# Matrices over a finite field with given Jordan normal form over the algebraic closure

Can one describe the (conjugacy classes of) square matrices over a finite field such that over the algebraic closure of this finite field their Jordan normal form consists of one Jordan block? (Such matrices correspond to absolutely indecomposable representations of the quiver with one vertex and one loop, which are important for Kac's conjecture).

## migrated from mathoverflow.netMar 6 '15 at 12:46

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Let $F$ be the finite field and $K$ be its algebraic closure. The first thing to decide seems to be when we can have $(x-\lambda)^{n} \in F[x]$ for $\lambda \in K \backslash F.$ If we write $n = p^{a}b$ with $p \not\mid b,$ then we must have $a >0$ and $[(x-\lambda)^{b}]^{p^a} \in F[x].$ Since $F$ is finite, this forces $(x-\lambda)^{b} \in F[x],$ and then we contradict $\lambda \not \in F$ as $p \not\mid b.$ Hence we must have $\lambda \in F$ after all. Since the matrix has a single Jordan block, it follows that there is one conjugacy class of such matrices for each $\lambda \in F.$
It is a general fact for square matrices with entries in a field$~F$, that if they are similar over an extension field$~K$, they are already similar over$~F$. Since having the same Jordan normal form over the algebraic closure$~\overline F$ implies being similar over$~\overline F$, it follows that there is at most one such conjugacy class for any given eigenvalue$~\lambda\in\overline F$.
It remains to see for which such eigenvalues$~\lambda$ there is actually a representative matrix over$~F$ of that Jordan block. This requires that the minimal polynomial $(X-\lambda)^n$ of that matrix have entries in$~F$. Since finite fields are perfect, one knows that the minimal polynomial over$~F$ of the algebraic element $\lambda\in\overline F$ is separable, and given that $(X-\lambda)^n\in F[X]$ it must be that this minimal polynomial is $X-\lambda$. So in fact $\lambda\in F$, in which case of course the conjugacy class is that of the Jordan block, which is defined over$~F$.