Question

Let R =
(Z Z2)
(0 Z2).
Show that every right 0-divisor in R is a left 0-divisor.

Answer 1

Suppose $\alpha = \begin{pmatrix} m & \overline{a} \\ 0 & \overline{n} \end{pmatrix}$ is not a left 0-divisor. Then $m$ must be odd, for otherwise $\begin{pmatrix} m & \overline{a} \\ 0 & \overline{n} \end{pmatrix}$ is right annihilated by $\begin{pmatrix} 0 & \overline{1} \\ 0 & 0 \end{pmatrix}$ . In addition, we must have $\overline{n} = \overline{1}$ , for otherwise $\overline{n} = 0$ , and $\alpha$ would be right annihilated by $\begin{pmatrix} 0 & \overline{a} \\ 0 & \overline{1} \end{pmatrix}$ since $m\overline{a} + \overline{a} \in 2\mathbb{Z} \cdot \overline{a} = 0$ . But then $\alpha$ is also not a right 0-divisor, for $0 = \begin{pmatrix} x & \overline{z} \\ 0 & \overline{y} \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & \overline{1} \end{pmatrix} = \begin{pmatrix} 2x & \overline{z} \\ 0 & \overline{y} \end{pmatrix} \Rightarrow x = 0, \overline{y} = \overline{z} = 0$