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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).

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migrated from mathoverflow.net Mar 6 '15 at 12:46

This question came from our site for professional mathematicians.

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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 $\lambda \in F.$

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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$~F$.

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As was suggested to me by a colleague, the question about conjugacy classes of matrices over a finite field that have one Jordan block over the algebraic closure, is almost tautological. Namely, by subtracting a scalar matrix we can assume that the matrix is nilpotent, and by construction of Jordan normal form the matrix of change of base to the Jordan normal form for a nilpotent matrix over a finite field has entries again in this finite field (since the eigenvalues of the matrix are in this finite field). Hence all nilpotent matrices over a finite field are conjugate, and thus the answer to the original question is that the conjugacy class of the matrices in question depends only on their eigenvalues (which are all equal for a given matrix).

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    $\begingroup$ This is certainly wrong. If it were the case that "all nilpotent matrices over a finite field are conjugate" then they would all be zero (which is nilpotent). Also the scalar matrix you are subtracting early in the argument need not have its coefficients in the base field, unless you prove that separately. $\endgroup$ – Marc van Leeuwen Mar 7 '15 at 12:50

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