# Zero divisors in matrix rings [closed]

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 $0$.

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, Lee Mosher, dustin, TMM, graydad
If this question can be reworded to fit the rules in the help center, please edit the question.

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