# What is the Picard group of $z^3=y(y^2-x^2)(x-1)$?

I'm actually doing much more with this affine surface than just looking for the Picard group. I have already proved many things about this surface, and have many more things to look at it, but the Picard group continues to elude me.

One of the biggest problems seems to be that I'm not really sure what tools I have at my disposal to attempt such a problem. This surface has 4 singularities, one of which (the origin) is particularly nasty (the exceptional fiber over the origin when blowing-up is an elliptic curve).

Let $X=\mathcal{Z}(z^3-y(y^2-x^2)(x-1))$ be the surface. I know that the divisor class group of the surface is $\mathrm{Cl}(X)\cong (\mathbb{Z}/3\mathbb{Z})^{3}\oplus \mathbb{Z}^{2}$, and that the Picard group is (isomorphic to) a subgroup of this. If we let $p_i,i=1,2,3,4$ be the singular points of $X$, then there is an exact sequence

$0\rightarrow\mathrm{Pic}(X)\rightarrow\mathrm{Cl}(X)\rightarrow\bigoplus\mathrm{Cl}(\mathcal{\hat{O}}_{X,p_i})$,

though the hat (for completion) only really matters on the singularity at the origin. I have shown that the three generators for the torsion part of $\mathrm{Cl}(X)$ map to linearly independent elements of this last direct sum, so nothing in the torsion subgroup can be in the kernel of that map, which by exactness equals $\mathrm{Pic}(X)$.

This is where I get stuck. I don't really know what else I can do; most of the things I can find in the literature seems to be only for nonsingular surfaces, or surfaces where the singularities are more simple than the mess at $(0,0,0)$.

I'd like to thank in advance anyone who takes some time to help me out.

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I don't know what you want exactly. You already proved that Pic$(X)$ is isomorphic to $\mathbb Z^2$. – user18119 Feb 25 '13 at 8:49
@QiL'8 I don't know what the OP looked for, but I'd like to know how to compute the divisor class group of $X$ and which commutative ring (coordinate ring?) corresponds to the $X$ above (is this $\mathbb C[X,Y,Z]/(Z^3-Y(Y^2-X^2)(X-1))$?). Furthermore, the divisor class group of $X$ is the same as the divisor class group of the ring that corresponds to $X$? – user26857 Feb 25 '13 at 10:17
@YACP: yes the coordinate ring $O(X)$ of $X$ is what you write. When $X$ is normal (equivalently $O(X)$ is integrally closed), the divisor class group of $O(X)$ is the same as Cl$(X)$. I don't know a general approach to compute this group (in terms of what ?). – user18119 Feb 25 '13 at 20:30
@QiL'8 Thank you for your answer. I didn't ask for a general approach. I was eager to see how can use geometric arguments to compute the divisor class group for this $X$ which the OP says that is $\mathrm{Cl}(X)\cong (\mathbb{Z}/3\mathbb{Z})^{3}\oplus \mathbb{Z}^{2}$. (If you like, I can post this as a different question.) – user26857 Feb 25 '13 at 21:42
@YACP: I don't know even for this particular case. You may post on mathoverflow. – user18119 Feb 25 '13 at 21:54