# 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.

• I don't know what you want exactly. You already proved that Pic$(X)$ is isomorphic to $\mathbb Z^2$.
– user18119
Feb 25, 2013 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, 2013 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, 2013 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, 2013 at 21:42

The Picard group of $$X$$ is the set of divisor class group elements that map to 0 in $$\bigoplus\mathrm{Cl}(\mathcal{\hat{O}}_{X,p_i})$$. In other words, it is the set of divisors on $$X$$ whose restriction to each local ring $$\mathcal{O}_{X,p_i}$$ is principal.

Since the torsion subgroup of $$\mathrm{Cl}(X)$$ maps injectively into $$\bigoplus\mathrm{Cl}(\mathcal{\hat{O}}_{X,p_i})$$, we know that the torsion subgroup of $$\mathrm{Pic}(X)$$ is trivial. Therefore, $$\mathrm{Pic}(X)$$ is a free abelian group with basis given by the restriction of any set of generators for $$\mathrm{Cl}(X)$$ to $$X$$.

Now we just need to find a set of generators for $$\mathrm{Cl}(X)$$. We know that the map $$\mathrm{Cl}(X)\to\bigoplus\mathrm{Cl}(\mathcal{\hat{O}}_{X,p_i})$$ is surjective, so it suffices to find a set of generators for each $$\mathrm{Cl}(\mathcal{\hat{O}}_{X,p_i})$$.

At the nonsingular points $$p_1, p_2, p_3$$, we have $$\mathrm{Cl}(\mathcal{\hat{O}}_{X,p_i})\cong\mathbb{Z}$$, so any element of $$\mathrm{Cl}(X)$$ whose restriction to each of these points is principal will generate $$\mathrm{Cl}(X)$$.

At the singular point $$p_4$$, we have $$\mathrm{Cl}(\mathcal{\hat{O}}_{X,p_4})\cong\mathbb{Z}\oplus\mathbb{Z}/3\mathbb{Z}$$, so we need to find an element of $$\mathrm{Cl}(X)$$ whose restriction to $$p_4$$ is principal and has order 3.

It turns out that the divisor class of the line $$y=0$$ has this property. To see this, note that the local equation for $$X$$ at $$p_4$$ is $$z^3=x(x-1)$$, so a generator for $$\mathrm{Cl}(\mathcal{\hat{O}}_{X,p_4})$$ is given by the restriction of the divisor class of the line $$x=0$$. The restriction of the divisor class of the line $$y=0$$ to $$\mathcal{\hat{O}}_{X,p_4}$$ is then $$(1/3)(0,0,1)-(1/3)(0,0,-1)=0$$, so it is principal.

Now we just need to check that the divisor class of the line $$y=0$$ has order 3 in $$\mathrm{Cl}(X)$$. To do this, note that the local equation for $$X$$ at the origin is $$z^3=y^2(x-1)$$, so a generator for $$\mathrm{Cl}(\mathcal{\hat{O}}_{X,0})$$ is given by the restriction of the divisor class of the line $$x=1$$. The restriction of the divisor class of the line $$y=0$$ to $$\mathcal{\hat{O}}_{X,0}$$ is then $$(1/3)(0,1,1)-(1/3)(0,-1,1)=0$$, so it is principal.

Therefore, the divisor class of the line $$y=0$$ generates $$\mathrm{Cl}(X)$$, and the Picard group of $$X$$ is the free abelian group with basis given by the restriction of this divisor class to $$X$$.