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Let $k$ be a field of characteristic $p$ and $V$ be a $k^p$ vector space. Denote by $k_s$ the separable closure of $k$ and set $G_k := Gal(k_s|k)$. Prove that

$$ H^0(G_k, V \otimes_{k^p} k_s^p) = V \\ H^n(G_k, V \otimes_{k^p} k_s^p) = 0, \: r > 0. $$

The case $n = 0$ is easy, since $(V \otimes k_s^p)^{G_k} = V^{G_k} \otimes (k_s^p)^{G_k} = V \otimes k^p = V$.

I am unsure, if my reasoning in the case $n > 0$ is correct:

We can view $V \otimes k_s^p$ as a direct sum of copies of $k^p_s$. If $H^n(G_k, k^p_s) = 0, n>0$ and $H^n(.)$ commutes with the direct sum, the claim is proven.

So basically my proof assumes, that $H^n(G_k, k_s) = 0$ induces $H^n(G_k, k^p_s) = 0$. Is that correct?

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    What is your $r$? In addition, this seems to be exactly what [this book](http://www.amazon.com/Algebras-Cohomology-Cambridge-Advanced-Mathematics/dp/0521861039) says!2012-12-28
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    Sorry, I have overlooked your comment. I fixed the typo and will look at the book. Thanks2013-01-13
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    I looked at the book and couldn't find anything except [Lemma 4.3.11](http://www-fourier.ujf-grenoble.fr/~marin/une_autre_crypto/Livres/Gille-Szamuely-Central-simple-alg.pdf.pdf) which says: $H^n(G_k,k_s) = 0, n > 0$. I need $H^n(G_k, k_s^p) = 0$. Or did you mean something else?2013-01-14
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    I mean: at page 40, it is asserted that the tensor product is a direct sum of copies of K(Indeed the sum is different from what you wrote, but I was not remembering well...)And if $k^p$ is separable, then the theorem of Hilbert asserts your statement, right? Or are there some mistakes in the arguments? In any case, thanks for your attention.2013-01-14

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