aa r X i v : . [ m a t h . A C ] S e p ISOLATED POINTS OF THE ZARISKI SPACE
DARIO SPIRITO
Abstract.
Let D be an integral domain and L be a field contain-ing D . We study the isolated points of the Zariski space Zar( L | D ),with respect to the constructible topology. In particular, we com-pletely characterize when L (as a point) is isolated and, underthe hypothesis that L is the quotient field of D , when a valuationdomain of dimension 1 is isolated; as a consequence, we find allisolated points of Zar( D ) when D is a Noetherian domain. We alsoshow that if V is a valuation domain and L is transcendental over V then the set of extensions of V to L has no isolated points. Introduction
Let D be an integral domain with quotient field K , and let L be a fieldcontaining K . The Zariski space of L over D , denoted by Zar( L | D ), isthe set of all valuation rings containing D and having quotient field L .O. Zariski introduced this set (under the name abstract Riemann sur-face ) and endowed it with a natural topology (later called the Zariskitopology ) during its study of resolution of singularities; in particular,he used the compactness of the Zariski space to reduce the problem ofgluing infinitely many projective models to the gluing of only finitelymany of them [32, 33]. Later on, it was showed that Zar( L | D ) enjoyseven deeper topological properties: in particular, it is a spectral space ,meaning that there is always a ring R such that Spec( R ) (endowed withthe Zariski topology) is homeomorphic to Zar( L | D ), and an example ofsuch an R can be find using the Kronecker function ring construction[5, 6, 8]. Beyond being a very natural example of a spectral space “oc-curring in nature”, the Zariski topology can also be used, for example,to study representation of integral domains as intersection of overrings[20, 21, 22], or in real and rigid algebraic geometry [15, 26].As a spectral space, two other topologies can be constructed onZar( L | D ) starting from the Zariski topology: the inverse and the con-structible (or patch ) topology. Both of them give rise to spectral spaces(in particular, they are compact); furthermore, the constructible topol-ogy gains the property of being Hausdorff, and plays an important rolein the topological characterization of spectral spaces (see for example Date : September 24, 2020.2010
Mathematics Subject Classification.
Key words and phrases.
Zariski space; constructible topology; Cantor space; iso-lated points; perfect spaces; extensions of valuations.
Hochster’s article [14]). The constructible topology can also be stud-ied through ultrafilters [7], and this point of view allows to give manyexamples of spectral spaces, for example by finding them inside otherspectral spaces (see [22, Example 2.2(1)] for some very general construc-tions, [29] for examples in the overring case, and [10, 9] for examplesin the semistar operations setting).In this paper, we want to study the points of Zar( L | D ) that areisolated, with respect to the constructible topology. Our starting pointis a new interpretation of a result about the compactness of spaces inthe form Zar( D ) \ { V } [28, Theorem 3.6]: in particular, we show thatif V is isolated then V is the integral closure of D [ x , . . . , x n ] M , where x , . . . , x n ∈ L and M is a maximal ideal of D [ x , . . . , x n ] (Theorem3.4). Through this result, we characterize when L is an isolated point ofZar( L | D ) cons (i.e., Zar( L | D ) endowed with the constructible topology;Proposition 4.1) and, under the hypothesis that L = K is the quotientfield of D , when the one-dimensional valuation overrings are isolated(Theorem 5.2).In Section 6, we study the isolated points of the constructible topol-ogy when D is a Noetherian domain and L = K is its quotient field.Theorem 6.4 gives a complete characterization: V ∈ Zar( K | D ) =Zar( D ) is isolated if and only if the center P of V on D has heightat most 1 and P is contained in only finitely many minimal primes; inparticular, this cannot happen if D is local and of dimension at least3. In the countable case, we also give a complete characterization ofwhen Zar( D ) cons ≃ Zar( D ′ ) cons under the hypothesis that D and D ′ are Noetherian and local (Theorem 6.12).The last two sections of the paper explore the case of extension ofvaluations. Section 7 studies the case where D itself is a field: in partic-ular, we show that if the transcendence degree of L over D is at least 2then Zar( L | D ) cons has no isolated points, improving [3, Theorem 4.45].In Section 8, we show that if V is a valuation domain that is not a fieldand K ( X ) is the field of rational functions, then the set of extensionsof V to K ( X ) has no isolated points (Theorem 8.2), and as a conse-quence we further extend [3, Theorem 4.45] to Zar( L | D ) cons when D isan arbitrary integral domain (Theorem 7.4 and Corollary 8.7).2. Notation and preliminaries
Throughout the paper, all rings will be commutative, unitary and willhave no zero-divisors (that is, they are integral domains). We usuallydenote by D such a domain and by K its quotient field; we use D todenote the integral closure of D in K .A valuation domain is an integral domain V such that, for every x = 0 in the quotient field of V , at least one of x and x − is in V .Any valuation domain is local; we denote the maximal ideal of V by m V . If L is a field containing the quotient field K of V , an extension SOLATED POINTS OF THE ZARISKI SPACE 3 of V to L is a valuation domain W having quotient field L such that W ∩ K = V . We denote the set of extension of V to L by E ( L | V ); thisset is always nonempty (see e.g. [12, Theorem 20.1]).If D is an integral domain and L is a field containing D , the Zariskispace (or
Zariski-Riemann space ) of L over D , denoted by Zar( L | D ), isthe set of all valuation domains containing D and having quotient field L . The Zariski-Riemann space Zar( L | D ) is always non-empty; more-over, if P is a prime ideal of D , there is always some V ∈ Zar( L | D )such that P = m V ∩ D . When L is the quotient field of D , we de-note Zar( L | D ) simply by Zar( D ), and we call its elements the valu-ation overrings of D . If D ′ is the integral closure of D in L , thenZar( L | D ) = Zar( L | D ′ ); in particular, Zar( D ) = Zar( D ). A valuationring in Zar( L | D ) is minimal if it is minimal with respect to contain-ment.The Zariski-Riemann space Zar( L | D ) can be endowed with a naturaltopology, called the Zariski topology , which is the topology generatedby the basic open sets B ( x , . . . , x n ) := { V ∈ Zar( L | D ) | x , . . . , x n ∈ V } , as x , . . . , x n range among the elements of K . Under this topology,Zar( L | D ) is a compact T space that is (usually) not Hausdorff [34,Chapter VI, Theorem 40]; moreover, it is a spectral space , i.e., there isa ring R (which can be constructed explicitly) such that the spectrumof R (endowed with the Zariski topology) is homeomorphic to Zar( L | D )[6, 5]. The inverse topology on Zar( L | D ) is the topology such that the B ( x , . . . , x n ) are a subbasis of closed sets; a set X ⊆ Zar( L | D ) is closedwith respect to the inverse topology if and only if it is compact withrespect to the Zariski topology and it is closed by generization (i.e.,whenever V ∈ X W ∈ Zar( L | D ) satisfies V ⊆ W then W ∈ X ) [8,Remark 2.2 and Proposition 2.6].The constructible topology (or patch topology ) of Zar( L | D ) is thecoarsest topology such that the sets B ( x , . . . , x n ) are both open andclosed; that is, it is the topology generated by all the B ( x , . . . , x n )and their complements. We denote Zar( L | D ), endowed with the con-structible topology, by Zar( L | D ) cons . The constructible topology is finerthan both the Zariski and the inverse topology. The Zariski space, en-dowed with the constructible topology, is again compact and spectral;furthermore, it is Hausdorff [4, Proposition 1.3.13 and Theorem 1.3.14]and zero-dimensional (i.e., it has a basis of clopen subsets).The constructible topology can also be defined on the spectrumSpec( R ) of a ring R : in this case, it is defined as the coarsest topol-ogy such that every subbasic open set D ( r , . . . , r n ) is both open andclosed. An overring of D is, more generally, a ring contained between D and its quotientfield. DARIO SPIRITO If X ⊆ Zar( L | D ) is a subset closed in the constructible topology,then it is compact in the Zariski topology. Conversely, if X is compactin the Zariski topology, and X = X ↑ = { V ∈ Zar( L | D ) | V ⊇ W forsome W ∈ X } then X is closed in the inverse topology and thus alsoin the constructible topology.For every field L , we can define a map γ : Zar( L | D ) −→ Spec( D ) ,V m V ∩ D, which is called the center map . Under the Zariski topology (on bothsets), γ is continuous ([34, Chapter VI, §
17, Lemma 1] or [5, Lemma2.1]), surjective (this follows, for example, from [1, Theorem 5.21] or[12, Theorem 19.6]), closed [5, Theorem 2.5] and spectral (i.e, γ − (Ω)is compact for every compact set Ω of Spec( D ); [5, Theorem 4.1]). As aspectral map between spectral spaces, γ remains continuous even whenZar( L | D ) and Spec( D ) are endowed with the constructible topology [4,Theorem 1.3.21].If L ⊆ L ′ is a field extension, then we have a restriction map ρ : Zar( L ′ | D ) −→ Zar( L | D ) ,V V ∩ L, which is continuous in both the Zariski and the constructible topology;therefore, it is closed with respect to the constructible topology (onboth sets).If X is a topological space, a point p ∈ X is isolated in X if { p } isan open set. If X has no isolated points, then X is said to be perfect .3. General results
We begin by establishing some general criteria to determine whichvaluation domains are isolated in Zar( D ).Let D be an integral domain: we say that a prime ideal P of D is almost essential if there is a unique valuation overring of D havingcenter P ; equivalently, P is almost essential if and only if the integralclosure of D P is a valuation domain. If D is integrally closed, this isequivalent to saying that D P is a valuation ring, and if it happenswe say that D P is an essential valuation overring of D . In this case,isolated valuation rings correspond to isolated prime ideals. Proposition 3.1.
Let D be an integral domain, and let P be an almostessential prime ideal of D ; let V be the valuation overring with center P . Then, V is isolated in Zar( D ) cons if and only if P is isolated in Spec( D ) cons .Proof. Let γ : Zar( D ) −→ Spec( D ) be the center map; then, γ iscontinuous with respect to the constructible topology and thus it is SOLATED POINTS OF THE ZARISKI SPACE 5 also closed (since it is a map between two compact Hausdorff spaces).By hypothesis, γ − ( P ) = { V } .Therefore, if P is isolated in Spec( D ) cons then { P } is open andthus { V } = γ − ( { P } ) is open in Zar( D ) cons , i.e., V is isolated. Con-versely, if V is isolated then Zar( D ) \ { V } is closed, with respectto the constructible topology, and thus γ (Zar( D ) \ { V } ) is closed inSpec( D ) cons . However, γ is surjective and γ − ( P ) = { V } , so that γ (Zar( D ) \{ V } ) = Spec( D ) \{ P } . Hence, { P } is open and P is isolatedin Spec( D ) cons , as claimed. (cid:3) Corollary 3.2.
Let D be a Pr¨ufer domain, and let V be a valuationoverring of D with center P . Then, V is isolated in Zar( D ) cons if andonly if P is isolated in Spec( D ) cons . In particular, Zar( D ) cons is perfectif and only if Spec( D ) cons is perfect.Proof. Since D is a Pr¨ufer domain, every valuation overring is essential.The claim follows from Proposition 3.1. (cid:3) In general, almost essential valuation overrings are rare; for example,if D is Noetherian, no prime ideal of height 2 or more can be almostessential. For this reason, we need more general results; the first stepis connecting isolated valuation rings with compactness. Proposition 3.3.
Let D be an integral domain, L a field containing D , and let V be a minimal element of Zar( L | D ) . Then, the followingare equivalent:(i) V is isolated in Zar( L | D ) cons ;(ii) Zar( L | D ) \ { V } is compact, with respect to the Zariski topology;(iii) Zar( L | D ) \ { V } is closed in the inverse topology.Proof. Let X := Zar( L | D ) \ { V } .The equivalence of (ii) and (iii) follows from the fact that, since V is minimal, X is closed by generization.If (i) holds, then { V } is an open set in the constructible topology,and thus X is closed; since Zar( L | D ) cons is compact, it follows that X is compact in the constructible topology and thus also in the Zariskitopology (which is coarser). Thus, (ii) holds.Conversely, if (iii) holds, then X is closed in the inverse and thus inthe constructible topology; hence, { V } is open and V is isolated. (cid:3) The advantage of the previous proposition is that the property ofZar( L | D ) \ { V } being compact has very strong consequences. Theorem 3.4.
Let D be an integrally closed domain and let V ∈ Zar( L | D ) . Then, the following are equivalent.(i) V is isolated in Zar( L | D ) cons ;(ii) there are x , . . . , x n ∈ L and a maximal ideal M of D [ x , . . . , x n ] such that V is the integral closure of D [ x , . . . , x n ] M and M isisolated in Spec( D [ x , . . . , x n ]) cons ; DARIO SPIRITO (iii) there are x , . . . , x n ∈ L and a prime ideal P of D [ x , . . . , x n ] such that V is the integral closure of D [ x , . . . , x n ] P and P isisolated in Spec( D [ x , . . . , x n ]) cons ;Proof. By [30, Proposition 3.3], without loss of generality we can sup-pose that L is the quotient field of D .(i) = ⇒ (iii) Since V is isolated, there are x , . . . , x k , y , . . . , y m ∈ L such that { V } = B ( x , . . . , x n ) ∩ B ( y ) c ∩ · · · ∩ B ( y m ) c . In particular, { V } = Zar( D [ x , . . . , x k ]) ∩ B ( y ) c ∩ · · · ∩ B ( y m ) c , andthus V is a minimal valuation overring of D [ x , . . . , x k ]. By Proposition3.3, Zar( D [ x , . . . , x k ]) \ { V } is compact, with respect to the Zariskitopology; therefore, by [28, Theorem 3.6], there are x k +1 , . . . , x n ∈ L such that V is the integral closure of D [ x , . . . , x k ][ x k +1 , . . . , x n ] M = D [ x , . . . , x n ] M , for some maximal ideal M of D [ x , . . . , x n ]. Moreover, M is almost essential in D [ x , . . . , x n ], and thus by Proposition 3.1 M must be isolated in Spec( D [ x , . . . , x n ]) cons . Thus, (ii) holds.(ii) = ⇒ (iii) is obvious.(iii) = ⇒ (i) Let A := D [ x , . . . , x n ]. The set Zar( A ) = B ( x , . . . , x n ) isopen in the constructible topology, and thus V is isolated in Zar( D ) cons if and only if it is isolated in Zar( A ) cons . By hypothesis, P is almostessential for D [ x , . . . , x n ], and thus by Proposition 3.1 the integralclosure V of D [ x , . . . , x n ] P is isolated, as claimed. (cid:3) Dimension 0
In this section, we study when the field L is isolated in Zar( L | D ) cons .If L is the quotient field of D , then L is an essential valuation overringof D , and thus one can reason through Proposition 3.1; however, it ispossible to use a more general approach.A domain D with quotient field K is said to be a Goldman domain (or a
G-domain ) if K is a finitely generated D -algebra, or equivalentlyif K = D [ u ] for some u ∈ K . Proposition 4.1.
Let D be an integral domain with quotient field K ,and let L be a field extension of K . Then, L is isolated in Zar( L | D ) cons if and only if D is a Goldman domain and K ⊆ L is an algebraicextension.Proof. Suppose first that the two conditions hold. Then, K = D [ u ];hence, B ( u ) = Zar( L | K ) = { L } since K ⊆ L is algebraic. Hence, L isisolated in Zar( L | D ) cons .Conversely, suppose that L is isolated. By Theorem 3.4, there are x , . . . , x n ∈ L such that L is the integral closure of a localizationof D [ x , . . . , x n ] at a maximal ideal M ; since M must have height 0, F := D [ x , . . . , x n ] must be a field such that F ⊆ L is algebraic. SOLATED POINTS OF THE ZARISKI SPACE 7
Suppose that F is transcendental over K : then, we can take a tran-scendence basis y , . . . , y k of F over L . By construction, F is algebraicover D [ y , . . . , y k ]; since F is a field, it is a Goldman domain, and thusby [17, Theorem 22] so should be D [ y , . . . , y k ], against [17, Theorem21]. Thus F is algebraic over K . Applying again [17, Theorem 22]to the algebraic extension D ⊂ F , we see that D is a Goldman do-main; furthermore, L is algebraic over F and thus over K . The claimis proved. (cid:3) The previous result can be used to give some necessary conditionsfor V to be isolated. We premise a lemma. Lemma 4.2.
Let D be an integral domain, L be a field containing D ,and let W ∈ Zar( L | D ) . Let π : W −→ W/ m W be the quotient map.Then, the map { Z ∈ Zar( L | D ) | Z ⊆ W } −→ Zar( W/ m W | D/ ( m W ∩ D )) ,Z π ( Z ) is a homeomorphism, when both sets are endowed with either the Zariskiand the constructible topology.Proof. Let M := m W ∩ D .For a ring extension A ⊆ B , let Over( B | A ) be the set of rings between A and B . With the same proof of [30, Lemma 3.2], we see that the map e π : Over( W | D ) −→ Over( W/ m W | D/M ) ,T π ( T )is a homeomorphism in the Zariski topology.If Z ∈ Zar( L | D ) is contained into W , then π ( Z ) = e π ( Z ) = Z/ ( m W ∩ Z ) is a valuation ring. Take now any θ ∈ m Z \ m V . If φ ∈ W/ m W , we canfind φ ∈ V such that φ = π ( φ ): then, θφ ∈ Z , and so π ( θ ) φ ∈ π ( Z ).Thus, φ = ( π ( θ ) φ ) /π ( θ ) is in the quotient field of Z/ m Z , and so π ( Z ) ∈ Zar( W/ m W | D/M ).Conversely, if Z ′ ∈ Zar( W/ m W | D/M ), then Z = π − ( Z ′ ) is the pull-back of Z ′ along the quotient W −→ W/ m W . Then, Z ′ is a valuationdomain by [11, Proposition 1.1.8(1)], and its quotient field is L by [11,Lemma 1.1.4(10)]. Thus, Z ′ ∈ e π ( { Z ∈ Zar( L | D ) | Z ⊆ W } ). Therefore, e π establishes a homeomorphism between { Z ∈ Zar( L | D ) | Z ⊆ W } andZar( W/ m W | D/M ) in the Zariski topology.In particular, e π is a spectral map between the two sets, and thus itis also a homeomorphism in the constructible topology. (cid:3) Proposition 4.3.
Let V ∈ Zar( D ) be a valuation domain with cen-ter P on D . If V is isolated in Zar( D ) cons , then the field extension D P /P D P ⊆ V / m V is algebraic. DARIO SPIRITO
Proof.
Consider ∆ := { W ∈ Zar( D ) | W ⊆ V } . Since m V ∩ D = P by definition, by Lemma 4.2, the quotient map V −→ V / m in-duces a homeomorphism between ∆ and Zar( V / m V | D/P ) (both in theZariski and in the constructible topology), and thus if V is isolatedin Zar( D ) cons then V / m is isolated in Zar( V / m V | D/P ). By Proposition4.1, if F is the quotient field of D/P then the extension F ⊆ V / m V mustbe algebraic. The claim now follows from the equality F = ( D/P ) P/P = D P /P D P . (cid:3) Corollary 4.4.
Let D be an integral domain, and let γ : Zar( D ) −→ Spec( D ) be the center map. If V is isolated in Zar( D ) cons , then V ∈ Zar( D ) is minimal in γ − ( γ ( V )) .Proof. Let P := γ ( V ). If V is not minimal, then V / m V is not minimal inZar( V / m V | D P /P D P ); hence, the extension D P /P D P ⊆ V / m V cannotbe algebraic, against Proposition 4.3. (cid:3) Dimension 1
We now analyze the case where the valuation ring V has dimension1; however, we work only for valuation overrings of D , i.e., we studyZar( K | D ) only when K is the quotient field of D , so that Zar( K | D ) =Zar( D ). Note that, unlike the proof of Theorem 3.4, we cannot supposein general that K is the quotient field of D , since the construction of[30, Proposition 3.3] changes the dimension of the valuation domaininvolved.The idea of this section is to study the maximal ideals of the finitelygenerated algebras D [ x , . . . , x n ]. Proposition 5.1.
Let ( D, m ) be an integrally closed local domain, andlet T = D be a finitely generated D -algebra contained in the quotientfield K of D . If m T = T , then no maximal ideal of T above m hasheight .Proof. Let T := D [ x , . . . , x n ]; we proceed by induction on n .Suppose n = 1, and let x := x ; without loss of generality, x / ∈ D . If x − ∈ D , then it is contained in m , and thus m T = T , a contradiction.Hence x, x − / ∈ D . By [27, Theorem 6], the ideal p := m T is prime butnot maximal; since every maximal ideal of T above m must contain p ,it follows that no such maximal ideal can have height 1.Suppose that the claim holds up to n −
1; let A := D [ x , . . . , x n − ],so that T = A [ x n ]; without loss of generality, A = D and x n / ∈ A . Let M be a maximal ideal of T above m . If x n is integral over A , then T isintegral over A , and thus then the height of M is equal to the heightof M ∩ A , which is not equal to 1 by induction.Suppose that x n is not integral over A . Let A ′ be the integral closureof A ; then, T ⊆ A ′ [ x n ] is an integral extension, and since x n is notintegral over A it follows that A ′ ( A ′ [ x n ]. Let M be a maximal ideal SOLATED POINTS OF THE ZARISKI SPACE 9 of T such that M ∩ D = m , and take a maximal ideal M ′ of A ′ [ x n ]above M . Let N := M ′ ∩ A ′ ; then, N is a nonzero prime ideal of A ′ ,and thus A ′′ := ( A ′ ) N is a local integrally closed domain with maximalideal N ( A ′ ) N = (0). Then, the ring A ′′ [ x n ] is the localization of A ′ [ x n ]at the multiplicatively closed set A ′ [ x n ] \ N , the set M ′′ := M ′ A ′′ [ x n ]is a maximal ideal, and N ( A ′ ) N ⊆ M ′′ . Applying the case n = 1 to A ′′ and A ′′ [ x n ], it follows that the height of M ′′ is not 1; since the heightof M ′′ is the same of the height of M ′ and of M , it follows that theheight of M is not 1, as claimed. (cid:3) Theorem 5.2.
Let D be an integral domain, and let V ∈ Zar( D ) be avaluation overring of dimension . Then, V is isolated in Zar( D ) cons ifand only if V is a localization of D and its center on D is isolated in Spec( D ) cons .Proof. Since Zar( D ) = Zar( D ), we can suppose without loss of gener-ality that D is integrally closed.If the two conditions hold, then V is isolated by Proposition 3.1.Suppose that V is isolated in Zar( D ) cons . Let P be the center of V on D , and suppose that V = D P . Since V is also isolated in Zar( D P ) cons ,by Theorem 3.4, there are x , . . . , x n ∈ K \ D P such that V is theintegral closure of D P [ x , . . . , x n ] M , where M is a maximal ideal of D P [ x , . . . , x n ]. However, m V ∩ D P [ x , . . . , x n ] = M , and thus M ∩ D P = P D P , so that P D P · D P [ x , . . . , x n ] = D P [ x , . . . , x n ]; by Proposition5.1, M cannot have height 1. However, the dimension of the integralclosure of D P [ x , . . . , x n ] M is exactly the height of M ; hence, this con-tradicts the fact that V has dimension 1. Thus, V = D P . The fact that P is isolated in Zar( D ) cons now follows from Proposition 3.1. (cid:3) Corollary 5.3.
Let D be an integral domain, and let V ∈ Zar( D ) be aminimal valuation overring of D . If dim( V ) = 1 and Zar( D ) \ { V } iscompact, then the center of V on D has height .Proof. If Zar( D ) \{ V } is compact, then by Proposition 3.3 V is isolatedin Zar( D ) cons . The claim now follows from Theorem 5.2. (cid:3) Theorem 5.2 does not work when V has dimension 2 or more, as thenext example shows. Example 5.4.
Let F be a field, take two independent indeterminates X and Y , and consider D := F + XF ( Y )[[ X ]], i.e., D is the ringof all power series with coefficients in F ( Y ) such that the 0-degreecoefficient belongs to F . Then, D is a one-dimensional local integrallyclosed domain (its maximal ideal is XF ( Y )[[ X ]]), and its valuationoverrings are its quotient field, F ( Y )[[ X ]] and the rings in the form W + XF ( Y )[[ X ]], where W belongs to Zar( F ( Y ) | F ) \ { F ( Y ) } , i.e., W is either F [ Y ] ( f ) for some irreducible polynomial f ∈ F [ Y ]) or W = F [ Y − ] ( Y − ) . Each of these W + XF ( Y )[[ X ]] is isolated in Zar( D ) cons , since each W is isolated in Zar( F ( Y ) | F ) (this follows, for example, by applyingTheorem 6.4 below to F [ Y ] or to F [ Y − ]). However, since every W + XF ( Y )[[ X ]] has dimension 2, it can’t be a localization of D = D .6. The Noetherian case
In this section, we want to use Theorem 5.2 to characterize the iso-lated points of Zar( D ) cons when D is a Noetherian domain. If D isintegrally closed, this is rather immediate; to extend it to also the non-integrally closed case, we need a few lemmas. (Note that the integralclosure of a Noetherian domain is not necessarily Noetherian; see e.g.[19, Example 5, page 209] for an example.)The following two lemmas are essentially “dual” one of each other. Lemma 6.1.
Let D be an integral domain. Let V ∈ Zar( L | D ) and let ∆ ⊆ Zar( L | D ) . If V = S { W | W ∈ ∆ } , then V ∈ Cl cons (∆) .Proof. Let Ω be an open set of Zar( D ) cons containing V ; we have toshow that Ω meets ∆. Without loss of generality, we can suppose thatΩ is a subbasic open set, i.e., that it is either equal to B ( x ) or to B ( x ) c for some x ∈ L .If Ω = B ( x ), then x ∈ V , and thus by hypothesis there is a W ∈ ∆such that x ∈ W , i.e., W ∈ Ω. If Ω = B ( x ) c , then x / ∈ W for every W ∈ ∆, and thus ∆ ⊆ Ω. In both cases, Ω ∩ ∆ = ∅ and V ∈ Cl cons (∆). (cid:3) Lemma 6.2.
Let D be an integral domain. Take a prime ideal P anda subset ∆ ⊆ Spec( D ) . If P = T { Q | Q ∈ ∆ } , then P ∈ Cl cons (∆) .Proof. Let Ω be an open set of Spec( D ) cons containing P ; we have toshow that Ω meets ∆. Without loss of generality, we can suppose thatΩ is a subbasic open set, i.e., that it is either equal to V ( I ) for someideal I or to D ( J ) for some finitely generated ideal J .If Ω = V ( I ), then I ⊆ P , and thus I ⊆ Q for every Q ∈ ∆. IfΩ = D ( J ), then J * P , and thus there must be a Q ∈ ∆ such that J * Q ; thus, Q ∈ D ( J ). Hence, Ω ∩ ∆ = ∅ and P ∈ Cl cons (∆). (cid:3) Lemma 6.3.
Let A ⊆ B be an integral extension, and let P ∈ Spec( A ) , Q ∈ Spec( B ) be such that Q ∩ A = P . If T { P ′ ∈ Spec( A ) | P ′ ) P } = P , then T { Q ′ ∈ Spec( B ) | Q ′ ) Q } = Q .Proof. Let I := T { Q ′ ∈ Spec( B ) | Q ′ ) Q } , and suppose I = Q ; then, Q ( I and V ( I ) = V ( Q ) \ { Q } . Consider the canonical map of spectra φ : Spec( B ) −→ Spec( A ): then, φ is closed (with respect to the Zariskitopology) [2, Chapter V, §
2, Remark (2)], and thus φ ( V ( I )) is closedin Spec( A ).By the lying over and the going up theorems, every P ′ ) P belongsto φ ( V ( I )), while P / ∈ φ ( V ( I )); hence, φ ( V ( I )) = V ( P ) \ { P } . However,the condition T { P ′ ∈ Spec( A ) | P ′ ) P } = P shows that V ( P ) \ { P } SOLATED POINTS OF THE ZARISKI SPACE 11 is not closed (its closure is V ( P )), a contradiction. Hence, I = Q , asclaimed. (cid:3) Theorem 6.4.
Let D be a Noetherian domain, and let V ∈ Zar( D ) ;let P be the center of V on D . Then, V is isolated in Zar( D ) cons if andonly if h ( P ) ≤ and V ( P ) is finite.Proof. Suppose first that V is isolated in Zar( D ) cons .If dim( V ) >
1, then V is not Noetherian. By Theorem 3.4, V is theintegral closure of D [ x , . . . , x n ] M , for some x , . . . , x n ∈ V and somemaximal ideal M . However, D [ x , . . . , x n ] is Noetherian, and thus sois D [ x , . . . , x n ] M ; hence, its integral closure is a Krull domain, whichcan’t be a non-Noetherian valuation domain, a contradiction.If dim( V ) = 0, then V = K . By Proposition 4.1, D must be aGoldman domain; by [17, Theorem 146], V ( P ) is finite.If dim( V ) = 1, then by Theorem 5.2 D is the localization of D at aprime ideal of Q of height 1; let P := Q ∩ D . If V ( P ) is infinite, then P is the intersection of all the prime ideals properly containing it (since D/P is not a Goldman domain); by Lemma 6.3, the same propertyholds for Q , and thus by Lemma 6.2 Q is not isolated in Spec( D ) cons .Let γ : Zar( D ) −→ Spec( D ) be the center map: then, γ is continuousin the constructible topology, and γ − ( Q ) = D Q = V since D is a Krulldomain and Q has height 1. Hence, V is not isolated, a contradiction;thus V ( P ) must be finite.Conversely, suppose that the two conditions hold, and let V ( P ) = { P, Q , . . . , Q n } . For each i , let y i ∈ Q i \ P and let x i := 1 /y i : then, A := D [ x , . . . , x n ] is a Noetherian domain such that P A is a maximalideal of A of height ≤
1, and Zar( A ) = B ( x , . . . , x n ) is an open set ofZar( D ) cons ; hence, it is enough to prove that V is isolated in Zar( A ) cons .If P has height 0, then A = K = V and thus V is isolated. Supposethat h ( P ) = 1.Since A is Noetherian, { P A } is an open subset of Spec( A ), withrespect to the constructible topology; hence, γ − A ( P A ) is an open subsetof Zar( A ) cons , where γ A : Zar( A ) −→ Zar( D ) is the center map relativeto A . However, γ − A ( P A ) is the set of valuation overrings of A P A = D P centered on ( P A ) A P A = P D P ; since P has height 1, D P has dimension1, and thus γ − A ( P A ) is in bijective correspondence with the maximalideals of the integral closure B of D P , which is Noetherian by [17,Theorem 93]. In particular, since P is in the Jacobson radical of B ,they are finite; since Zar( A ) cons is Hausdorff, it follows that all thepoints of γ − A ( P A ) (and, in particular, V ) are open, i.e., isolated inZar( A ) cons and thus in Zar( D ) cons . (cid:3) Corollary 6.5.
Let D be a Noetherian local domain of dimension atleast . Then, Zar( D ) cons is perfect. Proof.
Suppose V is isolated in Zar( D ) cons . By Theorem 6.4, its center P must have height 1 and V ( P ) must be finite. However, since P hasheight 1 and the maximal ideal m of D has height at least 3, there areprime ideals between P and m , and since D is Noetherian there mustbe infinitely many of them [17, Theorem 144], a contradiction. Henceno V can be isolated, and Zar( D ) cons is perfect. (cid:3) We now want to show that, when D is countable, there are fewpossible topological structures for Zar( D ) cons . The one-dimensional caseis very easy. Proposition 6.6.
Let ( D, m ) and ( D ′ , m ′ ) be two Noetherian local do-mains of dimension . The following are equivalent:(i) | Max( D ) | = | Max( D ′ ) | ;(ii) Zar( D ) ≃ Zar( D ′ ) ;(iii) Zar( D ) cons ≃ Zar( D ′ ) cons .Proof. Since D is Noetherian and one-dimensional, D is a principalideal domain with finitely many maximal ideals; hence, Zar( D ) =Zar( D ) ≃ Spec( D ), and the homeomorphism holds both in the Zariskiand in the constructible topology.Hence, if | Max( D ) | = | Max( D ′ ) | then Spec( D ) ≃ Spec( D ′ ) and thusZar( D ) and Zar( D ′ ) are homeomorphic in both the Zariski and theconstructible topology. Conversely, if Zar( D ) ≃ Zar( D ′ ) (in any of thetwo topologies) then in particular they have the same cardinality, whichis equal to | Max( D ) | +1 = | Max( D ′ ) | +1; thus | Max( D ) | = | Max( D ′ ) | .The claim is proved. (cid:3) For larger dimension, we need to join the previous theorems with thetopological characterization of the Cantor set. We isolate a lemma.
Lemma 6.7.
Let D be a countable domain. Then, Zar( D ) cons is metriz-able.Proof. The space Zar( D ) cons is compact and Hausdorff, hence normal[31, Theorem 17.10] and in particular regular. Furthermore, the familyof sets B ( t ) and B ( t ) c (as t ranges in the quotient field of D ) forma subbasis of Zar( D ) cons , and thus Zar( D ) cons is second countable.By Urysohn’s metrization theorem [31, Theorem 23.1], Zar( D ) cons ismetrizable. (cid:3) Proposition 6.8.
Let ( D, m ) and ( D ′ , m ′ ) be two countable Noetherianlocal domains of dimension at least . Then, Zar( D ) cons ≃ Zar( D ′ ) cons .Proof. Both Zar( D ) cons and Zar( D ′ ) cons are totally disconnected (sincethey are zero-dimensional), compact and perfect (Corollary 6.5). ByLemma 6.7, they are also metrizable.By [31, Theorem 30.3], any two spaces with these properties arehomeomorphic; hence, Zar( D ) cons ≃ Zar( D ′ ) cons . (cid:3) SOLATED POINTS OF THE ZARISKI SPACE 13
To study the case of dimension 2, we need two further lemmas.
Lemma 6.9.
Let ( D, m ) be a local Noetherian domain with dim( D ) > . If D is countable, then D has exactly countably many prime idealsof height .Proof. By [17, Theorem 144], there are infinitely many prime idealsbetween (0) and m , and thus D has infinitely many prime ideals ofheight 1.Moreover, every prime ideal is generated by a finite set, and thusthe number of prime ideals of height 1 is at most equal to the numberof finite subsets of D . Since D is countable, so is the set of its finitesubsets; the claim is proved. (cid:3) Lemma 6.10.
Let ( D, m ) be a local Noetherian domain of dimension with quotient field K , and let X be the set of isolated points of Zar( D ) cons . Then:(a) X is nonempty and compact, with respect to the Zariski topol-ogy;(b) if D is countable, then X is countable;(c) Cl cons ( X ) = X ∪ { K } .Proof. (a) Let γ : Zar( D ) −→ Spec( D ) be the center map, and let X be the set of height 1 prime ideals of D . We claim that V ∈ X if andonly if γ ( V ) ∈ X : indeed, if V ∈ X then by Theorem 6.4 the heightof γ ( V ) is at most 1, but it can’t be 0 since V ((0)) is infinite. On theother hand, if γ ( V ) ∈ X then V ( γ ( V )) = { γ ( V ) , m } is finite, andthus V ∈ X again by Theorem 6.4. Therefore, X = γ − ( X ), and inparticular X is nonempty.Since D is a Noetherian ring, Spec( D ) is a Noetherian space withrespect to the Zariski topology (i.e., all its subsets are compact) (see [4,Theorem 12.4.3] or [1, Chapter 6, Exercises 5–8]). Since γ is a spectralmap, the counterimage of any compact subset of Spec( D ) is compact;therefore, X = γ − ( X ) is compact with respect to the Zariski topology,as claimed.(b) By Lemma 6.9, X is countable; furthermore, for every P ∈ X , γ − ( P ) is finite, since it is in bijective correspondence with the set ofmaximal ideals of the integral closure of D P . Since X = γ − ( X ), itfollows that X is countable.(c) Since X is compact, the set X ↑ := { V ∈ Zar( D ) | V ⊇ W for some W ∈ X } is closed in the inverse topology, and thus in theconstructible topology; since every element of X is a one-dimensionalvaluation ring, furthermore, X ↑ = X ∪ { K } . Hence, Cl cons ( X ) ⊆ X ∪{ K } .If they are not equal, then it should be Cl cons ( X ) = X . However, X isinfinite (since X is infinite, by Lemma 6.9) and discrete (by definition,all its points are isolated) and thus it is not compact with respect to the constructible topology; this is a contradiction, since a closed set ofa compact set is compact. Thus, Cl cons ( X ) = X ∪ { K } , as claimed. (cid:3) Note that the set X is not compact with respect to the constructibletopology, as it is discrete and infinite. Proposition 6.11.
Let ( D, m ) and ( D ′ , m ′ ) be two countable Noether-ian local domains of dimension . Then, Zar( D ) cons ≃ Zar( D ′ ) cons .Proof. Denote by
K, K ′ the quotient fields of D and D ′ , respectively.Let X and C be, respectively, the set of isolated and nonisolatedpoints of Zar( D ) cons . Then, C is closed (with respect to the constructibletopology) since it is the intersection of the closed sets Zar( D ) \ { V } as V ∈ X . Define in the same way X ′ and C ′ inside Zar( D ′ ); then, C ′ isclosed.As in the proof of Proposition 6.8, we see that C and C ′ are to-tally disconnected, perfect, compact and metrizable, and thus they arehomeomorphic. In particular, they are homeomorphic to the Cantor set[31, Corollary 30.4], which is homogeneous [31, Problem 30A]; hence,we can find a homeomorphism φ C : C −→ C ′ such that φ ( K ) = K ′ .The sets X and X ′ are both discrete and both countable (Lemma6.10(b)); hence, any bijective map from X to X ′ is a homeomorphism.Choose one and call it φ X .Define φ : Zar( D ) cons −→ Zar( D ′ ) cons ,V ( φ C ( V ) if V ∈ C,φ X ( V ) if V ∈ X. By construction, φ is bijective.Consider the restriction of φ to X := X ∪ { K } : since φ ( K ) = K ′ ,its image is X ′ := X ′ ∪ { K ′ } . By Lemma 6.10(c), K is the uniquelimit point of X and K ′ is the unique limit point of X ′ ; thus, φ | X is ahomeomorphism between X and X ′ (with respect to the constructibletopology). Therefore, the restriction of φ to the closed sets X and C (whose union is the whole Zar( D )), is continuous; by [31, Theorem 7.6], φ is continuous on Zar( D ) cons . By the same reasoning, also the inverse φ − : Zar( D ′ ) cons −→ Zar( D ) is continuous; hence, φ is a homeomor-phism. In particular, Zar( D ) cons ≃ Zar( D ′ ) cons . (cid:3) We summarize the previous results in the following theorem.
Theorem 6.12.
Let ( D, m ) and ( D ′ , m ′ ) be two countable Noetherianlocal domains. Then, Zar( D ) cons ≃ Zar( D ′ ) cons if and only if one of thefollowing conditions hold:(1) dim( D ) = dim( D ′ ) = 1 and | Max( D ) | = | Max( D ′ ) | ; A topological space Y is homogeneous is, for every y , y ∈ Y there is a home-omorphism f : Y −→ Y such that f ( y ) = y . SOLATED POINTS OF THE ZARISKI SPACE 15 (2) dim( D ) = dim( D ′ ) = 2 ;(3) dim( D ) ≥ and dim( D ′ ) ≥ .Proof. If D and D ′ satisfy one of the conditions, then Zar( D ) cons ≃ Zar( D ′ ) by, respectively, Proposition 6.6, Proposition 6.11 and Propo-sition 6.8.Suppose now that Zar( D ) cons ≃ Zar( D ′ ) cons .If dim( D ) = 1, then Zar( D ) is finite, and thus so must be Zar( D ′ );hence, dim( D ′ ) = 1, and | Max( D ) | = | Max( D ′ ) | by Proposition 6.6.If dim( D ) = 2, then Zar( D ) cons has isolated points; since Zar( A )is perfect if A is Noetherian and local with dim( A ) ≥
3, it must bedim( D ′ ) = 2. In the same way, if dim( D ) ≥ D ) cons is perfectand thus Zar( D ′ ) ≥ D ′ in the place of D , theclaim is proved. (cid:3) When D is a field In this section we analyze what happens if we consider the Zariskispace Zar( L | D ) when also D is a field, which in the following we denoteby K . Note that if L is algebraic over K , then Zar( L | K ) is just a point( L itself); thus, the only interesting case is when trdeg( L/K ) ≥ L | D ) cons and theisolated points of Zar( L ′ | D ), where L ′ is an algebraic extension of L . Proposition 7.1.
Let V be a valuation domain, L ⊆ L ′ be an algebraicextension, and let W ∈ Zar( L ′ | V ) . Let X ⊆
Zar( L ′ | V ) be a subsetcontaining all extensions of W ∩ L to L ′ . Then, if W is isolated in X cons then W ∩ L is isolated in X ∩ L := { Z ∩ L | Z ∈ X } , with respect tothe constructible topology. In particular, this happens if X = Zar( L ′ | V ) or X = E ( L ′ | V ) .Proof. Suppose that W is isolated in X : then, there are some elements x , . . . , x n , y , . . . , y m ∈ L ′ such that { W } = B ( x , . . . , x n ) ∩ B ( y ) c ∩· · · ∩ B ( y m ) c ∩ X . Let L ′′ be the field generated by the x i and the y j over L , and let W := W ∩ L ′′ : then, L ′′ is a finite extension of L ,and { W } = B L ′′ ( x , . . . , x n ) ∩ B L ′′ ( y ) c ∩ · · · ∩ B L ′′ ( y m ) c ∩ ( X ∩ L ′′ ); inparticular, W is isolated in X ∩ L ′′ .Let now F be the normal closure of L ′′ over L ; then, F is finite over L ′′ and over L , and thus the set X of extensions of W to F is finite[12, Theorem 20.1]. Let X be the set of extensions of the elements of X ∩ L to F . Then, X ⊆ X , and X = B F ( x , . . . , x n ) ∩ B F ( y ) c ∩ · · · ∩B F ( y m ) c ∩ X ; hence, X is open in the constructible topology. Since X is finite, it follows that it is discrete.Let X be the set of extensions of W ∩ L = W ∩ L to F . Since L ⊆ F is normal, for every Z ∈ X and every Z ∈ X there is an L -automorphism σ of F such that σ ( Z ) = Z [12, Corollary 20.2]. By construction, σ can be viewed as a map from X to itself; then, σ iscontinuous, with respect to the constructible topology, and thus a self-homeomorphism of X cons1 . Since the elements of X are isolated, X isdiscrete, and in particular it is open in X cons1 .Let ρ : Zar( F | V ) cons −→ Zar( L | V ) cons be the restriction map: then, ρ is a closed map, and by definition ρ ( X ) = X ∩ L and ρ − ( W ∩ L ) = X . Take a closed set C of Zar( F | V ) cons such that C ∩ X = X \ X :then, ρ ( C ) ∩ ( X ∩ L ) = ( X ∩ L ) \ { W ∩ L } . Therefore, { W ∩ L } isopen in ( X ∩ L ) cons , i.e., W ∩ L is isolated, as claimed.The “in particular” statement follows immediately. (cid:3) Corollary 7.2.
Let V be a valuation domain and L ⊆ L ′ be an alge-braic extension. Let X ⊆
Zar( L ′ | V ) be a subset such that, for every W ∈ X , all extensions of W ∩ L to L ′ belong to X . If X ∩ L is perfectand |X ∩ L | > , then X is perfect.Proof. Suppose that X is not perfect: then, there is a W ∈ X that isnot isolated. By Proposition 7.1, it would follow that W ∩ L is isolated.Since X ∩ L has more than one point, this is impossible, and so X isperfect. (cid:3) Corollary 7.3.
Let V be a valuation domain and L ⊆ L ′ be an alge-braic extension; suppose that L is transcendental over V . If Zar( L | V ) cons (respectively, E ( L | V ) cons ) is perfect, then Zar( L ′ | V ) cons (resp., E ( L ′ | V ) cons )is perfect.Proof. It is enough to apply Corollary 7.2 to X = Zar( L ′ | V ) or X = E ( L ′ | V ), using the hypothesis that L is transcendental over V to guar-antee that |X ∩ L | > (cid:3) The following result completely settles the problem of finding theisolated points when trdeg(
L/K ) ≥
2, generalizing [3, Theorem 4.45]and solving the authors’ Conjecture A (in an even more general formu-lation). Note that the final part of the proof (the fact that there are noisolated points in Zar( L ′′ | K ) cons ) is exactly [3, Theorem 4.45], but weobtain it in a different way as a consequence of Theorem 6.4. Theorem 7.4.
Let K ⊆ L be a field extension with trdeg( L/K ) ≥ .Then, Zar( L | K ) cons is perfect.Proof. Suppose V ∈ Zar( L | K ) is isolated, with respect to the con-structible topology; then, there are a , . . . , a j , b , . . . , b k ∈ L such that { V } = B ( a , . . . , a j ) ∩ B ( b ) c ∩ · · · ∩ B ( b k ) c . Let L ′ is the extensionof K generated by a , . . . , a j , b , . . . , b k and two algebraically indepen-dent elements (so that trdeg( L ′ /K ) ≥ V ∩ L ′ is isolated inZar( L ′ | K ) cons .Let x , . . . , x n be a transcendence basis of L ′ over K , and let L ′′ := K ( x , . . . , x n ); then, L ′′ ⊆ L ′ is algebraic, and thus by Proposition 7.1 W := V ∩ L ′′ is isolated in Zar( L ′′ | D ) cons . SOLATED POINTS OF THE ZARISKI SPACE 17
For each i , at least one of x i and x − i that belongs to W ; let it be t i .Then, V ∈ Zar( D [ t , . . . , t n ]), and so V is isolated in Zar( K [ t , . . . , t n ]) cons .Let P be the center of W on K [ t , . . . , t n ]; by Theorem 6.4, P has height1. Since K [ t , . . . , t n ] is isomorphic to a polynomial ring, every maximalideal of K [ t , . . . , t n ] has height n > P is not maximal. However, K [ t , . . . , t n ] is an Hilbert ring, andthus every non-maximal prime ideal is the intersection of the maximalideals containing it [17, Theorem 147]; in particular, this happens for P ,and thus V ( P ) must be infinite. This contradicts Theorem 6.4; hence, W cannot be isolated and so neither can V . Since V was arbitrary,Zar( L | D ) cons is perfect. (cid:3) When the transcendence degree of L over K is 1, the picture is lesssatisfying. We start from the case L = K ( X ). Compare the next tworesults with [28, Corollary 5.5(a)] and [30, Proposition 4.2]. Proposition 7.5.
Let K be a field. Then all points of Zar( K ( X ) | K ) ,except K ( X ) , are isolated with respect to the constructible topology.Proof. The points of Zar( K ( X ) | K ) are K ( X ), K [ X ] ( X − ) and the rings K [ X ] ( f ( X )) , where f ( X ) is an irreducible polynomial of K [ X ]. The firstone is not isolated by Proposition 4.1; on the other hand, { K [ X ] ( f ( X )) } = B ( f ( X ) − ) c and { K [ X − ] ( X − ) } = B ( X ) c , and thus these domains areisolated, as claimed. (cid:3) Lemma 7.6.
Let K be a field, and let L be an extension of K suchthat trdeg( L/K ) = 1 . Let V ∈ Zar( L | K ) , V = L . Then, V is isolatedin Zar( L | K ) cons if and only if there is a finitely generated extension L ′ ⊆ L of K such that V is the unique extension of V ∩ L ′ to L .Proof. If V is isolated, then V = B ( x , . . . , x n ) ∩ B ( y ) c ∩ · · · ∩ B ( y m ) c ,for some x , . . . , x n , y , . . . , y m ∈ L . The claim follows by taking L ′ := L ( x , . . . , x n , y , . . . , y m ).Conversely, suppose there is such an L ′ ; then, L ′ must be transcen-dental over K , and thus L ′ is finite over K ( X ) for some X ∈ L ′ thatis transcendental over K . By Proposition 7.5, W := W ∩ K ( X ) isisolated in Zar( K ( X ) | K ) cons . Since K ( X ) ⊆ L ′ is finite, W has onlyfinitely many extensions to L ′ , and their set is open (as the restric-tion map from L ′ to K ( X ) is continuous); hence, each of them is iso-lated, i.e., W = B L ′ ( x , . . . , x n ) ∩ B L ′ ( y ) c ∩ · · · ∩ B L ′ ( y m ) c , for some x , . . . , x n , y , . . . , y m ∈ L ′ . Since W is the only extension of W , wehave W = B L ( x , . . . , x n ) ∩ B L ( y ) c ∩ · · · ∩ B L ( y m ) c and thus W isisolated, as claimed. (cid:3) Proposition 7.7.
Let K be a field, and let L be an extension of K suchthat trdeg( L/K ) = 1 . Let X := Zar( L | K ) \ { L } . Then, the followingare equivalent: (i) X cons is perfect;(ii) for every X ∈ L , transcendental over K , every valuation do-main V of K ( X ) , V = K ( X ) , has only finitely many extensionsto L ;(iii) there is an X ∈ L , transcendental over K , such that everyvaluation domain V of K ( X ) , V = K ( X ) , has only finitelymany extensions to L .Proof. (i) = ⇒ (ii) Let V ∈ Zar( K ( X ) | K ), V = K ( X ). Then, there is a W ∈ X that extends V . Since W is isolated, by Lemma 7.6 there is afinitely generated extension L ′ of K such that W ∩ L ′ has exactly oneextension to L ; furthermore, without loss of generality we can suppose X ∈ L ′ . The extension K ( X ) ⊆ L ′ is finite, and thus W ∩ K ( X ) = V has only finitely many extensions to L ′ ; therefore, V has only finitelymany extensions to L , as claimed.(ii) = ⇒ (iii) is obvious.(iii) = ⇒ (i) Let W ∈ X , and let V := W ∩ K ( X ). Then, V = K ( X ),and thus V has only finitely many extensions to L ; in particular, wecan find a finite extension L ′ of K ( X ) and an extension V ′ of V to L ′ such that W is the unique extension of V ′ to L . Then, L ′ satisfiesthe hypothesis of Lemma 7.6, and thus W is isolated. Since W wasarbitrary, X cons is perfect, as claimed. (cid:3) Corollary 7.8.
Let K be a field and let L be a finitely generated ex-tension of K such that trdeg( L/K ) = 1 . Then, (Zar( L | K ) \ { L } ) cons isperfect.Proof. It is enough to apply Proposition 7.7. (cid:3)
Remark 7.9.
It is possible for a valuation domain V ∈ Zar( L | K ) cons to be isolated even if V ∩ K ( X ) has infinitely many extensions to L (where X ∈ L is transcendental over K ). For example, suppose that W is an extension of V to K ( X ) that is a discrete valuation domain.Using [18] (see also [13, Section 3]), it is possible to construct a chain K ( X ) ⊂ F ⊂ F ⊂ · · · such that: • W has two extensions to F , say W and W ; • W has only one extension to F i , for each i > • each extension of W to F i has more than one extension to F i +1 .Let L := S i ≥ F i . Then, W has a unique extension W to L , while W (and so W ) has infinitely many extensions; however, W is an extensionof W that is isolated in Zar( L | V ) cons , since if x ∈ m W \ m W then { W } = B ( x − ) c . 8. Extensions of valuations
In this section, we want to extend the results of the previous sectionby taking D = V to be a valuation domain; in particular, we study theset E ( L | V ) of extensions of V to L . SOLATED POINTS OF THE ZARISKI SPACE 19
The most important case is when L = K ( X ) is the field of ratio-nal functions. If V is a valuation domain with quotient field K and s ∈ K (or, more generally, if s belongs to the algebraic closure of thecompletion of K ), we set V s := { φ ∈ K ( X ) | φ ( s ) ∈ V } . Then, V s is an extension of V to K ( X ), and it is possible to analyzeits algebraic properties (see for example [23, Proposition 2.2] for a de-scription when V has dimension 1).The following lemma is a partial generalization of [23, Theorem 3.2],of which we follow the proof. Lemma 8.1.
Let V be a valuation domain with quotient field K , andlet U be an extension of V to the algebraic closure K . Let s, t ∈ K .Then, U s ∩ K ( X ) = U t ∩ K ( X ) , if and only if s and t are conjugatedover K .Proof. If s, t are conjugated, there is a K -automorphism σ of K sending s to t . Setting e σ ( P i a i X i ) := P i σ ( a i ) X i , we can extend σ to a K ( X )-automorphism e σ of K ( X ) such that e σ ( φ )( t ) = σ ( φ ( s )) for every φ ∈ K ( X ); in particular, if φ ∈ K ( X ) then e σ ( φ ) = φ and thus φ ( s ) ∈ V ifand only if φ ( t ) ∈ V , i.e., φ ∈ U s ∩ K ( X ) if and only if φ ∈ U t ∩ K ( X ).Therefore, U s ∩ K ( X ) = U t ∩ K ( X ).Conversely, suppose that s and t are not conjugate, and let p ( X ) bethe minimal polynomial of s over K : then, p ( t ) = 0, and thus thereis a c ∈ K such that v ( c ) > u ( p ( t )) (where v and u are, respectively,the valuations with respect to V and U and u | K = v ). Then, q ( X ) := p ( X ) c ∈ K ( X ) belongs to U s (since q ( s ) = 0 ∈ V ) but not to U t (since u ( q ( t )) = u ( p ( t )) − v ( c ) < U s ∩ K ( X ) = U t ∩ K ( X ), asclaimed. (cid:3) Theorem 8.2.
Let V be a valuation domain that is not a field. Then, E ( K ( X ) | V ) cons is perfect.Proof. Suppose first that K is algebraically closed. By [25, Theorem7.2], for all extensions W of V to K ( X ) there is a sequence E = { s ν } ν ∈ Λ such that W = V E = { φ ∈ K ( X ) | φ ( s ν ) ∈ V for all large ν } . has the form V E for some pseudo-monotone sequence E = { s ν } ν ∈ Λ [25,Theorem 7.2] (see [25, Section 2] for the definition of pseudo-monotonesequence); in particular, the elements φ ( s ν ) are either definitively in V or definitively out of V (by [25, Proposition 3.2]; see also the proof ofTheorem 3.4 therein). Hence, if W ∈ B ( ψ ) then it must be ψ ( s ν ) ∈ V definitively, and thus B ( ψ ) contains some V s ν ; on the other hand, if W ∈ B ( ψ ) c then ψ ( s ν ) / ∈ V definitively, and thus B ( ψ ) c contains some V s ν . Since the B ( ψ ) and the B ( ψ ) c form a basis for the constructibletopology, it follows that W = V E is in the closure of E (and, indeed, that it is the limit of the sequence { V s ν | ν ∈ Λ } ; see [24, Proposition6.9] for a more general result when V has dimension 1). In particu-lar, each open neighborhood of W contains some V s ν = W , and thus E ( K ( X ) | V ) cons is not perfect.Suppose now that K is arbitrary, and let W ∈ E ( K ( X ) | V ). Supposethat W is isolated in E ( K ( X ) | V ) cons . Let W ′ be an extension of W to K ( X ), and let U := W ′ ∩ K ; set ρ : E ( K ( X ) | U ) −→ E ( K ( X ) | V ) to bethe restriction map. Since W is isolated and ρ is continuous, ρ − ( W )is open; by the previous part of the proof, no point of ρ − ( W ) is open,and since E ( K ( X ) | V ) cons is Hausdorff it follows that ρ − ( W ) containsinfinitely many valuation rings in the form U t (with t ∈ K ), and inparticular all of them restrict to W . However, by Lemma 8.1, this canhappen to only finitely many of them (since every t ∈ K has finitelymany conjugates over K ); this is a contradiction, and thus W cannotbe isolated. It follows that E ( K ( X ) | V ) cons is perfect, as claimed. (cid:3) This theorem allows to determine the isolated points of Zar( K ( X ) | D ) cons for every integral domain D . We first need a lemma. Lemma 8.3.
Let D be an integral domain. If there are φ , . . . , φ n ∈ K ( X ) such that K ⊆ D [ φ , . . . , φ n ] , then D is a Goldman domain.Proof. If every φ i belongs to K , then the claim follows from [17, The-orem 18]. Suppose that one of them is transcendental, and let F bethe quotient field of A := D [ φ , . . . , φ n ]. By L¨uroth’s theorem (see e.g.[16, § ψ ∈ F such that F = K ( ψ ), and without loss ofgenerality we can suppose ψ ∈ A . Then, D [ ψ ] has quotient field F .Since A contains K , it is an overring of K [ ψ ], and, since K [ ψ ] ≃ K [ X ]is a principal ideal domain, A is a localization of K [ ψ ]. Furthermore, K [ ψ ] is the localization of D [ ψ ] at D \{ } , and thus A is a localization of D [ ψ ]. Therefore, for every i there is an s i ∈ D [ ψ ] such that s i φ i ∈ D [ ψ ]and s − i ∈ A . Let s := s · · · s n : then, A = D [ ψ ][1 /s ] = D [ ψ, /s ].For every nonzero prime ideal P of D , we have P D [ ψ ] = D [ ψ ] and P A = A , and so s must belong to P D [ ψ ]. However, since D [ ψ ] ≃ D [ X ]is a polynomial ring, this implies that the intersection of all such P is nonzero. This intersection is exactly the set of polynomials whosecoefficients are in the intersection of all the P , which thus must benonzero; hence, D is a Goldman domain. (cid:3) Proposition 8.4.
Let D be an integral domain that is not a field, andlet J be the intersection of the nonzero prime ideals of D .(i) If J = (0) , then Zar( K ( X ) | D ) cons is perfect.(ii) If J = (0) , then the only isolated points of Zar( K ( X ) | D ) cons are K [ X ] ( f ( X )) (where f ( X ) is an irreducible polynomial of K [ X ] )and K [ X ] ( X − ) . SOLATED POINTS OF THE ZARISKI SPACE 21
Proof.
Let W ∈ Zar( K ( X ) | D ). If V := W ∩ K = K , then W ∈E ( K ( X ) | V ), which is perfect (when endowed with the constructibletopology) by Theorem 8.2. In particular, W is not isolated in Zar( K ( X ) | D ) cons .Suppose now that W ∩ K = K . If W = K ( X ), then W is notisolated by Proposition 4.1, since K is not algebraic over K . Let thus W = K ( X ).Suppose that J = (0), and suppose that W is isolated: let x , . . . , x n , y , . . . , y m be such that { W } = B ( x , . . . , x m ) ∩ B ( y ) c ∩ · · · ∩ B ( y m ) c =:Ω. Since J = (0), by Lemma 8.3 the domain D [ x , . . . , x n ] does not con-tain K , and thus there is a valuation domain W ′ ∈ Zar( K ( X ) | D [ x , . . . , x n ])not containing K . However, by construction W ′ = W and W ′ ∈ Ω,against the fact that W is isolated.Suppose now that J = (0), and let j ∈ J . Then, D [ j − ] = K ,and thus B ( j − ) = E ( K ( X ) | K ) = Zar( K ( X ) | K ) is a clopen sub-set of Zar( K ( X ) | D ) cons ; in particular, W ∈ E ( K ( X ) | K ) is isolatedin Zar( K ( X ) | D ) cons if and only if it is isolated in Zar( K ( X ) | K ) cons .The claim now follows from Proposition 7.5. (cid:3) To conclude the paper, we extend Theorem 7.4 to valuation domains.
Theorem 8.5.
Let V be a valuation domain with quotient field K ,and let L be a field extension of K such that trdeg( L/K ) ≥ . Then, E ( L | V ) cons and Zar( L | V ) cons are perfect.Proof. Let W ∈ Zar( L | V ), and set V := W ∩ L . Suppose that W isisolated in E ( L | V ) cons .Let x, z , z , . . . be a transcendence basis of L over K . If m ∈ m V ,then one of m + x and x − belongs to m V ; let z be that element. Then, z , z , . . . is also a transcendence basis of L .Let L ′ := K ( z , z , . . . ) be the extension of K obtained adjoiningall the element of the basis save for z . Then, L ′ ( z ) ⊆ L is an alge-braic extension; by Proposition 7.1 W := W ∩ L ′ ( z ) is isolated in E ( L ′ ( z ) | V ) cons .However, W ′ := W ∩ L ′ is not equal to L ′ (since it does not con-tain z − ), and thus by Theorem 8.2 E ( L ′ ( z ) | W ′ ) cons is perfect. Since E ( L ′ ( z ) | W ′ ) is infinite, this contradicts the fact that W is isolated.Hence, W cannot be isolated in E ( L | V ) cons , and thus E ( L | V ) cons isperfect. Furthermore, since this holds for every V , it follows that alsoZar( L | V ) cons is perfect. (cid:3) Corollary 8.6.
Let V be a valuation domain with quotient field K ,suppose V = K , and let L be a transcendental field extension of K .Then, E ( L | V ) cons is perfect.Proof. If trdeg(
L/K ) ≥ L/K ) =1, let X ∈ L be transcendental over K . By Theorem 8.2, E ( K ( X ) | V ) cons is perfect; by Corollary 7.3, also E ( L | V ) cons is perfect. (cid:3) Corollary 8.7.
Let D be an integral domain, and let L be a transcen-dental extension of the quotient field K of D . If trdeg( L/K ) ≥ , then Zar( L | D ) cons is perfect.Proof. Any W ∈ Zar( L | D ) belongs to E ( L | V ) for some V ∈ Zar( D ).By Theorem 8.5, all E ( L | V ) cons are perfect, and thus no W is isolated.Hence, Zar( L | D ) cons is perfect. (cid:3) References [1] M. F. Atiyah and I. G. Macdonald.
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Dipartimento di Matematica “Tullio Levi-Civita”, Universit`a degliStudi di Padova, Padova, Italy
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