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 $GL(k,\mathbb{C}[z_{1},z_{2}, \ldots ,z_{n}])$?

  • $\begingroup$ Jordan decomposition no longer holds because ${\mathbb C}[z_1, \ldots ,z_n]$ is not an algebraically closed field- in fact, it's not even a field. This suggests the following (seemingly easier) question : what if we replace ${\mathbb C}[z_1, \ldots ,z_n]$ by $\mathbb Z$ ? By $\mathbb Q$ ? If a reasonably simple answer exists for ${\mathbb C}[z_1, \ldots ,z_n]$, there should also be one for $\mathbb Z$ or $\mathbb Q$. $\endgroup$ Jun 17 '10 at 19:58
  • $\begingroup$ Clay, are you sure you 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 21:26
  • $\begingroup$ For conjugacy classes of integral matrices there is no straightforward description of what happens, but if you focus on integral matrices with a fixed irreducible characteristic 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
  • $\begingroup$ Hi Keith. Thanks for your notes. I believe I am asking the correct question for the application I have in mind. It is interesting that it is so difficult. I am curious to know whether it is any easier if I restrict the size of the matrix to say 2x2 traceless matrices with polynomial entries. Are the possible conjugacy classes by SL_2 known? $\endgroup$ Jun 17 '10 at 22:05
  • $\begingroup$ Even 2 x 2 is tricky over the integers: all the examples in the URL from my previous comment are 2 x 2 and this case is basically tantamount to trying to understand ideal class groups of all orders in quadratic fields. Replacing conjugacy by GL_2 with SL_2 isn't going to simplify things a whole lot either, as it sort of amounts to replacing ordinary class groups by "narrow" class groups, but maybe that is too algebraic for you? If your question has a geometric origin (line bundles?) maybe you can say something more, and I'd expect your case n = 1 to be very different from n > 1. $\endgroup$
    – KConrad
    Jun 17 '10 at 22:26

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.


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 $R.$

  • $\begingroup$ Although when k = 2 this question is about the classification of C[z,w]-module structures on C[z]^k (let w act as your chosen matrix, which does commute with scaling by C[z]), there will be a polynomial over C[z] killing w by Cayley--Hamilton. Thus you are looking at C[z,w]/(f)-module structures on C[z]^k, where f is some monic polynomial in w. (more next) $\endgroup$
    – KConrad
    Jun 27 '10 at 1:34
  • $\begingroup$ With the analogy Z <---> C[z] in mind, considering that both are PIDs, a paper you might get something useful from is by Estes and Guralnick "Representations under ring extensions: Latimer-MacDuffee and Taussky correspondences," Adv. in Math. 54 (1984), no. 3, 302--313. (On MathSciNet, see MR0766270.) In your example from the 2 x 2 case, the matrix [0 1; z^p 0] has char. polynomial T^2 + z^p, which is separable, and that's the kind of thing Estes and Guralnick look at. Disclaimer: I haven't looked at the paper. $\endgroup$
    – KConrad
    Jun 27 '10 at 1:36
  • $\begingroup$ Thank you for the references, Keith! I understand what happens for matrices over $K[x] (n=1)$ when the char polynomial $f(x,\lambda)$ is irreducible and defines a non-singular (hence normal) affine curve, by the correspondence with $A=K[x,\lambda]/(f)$-modules, since $A$ is a Dedekind ring. It would be nice to see more general cases worked out in some detail. By the way, explicit matrices you quote were given by the OP, Clay Cordova. But I am not sure how much he is still interested in the question. $\endgroup$ Jun 28 '10 at 5:11

"Answer" deleted due to the inability of the author to read.

  • $\begingroup$ If I understand Wikipedia correctly then the Smith normal form is the analogue over a PID of the usual singular value decomposition. One considers the action: M->GMH Where M is a general endomorphism and G and H are invertible. In particular H is NOT the inverse of G so this is not the action by conjugation. $\endgroup$ Jun 17 '10 at 23:35
  • $\begingroup$ Ahh I see sorry, I misread the question. I'm going to withdraw my answer. $\endgroup$
    – GMRA
    Jun 17 '10 at 23:36

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