13
$\begingroup$

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 $B'$.

Improve this question
$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$
    – Fernando Muro
    Commented Nov 1, 2012 at 1:58
  • 3
    $\begingroup$ $\mathbb{Z}[t]/(t^2-1)$ 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. $\endgroup$
    – David E Speyer
    Commented Nov 1, 2012 at 2:36
  • $\begingroup$ One can see that the set of zero divisors of $R$ can be described as $(1\pm x)\mathbb{Z}$. $\endgroup$
    – Dima Pasechnik
    Commented Nov 1, 2012 at 3:41
  • $\begingroup$ Sure, I don't know what I was thinking of, I took $t^2-1$ as irreducible. $\endgroup$
    – Fernando Muro
    Commented Nov 1, 2012 at 8:55
  • $\begingroup$ 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. $\endgroup$
    – François Brunault
    Commented Nov 13, 2012 at 12:26

1 Answer 1

Reset to default
13
$\begingroup$

There is a general concept of Hermite Normal Form developed by Kaplansky [1] for associative rings with identity. His results were revived in [Appendix to §I.4 and Notes on Chapter I, 3] and [4]. (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 $n$ [2]. 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.

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

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .