SKETCHED SOLUTIONS TO EXERCISES OF CHAPTER 4
4.1. It is straightforward to verify that if
0 → Mi0 → Mi → Mi00 → 0
is a family of exact sequences of A-modules, where i ∈ I, then
Y
Y
Y
0→
Mi0 →
Mi →
Mi00 → 0
i∈I
i∈I
i∈I
and
0→
M
Mi0 →
i∈I
M
Mi →
i∈I
M
Mi00 → 0
i∈I
are exact as well.
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4.2. Since
is exact, it commutes with kernels and cokernels. Since coimages are
Ldefined via cokernels and coimages are isomorphic to images in abelian category, it follows that
also commutes
with images. We conclude that for two complexes
fi
gi
Xi0 → Xi → Xi00 ,
i = 1, 2
one has Ker(g1 ⊕ g2 ) = Ker g1 ⊕ Ker g2 and Im(f1 ⊕ f2 ) = Im f1 ⊕ Im f2 . Therefore the two
sequences are exact if and only if their direct sum is exact.
∼ M (I) for a set I and L is exact, it is clear that HomA (A(I) , −) is
4.3. (i) Since HomA (A(I) , M ) =
exact, i.e. A(I) is projective.
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(ii) “Only if” part: Let A(P ) = p∈P Ap be the free A-module indexed by P and ψ : A(P ) → p be
the A-homomorphism such that ψ(1p ) = p. Then ψ is surjective. If P is projective, then ψ has a
section, i.e. an A-homomorphism φ : P → A(P ) such that ψ ◦ φ = idP . This shows P is a direct
summand of A(P ) .
“If” part: If P ⊕ K is free, then by (i) P ⊕ K is projective, hence HomA (P ⊕ K, −) = HomA (P, −) ⊕
HomA (K, −) is exact. This implies HomA (P, −) is exact, hence P is projective.
(iii) It is easy to show that free modules are flat, hence by (ii) − ⊗A (P ⊕ K) = (− ⊗A P ) ⊕ (− ⊗A K)
is exact, which implies that − ⊗A P is exact, i.e. P is flat.
4.4. (i) Let M0 be an abelian group and M be a subgroup. We only need to show that any
homomorphism ϕ : M → Q/Z extends to M0 . Let S be the set of pairs (N, φ) where N is a
subgroup of M0 containing M and φ : N → Q/Z is a homomorphism such that φ|M = ϕ. Define
a partial order on S such that (N, φ) ≤ (N 0 , φ0 ) if N ⊂ N 0 and φ0 |N = φ. Clearly Zorn’s lemma
applies and we let (N, φ) be a maximal element of S. It suffices to show that N = M0 . Otherwise we
may take x ∈ M0 \ N . Consider Zx̄ ⊂ M0 /N , where x̄ = x + N . If Zx̄ ∼
= Z, we extend φ to N + Zx
∼
by defining φ(x) = 0; if Zx̄ = Z/n for some n ∈ N, then we pick up an arbitrary representative
r ∈ Q of φ(nx) ∈ Q/Z and extend ϕ to N + Zx by defining φ(x) = r/n mod Z. Then one can show
that in both cases φ is well-defined and extends ϕ to N + Zx. But this contradicts the maximality
of N . Hence N = M0 and the proof is finished.
(ii) Let ϕ ∈ HomZ (M, N ). By (i) we have the exact sequence
ϕ∨
N ∨ → M ∨ → (Ker ϕ)∨ → 0.
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If ϕ∨ = 0, then M ∨ ∼
= (Ker ϕ)∨ . Since ϕ = 0 on Ker ϕ, it follows that ϕ = 0 on M as well. This
shows that the map ϕ 7→ ϕ∨ is injective.
(iii) One has
HomA (−, P ∨ ) = HomA (−, HomAop (P, Q/Z)) ∼
= HomZ (− ⊗A P, Q/Z).
P is projective hence flat, together with (i) we see that the last functor above is a composition of
two exact functors − ⊗A P and HomZ (−, Q/Z), hence is exact. This shows P ∨ is left A-injective.
(iv) Clearly Z∨ = Q/Z. By (ii) M = HomZ (Z, M ) → HomZ (M ∨ , Q/Z) = M ∨∨ is injective.
Take a surjective Aop -homomorphism P → M ∨ where P is projective, which gives an injective
A-homomorphism M ∨∨ → P ∨ . By (iii) P ∨ is injective, and M → M ∨∨ → P ∨ is an injection.
4.5. We can make I a filtrant category by giving an arbitrary total order on I, which exists by
Zorn’s lemma. Define two functors α, β : I → C such that
Xi0 , i = i0
α(i) =
α(s) = 0 for any s : i ≤ j, i 6= j,
0,
otherwise,
β(i) = Xi ,
β(s) = 0 for any s : i ≤ j, i 6= j.
Then the
Lobvious morphism α → β is a monomorphism. Taking direct limit gives a monomorphism
Xi0 → i∈I Xi .
4.6. (i) “Only if” part is clear since HomC (W, −) is left exact. For the “if” part, the exactness of
the Hom complex implies that
HomC (W, X) ∼
= HomC (W, Ker g)
for any W ∈ C, where g is the morphism Y → Z. It follows from the Yoneda lemma that X ∼
= Ker g.
(ii) By passing to the opposite category, one can similarly show that X → Y → Z → 0 is exact if
and only if for any W ∈ C the complex
0 → HomC (Z, W ) → HomC (Y, W ) → HomC (X, W )
is exact.
4.7. (i) Applying HomC (W, −) to the diagram for any W ∈ C, by 4.6 (i) we may reduce to the case
C = Mod(Z). Let y ∈ Y . Because f is an epimorphism, we may find x ∈ X such that f (x) = g(y).
Hence (x, y) ∈ Ker(f − g) and we may find z ∈ V such that f 0 (z) = y, g 0 (z) = x. Hence f 0 is an
epimorphism.
(ii) By passing to the opposite category, we may reverse the arrows and switch from epimorphism
to monomorphism.
4.8. Applying HomC (W, −) to the diagram for any W ∈ C, by 4.6 (i) we may reduce to the case
C = Mod(Z). We need to show the first column is exact.
Exactness at X00 : if x00 ∈ X00 is mapped to zero in X10 , then it is mapped to zero in X1 . But
X00 → X0 → X1 is injective, hence x00 = 0.
Exactness at X10 : let x01 ∈ X10 be mapped to zero in X20 . Then its image x1 ∈ X1 is mapped to zero
in X2 , which implies x1 is the image of x0 ∈ X0 because the second column is exact. Since x1 is
mapped to zero in X100 , we see x0 is also mapped to zero in X100 . But X000 → X100 is injective, x0 is
mapped to zero in X000 . The first row is exact, hence x0 is the image of x00 ∈ X00 . Finally x00 has to
be mapped to x01 , because X10 → X1 is injective.
4.9. (i) It is clear that W/(x1 W + · · · + xj−1 W ) = k[xi , ∂i ]j≤i≤n [∂1 , . . . , ∂j−1 ] is a polynomial
algebra of j − 1 variables over the Weyl algebra k[xi , ∂i ]j≤i≤n ∼
= Wn−j+1 . The left multiplication
2
by xj is injective on k[xi , ∂i ]j≤i≤n , hence is also injective on W/(x1 W + · · · + xj−1 W ). This shows
ϕ is a regular sequence. It follows that H j (K • (W, ϕ)) is concentrated in degree n and
H n (K • (W, ϕ)) ∼
= W/(x1 W + · · · + xn W ) ∼
= k[∂1 , . . . , ∂n ],
which is isomorphic to a polynomial ring of n variables over k.
(ii) Similar to (i), one shows that W/(W ∂1 + · · · + W ∂j−1 ) = k[xi , ∂i ]j≤i≤n [x1 , . . . , xj−1 ] on which
the right multiplication by ∂j is injective, hence ψ is a regular sequence. Then H j (K • (W, ψ)) is
concentrated in degree n and
H n (K • (W, ψ)) ∼
= W/(W ∂1 + · · · + W ∂n ) ∼
= k[x1 , . . . , xn ].
(iii) We have Ker ∂1 ∩ · · · ∩ Ker ∂j−1 ∼
= k[xj , . . . , xn ], on which ∂j is surjective (integral of a
polynomial is a polynomial), hence λ is a coregular sequence. Then H j (K • (O, λ)) is concentrated
in degree 0 and
H 0 (K • (O, λ)) ∼
= k.
= Ker ∂1 ∩ · · · ∩ Ker ∂n ∼
4.10. Since ϕ1 is injective on A and ϕ2 is injective on A/ϕ1 (A) ∼
= k[x2 , ∂2 ][∂1 ], the sequence
ϕ = (ϕ1 , ϕ2 ) is regular. Hence H i (K • (A, ϕ)) is concentrated in degree 2 and H 2 (K • (A, ϕ)) =
A/(ϕ1 (A) + ϕ2 (A)) = k[∂1 , x2 ], which is isomorphic to a polynomial ring of 2 variables over k.
4.11. Since ϕ1 is injective and ϕ2 is surjective on M , one has
H 0 (K • (M, ϕ)) = 0,
H 2 (K • (M, ϕ)) = 0.
To calculate H 1 (K • (M, ϕ)), consider the exact sequence
ϕ2
H 0 (K • (M, ϕ1 )) → H 1 (K • (M, ϕ)) → H 1 (K • (M, ϕ1 )) → H 1 (K • (M, ϕ1 )).
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We compute that H 0 (K • (M, ϕ1 )) = 0, and H 1 (K • (M, ϕ1 )) = i≥1 kti , on which the kernel of ϕ2
is kt ∼
= k. It follows that H 1 (K • (M, ϕ)) ∼
= k.
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