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Let $R$ and $R'$ be two rings $R＝R'$ as sets, and they are also isom as rings. I would like to know the examples, in which, operator ＋,× of $R$ and $R'$ are(completely) different, but $R$ and $R'$ are isom as rings and sets.

For example, I occurred upon $(\Bbb{Z},＋,×),(\Bbb{Z},＋,・)$, where former is $a×b＝-ab$, the latter is $a・b＝ab$.

But this is damb example, I want to know surprising example.

Pont
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  • Oh, I am sorry, you just added a new line. I find this example with $R=\Bbb Z$ and $a\oplus b=a + b - 1$ and $a\odot b=a + b - ab$ not "damb". But this would be opinion-based, I think. You can find many other examples here online (only you can chose then one which you like or you find "surprising". So try searching yourself). – Dietrich Burde Apr 12 '22 at 08:44
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    A funny little example: Let $R$ be a ring, and $f:(R,+)\to (R,+)$ a bijective morphism of the underlying additive group. Define a ring $R' = (R,)$ by $ab = f(ab)$. Then $f:R\to R'$ is an isomorphism of rings. – Erik D Apr 12 '22 at 22:45

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