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

I have proven in the past 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, since every ideal is a power of $\frak{m}$.

Conversely, the Krull Intersection Theorem guarantees that if $A$ is a Noetherian local ring with maximal ideal $\mathfrak{m}$, then $\bigcap_{n>1}\mathfrak{m}^n = \{0\}$.

I'm wondering about the first theorem without the assumption that $\mathfrak{m}$ is principal, and is instead just finitely generated? Here is a demonstration of where my previous proof breaks down:

Let $x \in I$ for some ideal of $A$. Assuming (wlog) that $I$ is not maximal, we know that $I \subseteq \mathfrak{m}$. Choose $n$ so that it is the maximal power where $I \subseteq \mathfrak{m}^n$ since it would otherwise be trivial, by assumption.

If $\mathfrak{m}$ were principal, then $x = a \cdot m^n$, where $m$ is the generator of $\mathfrak{m}$. When $\mathfrak{m}$ is principal, I could then argue that $I = \mathfrak{m}^n$ since $a$ would then be a unit. However, I don't think this argument works here.

If this does not work, I was wondering if maybe using the fact that the natural homomorphism

$$\phi\colon A \to \prod_{i>0}A/I_i$$

is an injection, since $\ker(\phi) = \bigcap I_i = \{0\}$, but I'm unsure of how I can use this observation.

So: is the assumption that the maximal ideal is finitely generated sufficient here? If yes, is there any way to recover either of the "partial" proof ideas that I was concerned with?

• OP: if you replace the condition of intersection of powers being 0 i.e. m-adically separated with m-adically complete then it is true by Cohen structure theorem – Jay Oct 18 '16 at 19:41

The ring $B$ in Proposition 3.2 on page 6 is local with $2$-generated maximal ideal. Note 3.5(2) on page 7 says that the intersection of powers of the maximal ideal is zero. $B$ is not Noetherian because $\dim B = 3$ (so if $B$ were Noetherian then Krull's Principal Ideal Theorem would imply the maximal ideal needs at least $3$ generators).