Can $\Bbb Z_n$l ideals be generated by one element?

Question

Let's define the ring, $\Bbb Z_n$ (It is surely a well-known ring.)

set : $Z_n$ ={$[0]_{n}$, $[1]_{n}$, ... $[n-1]_{n} \}$
operation : addition or multiplication for the $mod n$

Say the $m \in \Bbb Z_n$, It is obvious that $\langle m \rangle$ is an ideal of $\Bbb Z_n$.

Then, $\forall$ ideal J of the $Z_n$, can the ideal J be expressed as the $\langle m \rangle$ ?

If said result on cyclic groups is unfamiliar then you can instead use the same one-line proof as in $\Bbb Z$ to show that every ideal $\neq 0$ is generated by its least positive element. — Bill Dubuque, Jul 23 '19 at 16:47
@DavidC.Ullrich As in the OP we presume the standard nonnegative reps $,0,1,\ldots, n!-!1\ \ $ — Bill Dubuque, Jul 23 '19 at 21:08

score 1 · Accepted Answer · answered Jul 30 '19 at 10:19

consider $I\subset \mathbb{Z}_n$ be ideal.Let $\eta:\mathbb{Z}\to \mathbb{Z}_n$ be natural projection then observe $\eta^{-1}(I)$ is ideal of $\mathbb{Z}$. So it is generated by a single element (say a) as $\mathbb{Z}$ being PID. Now it is easy to see $I$ is generated by $\eta(a)$.

score -1 · Answer 2 · answered Jul 24 '19 at 17:56

-1

Since $\mathbb{Z}_n\cong\frac{\mathbb{Z}}{\langle n\rangle}$ and $\mathbb{Z}$ is a PID, we have $\mathbb{Z}_n$ is also a PID.

answered Jul 24 '19 at 17:56

E.R

2,917

2

This is incorrect. Being a PID includes (as the name suggests) being a domain. – Tobias Kildetoft Jul 24 '19 at 17:58
Not to mention it is almost surely beyond the OP's knowledge level. – Bill Dubuque Jul 24 '19 at 21:41

Can $\Bbb Z_n$l ideals be generated by one element?

2 Answers2