# Do these matrix rings have non-zero elements that are neither units nor zero divisors?

Let $$RR$$ be a commutative ring (with $$11$$) and $$Rn×nR^{n \times n}$$ be the ring of $$n×nn \times n$$ matrices with entries in $$RR$$.

In addition, suppose that $$RR$$ is a ring in which every non-zero element is either a zero divisor or a unit [For example: take any finite ring or any field.] My question:

Is every non-zero element of $$Rn×nR^{n \times n}$$ a zero divisor or a unit as well?

We know that if $$A∈Rn×nA \in R^{n \times n}$$, then $$AC=CA=det(A)InAC=CA=\mathrm{det}(A)I_n$$ where $$CC$$ is the classical adjoint of $$AA$$ and $$InI_n$$ is the identity matrix.

This means that if $$det(A)\mathrm{det}(A)$$ is a unit of $$RR$$, then $$AA$$ is a unit of $$Rn×nR^{n \times n}$$ (since $$A−1=(det(A))−1CA^{-1}=(\mathrm{det}(A))^{-1}C$$). Also, the converse holds, if $$AA$$ is a unit of $$Rn×nR^{n \times n}$$, then $$det(A)\mathrm{det}(A)$$ is a unit.

I would like to know if one can show $$0≠A∈Rn×n0 \not= A \in R^{n \times n}$$ is a zero divisor if $$det(A)\mathrm{det}(A)$$ is zero or a zero divisor.

Things to consider:

1) This is true when $$R=FR=\mathbb{F}$$ a field. Since over a field (no zero divisors) and if $$det(A)=0\mathrm{det}(A)=0$$ then $$Ax=0Ax=0$$ has a non-trivial solution and so $$B=[x|0|⋯|0]B=[x|0|\cdots|0]$$ gives us a right zero divisor $$AB=0AB=0$$.

2) You can’t use the classical adjoint to construct a zero divisor since it can be zero even when $$AA$$ is not zero. For example:

$$A=[111000000]impliesclassicaladjoint=0A=\begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \qquad \mathrm{implies} \qquad \mathrm{classical\;adjoint} = 0$$
(All $$2×22 \times 2$$ sub-determinants are zero.)

3) This is true when $$RR$$ is finite (since $$Rn×nR^{n \times n}$$ would be finite as well).

4) Of course the assumption that every non-zero element of $$RR$$ is either a zero divisor or unit is necessary since otherwise take a non-zero, non-zero divisor, non-unit element $$rr$$
and construct the diagonal matrix $$D=diag(r,1,…,1)D = \mathrm{diag}(r,1,\dots,1)$$ (this is non-zero, not a zero divisor, and is not a unit).

Edit: Not totally unrelated…
https://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor

Edit: One more thing to consider…

5) This is definitely true when $$n=1n=1$$ and $$n=2n=2$$. It is true for $$n=1n=1$$ by assumption on $$RR$$. To see that $$n=2n=2$$ is true notice that the classical adjoint contains the same same elements as that of $$AA$$ (or negations):

$$A=[a11a12a21a22]⟹classicaladjoint=C=[a22−a12−a21a11] A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \qquad \Longrightarrow \qquad \mathrm{classical\;adjoint} = C = \begin{bmatrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{bmatrix}$$

Thus if $$det(A)b=0\mathrm{det}(A)b=0$$ for some $$b≠0b \not=0$$, then either $$bC=0bC=0$$ so that all of the entries of both $$AA$$ and $$CC$$ are annihilated by $$bb$$ so that $$A(bI2)=0A(bI_2)=0$$ or $$bC≠0bC \not=0$$ and so $$A(Cb)=det(A)bI2=0I2=0A(Cb)=\mathrm{det}(A)bI_2 =0I_2=0$$. Thus $$AA$$ is a zero divisor.

As you’ve demonstrated in 1), the question boils down to when $Ax=0$ has a non-trivial solution. It turns out that this is the case if and only if $\det A$ is a zero divisor. I’ve written this up in a separate post because it’s of interest in its own right: necessary and sufficient condition for trivial kernel of a matrix over a commutative ring. It follows that the answer to your question is yes, $A$ is a zero divisor if and only if $\det A$ is a zero divisor, and thus $R^{n\times n}$ inherits from $R$ the property that all non-zero elements are either units or zero divisors.