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.



Condition for a local ring whose maximal ideal is principal to be Noetherian
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 $R,M,M^2,\dots,M^n=0$.


