Galois Cohomology

Suppose $ L/K$ is a finite Galois extension of fields, and $ A$ is a module for $ \Gal (L/K)$. Put

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

Next suppose $ A$ is a module for the group $ \Gal (K^{\sep }/K)$ and for any extension $ L$ of $ K$, let

$\displaystyle A(L) = \{x \in A : \sigma(x) = x$    all $\displaystyle \sigma \in\Gal (K^{\sep }/L)\}.$

We think of $ A(L)$ as the group of elements of $ A$ that are ``defined over $ L$''. For each $ n\geq 0$, put

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

Also, put

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

where $ L$ varies over all finite Galois extensions of $ K$. (Recall: Galois means normal and separable.)

Example 11.4.1   The following are examples of $ \Gal (\overline{\mathbf{Q}}/\mathbf{Q})$-modules:

$\displaystyle \overline{\mathbf{Q}},\quad \overline{\mathbf{Q}}^*,\quad \overli...{\mathbf{Q}}),\quad
E(\overline{\mathbf{Q}})[n], \quad {\rm Tate}_{\ell}(E),

where $ E$ is an elliptic curve over  $ \mathbf {Q}$.

Theorem 11.4.2 (Hilbert 90)   We have $ \H^1(K,\overline{K}^*) = 0$.

Proof. See [Ser79]. The main input to the proof is linear independence of automorphism and a clever little calculation. $ \qedsymbol$

William Stein 2012-09-24