Let \(X\) be an integral scheme of finite type over a field \(k\), having function field \(K\). We say that a valuation of \(K/k\) has center \(x\) on \(X\) if its valuation ring \(R\) dominates the local ring \(\OO_{x, X}\).
If \(X\) is separated over \(k\), then the center of any valuation of \(K/k\) on \(X\), if it exists, is unique.
If \(X\) is proper over \(k\), then every valuation of \(K/k\) has a unique center on \(X\). Note: if \(X\) is a variety over \(k\), the criterion of (b) is sometimes taken as the definition of a complete variety.
Prove the converses of (a) and (b). Hint: while (a) and (b) follow easily from (4.3) and (4.7), their converses require some comparison of valuations in different fields.
If \(X\) is proper over \(k\), and if \(k\) is algebraically closed, show that \(\Gamma(X, \OO_X) = k\). This result generalizes (I, 3.4a). Hint: let \(a \in \Gamma(X, \OO_X)\) with \(a \not\in k\). Show that there is a valuation ring \(R\) of \(K/k\) with \(\inverseof{a} \in \mfm_R\), then use (b) to get a contradiction.
Solution
Let \[ \eta\in X \] be the generic point, so \[ \kappa(\eta)=K(X)=K. \]
1. Let \(R\subseteq K\) be a valuation ring containing \(k\).Giving a center \(x\in X\) of \(R\) is equivalent to giving a \(k\)-morphism\[ \widetilde\eta:\Spec R\longrightarrow X \]whose restriction to the generic point\[ \Spec K\subseteq\Spec R \]is the canonical morphism\[ \Spec K\longrightarrow X \]with image \(\eta\).
Suppose first that \(x\) is a center. By definition, \[ \mathcal O_{X,x}\subseteq R \] and \[ \mathfrak m_x = \mathcal O_{X,x}\cap\mathfrak m_R. \] Thus the inclusion \[ \mathcal O_{X,x}\longrightarrow R \] is a local homomorphism. A local homomorphism from the local ring at \(x\) to a local ring defines a unique morphism \[ \Spec R\longrightarrow X \] sending the closed point to \(x\). On fraction fields this is the identity map \[ K(X)=K\longrightarrow K, \] so the generic point maps canonically to \(\eta\).
Conversely, let such a morphism \(\widetilde\eta\) be given and let \(x\) be the image of the closed point of \(\Spec R\). The induced homomorphism \[ \mathcal O_{X,x}\longrightarrow R \] is local because morphisms of schemes are morphisms of locally ringed spaces. Its restriction on the generic point is the identity on \(K\), hence it identifies \(\mathcal O_{X,x}\) with a subring of \(R\). Therefore \(R\) dominates \(\mathcal O_{X,x}\), so \(x\) is a center.
2. If \(X\) is separated over \(k\), every valuation of \(K/k\) has at most one center on \(X\).
Let \(R\) be a valuation ring of \(K/k\). Two centers would, by 1, give two morphisms \[ \Spec R\rightrightarrows X \] whose restrictions to \[ \Spec K \] are both the canonical generic-point morphism.
The valuative criterion for separatedness says that a separated morphism has at most one such lift. Therefore the two morphisms, and hence their closed-point images, coincide.
3. If \(X\) is proper over \(k\), every valuation of \(K/k\) has a unique center on \(X\).
Let \(R\) be a valuation ring of \(K/k\). Consider the diagram \[ \begin{array}{ccc} \Spec K&\longrightarrow&X\\ \downarrow&&\downarrow\\ \Spec R&\longrightarrow&\Spec k. \end{array} \] The valuative criterion for properness gives a unique dotted lift \[ \Spec R\longrightarrow X. \] By 1 this lift is exactly a unique center of \(R\) on \(X\).
4. We use the refined Noetherian valuative criterion in the converse direction.For a finite-type morphism\[ T\longrightarrow S \]with \(S\) locally noetherian, separatedness and properness may be tested using only discrete valuation rings whose fraction field is the residue field of the generic point of an irreducible component of \(T\), with the generic point mapping canonically to that component.
This is the refined Noetherian valuative criterion; see the Stacks Project, Lemmas 32.15.2 and 32.15.3.
The comparison of valuations in different fields mentioned in Hartshorne’s hint is the content of the reduction to these generic-component valuation rings. Since the present \(X\) is integral, it has only one irreducible component and its generic residue field is exactly \(K(X)=K\).
5. Conversely, suppose every valuation of \(K/k\) has at most one center on \(X\).Then \(X\) is separated over \(k\).
Let \[ R\subseteq K \] be any discrete valuation ring containing \(k\) with fraction field \(K\). By hypothesis, \(R\) has at most one center on \(X\). By 1, the diagram \[ \begin{array}{ccc} \Spec K&\longrightarrow&X\\ \downarrow&&\downarrow\\ \Spec R&\longrightarrow&\Spec k \end{array} \] has at most one lift.
Thus the generic-component uniqueness condition of the refined valuative criterion 4 holds. Since \(X\) is of finite type over the field \(k\), that criterion implies \[ X\longrightarrow\Spec k \] is separated.
6. Conversely, suppose every valuation of \(K/k\) has a unique center on \(X\).Then \(X\) is proper over \(k\).
For every discrete valuation ring \[ R\subseteq K \] containing \(k\) and having fraction field \(K\), the hypothesis and 1 say that the generic-point valuative diagram has exactly one lift \[ \Spec R\longrightarrow X. \]
The existence-and-uniqueness condition in the refined valuative criterion 4 therefore holds. Since \(X\) is of finite type over the noetherian scheme \(\Spec k\), the criterion implies that \[ X\longrightarrow\Spec k \] is proper.
7. Parts (a)–(c) therefore give the equivalences\[ \boxed{ X\text{ separated over }k \iff \text{every valuation of }K/k\text{ has at most one center on }X, } \]and\[ \boxed{ X\text{ proper over }k \iff \text{every valuation of }K/k\text{ has exactly one center on }X. } \]
The forward implications are <1>2–<1>3, and the converses are <1>5–<1>6.
8. Assume now that \(X\) is proper over an algebraically closed field \(k\), and let\[ a\in\Gamma(X,\mathcal O_X). \]If \(a\notin k\), then \(a\) is transcendental over \(k\).
Because \(X\) is integral, restriction to the generic point gives an injection \[ \Gamma(X,\mathcal O_X)\hookrightarrow K(X)=K. \]
If \(a\) were algebraic over the algebraically closed field \(k\), then \(a\in k\). Thus \(a\notin k\) forces \(a\) to be transcendental.
9. Under the assumption \(a\notin k\), there is a valuation ring\[ R\subseteq K \]of \(K/k\) such that\[ a^{-1}\in\mathfrak m_R. \]
Since \(a\) is transcendental over \(k\), the subfield \[ k(a)\subseteq K \] is a rational function field. Consider the discrete valuation ring at infinity \[ A_0 = k[a^{-1}]_{(a^{-1})} \subseteq k(a)\subseteq K. \] Its maximal ideal is generated by \(a^{-1}\).
By the valuation-ring existence theorem, any local subring of a field is dominated by a valuation ring of that field. Thus there is a valuation ring \[ R\subseteq K \] with fraction field \(K\) which dominates \(A_0\); see the Stacks Project, Lemma 10.50.2.
Because \(k\subseteq A_0\subseteq R\), this is a valuation of \(K/k\). Domination gives \[ a^{-1}\in\mathfrak m_{A_0} \subseteq \mathfrak m_R. \]
10. The valuation ring from 9 cannot have a center on \(X\).
Suppose it had center \(x\). Then \[ \mathcal O_{X,x}\subseteq R. \] The global regular function \(a\) has a germ \[ a_x\in\mathcal O_{X,x}, \] so \[ a\in R. \]
But 9 also gives \[ a^{-1}\in\mathfrak m_R. \] Since \(a\in R\) is the inverse of \(a^{-1}\), the element \(a^{-1}\) is a unit of \(R\), contradicting membership in the maximal ideal.
11. Therefore every global regular function on \(X\) lies in \(k\):\[ \boxed{\Gamma(X,\mathcal O_X)=k.} \]
If some \(a\in\Gamma(X,\mathcal O_X)\) lay outside \(k\), <1>9–<1>10 would produce a valuation of \(K/k\) with no center on \(X\). This contradicts properness and part (b), proved in 3.
The reverse inclusion \[ k\subseteq\Gamma(X,\mathcal O_X) \] is the structure map of the \(k\)-scheme \(X\). Hence equality holds.
12. Q.E.D.
Steps 2, 3, <1>5–<1>7, and 11 prove parts (a), (b), (c), and (d), respectively.