Let's say $a, b$ are zero divisors in a Ring, (i.e., there exist some $x,y \in R$ s.t. $ax=0, by=0$). I feel that $a$ is a zero divisor of $xy$ (as $axy=0y=0$), but is $b$ a zero divisor of $xy$? If I take a look at $bxy$, I know I can't commute $b$ and $x$, but can $b$ be a zero divisor of $(xy)$?


Your intuition is right. Indeed, $axy=0$, but we cannot say that $bxy=0$.

For example, consider $M_2(\mathbb{Z})$, the $2\times2$ matrices with integer coefficients. We have that

\begin{align*} \begin{pmatrix} 1&0\\0&0 \end{pmatrix} \begin{pmatrix} 0&0\\1&1 \end{pmatrix}=\begin{pmatrix} 0&0\\0&0 \end{pmatrix} \end{align*}


\begin{align*} \begin{pmatrix} 0&1\\0&0 \end{pmatrix} \begin{pmatrix} 1&1\\0&0 \end{pmatrix}=\begin{pmatrix} 0&0\\0&0 \end{pmatrix} \end{align*}

It is clear that

\begin{align*} \begin{pmatrix} 1&0\\0&0 \end{pmatrix} \begin{pmatrix} 0&0\\1&1 \end{pmatrix} \begin{pmatrix} 1&1\\0&0 \end{pmatrix}=\begin{pmatrix} 0&0\\0&0 \end{pmatrix} \end{align*}


\begin{align*} \begin{pmatrix} 0&1\\0&0 \end{pmatrix} \begin{pmatrix} 0&0\\1&1 \end{pmatrix} \begin{pmatrix} 1&1\\0&0 \end{pmatrix}=\begin{pmatrix} 1&1\\0&0 \end{pmatrix} \end{align*}

  • 1
    $\begingroup$ Beautiful. Is there an algorithmic way to finding such counter-examples, or did you just look for a small non-commutative ring and try things out? $\endgroup$
    – bliipbluup
    Oct 22 at 4:38
  • 1
    $\begingroup$ @Bliipbluup No, no algorithm. I just thought of a simple non-commutative ring and tried the first non-zero divisors that came to mind. $\endgroup$
    – Bonnaduck
    Oct 22 at 5:16

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