A commutative ring R is an integral domain $\iff a,b,c \in R, a \neq0, > ab=ac \implies b=c$.

I am relatively new to rings and I am struggling with shaking off the habit of taking cancellation laws of groups. So, I would like some validation for the following proof which came out to be deceptively straightforward:


Let $R$ be a commutative ring which is also an integral domain. Let $a,b,c \in R$ be s.t. $a \neq 0, ab=ac$. Then $ab=ac \implies ab-ac=ac+(-ac)\implies ab-ac=0 \implies a(b-c)=0$

By hypothesis $a\neq 0$ so $b-c=0$ since $R$ is an integral domain. Hence, $b=c$.


Let $R$ be a commutative ring s.t. the property $a,b,c \in R$ for $a \neq 0$ be s.t. $ab=ac \implies b=c$ holds.

Let $x,y \in R$ s.t. $xy=0$. $\exists p,q \in R$ s.t. $y=p+(-q)$.

Then $$ \begin{equation} xy = 0 \\ x(p+(-q)) =0 \\ x(p-q)=0 \\ xp-xq =0 \\ xp-xq +xq = 0+xq \\ xp=xq \implies p=q \implies (p-q) =0 \implies y=0 \end{equation} $$ So $R$ has no zero divisors and hence, is an integral domain. $\Box$

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    $\begingroup$ The converseseems difficult to justify, in my opinion. $p$ and $q$ come out of nowhere and are wholly unmotivated and unjustified. Why bother? If $xy=0$ and $x\neq 0$, then $xy=x0$. Go from there. (To see why they are unnecessary... what prevents you from taking $p=y$ and $q=0$)? $\endgroup$ Commented Apr 10, 2020 at 1:06
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    $\begingroup$ Hint: it's equivalent to $\ a\neq 0,\, a(b-c)=0\, \Rightarrow\, b-c = 0\ \ \ $ $\endgroup$ Commented Apr 10, 2020 at 1:12
  • $\begingroup$ @ArturoMagidin I see your point and I like your approach better. But for my clarification, can’t we still cook up such $p,q$ s.t. $y=p-q$? Just like in groups, we have $a\in G$, then $a=bc$ for some $b,c \in G$. $\endgroup$
    – Sun
    Commented Apr 10, 2020 at 1:12
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    $\begingroup$ Yes, you can, which amounts to interpolating the equivalent form I gave between the two - which makes it clearer. $\endgroup$ Commented Apr 10, 2020 at 1:14
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    $\begingroup$ @Sun: Yes, you can, as I said: take $p=y$ and $q=0$. Or take $p=0$ and $q=-y$. In fact, those may be the only possible choices (e.g., in the field of $2$ elements). So, again, what is the point? You introduce two unmotivated new variables that play absolutely no actual role in your proof. And you don’t prove they exist (and if you do... then just use the two specific ones!). $\endgroup$ Commented Apr 10, 2020 at 1:14

1 Answer 1


For the side $\Leftarrow$ I think that you prove is a little bit too much complicated :

Here is my proof :

$xy=0 \Rightarrow xy+x=x=x(y+1) $ if we assume that x is not zero then :

We can by assumption "eliminate x"


Then y is equal to zero

  • $\begingroup$ In my opinion, this is a little but too complicated. Easier to just say $xy=0\implies xy=x0$. Save yourself two steps. $\endgroup$ Commented Apr 11, 2020 at 0:45
  • $\begingroup$ @Arturo Magidin You right .:) $\endgroup$
    – user747167
    Commented Apr 11, 2020 at 9:02

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