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§5. Some Calculations with Galois Cohomology 153

§5. Some Calculations with Galois Cohomology

1

Our interest in H (G, A) will be for G = Gal(k /k) for some Galois extension k /k,
and in this case we will speak of Galois cohomology groups especially in the case of
1
H .
(5.1) Proposition. For a finite Galois extension k /k we have

H 1 Gal(k /k), GL n k = 0.

Proof. Let a s be a 1-cocycle, and for any u ∈ M n (k ) form the “Poincar´ e series”

b = a t t(u) ∈ M n k for G = Gal k /k .
t∈G

We calculate for s ∈ Gal(k /k) using the cocycle relation
−1 −1
s(b) = s (a t ) st(u) = a s a st st(u) = a s b.
t∈G t∈G
Hence a s is a coboundary if we can choose u such that b is invertible. For n = 1the
linear independence of automorphisms gives an element u with b = 0.
For n > 1 we consider the linear transformation B : k n → k n defined by the

matrix equation B(x) = a t t(x).
t∈G
n

Assertion: The B(x) generate the k -vector space k . For otherwise, there exists a
linear form f with f (B(x)) = 0 for all x, then for all scalars c,wehave

0 = f (B(x)) = a t f (t(c)t(x)) = t(c) f (a t t(x)) .
t∈G t∈G

This is a linear relation between all t(c), t ∈ G as c varies over k . The linear inde-
pendence of the elements t of G implies that each f (a t t(x)) = 0, and since the a t
are invertible, we deduce that f = 0 which is a contradiction.
Choose x 1 ,... , x n ∈ k n such that the B(x 1 ),... , B(x n ) are linearly indepen-

dent, and choose u ∈ M n (k ) with ue i = x i for i = 1,... , n. For the matrix

b = a t t(u), it follows that b(e i ) = B(x i ) which implies that b ∈ GL n (k ).
t∈G
This proves the proposition.
For a cyclic group G of order n with generator s and x ∈ A, the function G → A
defined inductively by the relations

a 1 = 1, a s = x,... , a i = x a i−1 ,...
s s
is a 1-cocycle on G with values in the abelian group A if and only if
2
1 = N(x) = x · s(x) · s (x)... s n−1 (x).
This observation and previous proposition leads to the next corollary.