Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 113 / 4

A morphism of schemes $i:Z\to X$ is a closed immersion if it identifies $Z$ homeomorphically with a closed subset of $X$ and the morphism of sheavesis surjective. Equivalently, every point of $X$ has an affine neighborhood $\mathrm{Spec}A$ on which $i$ is isomorphic to $\mathrm{Spec}(A/I)\to \mathrm{Spec}A$ for some ideal $I\subseteq A$.

A closed subscheme of $X$ is a scheme $Z$ supplied with a closed immersion $Z\hookrightarrow X$, considered up to isomorphism over $X$.

Solved by gpt-5.6-sol high.

On an affine open $W=\mathrm{Spec}A\subseteq X$, write $Z\cap W=V(I)$. Give $Z\cap W$ the reduced induced subscheme structure $\mathrm{Spec}(A/\sqrt{I})$. These constructions agree under localization and therefore glue to a reduced closed subscheme $Z\hookrightarrow X$.

Let $Z^{\prime }\hookrightarrow X$ be another closed immersion with the same underlying closed set. Affine-locally write $Z^{\prime }=\mathrm{Spec}(A/J)$. Since $V(J)=Z\cap W$, its vanishing ideal is $\sqrt{J}=\sqrt{I}$, and the inclusion $J\subseteq \sqrt{J}$ induces a quotient homomorphismContravariance of the spectrum of a commutative ring gives a factorizationThe quotient maps force these local factorizations to agree on overlaps, so they glue. They are also the only possible maps over $X$, which proves uniqueness.

Solved by gpt-5.6-sol high.

Write $X=\mathrm{Spec}B$, $Y=\mathrm{Spec}A$, and let the morphism correspond to a homomorphism $\phi :A\to B$. Put $I=\ker \phi$ andThe injection $A/I\hookrightarrow B$ gives $X\to Z$, and the quotient $A\to A/I$ gives a closed immersion $Z\hookrightarrow Y$, so $f$ factors through $Z$.

If $f$ factors through another closed subscheme $Z^{\prime }=\mathrm{Spec}(A/J)$, then $J\subseteq \ker \phi =I$. The resulting quotient $A/J\to A/I$ induces the unique factorization $Z\to Z^{\prime }\to Y$. Thus $Z$ is the scheme-theoretic image.

It remains to identify its underlying set. A principal open subscheme $D(a)\subseteq \mathrm{Spec}A$ misses $f(X)$ exactly when $D(\phi (a))$ is empty, equivalently when $\phi (a)$ is a nilpotent element. Hence the ideal of functions vanishing set-theoretically on $f(X)$ has radical $\sqrt{\ker \phi }$. The closure is thereforeas required.

Solved by gpt-5.6-sol high.