# A question on the Picard group

Let $X$ be a simply connected smooth projective variety, whose Picard group is generated by the classes of the irreducible codimension 1 loci $D_1, \ldots, D_k$. Let $E_1, \ldots, E_r$ be other irreducible codimension 1 loci, and suppose that $X^0$ is the complement in $X$ of the divisors $D_i$ and $E_j$.

Suppose now that $X_0$ is the complement of $n$ irreducible loci of codimension $1$ in $Y$, a smooth projective variety.

Question: Can I conclude that the Picard group of $Y$ has rank $n-r$?

I can answer the question affirmatively over $\mathbb{C}$, by using the long exact sequence with compact support associated with the inclusion $Y \setminus X^0 \to Y$, but I would like to know if there is an algebraic proof of this (valid over any algebraically closed field $k$).

EDIT: As pointed out in the answer, I am actually assuming that the Picard group of $X$ is FREELY generated by the $D_1, \ldots, D_k$.

• I'd guess that, by using etale cohomology, you can give an algebraic proof valid over any algebraically closed field (use the same argument you use over $\mathbb{C}$). – anon Jan 30 '13 at 7:59
• What I prove is that H^2(Y)=H^{1,1}(Y) has rank $n-r$: I have an exact sequence $$0 \to H^2(Y) \to \mathbb{Q}^n \to H^1(X^0)= \mathbb{Q}^r \to H^1(Y)$$ and I have to proove the surjectivity of $\mathbb{Q}^n \to \mathbb{Q}^r$ to conclude. To do so, I prove $\mathbb{Q}^r \to H^1(Y)$ is the zero map, by using the purity of the Hodge structures. – OldMacdonaldHadaForm Jan 30 '13 at 9:59
• The point is that I am not sure how the theory of (mixed) Hodge structures on the ètale cohomology groups works, and if it exists at all. – OldMacdonaldHadaForm Jan 30 '13 at 10:00
• You don't have a MHS on étale cohomology but you have a weight filtration which satisfies exactly the same properties as the weight filtration in mixed Hodge theory. The reference is Deligne's papers on the Weil conjectures. So your proof should work in the same way, by saying that $H^1(X_0)$ has weight $2$ and $H^1(Y)$ has weight $1$ since $Y$ is smooth and projective. You just have to put a subscript $\ell$ next to all of your $\mathbb Q$'s. ;) – Dan Petersen Jan 30 '13 at 10:22
• Thanks! I think my question is solved affermatively now. – OldMacdonaldHadaForm Jan 30 '13 at 15:09

## 2 Answers

I'm going to assume that $X^0$ is the same as $X_0$ and that the Picard group of $X$ is freely generated by the $D_i$, since without that the question doesn't make much sense. In this cases, the answer is yes. This is the same as Mohan's answer, but with a little more detail.

The relevant tool is that if $Z$ is a normal variety and $D$ is an irreducible divisor in $Z$, there's an exact sequence: $$0 \rightarrow \mathcal O(Z)^* \rightarrow \mathcal O(Z \setminus D)^* \rightarrow \mathbb Z \rightarrow \operatorname{Cl}(Z) \rightarrow \operatorname{Cl}(Z \setminus D) \rightarrow 0$$ Here $\mathcal O(Z)^*$ denotes the ring of global functions on $Z$. I'm writing $\operatorname{Cl}$ instead of $\operatorname{Pic}$ because this applies to the Weil divisor group of an arbitrary normal variety. The right part of this exact sequence is Proposition II.6.5 in Hartshorne and it's not hard to see that the left part is also exact.

What this means is that we can define the virtual class rank'' of a variety $Z$ over a field $K$ to be $$\operatorname{rank} (\operatorname{Cl}(Z)) - \operatorname{rank}(\mathcal O(Z)^* / K^*),$$ at least when these ranks are finite. Then, the exact sequence tells us that removing an irreducible divisor drops the virtual class rank by exactly 1. In your case, $X$ is projective, so it has no non-constant regular functions, so its virtual class rank is the rank of its class group, $k$. By induction, the virtual class rank of $X^0$ is $-r$ and for $Y$ it is $n-r$, which is again the rank of the class group of $Y$, because it is projective.

This only tells you about the rank of the torsion-free part of $\operatorname{Pic}(Y) = \operatorname{Cl}(Y)$. If part of your question was whether this Picard group is torsion-free, I'm not sure what you can say about that.

• Thank you very much. This is much more natural and elementary than the approach I was following. – OldMacdonaldHadaForm Jan 31 '13 at 10:14

Here is a fairly elementary proof (actually statement, and the proof is straightforward).

Let $X_0$ be a smooth quasi-projective variety over an algebraically closed field $k$, let $Y$ be a smooth projective completion of $X_0$ and assume that $Y-X_0$ is the union of $n$ distinct irreducible divisors. Then the following are equivalent.

1. $X_0$ is affine, $\mathrm{Pic}\,X_0$ is torsion and the group of units of $X_0$ modulo $k^*$ is an abelian group of rank $r$.

2. $\mathrm{Pic}\,Y$ has rank $n-r$ and the $n$ divisors of $Y-X_0$ generate the Picard group of $Y$ up to torsion.

• Thanks a lot for your answer. I picked the other one because I could understand better the details ;) – OldMacdonaldHadaForm Jan 31 '13 at 10:18