Lemma 59.63.2. Let $k$ be an algebraically closed field of characteristic $p > 0$. Let $V$ be a finite dimensional $k$-vector space. Let $F : V \to V$ be a frobenius linear map, i.e., an additive map such that $F(\lambda v) = \lambda ^ p F(v)$ for all $\lambda \in k$ and $v \in V$. Then $F - 1 : V \to V$ is surjective with kernel a finite dimensional $\mathbf{F}_ p$-vector space of dimension $\leq \dim _ k(V)$.

**Proof.**
If $F = 0$, then the statement holds. If we have a filtration of $V$ by $F$-stable subvector spaces such that the statement holds for each graded piece, then it holds for $(V, F)$. Combining these two remarks we may assume the kernel of $F$ is zero.

Choose a basis $v_1, \ldots , v_ n$ of $V$ and write $F(v_ i) = \sum a_{ij} v_ j$. Observe that $v = \sum \lambda _ i v_ i$ is in the kernel if and only if $\sum \lambda _ i^ p a_{ij} v_ j = 0$. Since $k$ is algebraically closed this implies the matrix $(a_{ij})$ is invertible. Let $(b_{ij})$ be its inverse. Then to see that $F - 1$ is surjective we pick $w = \sum \mu _ i v_ i \in V$ and we try to solve

This is equivalent to

in other words

The algebra

is standard smooth over $k$ (Algebra, Definition 10.137.6) because the matrix $(b_{ij})$ is invertible and the partial derivatives of $x_ j^ p$ are zero. A basis of $A$ over $k$ is the set of monomials $x_1^{e_1} \ldots x_ n^{e_ n}$ with $e_ i < p$, hence $\dim _ k(A) = p^ n$. Since $k$ is algebraically closed we see that $\mathop{\mathrm{Spec}}(A)$ has exactly $p^ n$ points. It follows that $F - 1$ is surjective and every fibre has $p^ n$ points, i.e., the kernel of $F - 1$ is a group with $p^ n$ elements. $\square$

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