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.



1 Answer 1


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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .