Rings, ID, and Fields

• Mar 24th 2010, 10:13 AM
bookie88
Rings, ID, and Fields
Assume that the ring R is isomorphic to the ring R'. Prove that if R is commutative, then r' is commutative.
• Mar 24th 2010, 10:24 AM
Haven
Let $\phi : R \rightarrow R'$
If R is commumative then
$\forall a,b \in R: ab = ba$
$\Rightarrow \phi(ab) = \phi(ba)$
Also $\phi(ab) = \phi(a)\phi(b)$ and $\phi(ba) = \phi(b)\phi(a)$
and $\phi$ is bijective.

This should be enough for you to see why that R' is communative
• Mar 24th 2010, 10:31 AM
hatsoff
Quote:

Originally Posted by bookie88
Assume that the ring R is isomorphic to the ring R'. Prove that if R is commutative, then r' is commutative.

Let $\phi:R\to R'$ denote an isomorphism, and $x,y\in R'$. Then there are $a,b\in R$ with $\phi(a)=x$ and $\phi(b)=y$. Furthermore, $xy=\phi(a)\phi(b)=\phi(ab)=\phi(ba)=\phi(b)\phi(a) =yx$. So $R'$ is commutative.

EDIT: Looks like someone beat me to it. Oh well.