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Math Help - example of a noncommutative ring R for which R is not isomorphic to R^op

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    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.
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    Quote Originally Posted by dori1123 View Post
    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}.
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