Proving that $\mathbb{Z}$ is a Noetherian ring

Question

So by definition I know that for all increasing chain of ideals: $I_1\subseteq I_2\subseteq \ldots I_k\subseteq I_{k+1}\subseteq \ldots$ in $\mathbb{Z}$, $\exists n\in \mathbb{N}, n \geq 1: I_n = I_{n+m} \forall m\in \mathbb{N}, m\geq 1.$ Or at least that was the definition I was given. But I am a little lost in how to work with Ideals, let a lone this sort of sequence of ideals. I thought of working with $\mathbb{Z}_p$ with p prime. But didn't get far. Can anyone guide me through this? explain everything. That would be greatly appreciated.

Answer 1

You can show that any PID is Noetherian (hint: suppose an infinite sequence of ideals exists, and take their union $I$ and verify it's an ideal in a PID $I = (a)$, and find $I_n$ where $a \in I_n$ since $a \in I$).

To show that $\mathbb{Z}$ is noetherian, just note that $\mathbb{Z}$ is a PID.

Answer 2

$\newcommand{\Z}{\mathbb{Z}}$ There's a lot of ways to go about approaching this. You can show for yourself (the very useful fact) that if $R$ is a commutative ring then following are equivalent

1. Every ideal in $R$ is finitely generated.
2. Every ascending sequence of ideals stabilizes.
3. Every non-empty collection of ideals of $R$ has a maximal element with respect to containment.

As Anas mentioned if $R$ is a PID then obviously any ideal is finitely generated and by this equivalence is therefore Noetherian.

If you want to work directly with ideals of $\Z$, just remember that ideals of $\Z$ are exactly $n\Z$ for some $n \in \Z$. Therefore any strictly ascending chain of ideals has the form $$ (n_1)\subsetneq (n_2) \subsetneq \dots $$ But any such chain must terminate. The reason is that if there are only finitely many ideals of $\Z$ which contain $(n_1)$ by the correspondence isomorphism theorem. Precisely there is a bijection between the ideals of $\Z$ containing $(n_2)$ and ideals of $\Z/(n_2)$, which is a finite ring.