= Solution
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> $k[x_0,\ldots,x_n]$ is a <unique factorization domain>. Hence any <Weil divisor> can be represented by a homogeneous rational expression
$$
F=\prod_jF_j^{m_j}
$$
of 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 $U_i\cap U_j$,
$$
\frac{f_i}{f_j}=\left(\frac{x_j}{x_i}\right)^d
$$
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 $\mathbb A^1$ gives
$$
\operatorname{Cl}(X\times\mathbb A^n)\cong\operatorname{Cl}(X).
$$
The <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 $\mathbb P^n_{K(X)}$ kills pullbacks from $X$ and sends $[H]$ to the generator of $\operatorname{Cl}(\mathbb P^n_{K(X)})\cong\mathbb Z$. Therefore the sum is direct, proving
$$
\operatorname{Cl}(X\times\mathbb P^n)
\cong\operatorname{Cl}(X)\oplus\mathbb Z.
$$
This is the <divisor class group of a projective-space bundle with trivial vector bundle>.
Back to article page
