Jordan Form Over a Polynomial Ring 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 $GL(k,\mathbb{C}[z_{1},z_{2}, \ldots ,z_{n}])$?
 A: "Answer" deleted due to the inability of the author to read.
A: The short answer is "no". It is not too difficult to construct *invariants, but the canonical forms are hard.a What follows is not a full answer, but a useful way to think about the question. 
The problem of classifying $k\times k$ matrices over a commutative ring $R$ up to conjugacy is equivalent to the problem of classifying all $R[\lambda]$-module structures on $R^k$, the free $R$-module of rank $k,$ up to isomorphism. Given $A\in M_k(R)$, let the variable $\lambda$ act on $R^k$ via $A$ and conversely, given an $R[\lambda]$-module structure on $R^k,$ the action of $\lambda$ is $R$-linear, hence yields a matrix. It's a good exercise to see that conjugacy of matrices $\leftrightarrow$ isomorphism of modules.
Now, when $R$ is a field, $R[\lambda]$ is a principal ideal domain, and all finitely-generated modules can be completely classified using the theory of elementary divisors.b However, if $R$ is even a bit more complicated, such as $\mathbb{Z}$ or $K[X]$, the question involves modules over a ring of Krull dimension $2$ or larger, and one cannot hope for an explicit easy solution (except $k=1$). Already for $n=1, R=K[x]$ we are looking at the classification of $K[x,y]$-modules.  See van der Waerden for classical treatment.

Footnotes 
a For example, the characteristic polynomial and the Fitting invariants of the matrix in b.
b In the context of the conjugacy problem, you can replace a $k\times k$ matrix $A$ over $R$ modulo conjugation by $GL_k(R)$ with a $k\times k$ matrix $A-\lambda I_k$ over $R[\lambda]$ modulo left and right multiplication by $GL_k(R).$ The "elementary" part refers to the fact that when $R=K$ is a field, the general linear group is generated by elementary transformations, and  "divisors" refers to the form of the answer, where the canonical form is diagonal and $d_i$ divides the entries in rows $1$ through $i$. Neither fact is true for a more general $R.$
