A Short Course on Galois Cohomology - William Stein - University of ...

Definition 14. A G-module A ∈ ModG is relatively injective if it is a direct
factor of a co-induced module, i.e., if A⊕B is co-induced for some B ∈ ModG.
Proposition 15. If A ∈ ModG is relatively injective then H q (G, A) = 0 for
all q ≥ 1.
Proof. It suffices to consider co-induced G-modules, since we can then deduce
the general case from the identity H q (G, A⊕B) = H q (G, A)⊕H q (G, B).
So assume that A is co-induced, that is, A ∼ = HomZ(Λ, X) for some X ∈ Ab.
For B ∈ ModG,
HomZ(B, X) ∼ = HomG(B, A), f ↦→ (b ↦→ (s ↦→ f(s.b))).
From the above it follows that HomG(Pi, A) = HomZ(Pi, X), so
H q (G, A) = Ext q
Z (Z, X) = 0
for q ≥ 1. Moreover, HomZ(Z, X) ∼ = X so the functor HomZ(Z, −) is exact.
Corollary 16 (Dimension shifting). If 0 → A → A ∗ → A ′ → 0 is exact
with A ∗ co-induced then
for all q ≥ 1.
H q (G, A ′ ) = H q+1 (G, A)
Remark 17. This shows that the functor H q+1 (G, −) is completely determined
by H i (G, −) for i ≤ q, giving the claimed uniqueness.
We now explicitly consider H q (G, A) as a quotient of cocycles by coboundaries.
We have the following free resolution of Z as a G-module
. . . → P2 d2
−→ P1 d1
−→ P0 d0
−→ Z → 0 (†)
where Pi = Z[G i+1 ] = Z[G × · · · × G] with G-action given by s.(g0, . . . , gi) =
(s.g0, . . . , s.gi) and d: (g0, . . . , gi) ↦→ i
j=0 (−1)j (g0, . . . , ˆgj, . . . , gi).
Lemma 18. Pi is a free G-module.
Proof. The set {(1, g1, g1g2, . . . , g1 · · · gi) : g1, . . . , gi ∈ G} is a basis for Pi
over Z[G].
Proposition 19. The free resolution (†) is exact.
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