Let $A$ be an $n\times n$-matrix over $\mathbb Z[t^\pm]$. In general doesn't exist $P,Q\in GL(n,\mathbb Z[t^\pm])$ such that $PAQ$ is a diagonal matrix (this happens cause $\mathbb Z[t^\pm]$ is not a princiapl domain). On the other hand, clearly exist $P,Q\in Sl(n,\mathbb Q[t^\pm])$ such that $PAQ$ is a diagonal matrix. Therefore taking an apropiate number $m$ we have that $P,Q\in Gl(n,\mathbb Z_{(m)}[t^\pm])$, clearly we can suppose that this element $m$ is square-free. Let $S(A)$ be the set of all elements $m\in\mathbb N$ with this property and $T(A):=\{m\in S(A):\not\exists d\in S(A)\text{ such that } d\mid m, d\not=m\}$.
Question: Is $T(A)$ a finite set? There is an easy way to compute it?
For example, if $A$ diagonal then $T(A)=\{1\}$
