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?