In rings left and right cancellation laws generally don't hold.

can anyone generalize some cases so that we are ensured when the cancellation laws hold in rings?(the case I found was in Integral domains they hold.)

  • $\begingroup$ That's just the class of "noncommutative domains" $\endgroup$
    – rschwieb
    Commented Sep 30, 2014 at 17:56

2 Answers 2


Suppose that $ab = ac$ implies $b = c$ for non-zero $a$. Then $a(b-c) = 0$ implies $b-c = 0$. Suppose $a$ is a left zero divisor, i.e. $ad = 0$ for some non-zero $d$. Then taking $b = c+d$ gives $a(b-c) = ad = 0$. So $c = b = c+d$, so $d = 0$, which is a contradiction. So left cancellation implies that the ring has no zero divisors.

Similiarly, right cancellation also implies that the ring has no zero divisors.

Conversely, if a ring has no zero divisors (i.e. it is a domain) then $ab = ac$ ($a \neq 0$) implies $a(b-c) = 0$ implies $b - c = 0$ as $a$ is non-zero and there are no zero divisors, hence $b - c$. Similiarly right cancellation holds.

Conclusion: the following are equivalent for a ring $R$: $R$ has no zero divisors; the left cancellation law holds in $R$; the right cancellation law holds in $R$.

  • 2
    $\begingroup$ :Your choice of $b,b=c+d$ does'nt seems to be honest as just before this assumption you chose $d$ to be non-zero,while just after this you chose $d=b-c$,which is zero. $\endgroup$
    – Picaso
    Commented Sep 2, 2017 at 18:11

Let $R $ be a ring with cancellation laws holding.

Let $a,b\in R$.

Now,if $ab=ac$ then by left cancellation law we get,$b=c$.

Similarly,if $ba=ca$ then by right cancellation law we get,$b=c$.

Let $ab=0$ then if $a\neq 0$.So,$ab=a0,$by left cancellation law $b=0$.

Therefore,if $ab=0$ then either $a=0$ or $b=0$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .