# Can matrix factorizations be nonsquare?

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

• 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. – 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 $$BA=3xI_1=xI_1$$.