Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 113 / 2

The fibre product of schemes $X{\times }_{S}Y$ comes with projections to $X$ and $Y$ having equal composites to $S$, and is universal with this property. A morphism is universally closed when every base change is closed on underlying topological spaces.

The structural morphism ${\mathbb{A}}_k^1\to \mathrm{Spec}k$ is closed because its target has one point, but it is not universally closed. After base change by ${\mathbb{A}}_k^1$, it becomes the projection ${\mathbb{A}}_k^2\to {\mathbb{A}}_k^1$; the closed hyperbola $V(xy-1)$ has image $D(x)$, which is not closed.

For a finite-type universally closed nonseparated example, glue two copies of ${\mathbb{P}}_k^1$ along the complement of one rational point, producing a projective line with a doubled point. It is of finite type. After every base change, each of its two projective-line charts maps closedly to the base, so the image of any closed subset, being the union of the two closed images, is closed. Thus the structural morphism is universally closed. It is not separated because the two doubled points violate uniqueness in the valuative criterion for separatedness, or equivalently because its diagonal is not closed.

Fix $x\in X$ and choose affine opens $x\in U\subseteq X$ and $f(x)\in W\subseteq S$ with $f(U)\subseteq W$. The open subset $U{\times }_{W}U$ of $X{\times }_{S}X$ contains $\Delta (x)$. If $U=\mathrm{Spec}A$ and $W=\mathrm{Spec}R$, its diagonal is induced by the surjectionso it is a closed immersion. Hence every diagonal morphism is locally a closed immersion into an open subset, and therefore is a locally closed immersion.

For ${\mathbb{A}}_k^1$, the product is ${\mathbb{A}}_k^2=\mathrm{Spec}k[x,y]$ and the diagonal is $V(x-y)$. Its complement is the principal affine open

Take instead the separated scheme $X={\mathbb{A}}_k^2$. The complement of its diagonal in $X{\times }_{k}X\cong {\mathbb{A}}_k^4$ is ${\mathbb{A}}_k^4\setminus {\mathbb{A}}_k^2$, where the removed diagonal has codimension two. Its global functions still form the polynomial ring of ${\mathbb{A}}_k^4$. Were the complement affine, its canonical morphism to $\mathrm{Spec}$ of this ring would identify it with all of ${\mathbb{A}}_k^4$, which is impossible. Thus this diagonal complement is not affine.

For a Noetherian separated integral scheme regular in codimension one, a Weil divisor is a finite integer combination of integral codimension-one closed subschemes. Principal divisors are the valuation divisors of nonzero rational functions, and the divisor class group is

For an open immersion $U\subseteq X$, restriction induces a surjectionevery prime divisor of $U$ closes to a prime divisor of $X$, while divisors supported in $X\setminus U$ form the kernel. Therefore $\mathrm{Cl}(X)=0$ implies $\mathrm{Cl}(U)=0$.

Affine schemes need not have trivial class group. For example,is a normal affine surface with $\mathrm{Cl}(X)\cong \mathbb{Z}/2\mathbb{Z}$. The height-one prime $(x,z)$ represents the nonzero class: twice this divisor is the principal divisor of $x$, but the divisor itself is not principal.