TITLE: Condition for a local ring whose maximal ideal is principal to be Noetherian
QUESTION [7 upvotes]: 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 $n\geq 1$.

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?
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.

REPLY [10 votes]: 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.

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