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

Proof. The map d: Pi+1 → Pi induces a map HomG(Pi, A) → HomG(Pi+1, A)
via φ ↦→ φ ◦ d,
(dφ)(g1, . . . , gn) = f d(1, g1, g1g2, . . . , g1 · · · gi+1)
= f
+
(g1, g1g2, . . . , g1 · · · gi+1)
i
j=1
(−1) j (1, g1, . . . , g1 ·
· · gj, . . . , g1 · · · gi+1)
+ (−1) i+1
(1, g1, g1g2, . . . , g1 · · · gi)
= g1φ(g2, g3, . . . , gi+1) +
+ (−1) i+1 φ(g1, . . . , gi)
i
j=1
(−1) j φ(g1, . . . , gjgj+1, . . . , gi+1)
We now make some explicit considerations in small cases. Begin with
the case i = 0. If φ ∈ HomG(P0, A) then there is an a ∈ A such that
φ() = a. Thus, (dφ)(g) = gφ() − φ() = ga − a. Now let i = 1 and take
φ ∈ HomG(P1, A). Then
Thus
H 0 (G, A) = A G
(dφ)(g1, g2) = g1φ(g2) − φ(g1g2) + φ(g1)
H 1 (G, A) = {φ: G → A : dφ = 0}/{dφ : φ ∈ HomG(P0, A)}
= {φ: G → A : φ(g1g2) = φ(g1) + g1φ(g2)}/{φa : G → A : a ∈ A}
where φa(g) = ga − a,
H 2 (G, A) = {φ: G × G → A : g1φ(g2, g3) − φ(g1g2, g3)
+ φ(g1, g2g3) − φ(g1, g2) = 0}/⋆
5 Morphisms of pairs
5.1 Morphism of pairs
A morphism of pairs (G, A) → (G ′ , A ′ ) together with homomorphisms f : G ′ →
G and g : A → A ′ such that g(f(s ′ ).a) = s ′ .g(a) for all s ′ ∈ G ′ , a ∈ A induces
morphisms H q (G, A) → H q (G ′ , A ′ ) for all q ≥ 0.
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