Is the determinant of a zero divisor zero? Suppose that $A$ is a zero divisor in the ring of $(n\times n)$-matrices over the ring $R$.
Is $\det(A) =0$ if $R$ is a field?
Is $\det(A) =0$ if $R$ is an integral domain?
It's not necessarily true if $R$ has zero-divisors. Take for example $R=\mathbf{Z}/2^{n+1}\mathbf{Z}$ and $A=\textrm{diag}(2,2,\ldots,2)$. Then $A^{n+1}=0$ but $\det A = 2^n \neq 0$. 
Of course, if $R$ is a reduced ring and $A$ is nilpotent, we have that $\det(A) =0$. In fact, $A^m=0$ for some $m>0$. Thus, $\det(A)^m = \det(A^m) =0$. Thus $\det(A) $ is nilpotent in $R$. Therefore, we have that $\det(A) =0$.
 A: For a field, $A\neq 0$ is a zero divisor if and only if it is not invertible.
If $A$ is not invertible, then let $P$ be the projection onto the eigenspace of $A$ corresponding to $0$. Then $P\neq 0$, since the eigenspace is nontrivial, but $AP=0$. And letting $W$ be any complement to $\mathrm{Im}(A)$ (which is nontrivial by the Rank-Nullity Theorem), and letting $Q$ be the projection onto $W$ along $\mathrm{Im}(A)$ gives $QA=0$, so $A$ is a zero divisor. Conversely if $A$ is a zero divisor, then it is not a unit.
Thus, $A$ is a zero divisor if and only if $A$ is not invertible, if and only if $\det(A)=0$.
For $R$ an integral domain, look at $A$ as a matrix over $Q(R)$, the fraction field of $R$. It's still a zero divisor over $Q(R)$, hence its determinant is zero by the argument above.
Added. The converse also holds over an integral domain (it holds in fields, as noted above): if $\det(A)=0$, then working over $Q(R)$ you can find a matrix $P$ and a matrix $Q$ with coefficients in $Q(R)$ such that $QA = AP = 0$. Though these matrices may not lie in $R$, multiplying them by a suitable scalar will make them lie in $R$: just express each entry as a fraction $\frac{a}{b}$ with $a,b\in R$, and multiply by the product of all denominators. Though there may not be unique expressions (even unique reduced expressions), simply pick one expression for each entry and use it. 
A: I assume the ring $R$ is commutative.
The formula $A.A^{adj}=A^{adj}.A=det(A).Id$ shows that if $det(A)\in R^\times$, that is if $det(A)$ is invertible in $R$, then the matrix $A$ will be invertible and thus certainly will not be a zero-divisor in $M_n(R)$.
By contraposition we deduce that if $A$ is a zero divisor in $M_n(R)$ and if $R$ is a field, then $det(A)=0$. 
If $R$ is only a domain and $A$ is a zero divisor in $M_n(R)$, the matrix $A$ will a fortiori be a zero divisor in $M_n(Frac(R))$, and so again $det(A)=0$ .
(Be sure to notice that the determinant of $A$ is the same whether you consider that $A$ has coefficients in $R$ or in $Frac(R)$ !)
