# a normal form for matrices over Z[x]/(x^2-1) ?

We are discussing, offline, modules over the $\mathbb{Z}$-group ring of the cyclic group of order 2, which is probably better known as the quotient ring $R=\mathbb{Z}[t]/(t^2-1)$. Is there any way to describe matrices over it, in a way similar to Smith Normal Form (SNF), or Hermite Normal Form (HNF)? That is, for $A\in R^{n\times m}$, find $X\in GL(n,R)$ and $Y\in GL(m,R)$, such that $XAY$ is "nice", e.g. diagonal (resp. upper-triangular), like one would get if SNF (resp. HNF) was possible for $R$.

I am aware of a similar question for $\mathbb{Z}[t]$, which looks harder. One immediate observation is that $A=B+tC$, for $B,C\in \mathbb{Z}^{n\times m}$, and so one can choose $X$, $Y$ to have integer entries, so that $XAY=B'+tC'$, where $C'$ is the SNF of $B'$.

-
Your (commutative) ring has (Krull) dimension 1 because it is the quotient of a 2-dimensional ring by a principal ideal. You can check that it's regular, so it's a Dedekind domain. Modules of finite type over Dedekind domains are classified. You can probably obtain a matrix nomar form from this. There's actually a paper by Henry Cohen on the Smith normal form for Dedekind domains. I may also be possible that your ring is a PID or even an Euclidean domain. In that case you would have the classical Smith normal form. – Fernando Muro Nov 1 '12 at 1:58
$\mathbb{Z}[t]/(t^2-1)$ is not a Dedekind domain because it is not a domain. Moreover, it is not even locally a Dedekind domain -- the localization to the prime $(2, t-1)$ is not a domain. – David Speyer Nov 1 '12 at 2:36
One can see that the set of zero divisors of $R$ can be described as $(1\pm x)\mathbb{Z}$. – Dima Pasechnik Nov 1 '12 at 3:41
Sure, I don't know what I was thinking of, I took $t^2-1$ as irreducible. – Fernando Muro Nov 1 '12 at 8:55
In the last sentence you probably mean that $C'$ is the SNF of $C$? In the case $m=n$, if $B$ or $C$ is invertible then you can reduce to the case $B$ or $C$ is the identity matrix so you're essentially looking at a free $\bf{Z}$-module $M$ equipped with an involution. These objects are completely described by the rank of the $\pm 1$-parts together with the cohomology groups $H^i(\bf{Z}/2\bf{Z},M)$ with $i=0,1$. In the general case I don't know, note that $R$ is a subring of index $2$ in $\bf{Z} \times \bz{Z}$ so I would start by looking at the SNF or HNF there. – François Brunault Nov 13 '12 at 12:26