Galois Cohomology

Let $K$ be a field and $L$ a finite Galois extension of $K$, so the set of field automorphisms of $L$ that fix $K$ equals the dimension of $L$ viewed as a $K$-vector space.

For any $\Gal (L/K)$-module $A$ and any $n\geq 0$, let

$$\H^n(L/K, A) = \H^n(\Gal (L/K), A).$$

If $M/L/K$ is a tower of Galois extensions of $K$ and suppose $\Gal (M/K)$ acts on $A$. Then inf defines a map

$$\H^n(L/K,A^L) \to \H^n(M/K,A).$$ (2.1.1)

Let $K^{\sep }$ denote a separable closure of $K$ and suppose $A$ is a (continuous) $\Gal (K^{\sep }/K)$-module. (Note - if $K$ has characteristic $0$, then a separable closure is the same thing as an algebraic closure.) For any subfield $L\subset K^{\sep }$ that contains $K$, let $A(L) = A^L$. Let

$$\H^n(K,A) = \varinjlim_{L/K\text{ finite Galois}} \H^n(L/K, A(L)),$$

where the direct limit is with respect to the maps (2.1.1). We can think of this direct limit as simply the union of all the groups, where we identify two elements if they are eventually equal under some map (2.1.1).

One can prove (see [Cp86, Ch. V]) that changing the choice of separable closure $K^{\sep }$ only changes $\H^n(K,A)$ by unique isomorphism, i.e., the construction is essentially independent of the choice of seperable closure.