Principal ideals in completions of Zariski rings

Let $A$ be a Noetherian ring and $\mathfrak{a}$ some ideal contained in the Jacobson radical of $A$. Now $A$ is endowed with the $\mathfrak{a}$-adic topology, i.e. $A$ is a Zariski ring.

If $\mathfrak{b} \subset A$ is an ideal so that $\mathfrak{b} \hat{A}$ is a principal ideal ($\hat{A}$ denotes the completion of A), then $\mathfrak{b}$ is a principal ideal. How can I prove this statement?

I know that my assumption $\mathfrak{a} \subset \operatorname{J}(A)$ is equivalent to the statement that all maximal ideals of $A$ are closed in the $\mathfrak{a}$-adic topology, but I don't think that this gets me any further. Any hints would be appreciated.

As $A\to \hat{A}$ is flat, we have $$\mathfrak b \hat{A}\otimes_{\hat{A}} (\hat{A}/\mathfrak a\hat{A})=(\mathfrak b\otimes_A \hat{A})\otimes_{\hat{A}} (A/\mathfrak a)=\mathfrak b\otimes_A A/\mathfrak a.$$ By hypothesis, the LHS is generated by one element, so is the RHS. By Nakayama's lemma (because $\mathfrak a$ is contained in the Jacobson radical of $A$), $\mathfrak b$ is generated by one element.
• @Paul: see en.wikipedia.org/wiki/Nakayama_lemma statement 3. If $b\in \mathfrak b$ generated the latter in $\mathfrak b\otimes_A A/\mathfrak a=\mathfrak b/\mathfrak a\mathfrak b$, then $M:=\mathfrak b/bA$ satisfies the condition $M=\mathfrak a M$.