# divisor class group of a product of schemes

The first part of this question is quite general: let $X$ and $Y$ be noetherian integral separated schemes which are regular in codimension one. Is there any relationship between the divisor class group $\text{Cl}(X \times Y)$ and the groups $\text{Cl}(X)$, $\text{Cl}(Y)$? For instance, I think there are natural maps on divisors $\text{Div}(X) \to \text{Div}(X \times Y)$ and $\text{Div}(Y) \to \text{Div}(X \times Y)$ which send $Z \mapsto Z \times Y$ and $Z \mapsto X \times Z$. Are these maps well-defined modulo linear equivalence? If so, are the induced maps on class groups injective?

This question occurred to me while working on Hartshorne exercise II.6.1, which claims that $\text{Cl}(X \times \mathbb{P}^n) \cong \text{Cl}(X) \times \mathbb{Z}$. In this case, taking $H_{\infty} \subset \mathbb{P}^n$ to be the hyperplane at infinity, $Z = X \times H_{\infty}$, and $U = X \setminus Z$, there is a natural surjection $\text{Cl}(X \times \mathbb{P}^n) \to \text{Cl}(U) \cong \text{Cl}(X)$ whose kernel is the image of the map $\mathbb{Z} \to \text{Cl}(X \times \mathbb{P}^n)$ which sends $n \mapsto n \cdot Z$. Why is this map injective?

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What do you mean by Div$(X)$ ? Cartier divisors ? – user18119 Nov 29 '11 at 22:44
I meant Weil divisors, hence all the hypotheses to ensure that they are well-behaved. – Justin Campbell Nov 29 '11 at 22:49
So you are talking about cycles of codimension 1. Note that the hypothesis on the regularity in codimension 1 doesn't help. So consider $S$-schemes $X, Y$. In general $Z\times_S Y$ needs not be of codimension 1 in $X\times_S Y$ (for example take $Y\subset S$ and look on what happens). When $Y\to S$ is faithfully flat, then things go better, you can go through linear equivalence. If moreover $Y\to S$ has a section, then $X\times_S Y\to X$ has a section and this implies that $\mathrm{Cl}(X)\to \mathrm{Cl}(X\times_S Y)$ is injective. – user18119 Nov 29 '11 at 23:45

For the general situation, you have to consider the fiber product of $X, Y$ over some base scheme, then see this answer in MO.
For the specific question, you probably ask why the map is surjective. Take a Weil divisor $D=\sum_i n_i Z_i$ on $U$, then the class of the Weil divisor $\overline{D}:=\sum_i n_i\overline{Z}_i$, where $\overline{Z}_i$ is the Zariski closure of $Z_i$ in $X\times \mathbb P^n$, is mapped to the class of $D$ in Cl$(U)$.
Add answer to the real question: suppose $kZ=0$ in Cl$(X\times \mathbb P^n)$. One can take $k\le 0$. Let $f$ be a rational function whose divisor is $kZ$. Then $f$ has no pole, hence $f\in O(X\times\mathbb P^n)=O(X)$ and the divisor of $f$ is of the form $D\times \mathbb P^n\ne kZ$ unless $k=0$.
Thanks, but although this MO question is related, I don't think it actually answers my general question. And I was asking why the map $\mathbb{Z} \to \text{Cl}(X \times \mathbb{P}^n)$ is injective. – Justin Campbell Nov 29 '11 at 16:04
Thanks! One point of clarification: when you say $f \in O(X \times \mathbb{P}^n)$, do you mean $f$ that is a global section of the structure sheaf? – Justin Campbell Nov 30 '11 at 17:37
@JustinCampbell: yes, but $O()$ I mean the sectios of the structural sheaf. – user18119 Nov 30 '11 at 20:37
Would either of you be able to explain the following points to me? 1. Why does $(f) = kZ$, $k \le 0$, imply $f$ has no pole? Hartshorne in fact defines a pole to occur along a divisor with negative coefficient. 2. What's the easiest way to see the equality $O(X \times \mathbb{P}^n) = O(X)$? 3. Finally, why does $f$'s being a global section imply the divisor has that form? Thanks for any help! – Joshua Ciappara Jul 20 at 2:02