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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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The first paragraph is an example of what is known as transport of structure. –  Arturo Magidin Jun 12 '12 at 6:13
    
Note, however, that the first paragraph is really irrelevant to your second paragraph: two rings $R$ and $S$ are isomorphic if and only if there exists a bijection $f$ such that $f(r+s) = f(r)+f(s)$ and $f(rs) = f(r)f(s)$. Applying the inverse function $f^{-1}$ to both sides of both equations we get your two displayed properties; the first displayed equation already implies that $f(0)=0$; and the fact that $f$ is onto and multiplicative implies that $f(1)$ is necessarily a unity, hence equal to $1$. –  Arturo Magidin Jun 12 '12 at 6:20
    
Thanks for these comments. I think the word shortcut was bad word choice on my part. –  Linda Cortes Jun 12 '12 at 6:31
    
(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 $f(1') = 1$.) –  Arturo Magidin Jun 12 '12 at 6:31
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up vote 1 down vote accepted

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.

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Thanks Cocopuffs. –  Linda Cortes Jun 12 '12 at 6:15
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