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Given an affine algebraic variety $V$ such that $\Gamma(V,\mathcal{O}_V)$ is a UFD, its sheaf of ring can be determined easily since one can show that:

$$\Gamma(D(f_1) \cup \cdots \cup D(f_n),\mathcal{O}_V) \simeq \Gamma(D(h),\mathcal{O}_V),$$ where $h=\mathrm{gcd}(f_1, \ldots, f_n)$.

It is then natural to ask whether we can characterize the set of affine algebraic varieties $V$ such that $\Gamma(V,\mathcal{O}_V)$ is a UFD? Are there any geometric interpretations?

Thanks you!

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2 Answers 2

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If and only if the Picard group is trivial. (Similar to in algebraic number theory).

Proof: Suppose $R$ is a UFD. Then every codimension 1 prime ideal is principal. Since these generate the divisor group, every divisor is principal. Suppose $R$ is not a UFD. Take an irreducible element that is not prime. This element must be contained in a codimension 1 prime ideal that is not principal (I think?). This would be a non-principal divisor.

There may be some assumptions I'm forgetting on the divisor group.

If you think the Picard group is geometric enough, we are done. (In particular, there are topological obstructions to having trivial Picard group.)

EDIT: Francois Brunault points out that these arguments are about the Weil divisor class group, not the Picard group. These concepts agree on nonsingular varieties but differ on singular ones. Sadly, the Weil group is less geometric than the Picard group.

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    $\begingroup$ Maybe one needs to replace the Picard group by the Weil divisor class group, see this previous question mathoverflow.net/questions/25758/… $\endgroup$
    – François Brunault
    Commented Dec 7, 2011 at 0:21
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    $\begingroup$ To Will: I would have said that the Weil class group is more geometric that the Picard group. Look at what happens for a nodal curve, or a curve with a cusp. $\endgroup$
    – Angelo
    Commented Dec 7, 2011 at 6:07
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    $\begingroup$ I completely agree with Angelo. Weil divisors are totally geometric: they are defined as (linear combinations of) subvarieties. Cartier divisors are on the other hand more algebraic as they are defined by their defining equations. Their geometric nature lies in their associated Weil divisor. $\endgroup$
    – Sándor Kovács
    Commented Dec 7, 2011 at 6:27
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    $\begingroup$ The Cartier group is geometric because it classifies line bundles, which are certainly geometric. While Weil divisors are geometric, the Weil class group is less so, being a quotient by an entirely algebraically-defined subgroup (though perhaps still more geometric than the Picard group? The distinctions seem silly at this point.) $\endgroup$
    – Will Sawin
    Commented Dec 7, 2011 at 6:43
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    $\begingroup$ Note also that the fact that $k[V]$ is a UFD does not necessarily imply that $V$ is smooth. This is true if $V$ is a curve, but in general something like $k[X,Y,Z,T]/(XY-Z^2-T^3)$ should give a counterexample. $\endgroup$
    – François Brunault
    Commented Dec 7, 2011 at 9:12
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A noetherian domain is a UFD if and only if it is normal and the divisor class group is zero. See, for example, exercise 1D here.

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    $\begingroup$ This is Prop. 6.2, p. 131 in Hartshorne's "Algebraic geometry". $\endgroup$
    – Damian Rössler
    Commented Dec 7, 2011 at 20:49
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    $\begingroup$ For what it's worth, this is equally true for Krull domains. $\endgroup$
    – Pete L. Clark
    Commented Feb 24, 2013 at 2:47

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