If $X$ is such that any étale $X$-scheme is locally factorial, is any smooth $X$-scheme locally factorial?

We say that a local ring is geometrically factorial if its strict henselization is a UFD, and that a scheme is geometrically factorial if all of its local rings are. If $$X$$ is a normal, locally Noetherian and geometrically factorial scheme, is any smooth $$X$$-scheme geometrically factorial?

$$\textbf{Edit:}$$ As Minsheon Shin pointed out in his answer, this question is equivalent to the one in the title by Proposition 1 of Danilov, $$\textit{On a conjecture of Samuel}$$. In the same article, Danilov conjectures that if $$R$$ is local, Noetherian and geometrically factorial, then $$R[[T]]$$ is geometrically factorial; and his main theorem is a special case of this conjecture. If $$R$$ is excellent, a positive answer to my question would imply the conjecture by https://stacks.math.columbia.edu/tag/07GC.

$$\textbf{Other equivalent formulations:}$$ By the local structure of smooth morphisms+induction, the question is equivalent to:

"If $$X$$ is a normal, Noetherian, geometrically factorial scheme, then $$\mathbb A^1_X$$ is geometrically factorial."

Working locally, this, in turn, reduces to:

"If $$R$$ is a strictly henselian local UFD, then the strict localizations of $$R[T]$$ at $$(\mathfrak m_R,T)$$ and at $$\mathfrak m_R$$ are UFDs."

The strict localization at $$(\mathfrak m_R,T)$$ has completion $$R[[T]]$$ so it is a UFD if Danilov's conjecture holds.

• I haven't thought it through, but a smooth morphism is locally factorizable as etale into an affine space over the base. So if you know that $\mathbb{A}^n_X$ has the property that its etale spaces are locally factorial then you win (at least for the question in the title). Feb 17, 2021 at 0:35

For (1), see Danilov, On a conjecture of Samuel (link), Proposition 1, which I reproduce here: Let $$A \to B$$ be a faithfully flat ring homomorphism. Then (a) if $$B$$ is a Krull ring, then $$A$$ is a Krull ring, and (b) if $$\operatorname{Pic}(A) = 0$$, then $$\operatorname{Cl}(A) \to \operatorname{Cl}(B)$$ is injective. The argument for (b) is: if an ideal $$\mathfrak{a}$$ of $$A$$ is such that $$\mathfrak{a} \otimes_{A} B$$ is a principal ideal, then $$\mathfrak{a}$$ is a finitely generated projective $$A$$-module of rank 1 (i.e. a line bundle) by faithfully flat descent; then $$\operatorname{Pic}(A) = 0$$ implies that $$\mathfrak{a}$$ is principal.
Then as Alex notes, you can conclude using that (i) if $$A$$ is a UFD then any localization is a UFD, and (ii) if $$A$$ is a UFD then the polynomial algebra $$A[t]$$ is a UFD. I was surprised to learn (in the same paper mentioned above) that the formal power series ring $$A[[t]]$$ need not be a UFD even if $$A$$ is.
(The property that $$R^{sh}$$ is a UFD is also called "geometrically factorial".)
• Thank you for your answer! I am confused about the "you can conclude" part though. If $X$ is locally factorial, then $\mathbb A^n_X$ is certainly locally factorial as well, but why would this remain true for geometric local factoriality? Or is this also from Danilov's article? (I cannot access it at the moment.) Edit: You were faster than me! Feb 17, 2021 at 3:13