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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 $0$.

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closed as off-topic by user26857, Lee Mosher, dustin, TMM, graydad Mar 15 '15 at 2:19

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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.

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  • $\begingroup$ A smart proof, thanks a lot. $\endgroup$ – Censi LI Mar 15 '15 at 9:28
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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 $PA=0$.

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Hint: Assume $a\det P=0$ with $a\ne 0$. Let $Q=\operatorname{diag}(a,1,\ldots,1)$. Then $\det(PQ)=0$.

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  • $\begingroup$ What is this hint supposed to prove? I can't see any reason to get $P$ a zero-divisor from $\det(PQ)=0$. (-1) $\endgroup$ – user26857 May 17 '15 at 9:19

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