The following got no answer on mathstackexchange. I believe it not to be hard, but maybe it is a little specialized?


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:
$$
\text{WCl}(\tilde S) \cong \epsilon^\* \text{Wcl}(S) \oplus \mathbb{Z} E
$$
$$
\text{Pic}(\tilde S) \cong \epsilon^\* \text{Pic}(S) \oplus \mathbb{Z} R \oplus \mathbb{Z} E
$$
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!



Divisor class group on blowup of nodal surface
This is a partial answer to the part concerning Weil divisors. 


I prefer to use the terminology of cycles of codimension $1$ instead of Weil divisors. The group of Weil divisors modulo rational equivalence is the Chow group $A^1$. 


Let $f : \tilde{S}\to S$ be any proper birational morphism of integral normal algebraic varieties (integral normal schemes of finite type over a field) of dimension $\ge 2$. Let $E\_1, \dots, E\_n$ be the irreducible components of codimension $1$ (in $\tilde{S}$) of the exceptional locus of $f$. 


As $f$ is proper, there is a canonical pushforward map $f\_\\*: A^1(\tilde{S})\to A^1(S)$ (see Fulton, Intersection Theory, Chapter 1) and it is a group homomorphism. As $S$ is normal, $f$ is an isomorphism outside of a codimension $\ge 2$ closed subset of $S$. In particular, $f\_\\*$ is surjective because for any irreducible cycle $\Gamma$ of codimension $1$ in $S$, $f\_\*$ maps the class of the strict transform $\tilde{\Gamma}$ to the class of $\Gamma$. However, in general $f\_\\*$ doesn't have a section. 


By construction, the kernel of $f\_\\*$ is generated by the classes of the $E\_i$. Let us show the map $\oplus\_{1\le i\le n} [E\_i]\mathbb Z \to A^1(\tilde{S})$
is injective. Let $g\in k(\tilde{S})$ be a rational function with divisor $\mathrm{div}\_{\tilde{S}}(g)$ supported in $\cup\_i {E\_i}$. Then the divisor $\mathrm{div}\_{S}(g)$ of $g$ as a rational function on $S$ is $0$. As $S$ is normal, this forces $g$ to be a unit in the ring of regular functions $O(S)$, hence $g$ is a unit in $O(\tilde{S})$ and $\mathrm{div}\_{\tilde{S}}(g)=0$. Conclusion, we have an exact sequence
$$0\to \oplus\_{1\le i\le n} [E\_i]\mathbb Z \to A^1(\tilde{S}) \to A^1(S) \to 0.$$ 


