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I recently proved to myself that if $R$ is a ring, and $R'$ a set in bijection with $R$, say by $f\colon R'\to R$, then one can turn $R'$ into a ring by defining $0'=f^{-1}(0)$, $1'=f^{-1}(1)$, $ r'+s'=f^{-1}(f(r')+f(s')),\qquad r's'=f^{-1}(f(r')f(s')), $ and then $f$ is a ring isomorphism.

Now suppose you put a new ring structure on $R$, say $(R,+,\cdot_u,0,u^{-1})$, where $a\cdot_u b=aub$. I want to use the above result as a shortcut to show $(R,+,\cdot, 0,1)$ is isomorphic to $(R,+,\cdot_u, 0,u^{-1})$ by exhibiting a bijection on $R$ which satisfies the four properties I listed above. I've had trouble thinking of what the map would look like. Does anyone see what the map would be?

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    (So using this "method" you end up doing *more* work than simply checking to see if you have a bijective ring homomorphism, which does not require checking $f(0')=0$ and $f(1') = 1$.)2012-06-12

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If $u$ is invertible, then $f: R \rightarrow R$, $r \mapsto ru$ is a bijection and satisfies the properties you want.

$f : R \rightarrow R$, $r \mapsto ur$ will also work.