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

A morphism of schemes is proper when it is of finite type, separated, and universally closed.

Each property is local on the target. Finite type is local by its affine definition. The restrictions of the diagonal morphism of $\phi$ over the open sets $U_i$ are closed immersions; since being a closed subset is local on an open cover, the diagonal itself is a closed immersion, so $\phi$ is separated. Finally, after any base change $Y^{\prime }\to Y$, the inverse images $U_i^{\prime }=Y^{\prime }{\times }_Y{U}_i$ cover $Y^{\prime }$. For every closed $C\subseteq X{\times }_Y{Y}^{\prime }$, its image has closed intersection with every $U_i^{\prime }$ because the restricted base-changed morphism is closed. The image is therefore closed in $Y^{\prime }$. Thus $\phi$ is universally closed and hence proper. This proves that properness is local on the target.

Write $f:U={\mathbb{A}}_{\mathbb{C}}^1\setminus \{1\}\to {\mathbb{A}}_{\mathbb{C}}^1$ for $f(t)=t^2$. For any closed point $a\in \mathbb{C}$, the fiber is the zero-dimensional closed subscheme of $U$ cut out by $t^2-a$. It is finite over $\mathbb{C}$, and every finite morphism is proper. This includes $a=1$, whose fiber in $U$ consists only of $t=-1$.

The morphism itself fails the valuative criterion for properness. Take the discrete valuation ring $R=\mathbb{C}[[s]]$ with fraction field $K=\mathbb{C}((s))$. The $K$-point $t=1+s$ lies in $U(K)$ and lies over the $R$-point of the target defined byAny extension $\mathrm{Spec}R\to {\mathbb{A}}^1$ of that $K$-point must still be given by $t=1+s$, whose closed point maps to the deleted point $t=1$. It therefore cannot factor through $U$. The required lift does not exist, so $f$ is not proper. This is an instance of the fact that proper closed-point fibers do not imply properness.

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.