Is there a (not so) generalized version of Hilbert's Theorem 90? I'm sorry if my following question doesn't make any sense. 
We know that if $L/k$ is a finite Galois extension then $H^{1}(\mathrm{Gal}(L/k),L^{*})=0$ (Hilbert's theorem 90). However I would like to know if there is some generalized version involving some field extension $M/L$ such that $H^{1}(\mathrm{Gal}(L/k),M^{*})=0$? Here note that $L$ and $M$ are not the same as in the usual version $H^{1}(\mathrm{Gal}(L/k),L^{*})$=0. Also I'm assuming that $M$ is already a $\mathrm{Gal}(L/k)$-module. 
(Edit: If $M/k$ is a Galois extension and $\sigma\in{\mathrm{Gal}(L/k)}$, then choose an automorphism $\tilde{\sigma}:M\rightarrow{M}$ extending $\sigma$ and define $\sigma\cdot{m}:=\tilde{\sigma}(m)$ for all $m\in{M}$. This could give $M$ a structure of $\mathrm{Gal}(L/k)$-module but I'm not sure!)
The motivation of this question is a proof regarding divisors on a smooth curve that I'm trying to understand. In that proof the author says that he will use a generalized version of Hilbert's theorem 90 but he doesn't say what version is. The context is a smooth curve $C/k$ with a generic point $\xi$, a finite Galois extension $L/k$ and a 1-cocycle $f:\mathrm{Gal}(L/k)\rightarrow{L(\xi)^{*}}$ and then the author says "by a generalization of Hilbert's theorem 90 this 1-cocycle is a 1-coboundary", so I'm assuming that the version is something like $H^{1}(\mathrm{Gal}(L/k),L(\xi)^{*})=0$ (whereas the usual version only implies $H^{1}(\mathrm{Gal}(L/k),L^{*})=0$). 
Could anyone tell me what this generalization could be? I know that there is a generalization involving étale cohomology but I don't think it should be that complicated. Any help is appreciated.
 A: Here, we have $L/k$ a finite Galois extension, and $M=L(\xi)$, with $\xi$ algebraic over $L$. You define the action of $G=Gal(L/k)$ over $M^{*}$
$G\times M^{*}$ $\rightarrow$ $M^{*}$ 
$(\sigma,x)\rightarrow \tilde \sigma(x)$, 
where $\tilde\sigma:M\rightarrow\bar M$ is the unique extension of $\sigma$ to $M$ such that $\tilde\sigma(\xi)=\xi$. It follows that, since $M=L(\xi)$, then $\tilde\sigma\in Aut(M/k)$, and that, by Artin's Lemma,since G is finite, and so $\tilde G$={$\tilde\sigma|\sigma\in G$} is finite,and since clearly $\tilde G$<$Aut(M/k)$, and Fix($\tilde G$)=k($\xi$), then M/k($\xi$) is a finite and Galois extension, and Gal($M/k(\xi)$)=$\tilde G$.
So, what you are really concerned with is $H^{1}(\tilde G, M^{*})$ with the standard action of $\tilde G$ over $M^{*}$ (that is, there is a commuting diagram involving this $\tilde G$-module and the G-module $H^{1}(G, M^{*})$, involving the injective group homomorphism $G\rightarrow \tilde G$,$\sigma\rightarrow\tilde\sigma$) , which is, by Hilbert theorem 90, null.  
