Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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.

share|cite|improve this answer
Thanks Cocopuffs. – Linda Cortes Jun 12 '12 at 6:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.