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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.

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up vote 6 down vote accepted

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.

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I have yet to understand your last step (application of NAK), but still, very nice. Thanks! (I'd be grateful if someone could expand on that in a comment) – Paul Dec 18 '11 at 21:40
@Paul: see 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 $M=\mathfrak a M$. – user18119 Dec 18 '11 at 21:47

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