Lemma 59.63.3. In the situation of Lemma 59.63.2 let $k'/k$ be an extension of algebraically closed fields. Set $V' = V \otimes _ k k'$ and $F' : V' \to V'$ be the Frobenius linear map induced by $F$. Then $\mathop{\mathrm{Ker}}(F - 1)$ maps isomorphically onto $\mathop{\mathrm{Ker}}(F' - 1)$.
Proof. Let $v_1, \ldots , v_ n$ be a basis of $V$ and say $F(v_ i) = \sum a_{ij}v_ j$. Then $v = \sum \lambda _ i v_ i$ is in $\mathop{\mathrm{Ker}}(F - 1)$ if and only if $\sum \lambda _ i^ pa_{ij} = \lambda _ j$ for $j = 1, \ldots , n$. By Lemma 59.63.2 these equations cut out a closed subscheme $Z$ of $\mathbf{A}^ n_ k = \mathop{\mathrm{Spec}}(k[\lambda _1, \ldots , \lambda _ n])$ finite over $\mathop{\mathrm{Spec}}(k)$. Then $Z(k) = Z(k')$ and we conclude. $\square$
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