# 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. Nov 1, 2012 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. Nov 1, 2012 at 2:36
• One can see that the set of zero divisors of $R$ can be described as $(1\pm x)\mathbb{Z}$. Nov 1, 2012 at 3:41
• Sure, I don't know what I was thinking of, I took $t^2-1$ as irreducible. Nov 1, 2012 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. Nov 13, 2012 at 12:26

There is a general concept of Hermite Normal Form developed by Kaplansky  for associative rings with identity. His results were revived in [Appendix to §I.4 and Notes on Chapter I, 3] and . (A quick Google search shows that other recent publications revolves around Kaplansky's definition.)

Let us suppose rings commutative for the sake of simplicity. A commutative ring $R$ with identity is said to be an elementary divisor ring if every matrix $A$ over $R$ admits a diagonal reduction, i.e., there are invertible matrices $P$, $Q$ over $R$ and elements $d_i \in R$ such that $PAQ = \operatorname{diag}(d_1, \dots, d_n)$ with $d_1 \, \vert \cdots \vert \, d_n$. A commutative ring $R$ with identity is a Hermite ring in the sense of Kaplansky, or concisely a K-Hermite ring (this is T. Y. Lam's naming), if every $1$-by-$2$ matrix admits a diagonal reduction, i.e., if for every $(a, b) \in R^2$ we can find an invertible matrix $Q$ such that $(a, b)Q = (d, 0)$ for some $d \in R$. It should be clear that a K-Hermite ring $R$ is a Bézout ring, i.e., the finitely generated ideals of $R$ are principal.

The ring $R = \mathbb{Z}[t]/(t^2 -1) = \mathbb{Z}[C_2]$ is not a Bézout ring since the image of the ideal $(2, t - 1)$ is not principal. Therefore we cannot expect matrices over $R$ to have a diagonal reduction in the sense of Kaplansky. However $R$ is a generalized Euclidean ring in the sense of P. M. Cohn, i.e., $SL_n(R)$ is generated by transvections for every $n$ . This fact is established using one of the obvious embeddings of $R$ into $\mathbb{Z}^2$ and some strong Euclidean property of $\mathbb{Z}$, that is $\mu(\mathbb{Z}) = \frac{1}{2}$, see [Lemma 4.1, 2]. This could be a starting point to study possible "nice" reduced forms for matrices over $R$. For instance, it is not difficult to show that the following holds: for every $(a, b) \in R^2$ there exists $E \in SL_2(R)$ such that $(a, b)E = (d, d')$ with $d d' = 0$; in addition we can take $(d, d') = (1, 0)$ if $(a, b)$ generates $R$.

In contrast, $\mathbb{Z}^2$ is trivially an elementary divisor ring.

 "Elementary divisors and modules", I. Kaplansky, 1949.
 "Generalized euclidean group rings", K. Dennis et al., 1984.
 "Serre's problem on projective modules", T. Y. Lam, 2006.
 "Euclidean pairs and quasi-Euclidean rings", A. Alahmadi et al., 2014.