Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 113 / 2 / c

A Cartier divisor on an integral scheme is an open cover $(U_i)$ together with nonzero rational functions $f_i$ such that every ratio $f_i/f_j$ is a regular unit on $U_i\cap U_j$.

Every prime Weil divisor on ${\mathbb{P}}_k^n$ is cut out by an irreducible homogeneous polynomial because the polynomial ring $k[x_0,\dots ,x_n]$ is a unique factorization domain. Hence any Weil divisor can be represented by a homogeneous rational expressionof some total degree $d$. On the standard chart $U_i=D_{+}(x_i)$, put $f_i=F/x_i^d$. This is a degree-zero rational function, and on $U_i\cap U_j$,is a regular unit. These local equations form a Cartier divisor whose associated Weil divisor is the original one.

Now put $P=X\times {\mathbb{P}}^n$ and let $H=X\times {\mathbb{P}}^{n-1}$ be the hyperplane divisor at infinity. Its complement is $X\times {\mathbb{A}}^n$. Iterating the given invariance under multiplication by ${\mathbb{A}}^1$ givesThe localization sequence for the divisor class group shows that every class on $P$ is a pullback of a class on $X$ plus an integer multiple of $[H]$. Restriction to the generic fiber ${\mathbb{P}}_{K(X)}^n$ kills pullbacks from $X$ and sends $[H]$ to the generator of $\mathrm{Cl}({\mathbb{P}}_{K(X)}^n)\cong \mathbb{Z}$. Therefore the sum is direct, provingThis is the divisor class group of a projective-space bundle with trivial vector bundle.