# Assume that the ring R is isomorphic to the ring R'. Prove that if R is commutative, then R' is commutative.

Assume that the ring R is isomorphic to the ring R.
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Delorenzoz

Definition of ring isomorphism:
Let R and R' denote two ring. A mapping $\varphi :R\to {R}^{\prime }$ is a ring isomorphism from R to R' provided the following conditions hold:
1. $\varphi$ is ont-to-ont correspondence (a bijection) from R to R'.
2. $\varphi \left(x+y\right)=\varphi \left(x\right)+\varphi \left(y\right)$ for all $x\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y\in R.$
3. $\varphi \left(x\cdot y\right)=\varphi \left(X\right)\cdot \varphi \left(y\right)$ for all $x\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y\in R.$
Proof:
Let R be an integral domain
Define a map $\varphi :R\to {R}^{\prime }$ such that $\varphi \left(x\right)={x}^{\prime }$
Let ${x}^{\prime },{y}^{\prime }\in {R}^{\prime }$. Since pho is onto, there exists $x,y\in R$ such that $\varphi \left(x\right)={x}^{\prime }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\varphi \left(y\right)={y}^{\prime }$
Now ${x}^{\prime }{y}^{\prime }=\varphi \left(x\right)\varphi \left(y\right)$
Since $\varphi$ is isomophism, from the definition of isomorphism,
$=\varphi \left(xy\right)$
Since R is a commutative ring,
$=\varphi \left(yx\right)$
$=\rho \left(y\right)\varphi \left(x\right)$
$={y}^{\prime }{x}^{\prime }$
Hence if R is a commutative, then R' is commutative

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