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

as follows:
N
A
A
N
H0(G, A)
∗
H0 (G, A)
ˆH0(G, A) = ker(N ∗
)
A[N]/IGA
ˆH 0 (G, A) = coker(N ∗ )
0
AG /N(A)
Proposition 36. If A is induced then ˆ H0 (G, A) = 0.
Proof. The assumption that A is induced means that A ∼ =
s∈G s.X ∼ =
Z[G] ⊗Z X, where X ≤ A is a subgroup. This is easy to see since G is finite.
Thus, each a ∈ A can be expressed uniquely as
a =
s.xs, xs ∈ X
s∈G
Now a ∈ A G if and only if all xs are equal if and only if a = N(xs). This
shows that A G = N(A), as claimed.
Remark 37. If A is induced, we also have that ˆ H0(G, A) = 0. The proof
of this statement will appear on the problem sheet.
Definition 38. We define the Tate cohomology groups as follows:
ˆH q (G, A) = H q (G, A), q ≥ 1
ˆH 0 (G, A) = A G /N(A)
ˆH −1 (G, A) = A[N]/IGA
ˆH −q (G, A) = Hq−1(G, A), q ≥ 2
Remark 39. The functor ˆ H 0 (G, −) is not left exact.
Proposition 40. Given a short exact sequence of G-modules 0 → A →
B → C → 0, we have the long exact sequence
· · · → ˆ H −2 (G, C) → ˆ H −1 (G, A) → ˆ H −1 (G, B) → ˆ H −1 (G, C)
→ ˆ H 0 (G, A) → ˆ H 0 (G, B) → ˆ H 0 (G, C)
→ ˆ H 1 (G, A) → ˆ H 1 (G, B) → ˆ H 1 (G, C) → ˆ H 2 (G, A) → · · ·
20