# $A$ local Noetherian ring with principal maximal ideal implies PIR?

Suppose that $A$ is a local Noetherian ring with principal maximal ideal. Can we prove that every ideal of $A$ is principal?

I tried to exploit the Noetherian property on the set of non-principal ideals obtaining that (if the statement is false) there must exist a prime non-principal ideal but I can't conclude nothing.

• I think you have to add "integral". – Captain Lama Jun 8 '16 at 17:59
• Much more is true: you can even drop the "local" condition. It's a famous theorem of Kaplansky that a commutative Noetherian ring is a principal ideal ring iff the maximal ideals are principal. – rschwieb Jun 8 '16 at 18:04
• I would try to look at the proof in the domain case (this should be in Atiyah-Macdonald, for example) and see where, if anywhere (I'm starting to think it's true), things break down. It looks like $A$ is $1$-dimensional. If $I$ is $\mathfrak{m}$-primary then everything should go through, but if not then it's not clear to me what to do. – Hoot Jun 8 '16 at 18:07
• @CaptainLama Not entirely sure which 'integral' you wanted to add, but see the first page of this – rschwieb Jun 8 '16 at 18:10
• Well, "drop the local condition" isn't really what I intended, I really mean that you can rephrase it to "all maximal ideals principal" and get more rings than local rings. – rschwieb Jun 8 '16 at 18:11

Let $I \subset (m)$ be a non-zero ideal. Define $n := \min \{ s \geq 1 | \exists x \in I \text{ such that } x \in (m)^s \setminus (m)^{s+1} \}$.
First, we have to show that $n < \infty$, i.e. we have to show that $I$ is not contained in $\bigcap\limits_{s \geq 1} (m)^s$, i.e. we have to show $\bigcap\limits_{s \geq 1} (m)^s=0$. This is clear with Krull's intersection theorem, but I don't want to invoke any theorem, so I will give an argument in this case:
Let $a$ be contained in that intersection, in particular $a=mb$ for some $b \in A$. I claim that $b$ is also contained in that intersection: If not, $b$ is non-zero in some $(m)^s/(m)^{s+1}$. Since this is a one-dimensional vector-space, we obtain that $b$ generates $(m)^s/(m)^{s+1}$. By Nakayama, $b$ generates $(m)^s$, i.e. $a$ generates $(m)^{s+1}$. In particular $a \notin (m)^{s+2}$, contradiction!
Thus, we have shown $(m) \bigcap\limits_{s \geq 1} (m)^s = \bigcap\limits_{s \geq 1} (m)^s$, i.e. $\bigcap\limits_{s \geq 1} (m)^s=0$ by Nakayama. This is where we need the noetherian hypothesis, because we need to guarantee that $\bigcap\limits_{s \geq 1} (m)^s$ is a priori finitely generated to invoke Nakayama.
Now it is very easy to show that $I=(m)^n=(m^n)$ holds: By the minimality of $n$, we have $I \subset (m)^n$ and we have some $x \in I$ with $x \notin (m)^{n+1}$. Again invoking Nakayama, we get $(x)=(m)^n$, i.e. $I \subset (m)^n=(x) \subset I$. The proof ends here.