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.

  • 3
    $\begingroup$ I think you have to add "integral". $\endgroup$ – Captain Lama Jun 8 '16 at 17:59
  • 1
    $\begingroup$ 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. $\endgroup$ – rschwieb Jun 8 '16 at 18:04
  • $\begingroup$ 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. $\endgroup$ – Hoot Jun 8 '16 at 18:07
  • 1
    $\begingroup$ @CaptainLama Not entirely sure which 'integral' you wanted to add, but see the first page of this $\endgroup$ – rschwieb Jun 8 '16 at 18:10
  • $\begingroup$ 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. $\endgroup$ – rschwieb Jun 8 '16 at 18:11

Here is an elementary proof (which is of course much easier than the proof of the deeper theorem of Kaplansky), which does not invoke any theorems in commutative algebra besides Nakayama:

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.