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Let $S\subseteq R$, where $R$ a commutative ring. Write $A\in S^{n\times n}(r)$ to mean that $A$ is an $n\times n$ matrix of rank $r$ with entries in $S$, with the rank understood in the sense of minors. Let $S^{n\times m}$ denote the set of all $n\times m$ matrices with entries in $S$.

Let us say that a matrix $A\in S^{n\times n}(r)$ has the full-rank factorization property (FRFP) over $S$ if $A$ can represented as the product $XY$ of matrices $X\in S^{n\times r}$ and $Y\in S^{r\times n}$.

If $R$ is a field, then it is easy to see that any $A\in R^{n\times n}(r)$ has the FRFP over $R$. E.g., let the columns of $X$ constitute a basis of the column space of $A$ and let the entries of $Y$ be the coefficients of the columns of $A$ in that basis.

Concerning the case $r=1$: It is not hard to see that any $A\in\N^{n\times n}(1)$ has the FRFP over $\N:=\{1,2,\dots\}$, as shown in Proposition 1. This result and its proof remain valid if $\N$ is replaced by $\Z\setminus\{0\}$ or $R\setminus\{0\}$, where $R$ is any unique factorization domain.

The case $r\ge2$ seems quite a bit more complicated in general; in what follows, let us indeed assume $r\ge2$. It is obvious that no matrix $A\in\N^{n\times n}(r)$ with at least one entry equal $1$ can have the FRFP over $\N$ -- because for all $X\in\N^{n\times r}$ and $Y\in\N^{r\times n}$ all the entries of $XY$ are $\ge2$. Also, for $\N_0:=\{0,1,\dots\}$, it is not hard to see that matrix $A=\left( \begin{array}{ccc} 3 & 2 & 0 \\ 5 & 4 & 4 \\ 0 & 1 & 6 \\ \end{array} \right)\in\N_0^{3\times3}$ does not have the FRFP over $\N_0$.

However, I have not been able to find a matrix $A\in\Z^{n\times n}(r)$ which does not have the FRFP over $\Z$.

So, a question is: Does such a matrix exist? More generally, are there good necessary and/or sufficient conditions on $S$ in order for all matrices $A\in S^{n\times n}(r)$ to have the FRFP?