Please check my proof

$\Leftarrow $ Suppose cancelation holds in $R$, but $R$ is not integral domain. There must exist zero divisor in $R$ We supoose $a$ is zero divisor and $b\neq c$ then suppose


cancel a form both side $b=c$ that is false then the if cancellation hold in $R$, $R$ can't has zero divisor or it's integral domain

$\Rightarrow $ Because $R$ is integral domain, there are no zero divisor in $R$

Consider $c=d$


cancel $a$ from both sides


that is true then cancellation holds if it is integral domain

  • 2
    $\begingroup$ How come "suppose" $\;ab=bc\;$ ?? Well, suppose not ...! This doesn't look right. Can you come up with a nice alternative? Your proof begins correctly. The other direction seems to have missed the point completely. $\endgroup$
    – DonAntonio
    Commented Dec 10, 2016 at 11:14

2 Answers 2


Take $a \in R$ and consider $\mu : R \to R$ given by $\mu(x)=ax$.

Then $\mu$ is a homomorphism of the additive group of $R$.

By definition, $a$ is a zero divisor iff $\ker\mu \ne 0$. Therefore, $R$ is a domain iff $\ker\mu = 0$ for all $a\ne0$.

It is well known that $\ker\mu = 0$ is equivalent to $\mu$ being injective, which is just another way of expressing cancellation.



$\;\Longleftarrow\;$ Suppose cancelation holds but $\;a\neq0\;$ is a zero divisor. Then there exists $\;b\neq0\;$ such that

$$0=a\cdot0=a\cdot b\;\;\ldots\text{and now finish}$$

$\;\implies\;$ Supose $\;R\;$ is an ID, and also that $\;a\neq0\;$ , so if

$$ab=ac\implies a(b-c)=0\;,\;\;\text{but...}$$

  • $\begingroup$ ok,thank for a hint. I will try to do itt $\endgroup$
    – Lingnoi401
    Commented Dec 10, 2016 at 11:21

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