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


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 $x=0$).


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 $$AB=x\cdot I_{n\times n}$$ $$BA=x\cdot I_{m\times m}$$

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    $\begingroup$ If you have such a 6-tuple $(R,m,n,A,B,x)$, then you also have one $(R',m,n,A',B',y)$ for some finite local ring $R'$ and $y^2=0$, by an easy argument. $\endgroup$ – YCor Jan 18 '20 at 16:59

Note that the equality of traces yields $|m-n|x=0$, and in particular this is impossible for $|m-n|=1$.

Still it is possible for $(m,n)=(1,3)$. Let $R$ be the ring $\mathbf{F}_2[a_1,a_2,a_3,b_1,b_2,b_3]/J$, where the ideal $J$ is generated by all $a_ib_j$ for $i\neq j$ and $a_1b_1-a_2b_2$, $a_2b_2-a_3b_3$. Write $x=a_1b_1=a_2b_2=a_3b_3$ in $R$. Note that $x\neq 0$. Indeed, grading $R$ with all generators of degree 1, $J$ is graded and $J_2$ is linearly generated by the given family, which is linearly independent, and we readily see $a_1b_1\notin J_2$.

Let $A$ be the $3\times 1$ matrix $(a_1,a_2,a_3)^\dagger$ and $B$ the $1\times 3$ matrix $(b_1,b_2,b_3)$. Then $AB=xI_3$, and $BA=3xI_1=xI_1$.


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