Local ring with principal maximal ideal and $\cap_{n>0}\mathfrak{m}^n=\{0\}$ is Noetherian?

Question

How can I show from scratch that, if $A$ is a local ring with maximal ideal $\mathfrak{m}$ so that $\mathfrak{m}$ is principal, and $\bigcap_{n>0}\mathfrak{m}^n = \{0\}$, then $A$ is necessarily Noetherian.

Answer 1

Let $\mathfrak{m} = mR$ for some $m \in R$.

For any $r \in R$, confirm that $r$ can be uniquely represented as $um^n$ for some $n \in \mathbb{N}$ and $u \in R$ a unit. (What's the largest power of $m$ that divides $r$? It must be finite because $\bigcap_n m^nR = 0$. And if $m^n$ divides $r$ but $m^{n+1}$ doesn't, then $\frac{r}{m^{n}} \notin mR = \mathfrak{m}$ and is therefore a unit because $R$ is local).

Now since every subset of $\mathbb{N}$ attains a minimum, it immediately follows that every ideal of $R$ is in fact principally generated.

So more than being Noetherian, the ring in question is a Discrete Valuation Ring.