Let $R$ be a commutative ring with unit. Let $M_A$ denote the submodule of $R^m$ generated by columns of a matrix $A$ with entries in $R$. Suppose we are given two matrices $A,B \in R^{m \times k}$. I want to know if the following statement is true:

- $M_A = M_B$ if and only if there exists an invertible matrix $P \in R^{k \times k}$ such that $A = BP$.

Note that one direction is super-easy: if $A = BP$ for invertible $P$, then the columns of both $A$ and $B$ generate the same submodule of $R^m$. So really the question is about the other direction. This is what I have tried for the "only if" direction: since $M_A = M_B$, we know that one can find matrices $P, Q \in R^{k \times k}$ such that $A = BP$ and $B = AQ$. From this we deduce that $A(I - QP)= B(I - PQ) = 0$. From here on, I don't really know what to do, but most likely this approach is just too naive.

However, I think the answer is true for a ring $R$ when $R$ is a principal ideal ring. Does anybody have a reference to this result (the particular case I want is when $R = \mathbb{Z}_d$, so a reference to this will also suffice)? More generally, I'd like to know for which rings does this property hold, and where can I find a proof of this result.

EDIT: Actually now I'm beginning to think that the result may not be true in general for PIRs. However, some special cases are true. I know the following facts:

(i) If you replace PIR by PID (principal integral domain), then the result is true.

(ii) If you have a PIR, and both $A,B$ have one row, then the result is true. Proof goes like this: Since a PIR is a Hermite ring, there exists an invertible matrices $K,K'$ such that $AK = (s,0,\dots,0)$ and $BK' = (s',0,\dots,0)$. Thus $s,s'$ generate the same ideal, and thus they must be associates (which is a property of PIRs by a result of I.Kaplansky). The result now follows.

(iii) If you are allowed to add zero columns to both $A,B$, then also the result is true, because of the existence of the Howell normal form. (see this and the references therein).

I'd like to stress that I'm interested in the case when the ring is $\mathbb{Z}_d$.

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