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Show that in the ring $\mathbb Z_n$ the divisors of zero are precisely those elements that are not relatively prime to $n$.

This is one of the problems in my quiz, I had proof but it's incomplete.

Proof: let $m$ be the element of $Z_n$ and $m\neq0$, and let the gcd of $m$ and $n$ be not equal to 1, say $d=\gcd(m,n) \neq1$. Then,

$m(\frac nd) = (\frac md)n$

thus $(\frac md)n$ gives 0 as a multiple of $n$. Hence $m(\frac nd) = 0$ in $Z_n$ this shows that neither $m$ nor $\frac nd$ is equal to zero, so $m$ is a 0 divisor.

This is where my proof ends, but my prof. was asking about the case when $\gcd(m,n) = 1$. I don't know what to do next I hope you could help me.

user26857
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sakura
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  • If $m$ and $n$ are relatively prime, then $gcd(m,n)=1$. So what you are showing is that if $d:=gcd(m,n)\not= 1 $, then it's a zero divisor. I assume your question should be, show that an element of $m$ of $\mathbb{Z}_n$ is relatively prime to $n$ iff $m$ is NOT a zero divisor in $\mathbb{Z}_n$. Or equivalently, $m$ is not relatively prime to $n$ iff $m$ is a zero divisor in $\mathbb{Z}_n$. – sqtrat Feb 22 '16 at 13:02
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    I feel obligated to point out that this proof is almost word for word identical to that in John B. Fraleigh's "A First Course in Abstract Algebra". – Dastur Aug 15 '19 at 03:10

2 Answers2

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If $\gcd(m, n) = g$, by Bézout's identity there are integers $r, s$ such that:

$\begin{align} r m + s n = g \end{align}$

Multiply by the integer $n / g$ to get:

$\begin{align} m \frac{r n}{g} + s \frac{n^2}{g} = n \end{align}$

This, interpreted modulo $n$ is:

$\begin{align} m \cdot \frac{r n}{g} \equiv 0 \pmod{n} \end{align}$

If $g \ne 1$, $r n / g \not\equiv 0 \pmod{n}$, and $m$ is a zero divisor.

If $g = 1$, Bezóut's identity modulo $n$ is:

$\begin{align} r m \equiv 1 \pmod{n} \end{align}$

and $r$ is an inverse for $m$, and $m$ can't be a zero divisor (in a finite ring, it is easy to prove that an element is a unit, i.e., has an inverse, or is a zero divisor).

vonbrand
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If $\gcd(m,n)=1$, then there exists a $k$ such that $km\equiv1\mod n$, and therefore $km$ cannot be a zero divisor.

vonbrand
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Daniel Robert-Nicoud
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