1. Valuative Criteria Specialization vs being closed
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1 1. Valuative Criteria 1.1. Specialization vs being closed Proposition 1.1 (Specialization vs Closed). Let f : X Y be a quasi-compact S-morphisms, and let Z X be closed non-empty. 1) For every z Z there exits a generic point x Z of Z such that x x. 2) The following are equivalent: i) f(z) Y is a closed subset. ii) The set f(z) is closed under specialization inside Y. iii) If i Z, f( i ) f(z) are generic points, then f({ i }) = {f( i )} in Y. To 1): Let U = Spec B X for any affine open neigbourhood of z. Then Z U X is a closed subset of X with z Z U (WHY). Hence Z U = V (a) for some reduced ideal a B (WHY), and z Z U is a prime ideal of B with z a. And if z Spec B is minimal satisfying z z a [such z always exist (WHY)], then z Z is a generic point of Z U (WHY), and z { z } (WHY). We claim that z Z is a generic point of Z. Indeed, let x Z satisfy x z. Then x lies in all open neighborhoods of x (WHY), hence x Z U = V (a), thus a x z (WHY). Hence we conclude that x = z (WHY). To 2): The implication i) ii) follows from the fact that closed sets are closed under specialization (WHY), and ii) iii) is clear (WHY). For the implication iii) i), notice that we w.l.o.g., we can suppose that X = Z (WHY), and that Y = Spec C is affine (WHY). Hence f being quasi compact implies that Z = f 1 (Y ) is quasi comapact (WHY), thus: Z = α I Z α, I finite, Z α affine open (WHY), hence f(z) = α f(z α ) (WHY). Since Z = α Z α (WHY), and f(z) = α f(z α ), one has: If f(z α ) Y is closed for all α I, then f(z) Y is closed (WHY). Hence w.l.o.g. we can suppose that I = 1, that is, Z = U with U = Spec B, and f : U Y defined by some S-morphimsm ϕ : C B (WHY). Moreover, f(z) Y = Spec C being a closed subset, there exists a reduced ideal a C such that f(z) = V (a) Y (WHY). Hence w.l.o.g., we can suppose that Y = f(z), or equivalently, a = (0), hence ϕ : C B is injective, and C is a reduce ring, i.e., (0) = (0). Lemma. Let ϕ : C B be an injective ring homomorphism. Then the corresponding map ϕ top : Specmin(B) Specmin(C) is surjective. Proof. ex. [Hint: Let Σ C C be the multiplicative system of the non-zero divisors, and R C := Σ 1 C C be the total ring of fractions of C. ThenSpecmax(R C ) = Spec(R C ) = Specmin(C) (WHY), and correspondingly for B. Further, one has S-embeddings C R C, B R B, and R C R B (WHY). Further, Specmax(R C ) = Spec(R C ) = Specmin(C), and correspondingly for B, and R C R B gives rise to a surjective map Spec(R B ) Spec(R C ) (WHY).] We now conclude the proof of assertion 2) as follows: Let y Y = f(z). Further, let y Specmin C be such that z y, hence y y in Y. Finally let i Specmin B satisfy 1
2 f( i ) = y, thus f( i ) = y y. Then y f(z) is a generic point of f(z) = Y (WHY), and i Z is a generic point of Z such that f( i ) = y, and y f({ i }) Closeness via Valuation rings Notations. O K is an S-valuation ring of an S-field K, and m are the generic point, respectively the closed point of T O := Spec O, thus ı : T O canonical. The valuative criteria for being separated and/or universally closed involve / are based on S-valuation rings O K of S-fields K, via diagrams of morphisms in Sch S of the form: ( ) f X f Y ı TO And given such commutative diagrams, the valuative criteria are about existence/uniqueness of morphisms f O : T O X, which make the resulting diagram below commutative: ( ) f X f Y f O ı TO We begin by announcing the following: Proposition 1.2 (Closeness vs Valuation rings). Let f : X Y be a quasi compact S-morphism. Then the following hold: 1) For y, y 0 Y the following are equivalent: i) y y 0 ii) There exists a local Y -domain R, p and a morphism Spec R Y, (0) y, p y 0. 2) For a closed subset Z X, set Σ := { i Z i Z, f( i ) f(z) are generic points }. Then the following are equivalent: i) f(z) Y is a closed subset. ii) For every ( ) with f () Z, there exists ( ) with f O (m) Z. iii) The same assertion, for all Y -isomorphisms f : i := f () with i Σ. 3) Finally one has: f : X Y is closed iff for every generic point i X and all diagrams ( ) in which f : i are Y -isomorphisms, diagrams ( ) exist. Proof. To 1): For the implication i) ii), let y y 0. Equivalently, for every open neighborhood V of y 0, one has y V. Hence if V = Spec C is affine, and y, y 0 Spec(C), one has: y y 0 iff y y 0 as ideals. Thus if O y0 = C y0 is the local ring of y 0, it follows that y Spec C Y lies actually in the image of the canonical S-morphism [which is a topological embedding (WHY)]: Spec O y0 Spec C Y. 2
3 Hence R := O y0 /y y0 endowed with p := (y 0 /y) y0 is a local domain, and C C y0 = O y0 R are S-morphisms giving rise to an S-morphism: Spec R Spec O y0 Spec C X top which is a topological embedding (WHY), such that (0) y, p y 0. For the implication ii) i), let R, p be a local domain, and f R : Spec R X with (0) y, p y 0. Then (0) p implies that y = f R (0) f R (p) = y 0 (WHY). To 2): For y, y 0 Y, by the discussion above above, one has: y y 0 iff there exists a local domain R, p and an S-morphism f R : Spec R Y with (0) y, p y 0. For y = f( i ), i Z generic point, consider the canonical Y -embedding f # i : κ(y) := Quot(R) κ( i ), and set R := f # i (R), p := f # i (p). Then R, p is a local Y -subring of κ( i ). Hence by Chevalley s Theorem, there exists a valuation Y -rin, m of the Y -field κ( i ) which dominates the given local subring R, p of κ( i ). Hence f # i : R R O is a Y -morphism of local rings (WHY), such that the corresponding Y -morphism f O : T O = Spec O Spec R Z satisfies: f ( f O ( i ) ) = y, f ( f O (m) ) = y 0. Finally apply Proposition 1.1, 2), etc. To 3): Let Z X be a closed subset. By assertion 2), one has: f(z) is closed, provided for all i Z, f( i) f(z) generic points, and every diagram ( ) in which f : i is a Y -isomorphism, diagrams ( ) exist. Now let such a situation be given, and i X be a generic point with i i. Consider any affine open subset U = Spec B X with i U. Then i U and i i as prime ideals of B (WHY), and one has a canonical Y -morphisms: B B/ i Quot(B/ i ) = κ( i ). Let i/ i B/ i κ( i ) be the images of i B under B κ( i ), and notice that one has canonical identifications κ( i ) = Quot(B/ i), κ( i) = κ( i/ i ) (WHY). Using Chevalley s Thm, let O 1, m 1 be a valuation ring of K := κ( i ) such that O 1, m 1 dominates B/ i, i/ i and the residue field κ(m 1 ) is an algebraic extension of κ( i) = κ( i/ i ). Then considering any prolongation O, m of O, m under the (algebraic) field extension κ( i) κ(m 1 ), one has: The preimage O, m of O, m under O 1 κ(m 1 ) is a valuation ring of K = ( i ) satisfying: - m 1 Spec O, and O 1 = O m1. - O/m 1 = O, m/m 1 = m. Therefore one has: - O O gives a closed immersion Spec O Spec O = T O with (0) m 1, m 0 m. Let Spec O = T O be the generic point, and f : i be the identity Y -map K κ( i ). Then we get a diagram ( ), and consider the resulting commutative diagram: ( ) f X f Y f O ı TO Since B B/ i O 1, the canonical S-morphisms Spec O 1 Spec B/ i Spec B = U satisfy: m 1 i/ i i (WHY), hence f O (m 1 ) = i Z (WHY). Hence m 1 m in T O, 3
4 implies that i = f O (m 1 ) f O (m). Thus Z being closed under specialization, i Z implies that f O (m) Z. Hence setting f O := f O (T O T O ), we get: f O ( ) = f O (m 1 ) = i Z, f O (m ) = f O (m) Z Valuative Criterion for Separated In the notation from the previous subsections one has: Theorem 1.3 (Valuative criterion for separated). For a quasi compact S-morphism f : X Y the following are equivalent: i) The morphism f is separated. ii) For every diagram ( ) as above, there exists at most one Y -morphism f O : T O X making the diagram ( ) commutative. iii) The same assertion for all Y -isomorphisms f : i with i X the generic points. Proof. i) ii): Suppose that f : X Y is separated, and a diagram ( ) be given. Let f, f : T O X be Y -morphisms making the diagram ( ) commutative. Then there exits a unique f : T O X Y X such that p X f = f, p X f = f (WHY). Further, since f () = f K () = f () (WHY), it follows that f() = f K () (WHY). Finally, recalling that p for all p T O, it follows that f() f(p) for all p T O (WHY). Since (X) X Y X is closed, and f() (X), by Proposition 1.1, above, it follows that f(p) (X). Hence im( f) (X), and therefore, f (p) = p X f(p) = p X f(p) = f (p), p T O, thus concluding that f = f (WHY). ii) iii): Clear. iii) i): We have to show that (X) X Y X is a closed subset. First notice that since f : X Y is quasi compact, so is : X X Y X (WHY). Hence by the Proposition 1.1 above, (X) X Y X is closed iff for every generic point i X, letting Z i := { i } X be the closure of i in X, one has: - (Z i ) X Y X is closed under specialization inside X Y X (WHY). Since the canonical embedding Z i Y Z i X Y X is a closed immersion (WHY), it suffices to prove that (Z i ) Z i Y Z i is closed. Equivalently, w.l.o.g., we can suppose that X = Z i is irreducible (WHY), with generic point X. Then (X) has generic point := ( X ), and consider the induced morphism of Y -fields # X : κ( ) κ( X ). Now let z inside X Y X. Let O z, m z be the local ring of z, and O, m κ( X ) be its image under # X. Proceeding as in the proof of Proposition 1.2, let O, m κ( X ) be a valuation ring which dominates O, m. Then the morphism of local Y -rings O z O O gives rise to a Y -morphism f : T O Spec O z X Y 4 X
5 such that f() = and f(m) = z. Then f := p X f and f := p X f are Y -morphisms T O X which satisfy: f () = X = f () (WHY). By the uniqueness of T O X as in diagram ( ), it follows that f = f, and therefore, p X(z) = p X f(z) = f (m) = f (m) = p X f(z) = p X(z) thus concluding that z (X). Hence finally (X) is closed, as claimed Valuative criterion for universally closed In the notation from the previous subsections one has: Theorem 1.4 (Valuative criterion for universally closed). For a quasi compact S-morphism f : X Y the following are equivalent: i) The morphism f is universally closed. ii) For every diagram ( ) as above, there exists at least one Y -morphism f O : T O X making the diagram ( ) commutative. iii) The same assertion for all Y -isomorphisms f : i with i X the generic points. Proof. i) ii): In the notations of the Theorem 1.4, let x := f (), hence f # : κ(x ) K is an Y -isomorphism. The Y -morphisms f : X, T O define a unique Y -morphism f : X Y T O such that p X f = f, p TO f = ı, as in the diagram below X f p X f ı X Y T O Then := f() satisfies: p X ( ) = x, and p TO ( ) =, thus one has Y -isomorphism of fields: T O p TO f # κ(x ) κ() p# T O κ( ) (WHY) Let Z X Y T O be the closure of Z := f() in X Y T O. Then Z is the generic point of Z (WHY), and Z := {z X Y T O Z z}. Hence f being universally closed, implies that p TO (Z) T O is closed (WHY). Since = p TO f () = p TO ( Z ) (WHY), it follows that p TO (Z) = T O (WHY). Let z Z be a preimage of m under p TO. Then Spec O Z,z Z satisfies: First, p TO (Spec O Z,z ) T O. Second, since Z Z is (by definition) the generic point of Z, one has Z Spec O Z,z (WHY). Hence O Z,z κ( Z ) is a local Y -subring of κ( Z ) with Quot(O Z,z ) = κ( Z ), and p TO : Spec O Z,z T O is defined by a Y -morphism of local rings p # T O : O O Z,z (WHY), giving rise to a fraction field Y -isomorphism Quot(O) = κ() p# T O κ(z ) = Quot(O Z,z ). Since O is a valuation ring, conclude that local ring morphism p # T O : O O Z,z must be an isomorphism (WHY). Finally, the inverse ı O : O Z,z O of p # T O gives rise to a Y -morphism f ıo : T O = Spec O Spec O Z,z Z X Y T O such that p TO f ıo = id TO. Hence f O := p X f TO : T O X is an Y -morphism as in diagram ( ) (WHY). 5
6 ii) iii): Clear. iii) i): Let g : T Y be an S-morphism, and setting X T := X Y T, we have to show that p T : X T T is closed, i.e., given any Z X T = X Y T closed subset, p T (Z) T is closed. Consider the following reduction steps: - Y = i I V i be a affine open covering of Y, and X i := f 1 (X i ) X, T i := g 1 (V i ) T. Then X i X, T i T are open subsets (WHY), and one has canonically X i Y T i = p 1 X (X i) p 1 T (T i) X Y T (WHY). Finally, Z i := Z (X i Y T i ) is closed in X i Y T i, and p T (Z) T i = p T (Z i ); and p T (Z) T is closed iff p T (Z i ) T i is closed (WHY). Hence w.l.o.g., Y is affine, and X = f 1 (Y ) implies: X is quasi compact (WHY). - p T (Z) T is closed iff p T (Z) U is closed in U for all affine open subsets U = Spec C of T (WHY). Hence w.l.o.g., T = Spec C is affine, thus quasi compact (WHY). Conclude that w.l.o.g., Y, T are affine S-schemes, X is a quasi-compact, hence X Y T is quasi compact, and p T, p X are quasi compact morphism (WHY). Finally, we have to show that p T : X Y T T is closed. Hence by Proposition 1.2, 3), we have to show that for every generic point i X Y T, and every commutative diagram of morphisms of T -schemes ( ) X Y T p T T f ı TO there exist f O : T O X Y T making the corresponding diagrams ( ) commutative. First notice that for the given i X Y T, the points i := p X ( i ) X, i := p T ( i ) T are generic points of X, respectively T (WHY). Hence setting : T O T p T Y, f : X Y T p X X, one has f () = i (WHY), and the diagram ( ) above gives rise to ( ) X f Y p X f ı TO g O Next recall that p X f : f i i gives rise to a Y -embedding of fields K := κ( i) κ( i ) κ() =: K. Hence letting m, O K be the preimages of m O under π, one has Y -morphisms p : T O = Spec O Spec O =: T O such that p() = and p(m) = m, and O, m is a valuation ring of K. Finally, i i factors uniquely as i and i is an isomorphism (WHY), and T O T Y 6
7 factors uniquely as T O T O Y (WHY). Conclude that ( ) gives rise canonically to ( ) f X f Y ı T O with f : i a Y -isomorphism. Hence since i X is (by construction) a generic point of X, by the hypothesis iii), there exists f O : T O X making the corresponding diagram ( ) commutative: ( ) f X f Y f O ı T O f O Therefore, the Y -morphism f O : T O T O X together with : T O T give rise to a unique Y -map f O : T O X Y T such that p T f O =, p X f O = f O. Hence we get a commutative diagram of T -morphims ( ) Department of Mathematics University of Pennsylvania DRL, 209 S 33rd Street Philadelphia, PA USA address: pop@math.upenn.edu URL: X Y T p T T f f O ı TO 7
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