# Hartshorne Exercise II.6.1 Proving that $\operatorname{Cl}(X\times\mathbb{P}^n)\cong \operatorname{Cl}(X)\times \mathbb{Z}$.

I have been stuck on this computation for some time (at least a year now since I started learning AG seriously).

The trick is to use Proposition II.6.5 and take the closed set $$Z$$ to be the hyperplane at $$\infty$$ to get a sequence $$\mathbb{Z}\rightarrow \operatorname{Cl}(X\times \mathbb{P}^n)\rightarrow \operatorname{Cl}(X\times\mathbb{A}^n)\cong \operatorname{Cl}(X)\rightarrow 0$$ with an inductive application of Proposition II.6.6. Now we prove that there is a section of the map on the right $$\operatorname{Cl}(X\times \mathbb{P}^n)\rightarrow \operatorname{Cl}(X\times\mathbb{A}^n)$$. To complete the proof, it then suffices to show that the left hand side map is injective. Here, I am stuck since I cannot seem to find a description of the closed set $$Z$$ and the generic point of $$X\times\mathbb{P}^n$$.

I've looked explanation like this but it flew over my head and wasn't really enlightening. Any thoughts / suggestions?

$$\def\ClGrp{\operatorname{Cl}} \def\PP{\mathbb{P}} \def\ZZ{\mathbb{Z}} \def\AA{\mathbb{A}}$$

To show $$\ClGrp(X\times\PP^n)\cong (\ClGrp X)\times\ZZ$$, we use the exact sequence of proposition II.6.5. Let $$i:X\times\PP^{n-1}\to X\times\PP^n$$ be the closed immersion with image $$X\times V(T_0)$$, and $$j:X\times\AA^n\to X\times\PP^n$$ be the open immersion with image $$X\times D(T_0)$$. By proposition II.6.5, we have an exact sequence $$\ZZ\cdot (X\times\PP^{n-1}) \to \ClGrp(X\times\PP^n) \to \ClGrp(X\times\AA^n) \to 0$$ and the third item is exactly $$\ClGrp(X)$$ by repeated applications of proposition II.6.6. It remains to show that the first map is injective and the exact sequence splits.

Let $$Z$$ denote the image of $$i:X\times\PP^{n-1}\to X\times\PP^n$$. If $$aZ$$ is a principal divisor, then $$aZ\cap \PP^n_{k(X)}$$ must also be a principal divisor: the nonzero function in $$k(X\times\PP^n)$$ cutting out $$aZ$$ is also a nonzero function in $$k(\PP^n_{k(X)})$$ which cuts out the divisor $$aZ\cap \PP^n_{k(X)}$$. So to show that $$aZ$$ cannot be principal for any $$a\neq 0$$, it suffices to show that $$aZ\cap \PP^n_{k(X)}$$ is not principal for any $$a$$.

Suppose that $$aZ\cap \PP^n_{k(X)}$$ is principal for some $$a\neq 0$$, and let $$f\in k(X\times\PP^n)$$ be a rational function with divisor $$aZ\cap \PP^n_{k(X)}$$. Looking at $$\AA^n_{k(X)} = D(T_0)\subset \PP^n_{k(X)}$$, we see that since the valuation of $$f$$ is zero at every codimension one point by assumption, we must have that $$f$$ and $$1/f$$ are both in the localization of $$k(X)[t_1,\cdots,t_n]$$ at every height one prime. Therefore by proposition II.6.3A, $$f$$ and $$1/f$$ belong to $$k(X)[t_1,\cdots,t_n]$$ and are units in that ring. As the only units in a polynomial ring over a field are the nonzero elements of the field, we have that $$f\in k(X)^\times$$, so its valuation in the local ring of $$V(T_0)$$ is zero. Thus $$aZ$$ is not a principal divisor for any $$a\neq 0$$ and the map $$\ZZ\cdot (X\times\PP^{n-1}) \to \ClGrp(X\times\PP^n)$$ is injective.

We can define a splitting of $$\ClGrp(X\times\PP^n) \to \ClGrp(X\times\AA^n)$$ by writing a map $$\ClGrp(X)\to \ClGrp(X\times\PP^n)$$ given on prime divisors by sending $$D_i$$ to $$D_i\times\PP^n$$ and composing with the isomorphism $$\ClGrp(X\times\AA^n) \to \ClGrp(X)$$ from proposition II.6.6. This map sends a prime divisor $$D_i\subset X$$ to $$D_i\times \PP^n$$ and then $$D_i\times\AA^n$$ and finally back to $$D_i$$ by construction, so this is a splitting, and by a general lemma from homological algebra this splitting gives that $$\ClGrp(X\times\PP^n)\cong \ClGrp(X)\times\ZZ$$ and we're finished.

• Okay, so I follow most of the steps except the one where $f$ has valuation zero at every codimension one point. What is the assumption we use here? Thank you for the detailed post. I'll be dissecting it :) Commented Jan 20, 2022 at 20:21
• That $aZ\cap\Bbb P^n_{k(X)}$ is (set-theoretically) $V(T_0)$ in $\Bbb P^n_{k(X)}$ and therefore does not intersect $D(T_0)$. Commented Jan 20, 2022 at 20:23
• That makes sense. Why does the nonzero function $f\in k(X\times \mathbb{P}^n)$ cutting out $aZ$ need be in $k(\mathbb{P}^n_{k(X)})$ and cutting out $aZ\cap \mathbb{P}^n_{k(X)}$? Is there some sort of inclusion of function fields I am missing here? Commented Jan 20, 2022 at 20:30
• The function fields $k(X\times\Bbb P^n)$ and $k(\Bbb P^n_{k(X)})$ are the same, and the local rings of $Z$ and $Z\cap\Bbb P^n_{k(X)}$ are the same. This is because they're just the local rings of the generic points of $X\times\Bbb P^n$ and $\Bbb P^n_{k(X)}$ and $Z$ and $Z\cap\Bbb P^n_{k(X)}$, since taking the fiber product with $k(X)$ is the same as just restricting the scheme structure of $X\times \Bbb P^n$ to the set-theoretic fiber of $X\times\Bbb P^n$ over $k(X)$. Commented Jan 20, 2022 at 20:44
• Okay that makes sense (and I was able to prove it more or less directly and using uniqueness of generic points for integral schemes). One final question (reality check question). The reason the valuation of $f$ on $V(T_0)$ is zero is just because it is a unit $f\in K(X)^\times$? Commented Jan 20, 2022 at 21:16