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Do you know rings without involutions, but auto-anti-isomorphic (isomorphic to their opposite)? In that case, what is the minimal example?

If a ring has an involution f, then f is an anti-automorphism; but there can be auto-anti-isomorphic ring without involutions.

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    $\begingroup$ By involution you probably mean an involutive self-map that is an anti-automorphism. So you want the existence of an anti-automorphism, but none being an involution (=being of order 1 or 2). $\endgroup$
    – YCor
    Mar 7, 2021 at 12:44
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    $\begingroup$ I don't see right away if any central simple algebra over $\mathbf{Q}$ might do the job. $\endgroup$
    – YCor
    Mar 7, 2021 at 12:45
  • $\begingroup$ 2x2 upper triangular matrices over a field are isomorphic to their opposite. I don't know if an obvious involution but maybe there is one $\endgroup$
    – Benjamin Steinberg
    Mar 7, 2021 at 13:04
  • $\begingroup$ I suppose you exclude commutative rings... $\endgroup$
    – Laurent Moret-Bailly
    Mar 7, 2021 at 14:29
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    $\begingroup$ This is answered here math.stackexchange.com/questions/3797004/… with a division algebra $\endgroup$
    – Benjamin Steinberg
    Mar 7, 2021 at 15:36

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