Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 20 / 4

A closed immersion $i:X\to Y$ is a morphism of schemes whose underlying map is a homeomorphism onto a closed subset and whose map of structure sheavesis surjective. Equivalently, $X$ is the closed subscheme defined by a quasi-coherent ideal sheaf of a closed subscheme on $Y$. The affine-local characterization is particularly useful: for every affine open subscheme $V=\mathrm{Spec}A\subseteq Y$,for an ideal $I\subseteq A$, with the map induced by the quotient homomorphism.

Work over an affine open subscheme $V=\mathrm{Spec}A$ of $Z$. Since $g$ is a closed immersion, its inverse image is $\mathrm{Spec}(A/I)$ for some ideal $I$. Since $f$ is a closed immersion, the inverse image of this affine scheme in $X$ is $\mathrm{Spec}((A/I)/J)$ for an ideal $J\subseteq A/I$. Let $\tilde{J}$ be the inverse image of $J$ in $A$. The composite on this open is induced by the surjectionThe affine-local characterization from part (a) therefore shows that $g\circ f$ is a closed immersion. In particular, closed-subscheme ideals compose by taking the inverse image of the second quotient ideal.

Over an affine open subscheme $\mathrm{Spec}A$ of the target, a closed immersion has the form $\mathrm{Spec}(A/I)\to \mathrm{Spec}A$. Pulling it back along an affine scheme map $\mathrm{Spec}B\to \mathrm{Spec}A$ givesThe identification follows directly by tensoring the quotient presentation; it does not require $B$ to be flat over $A$. Since $B\to B/IB$ is surjective, this is again a closed immersion. These affine calculations cover every base change of a morphism of schemes.

A morphism of schemes $f:X\to Y$ is a separated morphism if its diagonal morphismis a closed immersion. The diagonal morphism is induced by the pair of identity maps on $X$. Thus separatedness is a condition relative to the given target $Y$, not a requirement that $X$ be absolutely separated.

Since $g$ is a separated morphism, the diagonal morphism ${\Delta }_g:Y\to Y{\times }_{Z}Y$ is a closed immersion. The squareis cartesian. Indeed, on test schemes its pullback condition is $(f(x),y)=(t,t)$, forcing $y=t=f(x)$, and leaving exactly the choice of $x$. The graph morphism of schemes ${\Gamma }_f$ is therefore a base change of a morphism of schemes of ${\Delta }_g$. Part (c) shows that ${\Gamma }_f$ is a closed immersion.

Now assume additionally that $h=g\circ f:X\to Z$ is a closed immersion. The projection $p_Y:X{\times }_{Z}Y\to Y$ is its base change of a morphism of schemes along $g$, so is a closed immersion by part (c). Sincepart (b) shows that $f$ is a closed immersion. This covers the further assumption in the final unheaded paragraph.