Proof of $V\cong \overline{K} \otimes_{K} V_K$ using $H^1(G_{\overline{K}/K},\operatorname{GL}_n(K))＝0$

Question

This is from Silverman's book "The arithmetic of elliptic curves" (AEC), p.36, lemma 5.8.1.

Lemma 5.8.1 states

Let $V$ be a $\overline{K}$-vector space, and assume that $G_{\overline{K}/K}$ acts continuously on $V$ in a manner compatible with its action on $\overline{K}$. Then, $V\cong \overline{K} \otimes_{K} V_K$.

AEC reads there is a fancy proof which uses Hilbert's theorem 90, which says that $H^1(G_{\overline{K}/K},\operatorname{GL}_n(K))＝0$ (Exercise 2.13).

How can I apply Hilbert's theorem 90 to the proof of lemma 5.8.1?

Answer 1

Here's a sketch of the proof. I encourage you to fill in the details yourself. The definition of $V_K$ is $V_K=H^0(G_{\overline K/K},V)$. The key part of the proof is to show that $V$ has a $\overline{K}$ basis consisting of vectors in $V_K$. To find such a basis, start with an arbitrary basis $v_1,\ldots,v_n\in V$. Then each $\sigma\in G_{\overline K/K}$ gives a change of basis matrix $A_\sigma$ defined by writing $v_i^\sigma$ as a linear combination of $v_1,\ldots,v_n$. The map $$ G_{\overline K/K} \longrightarrow \operatorname{GL}_n(\overline K),\quad \sigma\longmapsto A_\sigma $$ is a $1$-cocycle. Using Hilbert Theorem 90, it follows that it is a coboundary, so there is a matrix $B\in\operatorname{GL}_n(\overline K)$ satisfying $A_\sigma=B^\sigma B^{-1}$. Now use $B$ (or maybe $B^{-1}$) to change the basis $v_1,\ldots,v_n$ to a new basis $w_1,\ldots,w_n$. If you've done it properly, the vectors $w_1,\ldots,w_n$ will be in $V_K$.