(R,Q) → HomZ(R,Q), (r, f) 7→ r ∗ f.

7 MAT 535 Algebra II Jason Starr Midterm 1, Thur. 2/11 Spring 2010 Problem Set 1 Due Thurs. 2/4

The claim is that this is a (left) R-module action. By the deﬁnition of addition of functions, the action distributes with respect to addition in f. Since multiplication sr distributes with respect to addition in the variable r, and since f commutes with addition, the action distributes with respect to addition in r. Clearly 1 ∗ f equals f. Finally, for elements r, r0 in R,

(r0 ∗ (r ∗ f))(s) = (r ∗ f)(sr0) = f((sr0)r) = f(s(r0r)) = ((r0r) ∗ f)(s).

Thus r0 ∗ (r ∗ f) equals (r0r) ∗ f, i.e., the last axiom for an R-module action is satisﬁed. Therefore

HomZ(R,Q) is an R-module via µ. In fact, it is straightforward to check that the rule Q 7→ H(Q) is a covariant (additive) functor

H : Z − Modules → R − Modules.

Next let M be a (left) R-module and let

i : M → Q be any group homomorphism. For every element m of M, consider the function

im : R → Q, t 7→ i(tm).

Since the R-module action of R on M distributes with addition of elements of R, im is a group homomorphism, i.e., an element of HomZ(R,Q). Thus there is an induced map

ηM,Q(i): M → HomZ(R,Q), m 7→ im. Using that the R-module action of R on M distributes with addition of elements of M, also j is a group homomorphism. Finally, for every m in M, and for every r, t in R,

irm(t) = i(t(rm)) = i((tr)m) = im(tr) = (r ∗ im)(t).

Therefore irm equals r ∗ im. So the group homomorphism ηM,Q(i) is an R-module homomorphism. This deﬁnes a map

ηM,Q : HomZ(M,Q) → HomR(M,H(Q)), which is easily seen to be a group homomorphism. There is an inverse map to ηM,Q as well. Consider the group homomorphism

θQ : HomZ(R,Q) → Q, θQ(f) = f(1).

Then for every R-module M, post-composing with θQ deﬁnes a group homomorphism

θM,Q : HomR(M,H(Q)) → HomZ(M,Q), u 7→ θQ ◦ u.

8 MAT 535 Algebra II Jason Starr Midterm 1, Thur. 2/11 Spring 2010 Problem Set 1 Due Thurs. 2/4

(Observe that θQ equals θH(Q),Q(IdH (Q)), which is too cumbersome to write repeatedly.) It is immediate to verify that θM,Q ◦ηM,Q is the identity map on HomZ(R,Q). Verifying that ηM,Q ◦θM,Q equals the identity comes down to the observation that for every R-module homomorphism,

u : M → H(Q) = HomZ(R,Q), m 7→ (um : R → Q), for every m in M and for every r in R,

um(r) = um(1r) = (r ∗ um)(1) = urm(1) = θQ ◦ urm = (θQ ◦ u)m(r).

Therefore, u = ηM,Q(θQ ◦ u) = ηM,Q(θM,Q(u)), so that ηM,Q ◦ θM,Q equals the identity. What this really means is that every R-module homomorphism α : M → H(Q) is of the form ηM,Q(i) for a unique group homomorphism i : M → Q. In particular, assume now that Q is injective. As above, let α : M → H(Q) be an R-module homomorphism coming from a group homomorphism i : M → Q. And let u : M → N be an injective R-module homomorphism. Then u is an injective group homomorphism. And since Q is injective, the homomorphism i : M → Q factors as j ◦u for some group homomorphism j : N → Q. Thus, setting β to be the R-module homomorphism ηN,Q(j), the identity i = j ◦u gives the identity α = β ◦ u. So α factors through an R-module homomorphism N → H(Q). Therefore H(Q) is an injective R-module. All of the above is really Exercise 15. Now let M be any R-module. By Corollary 37, there exists an injective Z-module Q and an injective group homomorphism i : M → Q. Deﬁne α : M → H(Q) to be ηM,Q(i). Since θQ ◦ α equals i, and since i is injective, also α is injective. And by the argument above, H(Q) is an injective R-module. Hence every R-module admits an injective R- module homomorphism into some injective R-module.