First, a disclaimer: This is a repost of a question I asked on stackexchange (no answer there).
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:
$$A=\begin{pmatrix} 1 & 1 & 1 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \end{pmatrix} \qquad \mathrm{implies} \qquad \mathrm{classical\;adjoint} = 0 $$
(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).
5) This is somewhat related to the question: Rings in which every non-unit is a zero divisor
6) 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):
$$ A = \begin{pmatrix} a_{11} & a_{12} \cr a_{21} & a_{22} \end{pmatrix} \qquad \Longrightarrow \qquad \mathrm{classical\;adjoint} = C = \begin{pmatrix} a_{22} & -a_{12} \\ -a_{21} & a_{22} \end{pmatrix} $$
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.
7) Apparently strange behavior can occur when $R$ is non-commutative (not surprising). Like a matrix can be both a left inverse and left zero divisor. [The determinant keeps this from happening in the commutative case.]
Best Answer
The answer given by David Speyer can be strengthened as follows. If $A$ is a non-invertible $n\times n$ matrix with entries in $R$ as described in the problem, then the linear maps $R^n \to R^n$ defined by either left or right multiplication are non-injective. In particular, $A$ is both a left-zero-divisor and a right-zero-divisor.
This is a consequence of McCoy's rank theorem. You can find a nice, brief account of this in Section 2 of this paper by Kodiyalam, Lam, and Swan. One consequence of the theorem is that for any commutative ring $R$, a square matrix $A$ over $R$ has linearly independent columns if and only if its determinant is not a zero-divisor, if and only if its rows are linearly independent.
So if every element of $R$ is either invertible or a zero-divisor, this means that every square matrix over $R$ defines a linear transformation that is either invertible or non-injective.