# Condition for a local ring whose maximal ideal is principal to be Noetherian

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.

• 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. Jan 24, 2014 at 19:58
• Jan 25, 2014 at 0:09
• Also related: mathoverflow.net/questions/36611/… Jan 25, 2014 at 12:09
• @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. Jan 25, 2014 at 13:42
• @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 $1$. Jan 25, 2014 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 $R,M,M^2,\dots,M^n=0$.

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.

• Thank you. Does this text have any information which might help with #2? (i.e. for local rings that aren't domains) Jan 24, 2014 at 21:06