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Find all the positive integers $m$ for which the zero divisors together with $0$ form an ideal in the ring $\mathbb{Z}/(m)$.

We need the set $\{ x \in \mathbb{Z}/(m) \mid x \text{ is a zero divisor}\} \cup \{ 0 \}$ to be an ideal in the ring $\left( \mathbb{Z}/(m), +, \cdot \right)=\left( \mathbb{Z}_m, +, \cdot \right)$.

I've been struggling with this problem, but I don't even know how to start the solution. I've tried some small examples, but that's it. Can anyone help me?

sicmath
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  • What have you tried? Have you checked the definitions? Have you tried some small examples? – Servaes Nov 05 '20 at 09:05
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    Hint: call the set $\cal Z.,$ Show $,ab\in \cal Z\Rightarrow a\in \cal Z,$ or $,b\in \cal Z,,$ so if $,\cal Z, $ is an ideal it is prime. $\ $ – Bill Dubuque Nov 05 '20 at 10:20
  • Further hint: the contrapositive of the prior inference is that cancellable (aka regular) elements are closed under product — which is clear (in fact they are also closed under divisors, i.e. they form a saturated monoid $,M,,$ i.e. $,ab\in M\iff a,b\in M).,$ Thus it follows that its complement $,\cal Z,$ is a union of prime ideals. That's overkill in this simple case, but useful to know for generalizations. – Bill Dubuque Nov 05 '20 at 10:41

2 Answers2

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Hint: The best way to start on a problem like this is to try to find examples. If you try the first few numbers, you should get that $\{1,2,3,4,5,7,8,9,11,\dots\}$ work, but $\{6,10,12,\dots\}$ don't. See if you can use the mini-proofs you come up with for these small cases to generalize to larger classes of numbers.

Carl Schildkraut
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Let $m$ be a positive integer with $m\ge 2$, and let $R$ be the ring $\mathbb{Z}/(m)$.

Let $S$ be the set of nonzero zerodivisors of $R$, and let $I= S\cup \{0\}$.

If $m$ is prime, then $R$ has no nonzero zerodivisors (why?), hence $I=\{0\}$ which is an ideal of $R$.

Next assume $m$ is composite, and suppose $I$ is an ideal of $R$.

If there are distinct primes $p,q$ such that $p,q$ are both divisors of $m$, then $p,q$ are zerodivisors (why?), hence $p,q\in I$, but then since $(p,q)=(1)$ (why?), we would have $1\in I$, contradiction (why?).

Hence $m$ must have the form $p^k$ where $p$ is prime and $k\ge 2$ (why?).

Thus for the case where $m$ is composite, we have a necessary condition on $m$.

To see that this condition is sufficient, suppose $m=p^k$ where $p$ is prime and $k\ge 2$. Now show that the zerodivisors of $R$ (including $0$) are precisely the elements of the ideal $(p)$. I'll leave that verification to you.

quasi
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