# Completions of non-Noetherian rings with respect to finitely generated ideal

All the standard counterexamples to flatness that I have seen involve completions with respect to non-finitely generated ideals. I am interested in the following two cases:

1. Let $$A$$ be a local ring with finitely generated maximal ideal $$\mathfrak{m}$$. Is the completion morphism $$A \rightarrow \hat{A} := \underset{n \in \mathbb{N}}{\mathrm{lim}} \text{ } \frac{A}{\mathfrak{m}^{n+1}}$$ conjectured to be injective? In other words, is Krull's intersection theorem expected to hold?

2. Let $$A$$ be a ring and let $$I$$ be a finitely generated ideal of $$A$$ such that $$A/I$$ is Noetherian. Is the completion with respect to $$I$$ expected to be flat?

Thank you in advance!

For question 2. notice that by Stacks 05GH your assumptions imply that $$\widehat{A}$$ is noetherian. Assume that $$A$$ is a local ring. If the map $$A\rightarrow \widehat{A}$$ was always flat, then the assumption that $$A$$ is local implies it is faithfully flat. But then $$A$$ is noetherian as well by faithfully flat descent. Thus, for a general ring $$A$$, using the same argument you would be able to show that all its stalks are noetherian, and hence $$A$$ is noetherian. Thus this is false in general.
For 1., take for $$A$$ the ring of germs of $$\mathscr{C}^{\infty}$$ functions at $$0\in \mathbb{R}$$. Then $$\mathfrak{m}=(x)$$ and $$\hat{A}=\mathbb{R}[[x]]$$, so the map $$A\rightarrow \hat{A}$$ is not injective.