Ideals in $\mathbb{Z}_n$

Question

For which $n > 1$, the set of all not invertible elements of the ring $\mathbb{Z}_n$ is an ideal?

This is where I got to:

The ideal, surely, should consists only of zero- divisors (because each not invertible element is a zero- divisor). Also, all zero- divisors in $\mathbb{Z}_n$ are the elements which are NOT coprime with $n$.

For example, in $\mathbb{Z}_6 = \{ \bar{0}, \bar{1},\bar{2},\bar{3},\bar{4},\bar{5}\}$, all zero-divisors are $\bar{0}, \bar{2},\bar{3},\bar{4}$, but obviously that's NOT an ideal. So, I get stuck on what should be those $n$, for which in $\mathbb{Z}_n$ all zero- divisors (not inversible elements) are an ideal ?

Consider what happens if $n$ has two different prime factors. In particular, what does Bezout's identity tell you about an ideal that contains both of them? — Troposphere, May 01 '21 at 14:12
I would suggest looking at the first several $\mathbb{Z}_n$, say for $n=2,3,...,10$ and see if that gives you any ideas. — paw88789, May 01 '21 at 14:14
@Troposphere Could you give me a little more details about this approach? — Simon Jachson, May 01 '21 at 17:05

score 1 · Answer 1 · answered May 01 '21 at 14:39

For a commutative ring $R$ the non-units form an ideal if and only if $R$ is local. Now $R=\Bbb Z_n$ is local iff $n$ is a prime power, see here:

Showing $R$ is a local ring if and only if all elements of $R$ that are not units form an ideal

Show that $\mathbb{Z}_n$ is local ring iff $n$ is a power of a prime number

Ideals in $\mathbb{Z}_n$

1 Answers1