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

Question

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!

Answer 1

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.

Answer 2

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.