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](http://en.wikipedia.org/wiki/Jordan_form)) 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 $GL(k,\mathbb{C}[z\_{1},z\_{2}, \ldots ,z\_{n}])$?



Jordan Form Over a Polynomial Ring
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 $R.$


