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

Let $$R$$ be a commutative ring (with $$1$$) and $$R^{n \times n}$$ be the ring of $$n \times n$$ matrices with entries in $$R$$.

In addition, suppose that $$R$$ 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 $$R^{n \times n}$$ a zero divisor or a unit as well?

We know that if $$A \in R^{n \times n}$$, then $$AC=CA=\mathrm{det}(A)I_n$$ where $$C$$ is the classical adjoint of $$A$$ and $$I_n$$ is the identity matrix.

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

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

Things to consider:

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

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

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

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

4) Of course the assumption that every non-zero element of $$R$$ is either a zero divisor or unit is necessary since otherwise take a non-zero, non-zero divisor, non-unit element $$r$$ and construct the diagonal matrix $$D = \mathrm{diag}(r,1,\dots,1)$$ (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 $$n=1$$ and $$n=2$$. It is true for $$n=1$$ by assumption on $$R$$. To see that $$n=2$$ is true notice that the classical adjoint contains the same same elements as that of $$A$$ (or negations):

$$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \qquad \Longrightarrow \qquad \mathrm{classical\;adjoint} = C = \begin{bmatrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{bmatrix}$$

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

• I am going to try reposting this on Mathoverflow. Commented Oct 11, 2011 at 13:36
• Thanks for catching the typo. Commented Dec 4, 2018 at 17:59

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.
For an artinian ring every element is either a unit or a zero divisor (yo can see this in any book about theory of rings). Moreover, if $R$ is an artinian ring, then $M_n(R)$ is also artinian. So, there are elements in $M_n(R)$ that are neither units nor zero divisor if the ring $R$ is not artinian.