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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8$\begingroup$ $\mathbb{Q}$-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. $\endgroup$– nafCommented Aug 23, 2011 at 15:32
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$\begingroup$ Thanks. But is it possible that a nodal-cuspidal plane curve is $\mathbb{Q}$-factorial? $\endgroup$– Fei YECommented Aug 24, 2011 at 5:38
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$\begingroup$ 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. $\endgroup$– nafCommented Aug 28, 2011 at 14:46
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$\begingroup$ 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. $\endgroup$– Fei YECommented Aug 28, 2011 at 15:41
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1$\begingroup$ 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. $\endgroup$– nafCommented Aug 29, 2011 at 5:48
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2 Answers
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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.
answered Aug 28, 2011 at 13:20
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$\begingroup$ 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? $\endgroup$ Commented Aug 28, 2011 at 13:42
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$\begingroup$ No, I meant the opposite. Let me make a quick edit, in case it's unclear. $\endgroup$ Commented Aug 28, 2011 at 14:07
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2$\begingroup$ @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. $\endgroup$– nafCommented Aug 28, 2011 at 15:08
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2$\begingroup$ @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 $\endgroup$ Commented Aug 28, 2011 at 15:19
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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.
answered Aug 28, 2011 at 19:57
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$\begingroup$ Can you give me some explanation on the first statement of your answer? See mathoverflow.net/questions/98919/… $\endgroup$– gioCommented Jun 11, 2012 at 14:26
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2$\begingroup$ The first statement follows from Theoreme XI.3.13 in SGA2 (As has been pointed out in the comments to your question.) $\endgroup$ Commented Jun 11, 2012 at 15:12
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$\begingroup$ Ok, thanks. Actually the statement applies to the local complete intersections. $\endgroup$– gioCommented Jun 13, 2012 at 13:34