Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 16 / 2

A proper morphism of schemes is a separated morphism of finite type which is a universally closed morphism: after every base change, the resulting map of underlying topological spaces takes closed subsets to closed subsets.

The valuative criterion for properness, in the Noetherian setting used here, says that a finite type morphism $f:X\to Y$ is a proper morphism if and only if the following existence and uniqueness condition holds. For every valuation ring $V$, its fraction field $K$, and compatible morphisms of schemes $a:\mathrm{Spec}K\to X$ and $b:\mathrm{Spec}V\to Y$ with $f\circ a=b{|}_{\mathrm{Spec}K}$, there is exactly one morphism of schemes $h:\mathrm{Spec}V\to X$ such that $h{|}_{\mathrm{Spec}K}=a$ and $f\circ h=b$. All valuation rings are allowed; no discreteness assumption is needed in this formulation. The following arguments also use that each of the three defining conditions for a proper morphism is preserved by base change.

A finite morphism is a finite type morphism. It is also a separated morphism: on an affine scheme chart $\mathrm{Spec}B\to \mathrm{Spec}A$, its diagonal morphism is represented by the surjective multiplication ring homomorphism $B{\otimes }_{A}B\to B$, and is therefore a closed immersion by Question 1.

We prove the lifting condition in the valuative criterion for properness. Choose an affine open subset of $Y$ containing the image of the closed point of $\mathrm{Spec}V$. Every point of $\mathrm{Spec}V$ is a generalization of its closed point, so the whole map factors through this affine open subset. Its inverse image under the finite morphism is $\mathrm{Spec}B$, with $B$ a finitely generated $A$-module. Every $b\in B$ is integral over $A$: apply the determinant trick to multiplication by $b$ on a finite set of module generators to obtain a monic polynomial which annihilates all generators and hence annihilates $1$.

In the given valuative square, we have compatible ring homomorphisms $A\to V$ and $B\to K$. The image in $K$ of each $b\in B$ satisfies a monic polynomial over $V$. To see directly why it belongs to $V$, suppose $x\in K\setminus V$. The defining property of a valuation ring gives $x^{-1}\in V$, and it lies in the maximal ideal since it cannot be a unit. A monic equation for $x$, multiplied by $x^{-n}$, would then express $1$ as a sum of elements in that maximal ideal, a contradiction. This proves that valuation rings are integrally closed and that $B\to K$ factors through $B\to V$. The factorization is unique because $V\hookrightarrow K$ is injective. Thus the lift exists uniquely, proving

The composite of two finite type morphisms is of finite type: on suitable affine open subsets, take the algebra generators of the first algebra together with the images of the algebra generators of the second. To check separatedness, factor the diagonal morphism of the composite asThe first arrow is a closed immersion, and the second is the base change of the closed immersion ${\Delta }_g:Y\to Y{\times }_{Z}Y$. Hence their composite is a closed immersion.

Finally, after any base change $Z^{\prime }\to Z$, the composite is $X^{\prime }\stackrel{f^{\prime }}{\to }{Y}^{\prime }\stackrel{g^{\prime }}{\to }{Z}^{\prime }$. Both arrows take closed subsets to closed subsets, so their composite does as well. This is universal closedness. All three conditions hold, and therefore

Use the graph morphism of schemesIt is a closed immersion: the square with ${\Delta }_g:Y\to Y{\times }_{Z}Y$ and the map $X{\times }_{Z}Y\to Y{\times }_{Z}Y$, $(x,y)\mapsto (f(x),y)$, is a fibre product of schemes square. The separatedness of $g$ makes ${\Delta }_g$ a closed immersion, and base change preserves a closed immersion.

The projection $\pi :X{\times }_{Z}Y\to Y$ is the base change of the proper morphism $gf:X\to Z$, so it is a proper morphism. The closed immersion ${\Gamma }_f$ is a finite morphism, hence a proper morphism by part (i). Since $f=\pi {\Gamma }_f$, part (ii) provesThis argument supplies finite type for $f$ as well; it need not be separately assumed.

We first justify surjectivity is preserved by base change for morphisms of schemes. Given $y^{\prime }$ over $y$ and a point $x$ over $y$, the fiber after the residue-field extension contains a point becauseIndeed, tensoring a nonzero vector space over a field with a field extension remains nonzero; a nonzero commutative ring has a prime ideal, giving the required point. Thus every base change of $f$ is surjective.

Now make an arbitrary base change $Z^{\prime }\to Z$. For a closed subset $C\subseteq Y^{\prime }=Y{\times }_Z{Z}^{\prime }$, its inverse image under $f^{\prime }:X^{\prime }\to Y^{\prime }$ is closed. The proper morphism $(gf{)}^{\prime }:X^{\prime }\to Z^{\prime }$ therefore has closed image on that inverse image. Surjectivity of $f^{\prime }$ givesso $g^{\prime }(C)$ is closed. This proves that $g$ is a universally closed morphism. Together with the given separatedness and finite type, it proves

Write $f$ as the ring homomorphism $A\to B$. By the permitted assumption, $B$ is a finitely generated $A$-algebra. We will prove that every element of $B$ is integral over $A$, including any nilpotent contribution. If $B=0$, there is nothing to prove.

We use the integral closure as an intersection of valuation rings in its general form: if an integral domain $R$ lies in a field $K$, then its integral closure inside $K$ is the intersection of all valuation rings of $K$ containing $R$. Here $K$ need not be the fraction field of $R$. One direction follows from valuation rings are integrally closed. For the other, if $0\ne x\in K$ is not integral, the ideal $x^{-1}R[x^{-1}]$ is proper: an expression for $1$ in that ideal, multiplied by a sufficiently large power of $x$, would be a monic equation for $x$. Choose a maximal ideal containing this ideal, and apply the valuation domination lemma to the corresponding local ring inside $K$. The resulting valuation ring $V$ contains $R$ and has $x^{-1}$ in its maximal ideal, so $x\notin V$. This proves the claimed intersection criterion. The valuation domination lemma is obtained by choosing, using Zorn's lemma, a maximal local overring dominating the given local ring; the usual alternative between adjoining $z$ and adjoining $z^{-1}$ forces this maximal overring to be a valuation ring.

The Noetherian ring $B$ has finitely many minimal prime ideals ${\mathfrak{q}}_1,\dots ,{\mathfrak{q}}_r$. SetTake any valuation ring $V$ of $K_i$ containing $R_i$. The maps $B\to D_i\to K_i$ and $A\to R_i\to V$ form a square for the valuative criterion for properness. Its lift gives $B\to V$, whose composite with $V\hookrightarrow K_i$ is the original map. Consequently the image $D_i$ is contained in $V$. The intersection criterion proves that every element of $D_i$ is integral over $R_i$.

Fix $b\in B$. For each $i$, choose a monic polynomial $P_i(T)\in A[T]$ whose value at the image of $b$ in $D_i$ is zero; coefficients over $R_i$ can be lifted to $A$. The product $P(T)={\prod }_i{P}_i(T)$ is monic and satisfiesThe nilradical of a Noetherian ring is nilpotent. Explicitly, choose finitely many generators $n_j$ with $n_j^{e_j}=0$; every product of $1+{\sum }_j(e_j-1)$ generators then contains a vanishing power. Thus $(\sqrt{(0)}{)}^N=0$ for some $N$, and $P(T{)}^N$ is a monic polynomial annihilating $b$. Hence $B$ is integral over $A$.

Finally choose algebra generators $b_1,\dots ,b_s$, with monic equations of degrees $d_1,\dots ,d_s$. Reducing powers in these equations shows that the finitely many monomialsspan $B$ as an $A$-module. Therefore $B$ is a finite $A$-module, proving