Completion of a local ring is noetherian (under some hypothesis)

Question

I was reading the proof of Lemma 10.12 in this paper. In the second sentence, the following fact is used implicitly:

Let $(R,\mathfrak{m})$ be a commutative local ring. Let $\widehat{R}$ be its $\mathfrak{m}$-adic completion, which is a local ring whose maximal ideal we denote by $\widehat{\mathfrak{m}}$. If $\widehat{\mathfrak{m}}$ is finitely generated, then $\widehat{R}$ is noetherian.

The first idea one could have is that in a local ring, it suffices to check that the maximal ideal is finitely generated to see that the ring is noetherian. However, this is not true, as the following example due to Clark shows. One could try to prove this under the stronger assumption that the local ring is complete with respect to its maximal ideal. However, we also don't know if $\widehat{R}$ is $\widehat{\mathfrak{m}}$-adically complete, for $\mathfrak{m}$ may not be finitely generated.

Does anyone know a proof for the claim in the box (in case it is true)?

Thank you for reading, any help will be highly appreciated :)

Remark. As Keerthi Madapusi recalls and follows from part of the proof of the Lemma, the vector space $\mathfrak{m}/\mathfrak{m}^2$ is finite-dimensional if and only if $\widehat{\mathfrak{m}}$ is finitely generated.

Answer 1

We can just follow the proof from the stacks tag linked from the one suggested by Keerthis Madapusi but under your hypothesis.

Let $a_1,\dots,a_r$ in $R$ generate $\mathfrak m/\mathfrak m^2$. Then the degree $k$ monomials in $a_1,\dots,a_r$ generate $\mathfrak m^k/\mathfrak m^{k+1}$.

Hence the homomorphism $R^{ \binom{k+r-1}{r-1}}\to \mathfrak m^k$ that sends each generator to a different monomial is surjective modulo $\mathfrak m$. Thus by the same lemma cited in the stacks tag above, except part (1) this time, the induced map on completions is surjective. The completion of $\mathfrak m^k$ is the kernel of $\hat{R} \to R/\mathfrak m^k$ and so the kernel of $\hat{R} \to R/\mathfrak m^k$ is generated by degree $k$ monomials in $a_1,\dots,a_r$ and thus is equal to $\hat{\mathfrak m}^k$ (of course it contains $\hat{\mathfrak m}^k$, and this shows it is generated by members of $\hat{\mathfrak m}^k$). So the limit topology is the $\hat{\mathfrak m}$-adic topology and thus $\hat{R}$ is complete for the $\hat{\mathfrak m}$-adic topology.

A comment on the tag references [Fujiwara. K and Kato. F, The foundation of Rigid Geometry I, Def. 7.2.6 and Prop. 7.2.7] as giving some kind of more general statement. I didn't check, but it's possible that general version is also what you need.