The statement "a local ring whose maximal ideal is principal is Noetherian" is (I think) false. The ring of germs about $0$ of $C^\infty$ functions on the real line seems to be a counterexample since $e^{-1/x^2}\in \left(x^n\right)$ for all $n\geq 1$.

  1. If I add to the hypothesis that the ring is a domain, then (I think) the statement is true. I'm trying to figure out if this must be true (I suspect not). Is there a nice example of a local Noetherian ring whose maximal ideal is principal that is not a domain?

  2. Is there a better, weaker condition to add to the hypothesis so that sufficiency holds? In other words, "if R is a local ring whose maximal ideal is principal, then R is Noetherian if and only if R is [what is the best thing to put here]?"

Here local rings are assumed to be commutative with unity.

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    $\begingroup$ Any valuation ring with a discretely ordered but nonarchimedean value group is a counterexample to the statement you claim to be true in 1. And for any field $K$, $K[x]/(x^2)$ is an example for the second part of 1. $\endgroup$ – Emil Jeřábek Jan 24 '14 at 19:58
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    $\begingroup$ Related: mathoverflow.net/questions/30715 , mathoverflow.net/questions/39866 $\endgroup$ – Torsten Schoeneberg Jan 25 '14 at 0:09
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    $\begingroup$ Also related: mathoverflow.net/questions/36611/… $\endgroup$ – Jesse Elliott Jan 25 '14 at 12:09
  • $\begingroup$ @EmilJeřábek: Is "discretely ordered but nonarchimedean value group" what you really want to say? I would have expected "value group that is discrete but not of rank 1". -- Btw, your answer is amazingly elegant. $\endgroup$ – Torsten Schoeneberg Jan 25 '14 at 13:42
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    $\begingroup$ @TorstenSchoeneberg: A totally ordered group $G$ is archimedean if for every positive $a,b\in G$, there exists an integer $n$ such that $a\le nb$. This is equivalent to having rank $1$. $\endgroup$ – Emil Jeřábek Jan 25 '14 at 14:03

Here is an elementary argument. Let $R$ be a local ring with a principal maximal ideal $M=mR$.

Since $1+M\subseteq R^*$, we have $ma\mid a$ only if $a=0$. Moreover, every nonunit is divisible by $m$, hence if there is a nonzero $a\in\bigcap_nM^n$, we can construct an increasing sequence of ideals $a_0R\subsetneq a_1R\subsetneq a_2R\subsetneq\cdots$ where $a_0=a$ and $ma_{n+1}=a_n$, in particular $R$ is not noetherian.

On the other hand, if $\bigcap_nM^n=0$, every nonzero element can be written as $um^n$ for some $u\in R^*$ and $n\in\omega$, so all nonzero ideals are of the form $M^n$. This makes $R$ a discrete valuation ring unless $m$ is nilpotent. Thus,

Fact: If $R$ is a local ring with a principal maximal ideal $M$, the following are equivalent.

  1. $R$ is noetherian.

  2. $\bigcap_{n\in\omega}M^n=0$.

  3. $R$ is a DVR, or there is $n\in\omega$ such that all ideals of $R$ are $R,M,M^2,\dots,M^n=0$.

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Let $R$ be a local domain whose maximal ideal $\mathfrak{m}$ is principal. Then, $R$ is noetherian if and only if its $\mathfrak{m}$-adic topology is separated. If $R$ is moreover not a field, then it is noetherian if and only if it is a discrete valuation ring. For proofs see Bourbaki's Algèbre commutative, VI.4.6 Proposition 9.

By considering the irreducible components one can get the following generalisation of the above:

Let $R$ be a local reduced ring with only finitely many minimal primes whose maximal ideal $\mathfrak{m}$ is principal. Then, $R$ is noetherian if and only if its $\mathfrak{m}$-adic topology is separated.

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  • $\begingroup$ Thank you. Does this text have any information which might help with #2? (i.e. for local rings that aren't domains) $\endgroup$ – Alexander Gruber Jan 24 '14 at 21:06

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