MAT 535 Algebra II Jason Starr Midterm 1, Thur. 2/11 Spring 2010 Problem Set 1 Due Thurs. 2/4 MAT 535 Problem Set 1 Solutions 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)i2I of R-modules, there is an isomorphism ! M ∼ M M ⊗R Ni = (M ⊗R Ni): i2I i2I (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 2 I, M qi : Ni ! Ni; i2I 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 i2I 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 ! i2I Ni)i2I is still a direct sum of (Ni)i2I , i.e., it satisfies the same universal property. Similarly, for the indexed family (M ⊗R Ni)i2I of Abelian groups, there is a direct sum which is an indexed family (ei)i2I of homomorphisms of Abelian groups M ei : M ⊗R Ni ! (M ⊗R Ni) ; i2I 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): i2I 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 i2I i2I 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 2 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) i2I 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)i2I , there exists a unique homomorphism of Abelian groups M M βm : Ni ! (M ⊗R Ni) i2I i2I such that for every i 2 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): i2I i2I Of course there is also the universal R-bilinear map ! ! M M γ : M × Ni ! M ⊗R Ni : i2I i2I The claim is that β is R-bilinear, and hence determines a unique group homomorphism ! M M b : M ⊗R Ni ! (M ⊗R Ni) i2I i2I 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)i2I , to prove that βm+m0 equals βm + βm0 , it suffices to prove for every i 2 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: i2I i2I 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 2 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)): i2I i2I 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): i2I i2I 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 2 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): i2I i2I 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 : i2I i2I By the universal property of (qi)i2I , this is equivalent to prove that for every index i 2 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)i2I , it suffices to prove that for every index i 2 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).