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Let $L\diagup K$ be a Galois extension of fields satisfying $\left[L:K\right] < \infty$. Let $B$ be a finite-dimensional (as a $K$-vector space) $K$-algebra. Then, the Galois group $G$ of $L\diagup K$ acts on the $L$-algebra $L\otimes_K B$ (although not by $L$-linear homomorphisms), thus also on its unit group $\left(L\otimes_K B\right)^{\times}$. Is it true that $H^1\left(G,\left(L\otimes_K B\right)^{\times}\right)$ is the one-element set?

For $B=K$, this would be Hilbert 90. More generally, for $B$ being a matrix algebra over $K$, this would be an extension of Hilbert 90 Milne claims to hold in his CFT. My main reason for generalizing to arbitrary $B$ is to prove the following fact, known as Noether-Deuring theorem:

Let $A$ be a $K$-algebra, and let $U$ and $V$ be two finite-dimensional representations of $A$ over $K$. Then, $U$ and $V$ are isomorphic representations if and only if the representations $L\otimes_K U$ and $L\otimes_K V$ are isomorphic representations of the algebra $L\otimes_K A$. Note that this holds not only for Galois extensions $L\diagup K$ but for arbitrary field extensions $L\diagup K$, but the (Galois) case of finite fields is the hardest. This generalizes the fact that two matrices over some field are similar if and only if they are similar over a field extension, which, in turn, is a particular case of "Conjugacy rank" of two matrices over field extension .

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up vote 13 down vote accepted

It's actually easier to go the other way around. Finite dimensional modules over an algebra $A$ fulfils the Krull-Remak-Schmidt theorem of being isomorphic to a direct sum of indecomposable modules with the indecomposable factors unique up to isomorphism. If now $L\bigotimes_KU$ and $L\bigotimes_KV$ are isomorphic as $L\bigotimes_KA$-modules they are also isomorphic as $A$-modules but as $A$-modules they are isomorpic to $U^n$ resp. $V^n$ where $n=[L:K]$ and by the KRS-theorem this implies that $U$ and $V$ are isomorphic. The generalised Thm 90 now follows as the cohomology set classifies $B$-modules whose scalar extension to $L$ are free of rank $1$.

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Ouch. Something tells me I should have seen that... – darij grinberg Jun 17 '10 at 9:49

Dear darij: yes, your $H^1$ is the one element set. A reference is Exercise 2 in Chapter X, §1 of Serre's Local Fields (page 152 in my edition). [The exercise is provided with hints!] And, yes, the case of finite fields is indeed the hardest.

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Thanks a lot. Somehow I used to think that Serre's Local Fields is only about local fields, and therefore never cared to open that book... – darij grinberg Jun 17 '10 at 9:50

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