Group action such that $H^1(G, L^{\times}) \neq \{1\}$.

Question

I am looking for an example of a field $L$ together with a group morphism $G \to \mathrm{Aut_{field}}(L)$ (i.e. a group $G$ acts by field automorphisms on $L$), and such that the cohomology group $H^1(G, L^{\times})$ is not trivial.

The typical example of such action is when $L$ is a Galois extension of some subfield $K$, and $G$ denotes the Galois group, but in that case Hilbert 90 asserts the vanishing of the first cohomology group.

(The motivation behind this question is an exercise in chap. I of Silverman's book "Arithmetic of elliptic curves", which states that $\mathbb P^n(L^G) = \mathbb P^n(L)^G$, where $L/K$ is Galois and $G$ the Galois group — this uses Hilbert 90 !).

Answer 1

A simple way to get such examples is to consider trivial actions. For instance, let $G=\mathbb{Z}$ and let $G$ act on any field $L$ trivially. Then $H^1(G,L^\times)\cong L^\times$. More generally, with $G=\mathbb{Z}$ acting possibly nontrivially on $L$, then $H^1(G,L^\times)$ is isomorphic to the coinvariants $L^\times_G$. (To verify these calculations you can use the very simple free resolution of the trivial module $\mathbb{Z}$ over the group ring $\mathbb{Z}[G]\cong\mathbb{Z}[t,t^{-1}]$: $0\to \mathbb{Z}[G]\stackrel{t-1}\to\mathbb{Z}[G]\to\mathbb{Z}\to 0.$)

Answer 2

In Eric Wofsey's answer, with $\Bbb{Z}$ acting trivially on $L^\times$,

The map is $a\in L^\times \to f_a\in H^1(\Bbb{Z},L^\times), f_a(\{n,m\})= a^{n-m}$ which is a cocycle because it satisfies $f_a(\{n,m\})=m.f_a(\{n-m,0\})$ and $f_a(\{n,m\}) f_a(\{m,l\})=f_a(\{n,l\})$.

We quotient those by the coboundaries, ie. the cocycles of the form $f(\{n,m\})= (n.b)/(m.b)=b/b=1$.