5
$\begingroup$

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 $(xy)$?

$\endgroup$
5
$\begingroup$

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*}

and

\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*}

but

\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*}

$\endgroup$
2
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.