This is the definition of zero divisor in Hungerford's Algebra:

A zero divisor is an element of $R$ which is BOTH a left and a right zero divisor.

It follows a statement: It is easy to verify that a ring $R$ has no zero divisors if and only if the right and left cancellation laws hold in $R$; that is, for all $a,b,c\in R$ with $a\neq 0$, $$ab=ac~~~\text{ or }~~~ba=ca~~~\Rightarrow~~~ b=c.$$

I think it is not true. But I can't find a counterexample.

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    $\begingroup$ Given that it is a "if and only if" statement, is there a direction where you think it is not true? And why do you think it is not true? $\endgroup$ – Thomas Andrews Jan 6 '16 at 18:00
  • $\begingroup$ If $a$ is a zero divisor, then $ab=a\cdot 0$ for some non-zero $b$, so there is no cancellation. Cancellation implies no zero divisors. So does no zero-divisors imply cancellation? No cancellation implies left- and right- zero divisors, but not necessarily both. $\endgroup$ – Thomas Andrews Jan 6 '16 at 18:05

Lemma: A ring has a left (or right) zero-divisor if and only if it has a zero divisor.

Proof: Assume $ab=0$ for $a,b\neq 0$.

If $ba=0$, you are done - $a$ is both a left and right zero divisor.

If $ba\neq 0$, then $a(ba)=(ab)a=0$ and $(ba)b=b(ab)=0$, so $ba$ is a left and right zero divisor.

Now it is much easier to prove your theorem.

If $ax=ay$ and $R$ has no zero-divisors, then $a(x-y)=0$. But, by the lemma, $R$ also has no left-zero divisors, so either $a=0$ or $x-y=0$.

Similarly for $xa=ya$.

On the other hand, if cancellation is true, then $a\cdot b=0=a\cdot 0$ means that either $a=0$ or $b=0$. So there can't be any left zero divisors, and thus no zero divisors.

  • $\begingroup$ Thanks! It is surprised the existence of one-sided zero divisor implies that two-sided. $\endgroup$ – bfhaha Jan 6 '16 at 18:52
  • $\begingroup$ Yeah, surprised Hungerford leaves that out, unless this is later dealt with in an exercise. $\endgroup$ – Thomas Andrews Jan 6 '16 at 18:53

Suppose $ab = 0$ with $a, b \ne 0$. Either $ba = 0$ (which means $a$ and $b$ are zero-divisors), or $ba \ne 0$, in which case $ba$ is a zero-divisor because $a(ba) = 0$ and $(ba)b = 0$.


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