Let $R$ be a commutative ring, $P \in M_n(R)$ and $\det(P)$ is a zero divisor of $R$. Must $P$ be a zero divisor of $M_n(R)$?
Here rings mean unital rings, $M_n(R)$ denotes the ring of square matrices over $R$ of order $n$, and zero divisor is understood to be nonzero. The difficulty lies in that the adjugate matrix of $P$ may well be $0$.
The answer is yes. First notice that it suffices to find a vector $x\neq 0$ with $Px=0$, because then the square matrix $Q$ formed by taking all of its columns to be $x$ will satisfy $PQ=0$ with $Q\neq 0$. The existence of such an $x$ is guaranteed precisely under the hypothesis of your question: See the answer at necessary and sufficient condition for trivial kernel of a matrix over a commutative ring:
A system of linear equations Ax=0 with a square n×n matrix A over a commutative ring R has a non-trivial solution if and only if its determinant (its only minor of dimension n) is annihilated by some non-zero element of R, that is, if its determinant is a zero divisor or zero.
By McCoy's theorem, the endomorphism associated to $P$ is injective if and only if $\det P$ is a non-zero divisor. Thus if $\det P$ is a zero divisor, $K=\ker P\neq\{0\}$. Consider the matrix $A$ of the projection onto $K$. Then $PA=0$.
Hint: Assume $a\det P=0$ with $a\ne 0$. Let $Q=\operatorname{diag}(a,1,\ldots,1)$. Then $\det(PQ)=0$.
