Let $R=\mathbb{Z}[\sqrt{-5}]$, which is well known to be a Dedekind domain but not a PID. Let $\mathrm{M}_{3}(R)$ be the set of $3\times3$ matrices over $R$. Does there exist a matrix $A\in\mathrm{M}_{3}(R)$ such that for all $g\in\mathrm{GL}_{3}(R)$ the $(1,3)$-entry of $gAg^{-1}$ is not $0$? In other words, does there exist a matrix $A\in\mathrm{M}_{3}(R)$ such that the $R$-similarity class of $A$ does not contain any matrix whose top right entry is zero?
I have asked the question in the simplest case where I don't know the answer, but one can of course generalise to $n\times n$ matrices and other rings.
If $R$ is a PID it is known that any $A\in\mathrm{M}_{3}(R)$ is $R$-similar to a matrix whose $(1,3)$-entry is zero. Namely, if $A=(a_{ij})$, $a_{ij}\in R$, then one can verify by direct computation that if $\begin{pmatrix}x & y\\ z & w \end{pmatrix}\in\mathrm{GL}_{2}(R)$ and $$ g=\begin{pmatrix}1 & 0 & 0\\ 0 & x & y\\ 0 & z & w \end{pmatrix}, $$ then the first row of $g^{-1}Ag$ is $(a_{11}\ \ \ a_{12}x+a_{13}z\ \ \ a_{12}y+a_{13}w)$. Thus, if $R$ is a PID, we can take $y,w\in R$ such that $y\gcd(a_{12},a_{13})=a_{13}$ and $w\gcd(a_{12},a_{13})=-a_{12}$ and $x,z\in R$ such that $xw-yz=1$ (which exist since $\gcd(y,w)=1$). This approach breaks down when $R$ is not a PID. On the other hand, if $R$ is an arbitrary commutative ring with identity and $(a_{12},a_{13})=(1)$, that is, if there exist $b_{1},b_{2}\in R$ such that $b_{1}a_{12}+b_{2}a_{13}=1$, then $\begin{pmatrix}x & y\\ z & w \end{pmatrix}=\begin{pmatrix}b_{1} & -a_{13}\\ b_{2} & a_{12} \end{pmatrix}$ will do. Thus, for this question, a starting point might be to consider matrices $A$ whose entries all lie in some maximal ideal of $\mathbb{Z}[\sqrt{-5}]$.
