Let $X$ be the set of $k\times k$ matrix with entries in $\mathbb{C}$, and let $M\in X$. The group $GL(k,\mathbb{C})$ acts on $X$ by conjugation, and according to the Jordan decomposition theorem (see e.g wikipedia) somewhere in the orbit containing $M$ is a block diagonal matrix with non-zero entries only on the diagonal and superdiagonal.

Suppose now we consider $k\times k$ matrices whose entries lie in the polynomial ring $\mathbb{C}[z_{1},z_{2}, \ldots ,z_{n}]$ and we study the action by conjugation of $GL(k,\mathbb{C}[z_{1},z_{2}, \ldots ,z_{n}])$. Then the Jordan decomposition theorem, as formulated above, clearly no longer holds. For example consider the matrix:

$ M=\left( {\begin{array}{cc} 0 & 1 \\\ z^{p}_{1} & 0 \end{array}} \right)$,

where in the above $p$ denotes a positive integer. If $p $ is odd, then $M$ cannot be diagonalized since the ring $\mathbb{C}[z_{1},z_{2}, \ldots ,z_{n}]$ does not contain the eigenvalues of $M$. On the other hand, if $p$ is even we still cannot diagonalize $M$ since when $z_{1}=0 $, $M$ is not diagonalizable.

My question is then what, if anything, remains of the Jordan decomposition in this case? Or equivalently given a $k\times k$ matrix $M$ with entries in $\mathbb{C}[z_{1},z_{2}, \ldots ,z_{n}]$ are there any particularly simple matrices related to $M$ via conjugation by an element of $GL(k,\mathbb{C}[z_{1},z_{2}, \ldots ,z_{n}])$?

sureyou want to work with GL_k over that polynomial ring? Your example is not invertible as a polynomial matrix. Such matrices would have to have constant (nonzero) determinant. So maybe "GL" is the wrong type of matrices and you just want matrices with nonzero determinant (i.e., polynomial matrices that are invertible over the rational functions in several variables over C). I think this is going to be pretty hard, even when n = 1. The case of C[z] matrices is analogous to integral matrices, and conjugacy classes in M_n(Z) by GL_n(Z) is sneaky! $\endgroup$ – KConrad Jun 17 '10 at 21:26irreduciblecharacteristic polynomial (OK, that won't have an analogue over C[z] very often) then there is a bijection between the conjugacy classes of such matrices and ideal classes in an order in a number field. This is a theorem of Latimer and MacDuffee. See math.uconn.edu/~kconrad/blurbs/gradnumthy/matrixconj.pdf. $\endgroup$ – KConrad Jun 17 '10 at 21:28