Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

All varieties will be over $\mathbb{C}$ and projective unless stated otherwise.

In Beauville - complex algebraic surfaces, the following is described: Let $S$ be a smooth surface and $p \in S$ a point. Let $\epsilon: \tilde S \rightarrow S$ be the blowup at $p$ and $E$ the resulting exceptional curve. Then $$ \text{Pic}(\tilde S) \cong \epsilon^* \text{Pic}(S) \oplus \mathbb{Z} E $$ With $\text{Pic}$ i mean either the group of invertible sheaves or those of Cartier divisors modulo equivalence.

Question 1: I was wondering about the situation when $p$ is a simple double singularity, a node, instead of being smooth and the rest of $S$ is smooth. Does the same formula hold? If not, is there is a similar formula that describes $\text{Pic}(S)$ as a direct summand of $\text{Pic}(\tilde S)$?

Question 2: Does anybody know a reference for this situation: relationship of Picard groups of singular surfaces (or varieties in general) with their smoothification?

My guess and thoughts so far:

My first guess was that the same would hold, but i think that is false. The question is treated locally in Hartshorne example 6.5.2, which examines $$ \text{Spec}(\mathbb{C}[x,y,z]/(xy - z^2)) $$ The Weil class group of this affine variety is $\mathbb{Z}/2\mathbb{Z}$ and is generated by a ruling of the cone. This ruling is not a Cartier divisor and the cartier divisor class group is trivial.

This makes me conjecture: \begin{align*} \text{WCl}(\tilde S) &\cong \epsilon^* \text{Wcl}(S) \oplus \mathbb{Z} E\\ \text{Pic}(\tilde S) &\cong \epsilon^* \text{Pic}(S) \oplus \mathbb{Z} R \oplus \mathbb{Z} E \end{align*} where $R$ corresponds to a ruling of the cone, which was a weil divisor on the singular variety but after the blowup corresponds to a Cartier divisor as well. $\text{WCl}$ means the group of Weil divsors modulo equivalence.

This is just an intuitive guess coming from Hartshorne's example so please please correct me if i'm wrong.

Thanks!

share|improve this question

1 Answer 1

up vote 1 down vote accepted

I think indeed, that the first guess does not quite hold, namely the group that you wrote can be only a subgroup of finite index in $\tilde S$. Consider the following example, the singular quadric in $\mathbb P^3$, $x_0^2+x_1^2+x_2^2=0$. Picard group of this quadric is $\mathbb Z$ and it is generated by $O(1)$, the corresponding Cartier divisor (generator of Picard) is a hyperplane section of the quadric, let us call it $C$. Clearly $C^2=2$. When you blow up the quadric for the exceptional curve $E$ you have $E^2=-2$, and $E\tilde C=0$. So $E$ and $\tilde C$ generate a subgroup of $Pic \tilde S$ such that the intersections of any two members in this subgroup is even. At the same time you can find curves on $\tilde S$ whose intersection is equal to $1$. So $E$ and $\tilde C$ don't generate whole $Pic \tilde S$.

Maybe the notes of Miles Reid might be of some help to you? http://arxiv.org/abs/alg-geom/9602006

share|improve this answer

Your Answer

 
discard

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.