Math 902

Solutions to Homework # 4

Except in Problem # 5, R denotes a commutative ring with identity.

1. Let M be a flat R-module. Prove the following conditions are equivalent. (If M satisfieseither of these conditions, M is said to be faithfully flat.)

(a) For every nonzero R-module N we have M ⊗R N 6= 0.

(b) For every maximal ideal m of R we have M 6= mM .

Solution: (a) =⇒ (b) : Letm be a maximal ideal of R. ThenM/mM ∼= M⊗RR/m 6=0. Thus, M 6= mM .

(b) =⇒ (a) : Let I be a proper ideal of R. Then I is contained in some maximal idealm. Hence, IM ⊆ mM ( M . Hence, M/IM 6= 0. Now let N be a nonzero R-moduleand let x ∈ N , x 6= 0. Then Rx ∼= R/I, where I = (0 :R x), As x 6= 0, I is a properideal. Thus, there exists an exact sequence 0 → R/I → N . Tensoring with M andusing that M is flat, we have an exact sequence

0→M ⊗R R/I →M ⊗R N.

Since M ⊗R R/I ∼= M/IM 6= 0, we conclude that M ⊗R N 6= 0.

2. Let N be an R-module. Prove that HomR(−, N) is left exact.

Solution: Let Af−→ B

g−→ C → 0 be exact. We need to show

0→ Hom(C,N)g∗−→ HomR(B,N)

f∗−→ HomR(A,N)

is exact. Let h ∈ HomR(C,N) and suppose g∗(h) = hg = 0. Since g is surjective, h = 0.Hence, g∗ is injective. Also, as f ∗g∗ = (gf)∗ = 0∗ = 0, we see that im g∗ ⊆ ker f ∗.Finally, suppose h ∈ ker f ∗. Then hf = 0. Then h(ker g) = 0. Define a map j : C → Nby j(c) = h(b) were b ∈ B such that g(b) = c. Since h(ker g) = 0, the map j is well-defined and so j ∈ HomR(C,N). Note that g∗(j) = jg = h. Thus, h ∈ im g∗ and thesequence is exact.

3. Let T be a (commutative) R-algebra and M an R-module.

(a) Prove that if M is projective, T ⊗R M is a projective T -module.

(b) Prove that if M is flat, T ⊗R M is a flat T -module.

Solution: Suppose M is a free R-module. Then M ∼= ⊕iR, where the sum is oversome index set I. Then T ⊗R M ∼= T ⊗R (⊕iR) ∼= ⊕i(T ⊗R R) ∼= ⊕iT . Thus, T ⊗R Mis a free T -module. Suppose now that M is a projective R-module. Then there existsan R-module Q such that M ⊕Q is free. Then T ⊗R (M ⊕Q) ∼= (T ⊗RM)⊕ (T ⊗RQ)

is a free T -module. As T ⊗RM is the direct summand of a free T -module, we see thatT ⊗R M is projective.

For (b), let 0 → Af−→ B be an exact sequence of T -modules. As every T -module is

an R-module (by restriction of scalars), we have that the sequence is also exact asR-modules. As M is a flat R-module, we have

0→ A⊗R Mf⊗1−−→ B ⊗R M

is exact. Now, as A is a T -module, A ∼= A⊗T T . By associativity of tensor products,A⊗R M ∼= (A⊗T T )⊗R M ∼= A⊗T (T ⊗R M). Similarly for B ⊗R M . Hence,

0→ A⊗T (T ⊗R M)f⊗1−−→ B ⊗T (T ⊗R M)

is exact. Thus, T ⊗R M is a flat T -module.

4. In the context of the Five-Lemma, prove that the middle map is surjective.

Solution: You can do this! ,

5. Let R be a (not necessarily commutative) ring. Prove that the following conditions areequivalent:

(a) R is semisimple.

(b) Every left R-module is projective.

(c) Every left R-module is injective.

Solution: (a) =⇒ (b) : Let M be a left R-module and φ : F → M a surjectivehomomorphism such that F is a free left R-module. Let K = kerφ. Since R issemisimple, F is semisimple and thus K is a direct summand of F . Hence, the shortexact sequence 0 → K → F → M → 0 splits. Then F ∼= K ⊕M and hence M isprojective.

(b) =⇒ (c) : Recall that a left R-module A is injective if and only if every shortexact sequence of left R-modules beginning with A splits. Similarly, a left R-moduleC is projective if and only if every short exact sequence ending with C splits. Byassumption, every left R-module is projective. Thus, every short exact sequence splitsand hence every left R-module is also injective.

(c) =⇒ (a) : To prove R is semisimple, it suffices to prove that every left ideal of I ofR is a direct summand of R; equivalently, the inclusion map I → R splits. But this isclear, as I is injective.