Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 113 / 2

A Weil divisor on $X$ is a finite formal sumover integral codimension-one closed subschemes $Z$. Since $X$ is regular in codimension one, the local ring at the generic point of each $Z$ is a discrete valuation ring. A nonzero rational function $f\in k(X{)}^{\times }$ therefore defines the principal divisor $\mathrm{div}(f)={\sum }_Z{v}_Z(f)[Z]$. The divisor class group is

The scheme ${\mathbb{P}}_k^3$ is Noetherian, integral, separated, and regular. Every open subscheme inherits these properties, so $U={\mathbb{P}}_k^3\setminus C$ satisfies $(\star )$. Because $C$ has dimension one, it has codimension two in ${\mathbb{P}}_k^3$ and contains no prime Weil divisor. The localization sequence for the divisor class group therefore makes restriction an isomorphismThe hyperplane divisor generates the class group of projective space, so

The assertion is false. For $n\ge 2$, take the affine hypersurfaceIt is an integral scheme, and its only possible singular point is the origin, which has codimension two. Hence it is regular in codimension one; as a hypersurface it satisfies Serre's condition $(S_2)$, so the Serre criterion for normality also makes it a normal scheme. The Divisor class group of an A-type surface singularity isgenerated by $V(x,z)$. Thus a closed affine subscheme satisfying $(\star )$ can have nonzero torsion in its class group.

The standard affine charts of ${\mathbb{P}}^3$ are copies of ${\mathbb{A}}^3$, and the product charts of ${\mathbb{P}}^1\times {\mathbb{P}}^1$ are copies of ${\mathbb{A}}^2$. Their local rings are localizations of polynomial rings over $k$ and hence are regular local rings. Both schemes are therefore regular.

On a regular integral scheme every Weil divisor is Cartier, so the divisor class group is naturally the Picard group. Pullback of line bundles along the Segre embedding therefore definesThe two groups areThe Segre coordinates are bihomogeneous of bidegree $(1,1)$, so ${\iota }^{\ast }{\mathcal{O}}_{{\mathbb{P}}^3}(1)=\mathcal{O}(1,1)$. In these bases the map is $m\mapsto (m,m)$ and