Let $A$ is a noetherian ring, $I\subseteq A$ is an ideal, and $\widehat{A}$ is the $I$-adic completion of $A$. If $I$ is contained in the Jacobson radical of $A$ then I have to show that $A\to \widehat{A}$ is faithfully flat.
My approach: First observe that every ideal $J$ of $A$ is closed. Since $A/J$ is finitely generated as $A$ module by Artin-Rees lemma we've $\bigcap_{n \geq 1} I^{n}(A/J)=0$, i.e., $J=\bigcap_{n\geq 1}(I^{n}+J)$, hence $J$ is closed. In particular $(0)$ is closed and thus $A \to \widehat{A}$ is injective. In order to show that $\widehat{A}$ is faithfully flat it is enough to show that $m\widehat{A} \neq \widehat{A}$ for all $m \in \mathrm{Max}(A)$, as we already know $\widehat{A}$ is flat $A$ module. (Probably we don't need that: $m\widehat{A}=\text{ the closure of } m \text{ in }\widehat{A}$.) Therefore $m \subset m\widehat{A} \cap A$, where $m\widehat{A}\cap A$ is closed ideal of $A$. From here how can I argue that it is in fact $m=m\widehat{A}\cap A$? I need some help. Thank you.