I originally asked this question on math stackexchange, but got no attention. As such, I am asking here as well.

$\textbf{Background}$

On page 49 of a 1980 paper by Eisenbud, he defines matrix factorizations. Let $R$ be a (unital) commutative ring. Given $x\in R$, Eisenbud defines a matrix factorization for $x$ as a pair of $R$-linear maps $(\varphi:F\to G,\psi:G\to F)$ where $G$ and $F$ are free $R$-modules such that $\varphi\circ\psi=x\cdot id_G$ and $\psi\circ\varphi=x\cdot id_F$. Note that some authors require in their definition that $\text{rank}(F)=\text{rank}(G)$, but I am trying to work with Eisenbud's definition. In the case that $F$ and $G$ have finite rank, the definition can be stated equivalently with matrices instead of $R$-linear maps.

It's not hard to see that $F$ and $G$ must have the same rank if $x$ is a non-zero divisor of $R$ since $x$ being a non-zero divisor implies $\varphi$ and $\psi$ must be injective. Eisenbud shows in Corollary 5.4 of his paper that $\text{rank}(F)=\text{rank}(G)$ whenever $R$ is Noetherian and $(x)/(x^2)$ is free over $R/(x)$ (although I think he is missing an assumption here since this appears to be false when $x=0$).

$\textbf{Question}$

Eisenbud does not include any examples of matrix factorizations with free modules of different ranks in his paper, nor have I encountered them elsewhere. As such, I am looking for an example of a matrix factorization for a nonzero element of a ring using free modules of different ranks. By the earlier statements, the matrix factorization would need to be for a zero divisor, $x\in R$, such that either $R$ is not Noetherian, or $(x)/(x^2)$ is not free over $R/(x)$. If possible, I would prefer for $R$ to be Noetherian and for the matrix factorization to be between modules of finite rank, so that the maps can be expressed as matrices.

Written in terms of matrices, I am looking for a Noetherian ring, $R$, a nonzero element, $x\in R$, an $n\times m$ ($n\neq m$) matrix, $A$, and an $m\times n$ matrix, $B$, such that $$AB=x\cdot I_{n\times n}$$ $$BA=x\cdot I_{m\times m}$$