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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.

Thanks.

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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
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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
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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
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2 Answers

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.

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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

$$x_0F+x_1G=0,$$

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.

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