# When is an algebraic variety $\mathbb{Q}$-factorial?

A variety is $\mathbb{Q}$-factorial if every global Weil divisor is $\mathbb{Q}$-Cartier. How bad singularities are allowed so that the algebraic variety is still $\mathbb{Q}$-factorial? Is a singular curve $\mathbb{Q}$-factorial? For example is a nodal-cuspidal plane curve $\mathbb{Q}$-factorial?

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$\mathbb{Q}$-factoriality is usually defined only for normal varieties since one wants the group of Cartier divisors to be a subgroup of the group of Weil divisors; a singular curve is never normal. –  ulrich Aug 23 '11 at 15:32
Thanks. But is it possible that a nodal-cuspidal plane curve is $\mathbb{Q}$-factorial? –  Fei YE Aug 24 '11 at 5:38
It depends on how you define $\mathbb{Q}$−factorial in the non−normal case. If you just say that a non−zero multiple of every Weil divisor lifts to a Cartier divisor then singular curves will be $\mathbb{Q}$-factorial. –  ulrich Aug 28 '11 at 14:46
Thanks. I was thinking of lifting a non-zero multiple of a Weil divisor to a Cartier divisor. Can you explain how to show that a singular curve is $\mathbb{Q}$-Cartier? I was thinking to lift a Weil divisor to the normalization and then maybe push forward to get something useful. But I don't have a clear picture in my mind. –  Fei YE Aug 28 '11 at 15:41
Any curve $C$ is quasi-projective. So assuming $C \subset \mathbb{P}^n$ you can find a hyperplane $D$ which intersects $C$ at a given singular point $x$ and no other singular point. Since any smooth point is a Cartier divisor and $D$ is also Cartier, it follows that some multiple of $x$ is the Weil divisor associated to a Cartier divisor. –  ulrich Aug 29 '11 at 5:48

I saw this a while ago, and I assumed that someone would give you an answer, but it doesn't look like anyone will. So I'll make an attempt. As Ulrich suggests, it is better to stick to normal varieties, so your nodal/cuspidal curves are immediately eliminated. So we should move to dimension at least two.

A very simple class of examples are cones. Let $V\subset \mathbb{P}^N$ be a nontrivial smooth variety, and let $R$ be the local ring of the vertex of the cone over $V$. This is normal and $\dim R\ge 2$. $R$ (or $Spec R$) is $\mathbb{Q}$-factorial exactly when $Cl(R)\otimes\mathbb{Q}=0$ where $Cl(R)$ is the (Weil) divisor class group. Now according to Lipman "Unique factorization in complete local rings", Alg. Geom, Arcata (1974) $$Cl(R)= Cl(V)/\langle\text{hyperplane}\rangle$$

Thus $R$ is $\mathbb{Q}$-factorial if and only if the $Cl(\mathbb{P}^N)\otimes \mathbb{Q}\to Cl(V)\otimes \mathbb{Q}$ is surjective.

This is generally not true. For example, if $V$ is a curve, $R$ is $\mathbb{Q}$-factorial if and only if $V$ has genus $0$.

Added Notes: (1) In the last sentence, I meant that a cone over a higher genus curve $V$ cannot be $\mathbb{Q}$-factorial because $\dim Cl(V)\otimes \mathbb{Q}=\infty$ in this case. (2) As Ulrich & Francesco point out $V$ should be projectively normal. For example, it would suffice to assume the $V$ is a hypersurface. (3) I forgot to say, that I'm working over an algebraically closed ground field.

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Could you clarify the last (counter)example? If V is genus 0, a hyperplane section rationally generates. Are you claiming there is a higher genus curve V for which a hyperplane slice also generates? –  Anton Geraschenko Aug 28 '11 at 13:42
No, I meant the opposite. Let me make a quick edit, in case it's unclear. –  Donu Arapura Aug 28 '11 at 14:07
@Donu: It is not true that a cone over a smooth projective variety $V \subset \mathbb{P}^N$ is always normal. Hartshorne, AG, Ex I 3.18, given a twisted quartic curve in $\mathbb{P}^3$ as an example; the point here is that this is not a linearly normal embedding. –  ulrich Aug 28 '11 at 15:08
@Ulrich: Right. In fact, it seems to me that the cone over $V \subset \mathbb{P}^n$ is normal if and only if $V$ is projectively normal –  Francesco Polizzi Aug 28 '11 at 15:19

A complete intersection $X$ in $\mathbb{P}^n$ is $\mathbb{Q}$ factorial if $\dim X_{sing}<\dim X-3$.

In general being $\mathbb{Q}$-factorial depends on the type of singularity you have, but also on the position of the singularities.

E.g., a degree $d$ threefold $X$ in $\mathbb{P}^4$ with only ordinary double points at $p_1,\dots,p_k$ is $\mathbb{Q}$-factorial if and only if the linear system of homogeneous polynomials of degree $2d-4$ polynomials through the singular points of $X$ has no defect (i.e., the codimension of this linear system is precisely $k$).

An example of a non-$\mathbb{Q}$-factorial threefold with only nodes is $f_1f_2+f_3f_4=0$, where $\deg(f_1)+\deg(f_2)=\deg(f_3)+\deg(f_4)$, and the $f_i$ are chosen sufficiently general.

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Can you give me some explanation on the first statement of your answer? See mathoverflow.net/questions/98919/… –  gio Jun 11 '12 at 14:26
The first statement follows from Theoreme XI.3.13 in SGA2 (As has been pointed out in the comments to your question.) –  Remke Kloosterman Jun 11 '12 at 15:12
Ok, thanks. Actually the statement applies to the local complete intersections. –  gio Jun 13 '12 at 13:34