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More precisely, if $X$ is a scheme such that for all $x\in X$, the ring $\mathscr{O}_{X,x}$ is a UFD, then is it true that we can cover $X$ by affine opens $U = \operatorname{Spec}A$ such that $A$ is a UFD?

Clearly the converse holds true; if we are given an affine scheme of the form $X=\operatorname{Spec}A$ for $A$ a UFD, then since the property of being a UFD is preserved by localization, $X$ is locally factorial. However, it is emphatically not true that given a locally factorial scheme $X$, any affine open $\operatorname{Spec}A$ has $A$ a UFD – for a counterexample, we need only consider $X=\operatorname{Spec}A$ where $A$ is a Dedekind domain with nontrivial ideal class group.

Therefore, my question reduces to the following problem in commutative algebra. Given a ring $A$ such that $A_\mathfrak{p}$ is a UFD for all $\mathfrak{p}\in\operatorname{Spec}A$, then for all $\mathfrak{p}\in\operatorname{Spec}A$, can we find some element $f\in A\setminus\mathfrak{p}$ such that $A_f$ is a UFD?

I don't see any clear answer to this question. My hunch is that it's not true, however.

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    $\begingroup$ No, it is not true. Any Dedekind domain is locally factorial, but if it is the co-ordinate ring of a smooth affine curve over say $\mathbb{C}$, then unless it is already a UFD, it can not be covered by open affines which are UFDs. $\endgroup$ – Mohan Apr 2 '17 at 0:37
  • $\begingroup$ However the answer should be positive, if $A$ is a Dedekind domain with finite ideal class group (i.e. integer rings of number fields), because we can always make one single ideal principal in a neighborhood of any point (generic freeness). If the ideal class is finite, we only have to do this for finitely many prime ideals to make sure that any ideal is principal, thus we find a common neighborhood, which makes all ideals principal (which results in a UFD). $\endgroup$ – MooS Apr 3 '17 at 11:35
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Let $X$ be a smooth connected projective curve over $\mathbb C$ of genus $g\gt 0$.
Since $X$ is smooth all local rings $\mathcal O_{X,x}$ are UFD's, but I claim that no affine open subset $U\subset X$ has a ring of regular functions $\mathcal O_X(U)$ which is a UFD.
Proof:
A subset $U\subset X$ is open and affine if and only it is obtained from $X$ by deleting a nonzero finite number of points from $X$ , namely $U=X\setminus \{x_1,\cdots,x_n\} \quad (n\geq 1)$.
The key tool is then the exact sequence $$\oplus _{i=1}^n\mathbb Z\cdot e_i \stackrel {u}\to \operatorname {Pic}(X) \stackrel {v}\to \operatorname {Pic}(U) \to 0 $$ in which the map $u$ sends the basis vector $e_i$ of the free module $\oplus _{i=1}^n\mathbb Z\cdot e_i$ to the line bundle $\mathcal O_X(x_i)$, while $v$ is just the restriction to $U$ of line bundles on $X$ . (Beware that $u$ nedn't be injective!)
This exact sequence can be found for $n=1$ in Hartshorne (Chapter II, Prop.6.5, page 133) and in Fulton's IntersectionTheory (Chap.1, Prop 1.8, page 21) for arbitrary $n$.

Since $\operatorname {Pic}(X)$ is non denumerable (for example because distinct points $x\in X$ give rise to distinct line bundles $\mathcal O_X(x)$ and $X$ is non denumerable since we are over $\mathbb C$) this proves that $\operatorname {Pic}(U)\neq0$.
And this allows us to conclude, as announced, that $\mathcal O_X(U)$ is not a UFD:
Indeed a noetherian domain $A$ is a UFD if and only if every prime ideal of height $1$ in $A$ is principal (Matsumura, Theorem 20.1, page 161) and that last condition translates for $A=\mathcal O(U)$ to every line bundle on $U$ being trivial.

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  • $\begingroup$ In Georges post the case $g=0$ must be excluded, because then $X$ is isomorphic to the projective space $\mathbb{P}^1_k$, which can be covered by two affine spaces $\mathbb{A}^1_k$. The Picard group in this case equals $\mathbb{Z}$. There is one point I don't understand: why are the line bundles $O_X(x)$ for two different points not isomorphic, in which case they would define the same element in the Picard group? Don't we need to involve Riemann-Roch somehow? $\endgroup$ – Hagen Knaf Apr 2 '17 at 8:51
  • $\begingroup$ Dear @Hagen: no, we don't need Riemann-Roch. If we had $\mathcal O(x)=\mathcal O(y)$ for distinct points $x\neq y\in X$ we would deduce that $x$ and $y$ are linearly equivalent, which means that there exists a rational function $f\in Rat(X)$ whose divisor is $div(f)=x-y$. But then the associated map $ \overline f:X\to \mathbb P^1$ would be a morphism whose fibre over $0\in \mathbb P^1$ is the divisor $1.x$ . Thus that morphism would be of degree $1$ and thus an isomorphism $\overline f: X\stackrel {\sim}{\to} \mathbb P^1$. That contradicts the hypothesis $g(X)\gt 0$. $\endgroup$ – Georges Elencwajg Apr 2 '17 at 10:33
  • $\begingroup$ OK, You are right. $\endgroup$ – Hagen Knaf Apr 2 '17 at 11:31

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