http://ijmms.hindawi.com
© Hindawi Publishing Corp.
AN EXTENSION THEOREM FOR SOBER SPACES AND THE GOLDMAN TOPOLOGY
EZZEDDINE BOUACIDA, OTHMAN ECHI, GABRIEL PICAVET, and EZZEDDINE SALHI
Received 21 December 2002
Goldman points of a topological space are defined in order to extend the notion of primeG-ideals of a ring. We associate to any topological space a new topology called Goldman topology. For sober spaces, we prove an extension theorem of continuous maps. As an application, we give a topological characterization of the Jacobson subspace of the spectrum of a commutative ring. Many examples are provided to illustrate the theory.
2000 Mathematics Subject Classification: 13C15, 54A10, 54B35, 54F65, 57R30.
1. Introduction. We start by recalling several definitions and notations we will be using in this paper.
IfXis a topological space, we denote byᏻ(X)the set of all open subsets of X. Recall that a continuous mapg:Y→Zis said to be aquasihomeomorphism ifUg−1(U)defines a bijectionᏻ(Z)→ᏻ(Y )[9].
A subsetSof a topological spaceXis said to bestrongly denseinXifSmeets every nonempty locally closed subset ofX[9]. Thus a subsetSofXis strongly dense if and only if the canonical injectionSXis a quasihomeomorphism.
It is well known that a continuous mapq:X→Y is a quasihomeomorphism if and only if the topology ofXis the inverse image byqof that ofY and the subsetq(X)is strongly dense inY [9].
The notion of quasihomeomorphism is used in algebraic geometry and it has recently been shown that this notion arises naturally in the theory of some foliations associated to closed connected manifolds (see [2, 3]). It is worth noting that quasihomeomorphisms are also linked with sober spaces. Recall that a topological spaceXis said to besoberif any nonempty irreducible closed subset ofXhas a unique generic point. LetXbe a topological space andS(X) the set of all irreducible closed subsets ofX[9]. LetU be an open subset of Xand setU= {C∈S(X)|U∩C≠∅}. Then the collection (U, U is an open subset ofX) provides a topology onS(X)and the following properties hold [9].
(i) The map
ηX:X →S(X), x→ {x} (1.1)
is a quasihomeomorphism.
(ii) The set S(X)is a sober space.
(iii) Let f:X→Y be a continuous map and let S(f ):S(X)→S(Y )be the map defined byS(f )(C)=f (C),for each irreducible closed subsetCofX.Then S(f )is continuous.
(iv) The topological spaceS(X)is called the sobrification of X,and the as- signment S defines a functor from the category Topof topological spaces to itself.
(v) If q:X→Y is a continuous map, then the following diagram is commu- tative:
X
f
ηX
Y
ηY
S(X) S(f ) S(Y ).
(1.2)
The sobrification serves, sometimes, to give topological characterization of particular spaces (see, e.g., [6]).
Now, we make some observations.
Observation1(General extension theorem of H. Tietze). LetXandY be two topological spaces. A“general extension theorem for maps of closed subsets of XintoY” is a statement giving conditions onXandY, under which it is true that for every closedA⊂X, each continuousf:A→Y is extendable overX relative toY. General extension theorems are rare and usually have interesting topological consequences. One can mention then a well-known result of H.
Tietze.
Theorem1.1(H. Tietze). LetXbe Hausdorff. The following two properties are equivalent:
(1) Xis normal;
(2) for every closedA⊂X, each continuousf:A→[0,1]has a continuous extensionF:X→[0,1]. Furthermore, if|f (a)|< conA, thenF can be chosen so that|F(x)|< conX.
The Tietze extension theorem for normal spaces has been improved, by Dugundji [5], using the richer structure of metric spaces.
Theorem1.2(Dugundji [5]). LetX be an arbitrary metric space,A⊂Xa closed subset, andLan affine space of typem. Then each continuousf:A→L has a continuous extensionF:X→L, and in factF(X)⊂[convex hull off (A)].
Observation2. LetXbe a topological space andβXthe Stone-ˇCech com- pactification ofX. Then every continuous function fromXto a compact space Kcan be extended toβX.
The above two observations have a common point: they provide theorems of extension. Our goal is to give such extension theorems for sober spaces.
Thus we state the following natural questions.
Problem1.3. LetXandY be topological spaces, characterize the continu- ous mapsq:X→Y, such that for each sober spaceZand for each continuous mapf:X→Z, there exists one and only one continuous mapf:Y→Zmaking the following diagram commutative:
X
q
f
Y
f
Z.
(1.3)
Problem1.4. Characterize the collection of topological spacesZsuch that for each continuous mapq:X→Y satisfyingProblem 1.3and for each contin- uous mapf:X→Z, there exists one and only one continuous mapf:Y →Z such that diagram (1.3) commutes.
The second section deals entirely with the solution of Problems1.3 and 1.4: quasihomeomorphisms answerProblem 1.3and sober spaces answerPro- blem 1.4. Naturally, these answers must provide some interesting applications;
these are what we promise in Sections5and6.
A topology ᐀ on a set X is defined to bespectral (and(X,᐀) is called a spectral space) if the following conditions hold:
(i) ᐀is sober;
(ii) the quasicompact open subsets ofXform a basis of᐀;
(iii) the family of quasicompact open subsets ofX is closed under finite intersections.
In a remarkable paper [10], Hochster has proved that a topological space is homeomorphic to the prime spectrum of some ring if and only if it is a spectral space. In the same paper, Hochster characterizes the maximal prime spectrum of a commutative ring as a quasicompactT1-space. Two years later, Hochster gave a topological characterization of the minimal prime spectrum of a ring [11].
Goldman ideals are important objects of investigation in algebra mostly be- cause of their role in the study of graded rings and some applications to al- gebraic geometry. Thus it is important to pay attention to the Goldman prime spectrum of a ring. Recall that a prime ideal of a commutative ringRis said to be aGoldman ideal(G-ideal) if there exists a maximal idealMofR[X]such thatp=M∩R. IfR is an integral domain and(0)is aG-ideal,R is called a G-domain.
Over the years, mathematicians have focused attention onG-domains; for instance, Goldman [8] and Krull [13] usedG-ideals for a short inductive proof
of the Nullstellensatz. The set of allG-ideals of a commutative ringRis denoted by Gold(R)and called theGoldman prime spectrumof that ring.
Recall that a topological spaceXis said to be aTD-space if for eachx∈X, {x}is locally closed.
In [4], Conte has proved that Spec(R)is aTD-space if and only if every prime ideal ofRis aG-ideal. He has also proved that if Spec(R)is Noetherian, then Spec(R)is aTD-space if and only if it is finite.
Fontana and Maroscia [7] have also established, by topological methods, several properties of the set ofG-ideals of a commutative ring. They also dis- cussed in detail a topological approach to a classification of the class of the commutative rings in which every prime ideal is aG-ideal.
Note also that rings in which every prime ideal is aG-ideal have been studied by Picavet in [15].
By agoldspectral spacewe mean a topological space which is homeomorphic to some Gold(R). Using the notion of sobrification, Echi has given an intrinsic topological characterization of the Goldman prime spectrum of a commutative ring [6].
Picavet characterizedG-ideals by a topological property:pis aG-ideal ofRif and only if{p}is locally closed in Spec(R)(equipped with the Zariski topology) [14]. This characterization motivated us to introduce inSection 3the notion of aGoldman pointin a topological space (a locally closed point).
We also give some characterizations of Goldman points. For instance,x∈X is aG-point if and only if{x}is strongly irreducible. Define Gold(X)to be the set of all Goldman points of a topological spaceX. WhenXisT0and has a base of quasicompact open subsets, then Gold(X)is the smallest strongly dense subset ofX. Salhi [17] defined a property(∗)on topological spaces satisfied by spectra of rings: a topological space(X,T )satisfies(∗)ifT is compatible with a partial ordering and for every chainC⊆Xthere is somea∈Xsuch that {a} =C. IfXsatisfies(∗)and Gold(X)=X, thenXis partially well ordered (for the partial ordering induced onXby the topology). Moreover, if the topology onXis compatible with a total ordering andXsatisfies(∗), then Gold(X)=X if and only if each nonmaximal element ofXhas an immediate successor. All these results and others are reminiscents of the theory ofG-ideals.
Picavet defined a new topology on the spectrum of a ring and called it the Goldman topologyorG-topologyowing to its links withG-ideals. This defini- tion can be extended to an arbitrary topological space(X,T ), providing a new topology(X,G)called again the Goldman topology orG-topology. Moreover, the locally closed subsets in the original topology are a base of open sets for theG-topology. If the original topological space isT0, theG-topology isT2; if X is a sober space, the sober subspaces of(X,T )identify with theG-closed subsets.
InSection 4, we define and study theG-topology on a topological spaceX.
WhenXis the spectrum of a ringR, then(X,G)is homeomorphic to the space
of minimal prime ideals of a ringS. However, we do not know of any natural link betweenRandS. Actually, theG-topology on a spectrum has a definition which looks like the patch topology definition of Hochster [10]. More precisely, the G-topology is finer than the patch topology. On a spectral space, these topologies are the same if and only ifXis a Noetherian space (this is not the case for an arbitrary topological space).
Jacobson topological spaces are also used in algebraic geometry and are linked with quasihomeomorphisms. A topological spaceXis called aJacobson space if the set Ꮿ(X) of its closed points is strongly dense in X. If X is a topological space, we denote by Jac(X)the set{x∈X:{x} = {x}∩Ꮿ(X)}. It is obviously seen that Jac(X)is a Jacobson space; we call it theJacobson subspace ofX.
LetRbe a ring and Spec(R)its prime spectrum equipped with the Zariski topology. We denote by Jac(R)the Jacobson subspace of Spec(R). Following Picavet [14], a prime idealpofRlies in Jac(R)if and only ifpis the intersection of some maximal ideals ofR.
By ajacspectral space we mean a topological space homeomorphic to the Jacobson subspace of Spec(R)for some ring R. Section 5deals with a nice topological characterization of jacspectral spaces: Jacspectral spaces are ex- actly the quasicompact Jacobson sober spaces.
InSection 6, we collect some examples, showing that many of the results in the earlier sections are best possible.
Note that through this paper,⊂denotes proper containment and⊆denotes containment with possible equality. If≤is an ordering on a setXandx∈X, then[x↑[denotes the set{y∈X|x≤y}and]↓x]is the set{y∈X|y≤x}. Recall that a topologyT on a spaceXiscompatible with a partial ordering≤ onXif{x} =[x↑[.
2. An extension theorem for sober spaces. In this section, we look more closely at quasihomeomorphisms.
Lemma2.1. Letq:X→Y be a quasihomeomorphism. Then the following properties hold:
(1) ifXis aT0-space, thenqis injective;
(2) ifXis sober andY is aT0-space, thenqis a homeomorphism.
Proof. (1) Letx1andx2be two points ofXwithq(x1)=q(x2). Suppose thatx1≠x2, then there exists an open subsetU ofXsuch thatx1∈U and x2∉U. Since there exists an open subsetVofY satisfyingq−1(V )=U, we get q(x1)∈Uandq(x2)∉U, which is impossible. It follows thatqis injective.
(2) We start with the obvious observation that ifS is a closed subset ofY, thenS is irreducible if and only if is soq−1(S).
We prove thatqis surjective. For this end, lety∈Y. According to the above observation, q−1({y})is a nonempty irreducible closed subset of X. Hence
q−1({y})has a generic pointx. Thus we have the containments
{x} ⊆q−1q(x)
⊆q−1 {y}
= {x}. (2.1)
Thenq−1({q(x)})=q−1({y}). It follows, from the fact thatqis a quasihome- omorphism, that{q(x)} = {y}. SinceY is aT0-space, we getq(x)=y. This proves thatqis a surjective map, and thusq is bijective. One may see that bijective quasihomeomorphisms are homeomorphisms.
Theorem2.2. Letq:X→Y be a continuous map. Then the following state- ments are equivalent:
(a) qis a quasihomeomorphism;
(b) S(q)is a homeomorphism.
Proof. First, we remark that the following diagram is commutative:
X
q
ηX
Y
ηY
S(X) S(q) S(Y ).
(2.2)
(a)⇒(b). SinceηY◦q=S(q)◦ηX is a quasihomeomorphism, the mapS(q) is necessarily a quasihomeomorphism. Thus, followingLemma 2.1,S(q)is a homeomorphism.
(b)⇒(a). SinceηX=((S(q))−1◦ηY)◦qand(S(q))−1◦ηYare quasihomeomor- phisms, it is easily seen thatqis a quasihomeomorphism.
As a consequence of the previous theorem, we state the following one.
Theorem2.3(extension theorem for sober spaces). (1)LetZbe a topolog- ical space. Then the following conditions are equivalent:
(i) Zis a sober space;
(ii) for each quasihomeomorphismq:X→Y and each continuous map f:X→Z, there exists one and only one continuous mapF :Y →Z such thatF◦q=f.
(2) Letq:X→Y be a continuous map. Then the following conditions are equivalent:
(i) qis a quasihomeomorphism;
(ii) for each sober spaceZ and each continuous map f :X→Z, there exists one and only one continuous mapF:Y→Zsuch thatF◦q=f. Proof. (1) (i)⇒(ii). Suppose that suchF exists. Then we haveS(F)◦S(q)= S(f ). ByTheorem 2.2,S(q)is a homeomorphism, henceS(F)=S(f )◦(S(q))−1.
On the other hand, the diagram Y
F
ηY
Z
ηZ
S(Y ) S(F) S(Z)
(2.3)
commutes; that is to say,ηZ◦F=S(F)◦ηY. Consequently, F=
ηZ−1
◦S(F)◦ηY= ηZ−1
◦S(f )◦ S(q)−1
◦ηY. (2.4) Thus, it suffices to verify thatF=(ηZ)−1◦S(f )◦(S(q))−1◦ηY does the job.
Indeed, the following diagram is commutative:
Z
ηZ
X
f q
ηX
Y
ηY
S(Z) S(f ) S(X) S(q) S(Y ).
(2.5)
Hence,
F◦q= ηZ−1
◦S(f )◦ S(q)−1
◦ηY◦q
= ηZ−1
◦S(f )◦ S(q)−1
◦S(q)◦ηX
= ηZ−1
◦S(f )◦ηX
= µC−1
◦ηZ◦f=f .
(2.6)
(ii)⇒(i). There exists a unique continuous map g:S(Z)→Z such that the diagram
Z
ηZ
1Z
S(Z)
g
Z
(2.7)
is commutative. Thus the diagram Z
ηZ
ηZ
S(Z)
ηZ◦g
S(Z)
(2.8)
commutes. HenceηZ◦g=1S(Z) by the implication (i)⇒(ii). ThereforeηZ is a homeomorphism and consequentlyZis a sober space.
(2) According to part (1), it suffices to show the implication (i)⇒(ii).
We will prove thatq is a quasihomeomorphism. FollowingTheorem 2.2, it suffices to prove thatS(q)is a homeomorphism.
There exist two morphismsηX:Y →S(X)andg:S(Y )→S(X)such that the following diagrams commute:
X
q
ηX
Y
ηX
S(X),
Y
ηY
ηX
S(Y )
g
S(X).
(2.9) Henceg◦ηY◦q=ηX. On the other hand, the rectangle (2.2) is commutative.
Thus(g◦S(q))◦ηX=g◦ηY◦q=ηX. Hence, using part (1) (i)⇒(ii), we easily getg◦S(q)=1S(X).
Now,(S(q)◦g)◦(ηY◦q)=S(q)◦ηX=ηY◦q. To prove thatS(q)◦g=1S(Y ), an analogous reasoning to the previous one is not valid since we have not yet shown thatηY◦qis a quasihomeomorphism. But one can show it by noticing that the following diagrams commute:
X
q
ηY◦q
Y
ηY
S(Y ),
X
q
ηY◦q
Y
(S(q)◦g)◦ηY
S(Y ).
(2.10) Hence, using part (1) (i)⇒(ii), for the quasihomeomorphismq, we get(S(q)◦ g)◦ηY=ηY, that is, the diagram
Y
ηY
ηY
S(Y )
S(q)◦g
S(Y )
(2.11)
is commutative. Thus, sinceηY is a quasihomeomorphism, one immediately hasS(q)◦g=1S(Y ), by part (1) (i)⇒(ii). This shows thatS(q)is a homeomor- phism, so thatqis a quasihomeomorphism.
3. Goldman points. In this section, we define Goldman points in topological spaces in order to recover results aboutG-ideals in a more general setting.
Definition3.1. LetXbe a topological space andx∈X. Thenxis said to be aG-point (Goldman point) ofX if{x}is a locally closed subset ofX. We denote by Gold(X)the set of allG-points ofX.
Proposition3.2. LetXbe a topological space andx∈X, then (i) x∈Gold(X)if and only if the derived set{x}= {x}\{x}is closed;
(ii) x∉Gold(X)if and only if{x}= {x}.
Proof. It is well known that a subsetS of a topological spaceXis locally closed if and only if S\S is closed. In particular, x∈X is aG-point if and only if{x}is closed. Moreover, if{x}is not closed, we get{x}= {x}so that {x}= {x}.
Next we give a generalization of a Picavet’s result [14, Section I, Proposition 2]. To begin, we need a definition: letC be a closed subset of a topological spaceX; thenCis said to bestrongly irreducibleif for every family{Ci}i∈Iof closed subsets ofX, such thatC= ∪i∈ICi, there is somei∈Isuch thatC=Ci. Proposition3.3. LetXbe aT0-space andx∈X. Thenxis aG-point if and only if{x}is a strongly irreducible closed subset ofX.
Proof. LetXbe a topological space (not necessarilyT0) andx∈Gold(X).
Suppose that{x} = ∪i∈ICifor some family{Ci}i∈Iof closed subsets ofX. Then, assuming thatCi⊂ {x}for alli∈I, we getCi⊆ {x}so that∪i∈ICi⊆ {x}. Since{x} is closed byProposition 3.2, we get{x} = ∪i∈ICi⊆ {x}, contra- dictingx∈ {x}. Therefore,{x}is a strongly irreducible closed subset ofX.
Conversely, suppose that{x}is a strongly irreducible closed subset of aT0- space X. In view ofProposition 3.2, x∉Gold(X)gives {x}= {x}, whence {x} = ∪[{y};y∈ {x}]. It follows that{x} = ∪[{y};y∈ {x}], from which we deduce that there is somey0∈ {x}such that{y0} = {x}. SinceXis aT0- space, we getx=y0, contradictingy0∈ {x}. Therefore,xis aG-point ofX.
Remark3.4. TheT0hypothesis cannot be deleted in the above result. Here are two examples.
(1) Take any setXwith at least two elements and consider the trivial topology onX. Then Gold(X)= ∅. Nevertheless,{x}is a strongly irreducible closed subset ofXfor eachx∈X.
(2) LetX= {1,2,3,4}and consider the topology defined byT = {∅,{1,2}, X}. Then Gold(X)is empty and{x}is a strongly irreducible closed subset of Xfor eachx∈X.
(3) Define Cl(x)= {y∈X| {x}={y}}forx∈X. In fact, forProposition 3.3 to hold, we only need to suppose that{x} =Cl(x).
From the definition of strongly dense subsets, it is evident that we have the following proposition.
Proposition3.5. LetXbe a topological space andY andZtwo subspaces ofXsuch thatX=Y∪ZandY∩Z= ∅.
(1)The setGold(X)is contained inGold(Y )∪Gold(Z).
(2)If, in addition,Y andZare locally closed, then (i) Gold(X)=Gold(Y )∪Gold(Z),
(ii) Gold(X)is strongly dense inXif and only ifGold(Y )is strongly dense inY andGold(Z)is strongly dense inZ.
LetR be a ring and Gold(R)the set of all G-ideals of R. It is well known that Gold(R)is strongly dense in Spec(R). Nevertheless, ifX is an arbitrary topological space, Gold(X)may be empty (seeRemark 3.4andExample 6.2).
Next we provide examples of topological spaces such that Gold(X)is strong- ly dense inX.
Proposition3.6. LetXbe aT0-space.
(1)IfXis quasicompact, then there is some closed pointx∈X.
(2)IfXhas a base of quasicompact open subsets, then the following properties hold:
(i) Gold(X)is strongly dense inX;
(ii) Gold(X)is the smallest strongly dense subset ofX.
Proof. We show (1). LetᏯbe the family of nonempty closed subsets of X. SinceXis quasicompact, the ordered set(Ꮿ,⊇)is inductive, whence(Ꮿ,⊆) has a minimal element by Zorn’s lemma. LetSbe a minimal element of(Ꮿ,⊆), then {x} =S for eachx∈S. It follows that S= {x}sinceX is a T0-space.
Next we show (i). Let Y =U∩F be a nonempty locally closed subset of X, whereU is open andF is closed. Letx∈Y andQa quasicompact open sub- set such that x∈Q⊆U. ThenQ∩F is closed inQ, whenceQ∩F is qua- sicompact. According to (1), there exists some y ∈Q∩F such that{y} is closed inQ∩F. Thus {y}is locally closed inX so thaty ∈Gold(X). This proves that Gold(X)∩Y= ∅. Hence Gold(X)is strongly dense inX. We prove (ii). Let Y be a strongly dense subset of X. If x ∈Gold(X), then {x} is a nonempty locally closed subset, whence {x} ∩Y = ∅. Therefore, we have Gold(X)⊆Y.
Corollary3.7. LetXbe a NoetherianT0-space. ThenGold(X)is strongly dense inX.
We next investigate when Gold(X)= X. We start with a straightforward lemma.
Lemma3.8. LetXbe a topological space andY a strongly dense subset ofX.
ThenGold(X)⊆Y.
Proposition 3.9. Let X be a topological space. Then the following state- ments are equivalent:
(1) Gold(X)=X;
(2) the only strongly dense subset ofXisX.
Proof. By Proposition 3.6, we get (1)⇒(2). Conversely, assume that (2) holds and letx∈X; thenX\{x}is not strongly dense inX. Hence, there ex- ists some nonempty locally closed subsetY ofXsuch thatY∩(X\{x})= ∅. Therefore, we getY= {x}so thatx∈Gold(X).
Proposition3.10. IfXis a NoetherianT0-space such thatGold(X)is finite, thenXis finite andGold(X)=X.
Proof. Suppose that Gold(X)=X. Define Y to be the set of all closed points ofX. Then obviouslyY ⊆Gold(X)so thatY is closed. SinceX\Y is a nonempty Noetherian T0-space, there exists somex1∈X\Y such that{x1} is a closed point ofX\Y (seeProposition 3.6(1)). Hence,{x1}is locally closed inX, whencex1∈Gold(X). Since Gold(X)=X, we get that (X\Y )\{x1}is a nonempty open subset ofX\Y. Arguing as above, we exhibitx2∈(X\Y )\{x1} such that x2 ∈Gold(X), and so on. Thus there is an infinite sequence in Gold(X), contradicting the hypotheses.
Remark 3.11. Noetherian spectral spaces have properties which do not hold for an arbitrary Noetherian topological space.
(1) In [14, Section II, Corollary of Proposition 3], Picavet proved that when R is a ring with Noetherian spectrum, then Gold(R)=Spec(R)if and only if Spec(R)is finite. This is no longer true for an arbitrary topological space (see Example 6.3).
(2) In the same paper [14], we find that ifR is a ring such that Spec(R)= Gold(R), then(Spec(R),⊆)is partially well ordered (i.e., every chain in Spec(R) is well ordered). This is also no longer true for the partial ordering induced by the topology of an arbitraryT0-space (seeExample 6.3).
(3) Nevertheless, observe that ifXis aT0-space such that the closure of each point has finitely many elements, then Gold(X)=X.
Now, we are able to prove that(X,≤)is partially well ordered if, in addition to Gold(X)=X, we assume thatXsatisfies property(∗)defined below.
Let(X,T )be a topological space. We say that(X,T )satisfies property(∗)if T is compatible with a partial ordering≤, and for every chainC= {ai}i∈I⊆X, there exists somea∈Xsuch that{a} =C. Note that the idea of property(∗) is due to Salhi (see his work about the space of leaves classes of a foliation [17]). First, we have the following Proposition.
Proposition3.12. LetX be aT0-space and≤ be the partial ordering in- duced on X by the topology. Let C = {ai}i∈I be a chain in X anda∈X. If {a} =C, thena=inf(C).
Proof. From{a} =C, we geta≤aifor everyi∈I. Now letb∈Xsuch that b≤ai for eachi∈I so thatC⊆ {b}. It follows thatC= {a} ⊆ {b}, whence b≤a. Therefore, we havea=inf(C).
The following result gives an example of topological spaces satisfying(∗).
Proposition3.13. LetRbe a ring. Then the spacesX=Spec(R)verify(∗).
Proof. LetC= {pi}i∈Ibe a chain of prime ideals ofR, thenp= ∩[pi;i∈I]
is the infimum of the family{pi}i∈I[12, Theorem 9]. Forx∈A, consider the special open subsetD(x)= {q∈Spec(R)|x∉q}of Spec(R)containing p.
Sincex∉p, there is somej∈I such thatx∉pjso thatpj∈D(x)∩Candp lies inC.
Remark3.14. There exists a nonspectral topological space satisfying(∗) (seeExample 6.4).
Proposition 3.15. Let(X,T )be a topological space such that the topol- ogyT is compatible with a partial ordering≤. IfXsatisfies property(∗)and Gold(X)=X, thenXis partially well ordered.
Proof. LetC be a chain in(X,≤). SinceXsatisfies property(∗), there is somex∈Xsuch thatx=inf(C)and {x} =C byProposition 3.12. Then we havex∈C, if this does not hold, then for everyy∈C, we have x < y and C⊆ {x}. Nowx∈Gold(X)implies that{x}is closed byProposition 3.2. This leads to a contradiction:x∈C⊆ {x}. ThusCis well ordered.
Proposition 3.16generalizes a result of Picavet [14, Section II, Proposition 5] (see also Fontana and Maroscia [7], Ramaswamy and Viswanathan [16]).
A topology᐀onXis called aprincipal topology[18] or agood topology[1] if arbitrary intersections of open subsets ofXare open. We review some proper- ties of principal topological spaces studied in [1]. LetXbe a set equipped with a binary relationandx∈X; we denote byl(x)the subset of all elements y ∈X such thaty =x or there exist finitely many elements x1,x2,...,xn such thatx1=y,xn =x, andxixi+1fori∈ {1,...,n−1}. The collection {l(x)|x∈X}is a base for a topology onX called theleft--topology and is denoted byTl(). We have proved that a topological space(X,T )is prin- cipal if and only if there exists a binary relationon X such thatT is the left--topology [1]. The following properties hold:
(i) a subsetUofXis open in(X,Tl())if and only ifl(x)⊆Ufor every x∈U;
(ii) the closure{x}of{x}is the subsetr(x)= {y∈X|x∈l(y)}; Recall that a topologyTonXis said to be compatible with the binary relation
if{x} =r(x)for eachx∈X.
Proposition3.16. Let(X,T )be a topological space such that the topology T is compatible with a total ordering ≤. If X satisfies Property(∗), then the following statements are equivalent:
(1) Gold(X)=X;
(2) every nonmaximal element ofXhas an immediate successor;
(3) T is a principal topology onX.
Proof. (1)⇒(2). Letxbe a nonmaximal element ofX, then{x}is closed byProposition 3.2, but{x}is a nonempty chain inX. We get{x}= {y}for some elementyby property(∗). Thusyis the smallest element of{x}since {x}is closed. It follows thatyis the immediate successor ofx.
(2)⇒(3). Let x be an element of X. If x is a maximal element of X, then ]↓x]=Xis an open subset ofX. Whenxis not maximal, letybe the imme- diate successor ofx. We get{y} =[y↑[=X\]↓x]sinceTis compatible with
≤. This proves that]↓x]is an open subset ofX. Therefore,T is a principal topology.
(3)⇒(1). For allx∈X, we have{x} =]↓x]∩[x↑[=]↓x]∩ {x}. It follows thatX=Gold(X).
Remark 3.17. The total ordering hypothesis is essential in the previous proposition (seeExample 6.5).
Proposition3.18. Let(X,≤)be a partially ordered set such that each ele- ment ofx∈Xhas a finite heightht(x). Assume that{x∈X|ht(x)=n}has finitely many elements for each integern. ThenGold(X)=Xfor each topology T onXwhich is compatible with≥.
Proof. Denote byXnthe set of allx∈Xsuch that ht(x)≤n. ThenX0is finite andXn+1=Xn∪(Xn+1\Xn). Moreover,X0is closed since it is a union of finitely many closed points. Suppose thatXiis a closed subset when 1≤i≤n and set{x1,...,xp} =Xn+1\Xn. Then{x1,...,xp}is a union of closed points in the open set X\Xn. We getXn+1=Xn∪(Xn+1\Xn)=Xn∪ {x1,...,xp} = Xn∪{x1,...,xp}.
It follows thatXn+1is a closed subset ofX. By induction,Xnis closed for each integern. Letx∈Xsuch that ht(x)=n≥1. Then{x}is closed inX\Xn−1. In that case,xbelongs to Gold(X). It follows that Gold(X)=X.
4. Goldman topology. Picavet [14] introduced the Goldman topology (G- topology) on the spectrum of a ringR. This topology is defined as follows. Let Y be a subset of Spec(R). The closure ofY with respect to theG-topology is the set of all prime ideals ofRwhich are an intersection of some elements of Y. Then a prime idealPofRis aG-ideal if and only ifPisG-open.
The family{V (I)∩D(x)|I is an ideal of R, x∈R}is a base for theG- topology on Spec(R). We show that this topology can be defined on any topo- logical space.
Proposition4.1. LetXbe a topological space,ᏼ(X)the set of all its subsets, andA∈ᏼ(X). SetAG= {x∈X| {x} =A∩{x}}. The mapµ:ᏼ(X)→ᏼ(X)de- fined byµ(A)=AGis a Kuratowski operator closure (thus providing a topology onX, called the Goldman topology orG-topology, for short).
Proof. We show four properties:
(1) µ(∅)= ∅,
(2) A⊆µ(A)for everyA∈ᏼ(X),
(3) µ(µ(A))=µ(A)for everyA∈ᏼ(X),
(4) µ(A∪B)=µ(A)∪µ(B)for everyA,B∈ᏼ(X).
The first two properties are obvious. Next we prove (3). Using (2), we get µ(A)⊆µ(µ(A)). Conversely, let x∈µ(µ(A))so that {x} =µ(A)∩{x}. Let Ube an open subset ofXcontainingx. Fromx∈µ(A)∩{x}, we deduce that there is somey∈U∩µ(A)∩{x}so thaty∈ {x}andy∈A∩{y}. Therefore, U∩(A∩ {y}) is nonempty and so isU∩(A∩ {x})since {y} ⊆ {x}. It fol- lows thatx∈A∩{x}. Thus we get{x} =A∩{x}. This proves thatµ(µ(A))= µ(A). To end, we show (4). It is enough to see thatµ(A∪B)⊆µ(A)∪µ(B).
Let x∈µ(A∪B); then{x} =(A∪B)∩{x} =A∩{x} ∪B∩{x}yields{x} = A∩{x}or {x} =B∩{x}. Therefore, x lies in µ(A)∪µ(B), which ends the proof.
It is easy to see that theG-topology on Spec(R)is the same as theG-topology defined by Picavet [14]. A closed (resp., open) set for theG-topology is termed G-closed (resp., G-open). Therefore, the collection of all G-open subsets is {X\µ(A)|A∈ᏼ(X)}.
Recall that thegenerizationofY ⊆Xis the subsetg(Y )of allx∈X such thatY∩{x} = ∅. Moreover, we haveg(Y )= ∩[O;Y⊆O,Oopen].
Proposition4.2. LetXbe a topological space.
(1)IfA∈ᏼ(X)is open or closed, thenAisG-closed.
(2)The family of all locally closed subsets ofXis a base for theG-topology on X. More precisely,{U∩{x} |Uis open andx∈U}is a base for theG-topology.
(3)The setg(A)is equal tog(AG)andAG⊆g(A)∩Afor allA⊆X.
Proof. We first prove that every locally closed subsetAofXisG-open. We only need to show that ifAis open or closed, thenAisG-closed. Letx∈AGso that{x} =A∩{x}. IfAis closed, we have{x} =A∩{x}, whencex∈ {x} ⊂A;
if Ais open, thenA meets {x}and x∈A. In any cases, x∈Aand A=AG. LetO be aG-open subset ofX; we must prove thatO is the union of some locally closed subsets ofX. We setF=X\O; letxbe an element ofOso that x∉F=FG; sincex∉F∩{x}, there is some open subsetUxcontainingxand such thatUx∩F∩{x} = ∅, whenceUx∩{x} ⊆O. Therefore,O= ∪[Ux∩{x};
x∈O]. Now, (3) follows from (1) sinceA⊂OAG⊂Ofor all open subsetO.
Corollary4.3. LetXbe a topological space. A subsetY ofX is strongly dense inXif and only ifY isG-dense inX.
Corollary4.4. LetRbe a ring. Then theG-closure ofGold(R)isSpec(R) and every strongly dense subset ofSpec(R)containsGold(R).
Proof. The set Gold(R) is G-dense in Spec(R) by Corollary 4.3 since Gold(R)is strongly dense in Spec(R). AG-dense subsetYof Spec(R)is strongly dense. HenceY contains Gold(R)byProposition 3.6.
Remark 4.5. TheG-topology on a topological spaceX is finer than the initial topology. In particular, forA⊆X, we have the containments
A◦⊆◦GA⊆A⊆AG⊆A, (4.1)
whereA,◦ ◦GA,AG, andAare, respectively, the interior in the initial topology, the interior in theG-topology, the closure in theG-topology, and the closure in the initial topology.
Proposition 4.6. Let X be a topological space and a∈X. Then{a}G = Cl(a)= {b∈X| {a} = {b}}.
Proof. Letx∈ {a}G. From{x} = {a}∩{x}, we geta∈ {x}andx∈ {a}.
Thus we have{x} = {a}, that is to say,x∈Cl(a). The converse is straightfor- ward.
Proposition 4.7. Let(X,T )be a topological space. Denote by (X,G) the spaceXequipped with the Goldman topology induced byT. The following state- ments are equivalent:
(1) (X,T )is aT0-space;
(2) (X,G)is aT2-space;
(3) (X,G)is aT1-space;
(4) (X,G)is aT0-space.
Proof. We first show (1)⇒(2). Letx=ybe two elements ofX. SinceX is aT0-space, there exists an open subsetU ofX such that (x∈U andy∉U) or (y ∈U and x ∉U). Set F =X\U. We have U∩F = ∅ with (x ∈U and y∈F) or (x∈F andy∈U). Therefore, theG-topology is Hausdorff. Now we show (4)⇒(1). Letaand bbe two elements ofX such that{a} = {b}so that Cl(a)=Cl(b). In view ofProposition 4.6, we get{a}G= {b}G. Since(X,G)is a T0-space, we havea=b. Therefore,(X,T )is aT0-space.
Remark4.8. It may be checked easily that the following properties hold.
(1) If(X,T )is aT1-space, then(X,G)is a discrete topological space.
(2) The set(X,G)is a discrete topological space if and only ifX=Gold(X).
In particular, if(X,G)is Hausdorff and nondiscrete,(X,T )is aT0-space and not aT1-space.
(3) Every locally closed subsetAofXis a clopen set in(X,G)and satisfies A= ∪[Cl(a);a∈A].
Proposition4.9. Let(X,T )be a topological space. If theG-topology onX is quasicompact, then the topological space(X,T )is Noetherian.
Proof. LetUbe an open subset ofX. BecauseUisG-closed byProposition 4.2,UisG-quasicompact. ThenUis quasicompact since theG-topology onX is finer than the original topology onX. Therefore,X is a Noetherian space.
Remark4.10. The converse of the above proposition is not true (seeEx- ample 6.3). Nevertheless, things are nicer for a spectral space.
The family of all closed subsets and all quasicompact open subsets ofX= Spec(R)is a subbase of closed sets for a topology calledthe patch topologyon X [10]. The family of all quasicompact open subsets and their complements is an open subbase for the patch topology. Apatchin Xis a closed set for the patch topology. It is easily seen that theG-topology is finer than the patch topology.
Proposition4.11. LetXbe a spectral space. Then the following statements are equivalent:
(1) theG-topology onXis compact;
(2) Xis a Noetherian space;
(3) theG-topology onXcoincides with the patch topology onX.
Proof. Thanks to [14, Section I, Proposition 4], (2) and (3) are equivalent.
Then (1)⇒(2) follows fromProposition 4.9and (3)⇒(1) is a consequence of the compactness of the patch topology onX[10].
Proposition4.12. Let(X,T )be a topological space andY ⊆X. ThenY is strongly dense inYG(for the topologies induced byT).
Proof. LetZ=Z∩YG be a nonempty locally closed subset ofYG, where Z is a locally closed subset ofX. SinceZ is G-open, we get Z∩Y = ∅ by ZG=Z∩YG. HenceY is strongly dense inYG.
Proposition4.13. EveryG-closed subset of a spectral spaceXhas at least a minimal element for the natural order defined by the topology (more precisely, everyG-closed subset ofXis≥-inductive).
Proof. LetX=Spec(R), whereRis a ring, andY ⊆XaG-closed subset.
Let{pi|i∈I} ⊆Y be a nonempty chain andpthe intersection of allpi. Since YGis the set of all prime ideals ofRwhich are an intersection of some prime ideals belonging toY, we see thatp∈Y. Therefore(Y ,⊇)is inductive.
Remark 4.14. The above proposition does not hold for an arbitraryT0- space (seeExample 6.3).
Proposition4.15. LetRbe a ring. Every patch ofSpec(R)is an intersection of Zariski quasicompactG-open sets ofSpec(R).
Proof. According to [10, Section 7], a patch set is an intersection of con- structible subsets. Moreover, a constructible subset is a union of finitely many locally closed quasicompact subsets.
Theorem4.16. LetXbe aT0-space.
(1)IfXis a sober space, everyG-closed subset ofXis sober.
(2)Every sober subspace ofXis aG-closed subset ofX.
Proof. We show (1). LetF be an irreducible closed subset of aG-closed subsetY ofX. We haveF =Y∩F. Since F is irreducible inY, so isF in X.
HenceF is an irreducible closed subset ofX. Therefore,F has a generic point x, that is to say,F= {x}becauseX is sober. It follows thatF =Y∩ {x}and F = {x} =Y∩{x}. This implies that x∈Y byG-closeness of Y. Moreover, we havex∈Y∩F =F, which proves that xis the generic point ofF (in the subspaceY). Now we prove (2). LetY be a sober subspace ofX andx∈YG, that is to say,{x} =Y∩{x}. SinceY∩{x}is irreducible, so isY∩{x}. Hence Y∩ {x}has a generic pointy∈Y∩ {x}so thatY∩ {x} =Y∩ {y}. Moreover, xandybelong toYG. This yields{x} = {y}. It follows thatx=y∈Y since Xis aT0-space. Therefore,Y is aG-closed subset ofX.
Remark4.17. In view ofTheorem 4.16, theG-closed subsets ofXare the sober subspaces ofXwhenXis a sober space. Nevertheless, whenXis aT0- space, aG-closed subset ofXneed not be sober (seeExample 6.3).
Proposition4.18. LetXbe aT0-space andY aG-dense subset. Then, for eachx∈X,x=inf({x}∩Y )(Xis equipped with the partial ordering≤induced by the topology).
Proof. For eachy∈ {x} ∩Y, we havey∈ {x} =[x↑[= {z∈X/x≤z}, whencex≤y. Letz∈X, such thatz≤yfor anyy∈Y∩{x}, and assume that x∉{z}. Then{x}∩(X\{z})is nonempty and locally closed. SinceYis strongly dense, we get{x} ∩(X\{z})∩Y = ∅. Hence, there exists some y∈ {x} ∩Y such thaty∈X\{z}, that is to say,zy; this is a contradiction. It follows thatz≤x, proving thatx=inf({x}∩Y ).
We recall some topological definitions and link them with theG-topology.
Let Min(R)denote the set of all minimal prime ideals of a ringR equipped with the relativization of the Zariski topology. It is well known that Min(R)is Hausdorff.
According to Hochster, a topological spaceX is said to be minspectral if there is some ringRsuch thatXis homeomorphic to Min(R)[11]. As in [11], an m-baseᏮfor a Hausdorff spaceXis a base of open sets such that every subset SofᏮwhich has the finite intersection property has nonempty intersection.
A topology is said to bescatteredif every point has a base of clopen neigh- borhoods. Such a topology is totally disconnected. A Hausdorff spaceXis said to becompletely regular(orTychonoff) if for each pointx∈Xand every closed subsetV ofXsuch thatx∉V, there is a continuous mapϕ:X→[0,1]such thatϕ(x)=1 andϕ(a)=0 for everya∈V.
Proposition4.19. LetXbe a topological space.
(1)IfXis a spectral space, then theG-topology onXis minspectral.
(2)IfXis aT0-space, then theG-topology onXis scattered.
(3)Every scattered topology is completely regular. In particular theG-topology on aT0-space is completely regular.
Proof. With the hypothesis of (1) being granted, theG-topology on X is Hausdorff sinceXis T0. LetᏮ be the family of all subsetsY ofXsuch that Y =F∩O, whereF is closed andO is quasicompact open. ThenᏮis a base for theG-topology and its elements are patches. Since the patch topology is compact [10], Ꮾis anm-base. Thanks to Hochster’s result [11, Theorem 1], theG-topology onXis minspectral. We show (2). Every locally closed subset isG-open andG-closed. Letx∈X, then the family of locally closed subsets of X containing x is a base ofG-clopen neighborhoods ofx. Now, (3) is a consequence of the following considerations. LetX be a scattered topologi- cal space, x∈X, andV a closed subset ofX such thatx∉V. There exists a clopen subsetO ofX containingx such thatO⊆X\V. Since {O,X\O}is an open covering ofX, we can define a continuous mapϕ:X→[0,1]such that ϕ(O) = {1} and ϕ(X\O) = {0}. We have thus checked out the three conditions.
Theorem4.20. LetXbe a sober topological space and Y⊆X. ThenYG is homeomorphic toS(Y ).
Proof. In view ofTheorem 4.16,YG is a sober space so thatYGS(YG).
On the other hand, the inclusionY→YGis a quasihomeomorphism (seePropo- sition 4.12). HenceS(Y )S(YG)byTheorem 2.2. ThereforeYGis homeomor- phic toS(Y ).
Proposition4.21. LetXbe a NoetherianT0-topological space.
(1)The canonical injectionX→S(X)identifiesXwith a subspace ofS(X),X is strongly dense inS(X)andS(X)is a Noetherian spectral space. In particular, theG-topology onS(X)is the patch topology and induces theG-topology onX.
(2)The setGold(X)is equal toGold(S(X)).
Proof. LetX be a Noetherian topologicalT0-space. Then its sobrification S(X)is a Noetherian spectral space. Indeed,S(X)is Noetherian wheneverX is Noetherian [9, Chapter 0, Section 3, Proposition 7.6]. In this case,S(X)is Noetherian and sober, whence a spectral space. ThenXcan be identified with a subspace ofS(X)andXis strongly dense inS(X)by [9, Chapter 0, Section 3, Proposition 7.1(b)]. We get Gold(S(X))⊆X byProposition 3.6becauseX is strongly dense. It follows that Gold(S(X))⊆Gold(X). Now Gold(S(X)) is strongly dense inXso that Gold(X)=Gold(S(X)).
5. A topological characterization of the Jacobson prime spectrum of a commutative ring. Recall that a topological spaceX is said to be aJacob- son spaceif the setᏯ(X)of all closed points ofX is strongly dense inX [9, Chapter 0, Section 3, Proposition 8.1]. Therefore,Xis a Jacobson space if and only ifX=Ꮿ(X)G. Obviously, whenXis a topological space, Jac(X)=Ꮿ(X)G is a Jacobson space; we call it theJacobson subspaceofX.
Let R be a ring. We denote by Jac(R)the Jacobson subspace of Spec(R).
Following Picavet [14], a prime idealpofRis in Jac(R)if and only ifpis the
intersection of all maximal idealsmofRsuch thatp⊆m. Ajacspectral space is defined to be a topological space homeomorphic to the Jacobson space of Spec(R)for some ringR.
We aim to give a topological characterization of jacspectral spaces by using our previous results. We need a lemma.
Lemma5.1. Letf:Y→Zbe a quasihomeomorphism. IfUis an open subset ofZ, then the following statements are equivalent:
(i) Uis quasicompact;
(ii) f−1(U)is quasicompact.
Denote byᏻ(X)the set of all open subsets of a topological spaceX. Then the proof ofLemma 5.1is an easy consequence of the following fact:Uf−1(U) defines a bijectionᏻ(Z)→ᏻ(Y ).
We now head towards an important result which completely characterizes jacspectral spaces.
Theorem5.2. LetX be a topological space. The following statements are equivalent:
(1) Xis a jacspectral space;
(2) Xis a quasicompact Jacobson sober space.
Proof. We first prove (1)⇒(2). We need only to show that the Jacobson space of a spectral space is a quasicompact Jacobson sober space. LetRbe a ring andX=Jac(R)the Jacobson space of Spec(R). As we have seen,X is a Jacobson space. SinceX=Max(R)G and Spec(R)is a sober space,Xis sober byTheorem 4.16. It is well known that Max(R)is quasicompact [10]. Moreover, the canonical injection Max(R)Max(R)G =X is a quasihomeomorphism since Max(R) is strongly dense in Max(R)G (see Proposition 4.12). Hence X is quasicompact byLemma 5.1. Next we show that (2)⇒(1). Suppose thatXis a quasicompact Jacobson sober space and letᏯ(X)be the set of all its closed points. We haveX=Ꮿ(X)G. Consequently, the canonical injectionᏯ(X)X is a quasihomeomorphism, whenceᏯ(X)is quasicompact byLemma 5.1. Ob- serve thatᏯ(X)is a T1-space. Therefore, there exists some ringRsuch that Ꮿ(X)is homeomorphic to Max(R)(see Hochster [10]). Letϕ:Ꮿ(X)→Max(R) be a homeomorphism andi: Max(R)→Jac(R)the canonical injection. Then f=i◦ϕ:Ꮿ(X)→Jac(R)is a quasihomeomorphism. In view ofTheorem 2.3, there exists a continuous extension ˜f:X→Jac(R). This extension is also a quasihomeomorphism. Now, sinceXis sober and Jac(R)is aT0-space, ˜f is a homeomorphism byLemma 2.1.
Our next concern will be the construction of jacspectral spaces from Jacob- son quasicompact spaces. First, we need some preliminary results.
Proposition5.3. (1)Letq:X→Y be an injective quasihomeomorphism, thenᏯ(Y )⊆q(Ꮿ(X)).
(2)LetXbe aT0-space. Thenq(Ꮿ(X))=Ꮿ(S(X)), whereq:X→S(X)is the injection ofXonto its sobrificationS(X).
(3)Letq:X→Y be a quasihomeomorphism andS a subset ofX. Then the following statements are equivalent:
(i) S is strongly dense inX;
(ii) q(S)is strongly dense inY. Proof. (1) The proof is straightforward.
(2) Following (1), the proof will be complete if we show that q(Ꮿ(X)) ⊆ Ꮿ(S(X)).
Letx∈Ꮿ(X). We claim that{q(x)} = {q(x)}.
LetF∈ {q(x)}. We must prove thatF=q(x)= {x} = {x}.
First, we observe thatx∈F. To see this, suppose thatx∉F, thenF∈U, whereU =X− {x}. Since F ∈ {q(x)}, q(x)∈U, hence q(x)∩U ≠∅. This yieldsx∈U, a contradiction. It follows thatx∈F.
Lety∈F. Suppose thaty≠x, thenF∩U≠∅, whereU=X−{x}. The rest of the proof runs as before, proving thaty=x. Thereforeq(x)∈Ꮿ(S(X)).
(3) The proof is straightforward.
Next, we derive a useful tool for constructing jacspectral spaces.
Corollary 5.4. Let X be a T0-space. Then the following statements are equivalent:
(i) Xis a quasicompact Jacobson space;
(ii) S(X)is a jacspectral space.
Proof. We start with the following two observations.
(i) Letq:X→S(X)be the injection ofXonto its sobrificationS(X). Fol- lowingProposition 5.3,Xis a Jacobson space if and only ifS(X)is.
(ii) The equivalence,Xis quasicompact if and only ifS(X)is, follows im- mediately fromLemma 5.1.
Therefore, ifXis a quasicompact Jacobson space, thenS(X)is a sober qua- sicompact Jacobson space, and according toTheorem 5.2,S(X)is a jacspectral space. Conversely, ifS(X)is jacspectral, then it is a quasicompact Jacobson space, and the above observations imply thatXis so.
Corollary5.5. LetX be a NoetherianT0-space. ThenS(Jac(X))is a jac- spectral space.
Corollary5.6. LetXandY be twoT0-spaces andq:X→Y a quasihome- omorphism. ThenXis a quasicompact Jacobson space if and only if is soY.
Corollary 5.7. Let X be a T0-space. Then the following statements are equivalent:
(i) Xis a quasicompact Jacobson space;
(ii) Xis injected by a quasihomeomorphism into a jacspectral space.
6. Examples. Recall that a topological space is said to beprincipalif every intersection of open subsets ofXis open [1,18]. Principal topological spaces provide examples ofG-points andG-topologies. The left topology associated to a partially ordered set is principal and every element ofX is a G-point.
Nevertheless, there exists a principal topological space which does not have a G-point.
Next, we describeG-points in principal topological spaces. It is not hard to verify the following proposition.
Proposition6.1. LetXbe a set equipped with a quasiorderinganda∈X.
The following statements are equivalent:
(1) ais aG-point in(X,Tl());
(2) a=bwheneverb∈Xis such thatabandba(i.e.,l(a)∩r(a)= {a}).
Example6.2(a principal topological space without anyG-point). TakeX= Zequipped with the binary relationdefined bymn(m≤n)or(nis even andn=m−1).
If we equip X with the left--topology, then, in view of Proposition 6.1, Gold(X)is empty.
Example6.3(a topological spaceXwhich is Noetherian, infinite, and such that Gold(X)=X). Consider the setX= {0}∪{1/n|n∈N∗}. ThenXequipp- ed with the left topology associated to the natural ordering is an infinite Noe- therian space such that Gold(X)=X.
(i) SinceF = {1/n|n∈N∗}has no smallest element,X is not partially well ordered.
(ii) We remark also that theG-topology is the discrete topology and since Xis infinite, theG-topology onXis not quasicompact.
(iii) We notice that(F,≥)is not inductive.
(iv) F is aG-closed irreducible subset ofXwithout generic point. HenceF is not sober.
Note that the next examples are spaces of leaves classes. Nevertheless, we describe them from a topological point of view.
Example6.4(a nonspectral topological space satisfying(∗)). LetXbe the subset ofR2defined byX= {a}∪S, wherea=(−2,0)andS= {(x,y)∈R2| x2+y2=1}.
Consider onX the topologyT such that X is the only open subset ofX containing aand such that for every b∈S, the family {B(b,ε)∩S |ε >0} is a base of neighborhoods of b (where B(b,ε)is the open ball of centerb and rayε). This topology is compatible with the partial ordering≤defined on X as follows: b≤b for everyb∈X and a≤b for everyb∈X. The space (X,T )satisfies property(∗) but is not spectral since(X,T )has no base of quasicompact open subsets.
Example6.5(a topological space showing that the total ordering hypothesis inProposition 3.16 is essential). (1) LetR be an integral domain with finite dimension andX=Spec(R). Assume thatRis not aG-domain. Then the second statement ofProposition 3.16holds while the first statement fails.
(2) LetX= {(0,1)} ∪ {(0,0)} ∪ {(1/n,0)|n∈Z∗}and seta=(0,1),aω= (0,0), andan=(1/n,0)ifn=0. We equipXwith the partial orderingx≤x andx≤afor eachx∈X.
The familyᏮ= {]↓x]}x∈X∼aω∪(∪n∈Z∗{aω} ∪ {ap| |p| ≥ |n|})is a base for a topologyTcompatible with the partial ordering≤. We have Gold(X)=X, but the topologyT is not the left topology.
References
[1] E. Bouacida, O. Echi, and E. Salhi,Topologies associées à une relation binaire et relation binaire spectrale[Topologies associated with a binary relation and spectral binary relation], Boll. Un. Mat. Ital. B (7)10(1996), no. 2, 417–439 (French).
[2] ,Nonfinite heights, Commutative Ring Theory (Fès, 1995), Lecture Notes in Pure and Appl. Math., vol. 185, Marcel Dekker, New York, 1997, pp. 113–
130.
[3] ,Feuilletages et topologie spectrale [Foliations and spectral topology], J.
Math. Soc. Japan52(2000), no. 2, 447–464 (French).
[4] A. Conte,Proprietà di separazione della topologia di Zariski di uno schema, Istit.
Lombardo Accad. Sci. Lett. Rend. A106(1972), 79–111 (Italian).
[5] J. Dugundji,An extension of Tietze’s theorem, Pacific J. Math.1(1951), 353–367.
[6] O. Echi,A topological characterization of the Goldman prime spectrum of a com- mutative ring, Comm. Algebra28(2000), no. 5, 2329–2337.
[7] M. Fontana and P. Maroscia,Sur les anneaux de Goldman, Boll. Un. Mat. Ital. B (5) 13(1976), no. 3, 743–759 (French).
[8] O. Goldman,Hilbert rings and the Hilbert Nullstellensatz, Math. Z.54(1951), 136–
140.
[9] A. Grothendieck and J. Dieudonné,Eléments de Géométrie Algébrique, Springer- Verlag, Heidelberg, 1971.
[10] M. Hochster,Prime ideal structure in commutative rings, Trans. Amer. Math. Soc.
142(1969), 43–60.
[11] ,The minimal prime spectrum of a commutative ring, Canad. J. Math.23 (1971), 749–758.
[12] I. Kaplansky,Commutative Rings, revised ed., The University of Chicago Press, Illinois, 1974.
[13] W. Krull, Jacobsonsche Ringe, Hilbertscher Nullstellensatz, Dimensionstheorie, Math. Z.54(1951), 354–387 (German).
[14] G. Picavet,Autour des idéaux premiers de Goldman d’un anneau commutatif, Ann. Sci. Univ. Clermont Math.57(1975), no. 11, 73–90 (French).
[15] ,Sur les anneaux commutatifs dont tout idéal premier est de Goldman, C. R.
Acad. Sci. Paris Sér. I Math.280(1975), no. 25, Ai, A1719–A1721 (French).
[16] R. Ramaswamy and T. M. Viswanathan,Overring properties ofG-domains, Proc.
Amer. Math. Soc.58(1976), 59–66.
[17] E. Salhi,Problème de structure dans les feuilletages de codimension un de classe C0, thèse d’Etat, IRMA, Strasbourg, France, 1984.
[18] A. K. Steiner,The lattice of topologies: structure and complementation, Trans.
Amer. Math. Soc.122(1966), 379–398.
Ezzeddine Bouacida: Département de Mathématiques, Faculté des Sciences de Sfax, Université de Sfax, BP 802, 3018 Sfax, Tunisia
E-mail address:ezzedine.bouacida@fss.rnu.tn
Othman Echi: Department of Mathematics, Faculty of Sciences of Tunis, University Tunis-El Manar, “Campus Universitaire,” 1092 Tunis, Tunisia
E-mail address:othman.echi@fst.rnu.tn
Gabriel Picavet: Laboratoire de Mathématiques Pures, Université Blaise Pascal, Com- plexe Scientifique des Cézeaux, 63177 Aubière Cedex, France
E-mail address:Gabriel.Picavet@math.univ-bpclermont.fr
Ezzeddine Salhi: Département de Mathématiques, Faculté des Sciences de Sfax, Uni- versité de Sfax, BP 802, 3018 Sfax, Tunisia
E-mail address:ezzeddine.salhi@fss.rnu.tn
