Past exam of the introduction to schemes course of the University of Oxford / 2023 / 2

For every valuation ring $D$ with fraction field $K$, consider compatible morphisms of schemesAn extension is a morphism of schemes $\ell :\mathrm{Spec}D\to X$ satisfying $\ell {|}_{\mathrm{Spec}K}=u$ and $f\ell =v$. The valuative criterion for universal closedness statesEquivalently, among quasi-compact morphisms, the existence condition characterizes universal closedness. A universally closed morphism is automatically a quasi-compact morphism; the converse direction of the criterion needs this compactness condition. All valuation rings are allowed, with arbitrary fraction fields; there is no Noetherian ring hypothesis and one cannot replace them by discrete valuation rings without additional assumptions.^[ref][ref]

Use the compatible pair $u,v$ and the notion of extension from (i). The valuative criterion for separatedness isA quasi-separated morphism means one whose diagonal morphism is a quasi-compact morphism. A separated morphism automatically has this property because its diagonal morphism is a closed immersion. The quasi-separated morphism hypothesis is needed in the converse valuative test for arbitrary schemes. Existence is not demanded.^[ref]

With the same compatible pair $u,v$, the valuative criterion for properness isEquivalently, one can replace finite type by being a quasi-compact morphism and locally of finite type. Necessity follows from the definition of a proper morphism and the existence and uniqueness criteria in (i) and (ii). Conversely, finite type supplies the compactness needed in (i), while being a quasi-separated morphism supplies the compactness of the diagonal morphism needed in (ii). Thus the extension condition gives both universal closedness and separatedness, and these together with finite type give a proper morphism. These hypotheses are part of the criteria, even though the question is phrased for unrestricted morphisms of schemes.^[ref]

Let $S^{\prime }\to S$ be a morphism of schemes and set $X^{\prime }=X{\times }_S{S}^{\prime }$, with projection $f^{\prime }:X^{\prime }\to S^{\prime }$. Given compatible morphisms of schemes $u^{\prime }:\mathrm{Spec}K\to X^{\prime }$ and $v^{\prime }:\mathrm{Spec}D\to S^{\prime }$, compose them with the projections to obtain a valuative problem for $f$. An extension $\ell :\mathrm{Spec}D\to X$ for that problem and the prescribed $v^{\prime }$ determine a unique morphism of schemesThe universal property of a fibre product shows that ${\ell }^{\prime }$ extends $u^{\prime }$ and lies over $v^{\prime }$. Conversely, every extension ${\ell }^{\prime }$ has such a first component $\ell$. Thus the sets of extensions in the two problems are in bijection.

If $f$ is a universally closed morphism, it is a quasi-compact morphism and satisfies existence. The allowed stability of quasi-compact morphisms under base change and (a)(i) prove that $f^{\prime }$ is a universally closed morphism. If $f$ is a separated morphism, it is a quasi-separated morphism and satisfies uniqueness. Stability of quasi-separated morphisms under base change and (a)(ii) prove that $f^{\prime }$ is a separated morphism.

If $f$ is a proper morphism, the same bijection supplies existence and uniqueness. The permitted stability under base change of quasi-compact morphisms, quasi-separated morphisms and locally of finite type supplies all remaining hypotheses of (a)(iii). Hence

Let $X\stackrel{f}{\to }Y\stackrel{g}{\to }S$ be morphisms of schemes. Start with $u:\mathrm{Spec}K\to X$ and $v:\mathrm{Spec}D\to S$ satisfying $gfu=v{|}_{\mathrm{Spec}K}$.

For existence, first apply the extension condition for $g$ to $fu$ and $v$, obtaining $w:\mathrm{Spec}D\to Y$. Then apply the extension condition for $f$ to $u,w$, obtaining $\ell :\mathrm{Spec}D\to X$. This $\ell$ solves the problem for $gf$. If both $f,g$ are universally closed morphisms, both are quasi-compact morphisms; their composite is a quasi-compact morphism by the allowed stability fact. Part (a)(i) therefore proves that $gf$ is a universally closed morphism.

For uniqueness, suppose ${\ell }_1,{\ell }_2$ are extensions for $gf$. Uniqueness for $g$ gives $f{\ell }_1=f{\ell }_2=w$, and uniqueness for $f$, now over this same $w$, gives ${\ell }_1={\ell }_2$. If $f,g$ are separated morphisms, their composite is a quasi-separated morphism by the allowed stability fact. Part (a)(ii) proves that $gf$ is a separated morphism.

Finally, if $f,g$ are proper morphisms, these arguments give both existence and uniqueness. Stability under composition of quasi-compact morphisms, quasi-separated morphisms and locally of finite type gives the other hypotheses of (a)(iii). Consequently

Let $x\in \mathrm{Frac}(D)$ satisfy a monic polynomial equationIf $x=0$, there is nothing to prove. Suppose $x\notin D$. The defining property of a valuation ring gives $y=x^{-1}\in D$. Multiplying the equation by $y^{n-1}$ yieldsa contradiction. ThereforeThis proof uses only the comparison of an element with its inverse; no discreteness or finite generation of the valuation is needed.

Let $D$ be a valuation ring, $K$ its fraction field, and let $u:\mathrm{Spec}K\to X$, $v:\mathrm{Spec}D\to S$ be a compatible pair. Since $D$ is a local ring, every open subset of $\mathrm{Spec}D$ containing its unique closed point is the whole space: a principal open subset containing the maximal ideal is defined by a unit. Choose an affine open subscheme $V=\mathrm{Spec}A\subseteq S$ containing the image under $v$ of that closed point. The entire $v$ factors through $V$.

Because $f$ is a finite morphism, $f^{-1}(V)=\mathrm{Spec}B$ with $B$ a finite $A$-module. The valuative problem is the compatible ring homomorphismsEvery $b\in B$ is an integral element over $A$. Mapping its monic polynomial equation to $K$ proves that $\psi (b)$ is an integral element over $D$. Part (c) gives $\psi (b)\in D$. Thus $\psi$ factors uniquely as a ring homomorphism $B\to D\hookrightarrow K$, producing the unique extension $\mathrm{Spec}D\to X$.

For completeness, a finite morphism is a quasi-compact morphism and of finite type: over $V$ its inverse image is an affine scheme, and finite $A$-module generators also generate it as an $A$-algebra. It is also a separated morphism. Affine-locally its diagonal morphism corresponds to the surjective multiplication ring homomorphismand hence is a closed immersion. In particular it is a quasi-separated morphism. The valuative criterion for properness now applies, giving

Write $X=\mathrm{Spec}B$, $S=\mathrm{Spec}A$ with $A,B$ integral domains, let $\alpha :A\to B$ be the ring homomorphism, and put $K=\mathrm{Frac}(B)$. We do not assume that $\alpha$ is injective. Write $A^{\prime }=\alpha (A)\subseteq B\subseteq K$.

Let $D\subseteq K$ be any valuation ring containing $A^{\prime }$, with fraction field $K$. The canonical morphism of schemes $\mathrm{Spec}K\to \mathrm{Spec}B$ and the ring homomorphism $A\to A^{\prime }\hookrightarrow D$ form a valuative problem for $f$. Since $f$ is a proper morphism, there is an extension. On coordinate rings it is a ring homomorphism $\beta :B\to D$ such thatis the original inclusion of $B$ into $K$. Therefore every element of $B$ lies in $D$. Since $D$ was arbitrary, the stated description of integral closure as an intersection of valuation rings givesThus every element of $B$ is an integral element over $A^{\prime }$, and hence over $A$: lift the coefficients of its monic polynomial along the surjection $A\to A^{\prime }$. The ring homomorphism $\alpha$ is therefore an integral extension in the sense of the question.

A proper morphism is of finite type, so in this affine situation $B=A[b_1,\dots ,b_s]$. If $b_i$ satisfies a monic polynomial of degree $n_i$, repeatedly reducing its powers shows that the finitely many monomialsgenerate $B$ as an $A$-module. Equivalently, apply the allowed equivalence between finite and integral plus finite type. Hence