Question

Let R =
(Z Z2)
(0 Z2).
Show that R is not isomorphic to its opposite ring R^op

Answer 1

First we notice the fact, that every right 0-divisor in $R$ is a left 0-divisor. If $R \sim R^{\wedge}$ op, mentioned fact would imply that every left 0-divisor in $R$ is a right 0-divisor. Now

$\beta = \left( \begin{array}{ll}2 & 0\\ 0 & \overline{1} \end{array} \right)$ is a left 0-divisor since $\left( \begin{array}{ll}2 & 0\\ 0 & \overline{1} \end{array} \right)\left( \begin{array}{ll}0 & \overline{1}\\ 0 & 0 \end{array} \right) = 0$ , but

$0 = \left( \begin{array}{ll}x & \overline{z}\\ 0 & \overline{y} \end{array} \right)\left( \begin{array}{ll}2 & 0\\ 0 & \overline{1} \end{array} \right) = \left( \begin{array}{ll}2x & \overline{z}\\ 0 & \overline{y} \end{array} \right)\Rightarrow x = 0,y = z = 0$

shows that $\beta$ is not a right 0-divisor—a contradiction.