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
(F - 1)(\sum \lambda _ iv_ i) = \sum \lambda _ i^ p a_{ij} v_ j - \sum \lambda _ j v_ j = \sum \mu _ j v_ j
This is equivalent to
\sum \lambda _ j^ p v_ j - \sum b_{ij} \lambda _ i v_ j = \sum b_{ij} \mu _ i v_ j
in other words
\lambda _ j^ p - \sum b_{ij} \lambda _ i = \sum b_{ij} \mu _ i, \quad j = 1, \ldots , \dim (V).
The algebra
A = k[x_1, \ldots , x_ n]/ (x_ j^ p - \sum b_{ij} x_ i - \sum b_{ij} \mu _ i)
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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