# What is a “unit of a zero-divisor” in a ring?

The following exercise is from Gallian's Contemporary Abstract Algebra:

Show that every nonzero element of ${\bf Z}_n$ is a unit of a zero-divisor.

A unit of a ring is an element which has a multiplicative inverse. A nonzero element $a$ is a commutative ring $R$ is called a zero-divisor if there is a nonzero element in $R$ such that $ab=0$.

Here is my question:

What is a unit of a zero-divisor in this exercise?

Edit:

According to the answers, this is a typo.

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@Bill: Hmm, it makes more sense. –  Jack Jul 4 '11 at 3:54
I deleted my comment and moved it to an answer, along with a link to a hint (after recalling that you were the author of the related prior question). –  Bill Dubuque Jul 4 '11 at 3:56

It should say "unit OR a zero-divisor". The proof follows immediately from your prior exercise and an application of Bezout's GCD identity to deduce that $\rm\ gcd(m,n)=1\ \Rightarrow\ m\$ is a unit in $\rm\:\mathbb Z/n\:.$

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Fair enough. Now I see. –  Jack Jul 4 '11 at 3:57

I think the question is stated wrongly. It should be:

Show that every nonzero element of $\mathbb{Z}_{n}$ is a $\text{unit or a zero-divisor}$.

Solution. Suppose that $a \in \mathbb{Z}_{n}$. If $\text{gcd}(a, n) = 1$, then we know that $a$ is a unit. Suppose that $\gcd(a, n) = d > 1$. Then $a(n/d)= (a/d)n = 0$, so $a$ is a zero-divisor.

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Why does gcd$(a,n)=1$ imply $a$ is a unit? Are you using the property of the congruence equation $ax\equiv b\pmod{n}$ that it has a unique solution (mod n) when gcd$(a,n)=1$? –  Jack Jul 4 '11 at 4:03
@Jack Hint: Bezout's identity. –  Bill Dubuque Jul 4 '11 at 4:05
@Bill: Hmm, Bezout's identity is more natural to find the inverse. I was thinking about the solution of the congruence equation $ax\equiv 1\pmod{n}$, which finally reduces to the issue of the identity you mentioned:) –  Jack Jul 4 '11 at 4:13
@Jack. Indeed. Note that Chandru's final sentence is the same as the hint I gave to your prior question. –  Bill Dubuque Jul 4 '11 at 5:31