# Thread: example of a noncommutative ring R for which R is not isomorphic to R^op

1. ## example of a noncommutative ring R for which R is not isomorphic to R^op

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

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

2. Originally Posted by dori1123
Give an example of a noncommutative ring $R$ such that $R$ and the opposite ring $R^{op}$ are not isomorphic.

I am thinking about the ring $Mat_n(R)$, where $R$ is any ring. But I have trouble showing that there is no isomorphism between $Mat_n(R)$ and $[Mat_n(R)]^{op}$. Some help please.
before giving you an example, i should explain the case $S=Mat_n(R).$ define $f: S \rightarrow Mat_n(R^{op})$ by $f(A)=A^{T}.$ see that $f$ is a ring isomorphism. so, if $R$ is commutative, then $S \cong S^{op}.$

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

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

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