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 satisfies
either 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) : Let m be a maximal ideal of R. Then M/mM ∼
= M ⊗R R/m 6=
0. Thus, M 6= mM .
(b) =⇒ (a) : Let I be a proper ideal of R. Then I is contained in some maximal ideal
m. Hence, IM ⊆ mM ( M . Hence, M/IM 6= 0. Now let N be a nonzero R-module
and let x ∈ N , x 6= 0. Then Rx ∼
= R/I, where I = (0 :R x), As x 6= 0, I is a proper
ideal. Thus, there exists an exact sequence 0 → R/I → N . Tensoring with M and
using 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.
f
g
Solution: Let A →
− B→
− C → 0 be exact. We need to show
g∗
f∗
0 → Hom(C, N ) −
→ HomR (B, N ) −→ 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 → N
by j(c) = h(b) were b ∈ B such that g(b) = c. Since h(ker g) = 0, the map j is welldefined and so j ∈ HomR (C, N ). Note that g ∗ (j) = jg = h. Thus, h ∈ im g ∗ and the
sequence 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 ∼
= ⊕i R, where the sum is over
∼
∼
some index set I. Then T ⊗R M = T ⊗R (⊕i R) = ⊕i (T ⊗R R) ∼
= ⊕i T . Thus, T ⊗R M
is a free T -module. Suppose now that M is a projective R-module. Then there exists
an R-module Q such that M ⊕ Q is free. Then T ⊗R (M ⊕ Q) ∼
= (T ⊗R M ) ⊕ (T ⊗R Q)
is a free T -module. As T ⊗R M is the direct summand of a free T -module, we see that
T ⊗R M is projective.
f
For (b), let 0 → A →
− 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 as
R-modules. As M is a flat R-module, we have
f ⊗1
0 → A ⊗R M −−→ 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,
f ⊗1
0 → A ⊗T (T ⊗R M ) −−→ 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 are
equivalent:
(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 surjective
homomorphism such that F is a free left R-module. Let K = ker φ. Since R is
semisimple, F is semisimple and thus K is a direct summand of F . Hence, the short
exact sequence 0 → K → F → M → 0 splits. Then F ∼
= K ⊕ M and hence M is
projective.
(b) =⇒ (c) : Recall that a left R-module A is injective if and only if every short
exact sequence of left R-modules beginning with A splits. Similarly, a left R-module
C is projective if and only if every short exact sequence ending with C splits. By
assumption, every left R-module is projective. Thus, every short exact sequence splits
and 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 of
R is a direct summand of R; equivalently, the inclusion map I → R splits. But this is
clear, as I is injective.