1. ## Rings

Prove that R is commutative if and only if R' is commutative.

2. What is R'?

3. Originally Posted by bookie88
Prove that R is commutative if and only if R' is commutative.
Originally Posted by Tinyboss
What is R'?
I'm going to take a wild guess. Let $\displaystyle R$ be a commutative ring and $\displaystyle \phi:R\to R'$ an epimorphism. Then, $\displaystyle R'$ is commutative. To see this let $\displaystyle \phi(r),\phi(s)\in R'$ then $\displaystyle \phi(r)\phi(s)=\phi(rs)=\phi(sr)=\phi(s)\phi(r)$

4. Originally Posted by Drexel28
I'm going to take a wild guess. Let $\displaystyle R$ be a commutative ring and $\displaystyle \phi:R\to R'$ an epimorphism. Then, $\displaystyle R'$ is commutative. To see this let $\displaystyle \phi(r),\phi(s)\in R'$ then $\displaystyle \phi(r)\phi(s)=\phi(rs)=\phi(sr)=\phi(s)\phi(r)$
It is an 'if and only if', so we would need an isomorphism (epimorphism gives the 'if', monomorphism gives the 'only if').

5. Originally Posted by Swlabr
It is an 'if and only if', so we would need an isomorphism (epimorphism gives the 'if', monomorphism gives the 'only if').
Why bother with the monomorphism bit? If $\displaystyle \phi:R\to R'$ is an isomorphism then it is an epimorphism and so the above proof works, and since $\displaystyle \phi^{-1}:R'\to R$ 1is an epimorphism too the conclusion follows.

6. Originally Posted by Drexel28
Why bother with the monomorphism bit? If $\displaystyle \phi:R\to R'$ is an isomorphism then it is an epimorphism and so the above proof works, and since $\displaystyle \phi^{-1}:R'\to R$ 1is an epimorphism too the conclusion follows.
One way is as easy as the other. I explained it in that way as I had maybe thought you had meant monomorphism as opposed to epimorphism, so I had thought about both cases.