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I am looking for an analogue of the Jordan normal form for nilpotent matrices over the polynomial ring ${\mathbb Z}[x_1, \dots, x_n]$. More precisely, is there a description for the orbits of action by conjugation of $GL_m({\mathbb Z}[x_1, \dots, x_n])$ on $M_{m \times m}({\mathbb Z}[x_1, \dots, x_n])$?

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I am very skeptical about the existance of a nice answer. Even if we take a field $K$ and consider the same problem for $K\left[X,Y\right]$ instead of $\mathbb Z\left[x_1,...,x_n\right]$, then a description of nilpotent matrices up to conjugation would yield a description of pairs of nilpotent commuting matrices over $K$ up to simultaneous conjugation by the same conjugator (in fact, a pair $\left(A,B\right)$ of two nilpotent commuting matrices yields a nilpotent matrix $AX+BY$ over $K\left[X,Y\right]$, and two such matrices are conjugate iff the pairs are simultaneously conjugate by the ... –  darij grinberg Aug 21 '10 at 18:30
    
... same conjugator). –  darij grinberg Aug 21 '10 at 18:31
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The case $n=1$ is as much as you can hope for. See also mathoverflow.net/questions/28560/… –  Victor Protsak Aug 21 '10 at 19:13
    
Thanks for the comments! –  Keivan Karai Aug 22 '10 at 12:44
    
This is a very ambitious question. The classification of conjugation orbits is well-understood in $M_n(k)$, $k$ a field (Frobenius normal forms). When the characteristic polynomial splits, this yields Jordan normal form. This classification comes from that of the orbits in $M_n(A)$ ($A$ a principal ideal domain) under the action of $GL_n(A)\times GL_n(A)$, $M\mapsto PMQ$ $PMQ$ is equivalent to $P$, instead of conjugated). This is the theory of elementary divisors. In your question, you keep conjugation, but your set of scalars is not a field, even not a P.I.D.! –  Denis Serre Sep 21 '10 at 9:07

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