MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given the equation of a cubic hypersurface $C\subset\mathbb{P}^{N}_{\mathbb{C}}$ ($N\geq 4$), there is an algorithm (or better a software) that allows to determine if $C$ is factorial (i.e., all of whose local rings are unique factorization domains, and hence there is no distinction between Cartier divisors and Weil divisors), and if $\mathrm{Pic}(C)=\mathbb{Z}\langle\mathcal{O}_C(1)\rangle$ ? Of course this is trivial if $C$ is smooth.


share|cite|improve this question
By the Grothendieck-Lefschetz theorem on Picard groups, the restriction map $Pic(X) \rightarrow Pic(D)$ is an isomorphism when $D$ is an ample effective divisor on $X$ and the dimension of $X$ is at least $4$. So the second part of your question is always the case. See Ample Subvarieties of Alg Var's by Hartshorne (Corollary IV.3.3) and Positivity in Alg. Geom. I Remark (3.1.26.) – Parsa Jun 17 '11 at 12:36
Parsa, this is true when $C$ is smooth. – Francesco Polizzi Jun 17 '11 at 12:39
Why do you need smoothness? In the case $D$ is reduced, you have the Lefschetz hyperplane theorem, and if $D$ is reduced means its exponential sequence is exact, so you use the Kodaira vanishings on $X$ to get the vanishings you need on $D$, and the 5-lemma on the long exact sequence associated to the exponential sequences of $X$ and $D$ gives you the isomorphism $H^1(X,\mathcal {O_X}^*) \cong H^1(D,\mathcal {O_D}^*)$. – Parsa Jun 17 '11 at 13:04
Sorry yiou are right, I misread your comment. Now I understand what you say. Well, usually "factorial" means that every codimension 1 subvariety is cut out by an hypersurface in the ambient space, and my answer shows that it is not always the case. But, as you remark, if C is not smooth this is not equivalent to $\textrm{Pic}(C)=Z$ . – Francesco Polizzi Jun 17 '11 at 13:26

First concerning your question: most people use $\operatorname{Pic}(C)$ for Cartier divisors. I interpret your question as follows: you want to determine whether every divisor on $C$ is linear equivalent with a Cartier divisor (this means factorial).

Now if $C$ is smooth this is true and I think this is also true if $\dim C-\dim C_{sing}>3$.

If $\dim C_{sing}\geq \dim C-3$ things are much more complicated. A necessary condition for being factorial is (roughly said) that the rank of $H^{N-2,N-2}(C,\mathbb{C}) \cap H^{2N-4}(C,\mathbb{Z})$ equals one. (If the MHS on H^{2N-2}$ does not have pure weight you have to be a bit more careful here.)

If $\Sigma=C_{sing}$ then you have an exact sequence $$H^{2N-5}(C)\to H^{2N-5}(C\setminus \Sigma)\to H^{2N-4}_\Sigma(C)\to H^{2N-4}(C).$$

If I remember correctly there should be a copy of $H^2(\Sigma) $ inside $H^{2N-4}_{\Sigma} (C)$. If this is all of $H^{2N-4}_\Sigma$ then you can relatively easily show that $H^{2N-4}(C)$ is one-dimensional and hence each divisor on $C$ is homologically equivalent to a Cartier divisor.

If $H^{2N-4}_\Sigma(C)$ is bigger then $H^2(\Sigma)$ things are getting complicated.

In the case that $\dim \Sigma=0$, i.e., $C$ has isolated singularities then the only interesting case is $N=4$. Now $H^4_\Sigma$ is the part of the cohomology of the Milnor fiber that is invariant under the monodromy. This can be calculated using Singular.

In some case you can actually calculate the cokernel $K$ of $H^3(C\setminus \Sigma)\to H^4_\Sigma(C)$. For this see e.g., Dimca's paper on Betti numbers and defects of linear systems. It turns out that $K$ is the primitive cohomology group $H^4(C,\mathbb{C})$

The formula Francesco mentioned is a special case of Dimca's approach.

Grooten-Steenbrink and Hulek-K. gave similar formula as Dimca for certain classes of nonisolated singularities.

share|cite|improve this answer

As Parsa explained in his comment, we always have $\textrm{Pic}(C)=\mathbb{Z}$ by Grothendieck.-Lefschetz. However, when $C$ is not smooth this does not mean that $C$ is factorial, that is that every Weil divisor is Cartier.

So we must understand when this happens.

I do not know whether there are satisfactory results in every dimension and for any type of singularities.

Let me give an answer for $N=4$, under the condition that $C$ has only isolated ordinary double points ("nodes").

Then there is the following result:

Theorem. Let $C \subset \mathbb{P}^4$ be a hypersurface of degree $d$ with at most ordinary double points as singularity. Let $\Sigma:=\textrm{Sing}(C)$. Then the following are equivalent:

  1. every divisor on the threefold $C$ is Cartier;
  2. every surface $S \subset C$ is cut out on $C$ by an hypersurface in $\mathbb{P}^4;$
  3. the set $\Sigma$ imposes independent linear conditions on linear forms of degree $2d-5$.

In other words, $C$ is factorial if and only if

$$H^1(\mathcal{O}_{\mathbb{P}^4}(2d-5) \otimes \mathcal{I}_{\Sigma})=0. \quad (\star)$$

If you have an explicit equation for $C$, you can easily check whether condition $(\star)$ holds by using Macauley2.

Cheltsov showed that that if $|\Sigma| <(d-1)^2$ then $C$ is factorial. For instance, a nodal cubic with at most $8$ nodes is factorial.

This result does not hold if $|\Sigma|=(d-1)^2$: in fact, any hypersurface of the form


with $F$ and $G$ general linear forms of degree $d-1$, is not factorial since it contains the $2$-plane $x_o=x_1=0$: notice that there are $(d-1)^2$ nodes on this plane.

For more details on these topics see [I. Cheltsov, Factorial Threefold hypersurfaces, J. Algebraic geometry 19 (2010), no. 4, 781–791] and the references given there.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.