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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 says! –  awllower Dec 28 '12 at 14:57
    
Sorry, I have overlooked your comment. I fixed the typo and will look at the book. Thanks –  p.kelly Jan 13 '13 at 21:55
    
I looked at the book and couldn't find anything except Lemma 4.3.11 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? –  p.kelly Jan 14 '13 at 14:33
    
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. –  awllower Jan 14 '13 at 15:07
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