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**dori1123** Give an example of a noncommutative ring $\displaystyle R$ such that $\displaystyle R$ and the opposite ring $\displaystyle R^{op}$ are not isomorphic.

I am thinking about the ring $\displaystyle Mat_n(R)$, where $\displaystyle R$ is any ring. But I have trouble showing that there is no isomorphism between $\displaystyle Mat_n(R)$ and $\displaystyle [Mat_n(R)]^{op}$. Some help please.

before giving you an example, i should explain the case $\displaystyle S=Mat_n(R).$ define $\displaystyle f: S \rightarrow Mat_n(R^{op})$ by $\displaystyle f(A)=A^{T}.$ see that $\displaystyle f$ is a ring isomorphism. so, if $\displaystyle R$ is commutative, then $\displaystyle S \cong S^{op}.$

an example of a ring $\displaystyle R$ for which $\displaystyle R \ncong R^{op}$ is the Klein four ring, i.e. $\displaystyle R=\{0,e_{11},e_{21},e_{11}+e_{21} \} \subset Mat_2(\mathbb{Z}/2\mathbb{Z}).$ here is why:

$\displaystyle e_{21}$ is the only non-zero nilpotent element of $\displaystyle R$ and thus if $\displaystyle g: R \rightarrow R^{op}$ is an isomorphism, then $\displaystyle g(e_{21})=e_{21}.$ thus $\displaystyle g(e_{11})= e_{11} \ \text{or} \ e_{11}+e_{21}$ and so we'll get the following contradiction:

$\displaystyle 0=g(e_{11}e_{21})=g(e_{21})g(e_{11})=e_{21}g(e_{11 })=e_{21}.$