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Suppose $R$ is a commutative ring. Show that $R$ is an integral domain iff for all $ x, y, z\in R, xy = xz $ implies $y = z$.


Proof:

$\Rightarrow $Let $x,y,z\in R$ such that $x(y-z)=0$ where $R$ is a commutative ring. Because $R$ is an integral domain, so if $x\neq 0$, then $y-z=0$. Thus if $y-z=0$, then $xy=xz$, hence $y=z$.

$\Leftarrow$ Suppose $R$ is a commutative ring. Let $x,y,z\in R$ such that $xy=xy$ and $y=z$. Then $xy-xz=x(y-z)=0$. So, it is either $x=0$ or $y-z=0$, thus $R$ is an integral domain.


Is this right? If not, can anyone give me a hit to write better one? Thanks

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    $\begingroup$ You need to assume also that $\,x\neq 0\,$ else the claim fails (your proof of $(\Rightarrow)$ implicitly assumes such). $\endgroup$ Jun 1, 2015 at 17:28

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In first $y-z=0 \implies y=z$.

In converse, Let $xy=0$, then $xy=0=x0$ and by hypothesis it implies that $y=0$ , so no proper zero divisors, and hence an integral domain.

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