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In the ring $\mathbb Z_n$, the divisors of zero are precisely those elements $m\in \mathbb Z_n$ such that $(m,n) > 1$.

Proof:

Let $d = (m,n)$ and note that $$m\cdot \frac nd = \frac md \cdot n \equiv 0 \ \ \ \ (\text{mod } n).$$

If $d > 1$, then $0 \not\equiv n/d \pmod n$ so that $m$ is a zero divisor. Conversely, if $d = 1$ and $ma \equiv 0\ (\text{mod }p)$, then $n|ma$ so that necessarily $n|a$. This implies $a ≡ 0 \ (\text{mod }p)$ so that $m$ is not a zero divisor.

Please help explain the proof in a more lucid language.

Asaf Karagila
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Shubham Goyal
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1 Answers1

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Well, you've got a $p$ sneaking in where you want $n$.

Then there are two parts to the argument based on whether $d = \gcd(m,n)$ is $1$ or more than $1$.

If $d > 1$, then you get a zero divisor because the product of $m$ and $n/d$ is $0$ in $\mathbb{Z}_n$ (and both $m$, $n/d < n$).

If $d = 1$, then let's look at a possible equation in $\mathbb{Z}_n$ showing that a number $m$ is a zero divisor, namely $$m a = 0 \pmod{n},$$ where neither $m$ nor $a$ are congruent to $0$ mod $n$.

Then $ma$ is a multiple of $n$ so $n \mid ma$ and because $n$ and $m$ are coprime, $n$ must actually divide $a$. But to be a zero divisor there must be an equation where neither $m$, $a$ are already $0$ mod $n$.

The proof is ok, apart from the erroneous $p$, but just a stream-of-consciousness proof that is not very well structured.

Gordon Royle
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