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](https://mathoverflow.net/questions/51942/analogue-of-smith-normal-form-for-matrices-over-mathbb-zt) 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 $B'$. 



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




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[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. 


