UFD property descends?

Question

Hi,

Let $k$ be a field and $A$ a local noetherian $k$-algebra. If its completion is a UFD, is it true that $A$ is a UFD? Proof?

Thanks

Answer 1

I think this is simpler than having to quote a "Theorem".

Let $\mathfrak a\subset A$ be an ideal such that $\hat{\mathfrak a}=\mathfrak a\hat A$ is principal, generated by $\hat f\in\hat A$. Let $f\in \mathfrak a$ be an element with the same residue mod $\hat{\mathfrak m}$, i.e., such that $\hat f\ \mathrm{mod}\ \hat{\mathfrak m} \leftrightarrow f\ \mathrm{mod}\ \mathfrak m$ via the isomorphism $\hat{\mathfrak a}/\hat{\mathfrak m}\hat{\mathfrak a}\simeq \mathfrak a/\mathfrak m\mathfrak a$. Then by Nakayama's lemma, $\hat{\mathfrak a}=f \hat A$. However, $\hat A$ is faithfully flat over $A$, so $\mathfrak a\hat A\cap A=\mathfrak a$, and hence we get that $\mathfrak a=fA$ is also principal.

For a noetherian ring, being a UFD is equivalent to the condition that the intersection of any two principal ideals is also principal (or having a greatest common divisor or a least common multiple). So let $\mathfrak a, \mathfrak b\subseteq A$ be two principal ideals. Then $\mathfrak a\hat A,\mathfrak b\hat A\subseteq \hat A$ are also principal (using the same generators). If $\hat A$ is a UFD, then $\mathfrak a\hat A\cap\mathfrak b\hat A$ is also principal, but $\mathfrak a\hat A\cap\mathfrak b\hat A = (\mathfrak a\cap \mathfrak b)\hat A$ (since $\hat A$ is flat over $A$), so by the above argument $\mathfrak a\cap \mathfrak b$ is principal, and hence $A$ is a UFD.