Zero divisors in a Ring

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*}

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*}

• 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? Oct 22 at 4:38
• @Bliipbluup No, no algorithm. I just thought of a simple non-commutative ring and tried the first non-zero divisors that came to mind. Oct 22 at 5:16