AN EXTENSION THEOREM FOR SOBER SPACES AND THE GOLDMAN TOPOLOGY

Goldman points of a topological space are defined in order to extend the notion of prime G-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.


Introduction.
We start by recalling several definitions and notations we will be using in this paper.
If X is a topological space, we denote by ᏻ(X) the set of all open subsets of X. Recall that a continuous map g : Y → Z is said to be a quasihomeomorphism if U g −1 (U ) defines a bijection ᏻ(Z) → ᏻ(Y ) [9].
A subset S of a topological space X is said to be strongly dense in X if S meets every nonempty locally closed subset of X [9].Thus a subset S of X is strongly dense if and only if the canonical injection S X is a quasihomeomorphism.It is well known that a continuous map q : X → Y is a quasihomeomorphism if and only if the topology of X is the inverse image by q of that of Y and the subset q(X) is strongly dense in Y [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 space X is said to be sober if any nonempty irreducible closed subset of X has a unique generic point.Let X be a topological space and S(X) the set of all irreducible closed subsets of X [9].Let U be an open subset of X and set U = {C ∈ S(X) | U ∩ C ≠ ∅}.Then the collection ( U, U is an open subset of X) provides a topology on S(X) and the following properties hold [9].
(i) The map 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 by S(f )(C) = f (C), for each irreducible closed subset C of X.Then S(f ) is continuous.
(iv) The topological space S(X) is called the sobrification of X, and the assignment S defines a functor from the category Top of topological spaces to itself.
(v) If q : X → Y is a continuous map, then the following diagram is commutative: The sobrification serves, sometimes, to give topological characterization of particular spaces (see, e.g., [6]).Now, we make some observations.
Observation 1 (General extension theorem of H. Tietze).Let X and Y be two topological spaces.A "general extension theorem for maps of closed subsets of X into Y " is a statement giving conditions on X and Y , under which it is true that for every closed A ⊂ X, each continuous f : A → Y is extendable over X relative to Y .General extension theorems are rare and usually have interesting topological consequences.One can mention then a well-known result of H. Tietze.
Theorem 1.1 (H.Tietze).Let X be Hausdorff.The following two properties are equivalent: (1) X is normal; (2) for every closed A ⊂ X, each continuous f : A → [0, 1] has a continuous extension F : X → [0, 1].Furthermore, if |f (a)| < c on A, then F can be chosen so that |F(x)| < c on X.
The Tietze extension theorem for normal spaces has been improved, by Dugundji [5], using the richer structure of metric spaces.Theorem 1.2 (Dugundji [5]).Let X be an arbitrary metric space, A ⊂ X a closed subset, and L an affine space of type m.Then each continuous f : A → L has a continuous extension F : X → L, and in fact F(X) ⊂ [convex hull of f (A)].
Observation 2. Let X be a topological space and βX the Stone-Čech compactification of X.Then every continuous function from X to a compact space K can be extended to βX.
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.
Problem 1.3.Let X and Y be topological spaces, characterize the continuous maps q : X → Y , such that for each sober space Z and for each continuous map f : X → Z, there exists one and only one continuous map f : Y → Z making the following diagram commutative: (1.3) Problem 1.4.Characterize the collection of topological spaces Z such that for each continuous map q : X → Y satisfying Problem 1.3 and for each continuous map f : X → Z, there exists one and only one continuous map f : The second section deals entirely with the solution of Problems 1.3 and 1.4: quasihomeomorphisms answer Problem 1.3 and sober spaces answer Problem 1.4.Naturally, these answers must provide some interesting applications; these are what we promise in Sections 5 and 6.
A topologyon a set X is defined to be spectral (and (X, -) is called a spectral space) if the following conditions hold: (i)is sober; (ii) the quasicompact open subsets of X form a basis of -; (iii) the family of quasicompact open subsets of X 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 quasicompact T 1 -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 because of their role in the study of graded rings and some applications to algebraic geometry.Thus it is important to pay attention to the Goldman prime spectrum of a ring.Recall that a prime ideal of a commutative ring R is said to be a Goldman ideal (G-ideal) if there exists a maximal ideal Over the years, mathematicians have focused attention on G-domains; for instance, Goldman [8] and Krull [13] used G-ideals for a short inductive proof of the Nullstellensatz.The set of all G-ideals of a commutative ring R is denoted by Gold(R) and called the Goldman prime spectrum of that ring.
Recall that a topological space X is said to be a T D -space if for each x ∈ X, {x} is locally closed.
In [4], Conte has proved that Spec(R) is a T D -space if and only if every prime ideal of R is a G-ideal.He has also proved that if Spec(R) is Noetherian, then Spec(R) is a T D -space if and only if it is finite.
Fontana and Maroscia [7] have also established, by topological methods, several properties of the set of G-ideals of a commutative ring.They also discussed in detail a topological approach to a classification of the class of the commutative rings in which every prime ideal is a G-ideal.
Note also that rings in which every prime ideal is a G-ideal have been studied by Picavet in [15].
By a goldspectral space we 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 characterized G-ideals by a topological property: p is a G-ideal of R if and only if {p} is locally closed in Spec(R) (equipped with the Zariski topology) [14].This characterization motivated us to introduce in Section 3 the notion of a Goldman point in a topological space (a locally closed point).
We also give some characterizations of Goldman points.For instance, x ∈ X is a G-point if and only if {x} is strongly irreducible.Define Gold(X) to be the set of all Goldman points of a topological space X.When X is T 0 and has a base of quasicompact open subsets, then Gold(X) is the smallest strongly dense subset of X. Salhi [17] defined a property ( * ) on topological spaces satisfied by spectra of rings: a topological space (X, T ) satisfies ( * ) if T is compatible with a partial ordering and for every chain C ⊆ X there is some a ∈ X such that {a} = C.If X satisfies ( * ) and Gold(X) = X, then X is partially well ordered (for the partial ordering induced on X by the topology).Moreover, if the topology on X is compatible with a total ordering and X satisfies ( * ), then Gold(X) = X if and only if each nonmaximal element of X has an immediate successor.All these results and others are reminiscents of the theory of G-ideals.
Picavet defined a new topology on the spectrum of a ring and called it the Goldman topology or G-topology owing to its links with G-ideals.This definition can be extended to an arbitrary topological space (X, T ), providing a new topology (X, G) called again the Goldman topology or G-topology.Moreover, the locally closed subsets in the original topology are a base of open sets for the G-topology.If the original topological space is T 0 , the G-topology is T 2 ; if X is a sober space, the sober subspaces of (X, T ) identify with the G-closed subsets.
In Section 4, we define and study the G-topology on a topological space X.When X is the spectrum of a ring R, then (X, G) is homeomorphic to the space of minimal prime ideals of a ring S.However, we do not know of any natural link between R and S. Actually, the G-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 if X is 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 space X is called a Jacobson 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 the Jacobson subspace of X.
Let R be 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 ideal p of R lies in Jac(R) if and only if p is the intersection of some maximal ideals of R.
By a jacspectral space we mean a topological space homeomorphic to the Jacobson subspace of Spec(R) for some ring R. Section 5 deals with a nice topological characterization of jacspectral spaces: Jacspectral spaces are exactly the quasicompact Jacobson sober spaces.
In Section 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 set X and x ∈ X, then [x ↑ [ denotes the set {y ∈ X | x ≤ y} and ] ↓ x] is the set {y ∈ X | y ≤ x}.Recall that a topology T on a space X is compatible with a partial ordering ≤ on X if {x} = [x ↑ [.

An extension theorem for sober spaces.
In this section, we look more closely at quasihomeomorphisms. Lemma 2.1.Let q : X → Y be a quasihomeomorphism.Then the following properties hold: (1) if X is a T 0 -space, then q is injective; (2) if X is sober and Y is a T 0 -space, then q is a homeomorphism.
Proof.(1) Let x 1 and x 2 be two points of X with q(x 1 ) = q(x 2 ).Suppose that x 1 ≠ x 2 , then there exists an open subset U of X such that x 1 ∈ U and x 2 ∉ U .Since there exists an open subset V of Y satisfying q −1 (V ) = U , we get q(x 1 ) ∈ U and q(x 2 ) ∉ U , which is impossible.It follows that q is injective.
(2) We start with the obvious observation that if S is a closed subset of Y , then S is irreducible if and only if is so q −1 (S).
We prove that q is surjective.For this end, let y ∈ Y .According to the above observation, q −1 ({y}) is a nonempty irreducible closed subset of X. Hence q −1 ({y}) has a generic point x.Thus we have the containments {x} ⊆ q −1 q(x) ⊆ q −1 {y} = {x}. (2.1) Then q −1 ({q(x)}) = q −1 ({y}).It follows, from the fact that q is a quasihomeomorphism, that {q(x)} = {y}.Since Y is a T 0 -space, we get q(x) = y.This proves that q is a surjective map, and thus q is bijective.One may see that bijective quasihomeomorphisms are homeomorphisms.
Theorem 2.2.Let q : X → Y be a continuous map.Then the following statements are equivalent: (a) q is a quasihomeomorphism; (b) S(q) is a homeomorphism.
Proof.First, we remark that the following diagram is commutative: S(Y ). ( (a)⇒(b).Since η Y • q = S(q) • η X is a quasihomeomorphism, the map S(q) is necessarily a quasihomeomorphism.Thus, following Lemma 2.1, S(q) is a homeomorphism.
As a consequence of the previous theorem, we state the following one.

Theorem 2.3 (extension theorem for sober spaces). (1)
Let Z be a topological space.Then the following conditions are equivalent: (i) Z is a sober space; (ii) for each quasihomeomorphism q : X → Y and each continuous map f : X → Z, there exists one and only one continuous map Let q : X → Y be a continuous map.Then the following conditions are equivalent: (i) q is a quasihomeomorphism; (ii) for each sober space Z and each continuous map f : X → Z, there exists one and only one continuous map On the other hand, the diagram Thus, it suffices to verify that Indeed, the following diagram is commutative: S(Y ). (2.5) Hence, There exists a unique continuous map g : commutes.Hence η Z • g = 1 S(Z) by the implication (i)⇒(ii).Therefore η Z is a homeomorphism and consequently Z is a sober space.
(2) According to part (1), it suffices to show the implication (i)⇒(ii).We will prove that q is a quasihomeomorphism.Following Theorem 2.2, it suffices to prove that S(q) is a homeomorphism.
There exist two morphisms η X : Y → S(X) and g : S(Y ) → S(X) such that the following diagrams commute: (2.9) Hence g • η Y • q = η X .On the other hand, the rectangle (2.2) is commutative.
To prove that S(q) • g = 1 S(Y ) , an analogous reasoning to the previous one is not valid since we have not yet shown that η Y • q is a quasihomeomorphism.But one can show it by noticing that the following diagrams commute: (2.10) Hence, using part (1) (i)⇒(ii), for the quasihomeomorphism q, we get is commutative.Thus, since η Y is a quasihomeomorphism, one immediately has S(q) • g = 1 S(Y ) , by part (1) (i)⇒(ii).This shows that S(q) is a homeomorphism, so that q is a quasihomeomorphism.

Goldman points.
In this section, we define Goldman points in topological spaces in order to recover results about G-ideals in a more general setting.Definition 3.1.Let X be a topological space and x ∈ X.Then x is said to be a G-point (Goldman point ) of X if {x} is a locally closed subset of X.We denote by Gold(X) the set of all G-points of X. Proposition 3.2.Let X be a topological space and x ∈ X, then (i) x ∈ Gold(X) if and only if the derived set {x} = {x}\{x} is closed; Proof.It is well known that a subset S of a topological space X is locally closed if and only if S \ S is closed.In particular, x ∈ X is a G-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: let C be a closed subset of a topological space X; then C is said to be strongly irreducible if for every family Proof.Let X be a topological space (not necessarily T 0 ) and x ∈ Gold(X).Suppose that {x} = ∪ i∈I C i for some family {C i } i∈I of closed subsets of X.Then, assuming that C i ⊂ {x} for all i ∈ I, we get C i ⊆ {x} so that ∪ i∈I C i ⊆ {x} .Since {x} is closed by Proposition 3.2, we get {x} = ∪ i∈I C i ⊆ {x} , contradicting x ∈ {x}.Therefore, {x} is a strongly irreducible closed subset of X. Conversely, suppose that {x} is a strongly irreducible closed subset of a T 0space X.In view of Proposition 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 some y 0 ∈ {x} such that {y 0 } = {x}.Since X is a T 0space, we get x = y 0 , contradicting y 0 ∈ {x} .Therefore, x is a G-point of X.
Remark 3.4.The T 0 hypothesis cannot be deleted in the above result.Here are two examples.
(1) Take any set X with at least two elements and consider the trivial topology on X.Then Gold(X) = ∅.Nevertheless, {x} is a strongly irreducible closed subset of X for each x ∈ X.
From the definition of strongly dense subsets, it is evident that we have the following proposition.Proposition 3.5.Let X be a topological space and Y and Z two subspaces of X such that X = Y ∪ Z and Y ∩ Z = ∅.
(2) If, in addition, Y and Z are locally closed, then (i) Gold(X) = Gold(Y ) ∪ Gold(Z), (ii) Gold(X) is strongly dense in X if and only if Gold(Y ) is strongly dense in Y and Gold(Z) is strongly dense in Z.
Let R 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, if X is an arbitrary topological space, Gold(X) may be empty (see Remark 3.4 and Example 6.2).
Next we provide examples of topological spaces such that Gold(X) is strongly dense in X. Proposition 3.6.Let X be a T 0 -space.
(1) If X is quasicompact, then there is some closed point x ∈ X.
(2) If X has a base of quasicompact open subsets, then the following properties hold: (i) Gold(X) is strongly dense in X; (ii) Gold(X) is the smallest strongly dense subset of X.
Proof.We show (1).Let Ꮿ be the family of nonempty closed subsets of X.Since X is quasicompact, the ordered set (Ꮿ, ⊇) is inductive, whence (Ꮿ, ⊆) has a minimal element by Zorn's lemma.Let S be a minimal element of (Ꮿ, ⊆), then {x} = S for each x ∈ S. It follows that S = {x} since X is a T 0 -space.Next we show (i).Let Y = U ∩ F be a nonempty locally closed subset of X, where U is open and (1), there exists some y ∈ Q ∩ F such that {y} is closed in Q ∩ F .Thus {y} is locally closed in X so that y ∈ Gold(X).This proves that Gold(X) ∩ Y = ∅.Hence Gold(X) is strongly dense in X.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 .Corollary 3.7.Let X be a Noetherian T 0 -space.Then Gold(X) is strongly dense in X.
We next investigate when Gold(X) = X.We start with a straightforward lemma.
Lemma 3.8.Let X be a topological space and Y a strongly dense subset of X.Then Gold(X) ⊆ Y .Proposition 3.9.Let X be a topological space.Then the following statements are equivalent: (1) Gold(X) = X; (2) the only strongly dense subset of X is X.
Proof.By Proposition 3.6, we get (1)⇒(2).Conversely, assume that (2) holds and let x ∈ X; then X\{x} is not strongly dense in X.Hence, there exists some nonempty locally closed subset Y of X such that Y ∩ (X\{x}) = ∅.Therefore, we get Y = {x} so that x ∈ Gold(X).Proposition 3.10.If X is a Noetherian T 0 -space such that Gold(X) is finite, then X is finite and Gold(X) = X.
Proof.Suppose that Gold(X) = X.Define Y to be the set of all closed points of X.Then obviously Y ⊆ Gold(X) so that Y is closed.Since X\Y is a nonempty Noetherian T 0 -space, there exists some x 1 ∈ X\Y such that {x 1 } is a closed point of X\Y (see Proposition 3.6(1)).Hence, {x 1 } is locally closed in X, whence x 1 ∈ Gold(X).Since Gold(X) = X, we get that (X\Y )\{x 1 } is a nonempty open subset of X\Y .Arguing as above, we exhibit x 2 ∈ (X\Y )\{x 1 } such that x 2 ∈ 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 if R 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 arbitrary T 0 -space (see Example 6.3).
(3) Nevertheless, observe that if X is a T 0 -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 that X satisfies 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 chain C = {a i } i∈I ⊆ X, there exists some a ∈ X such 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.Proposition 3.12.Let X be a T 0 -space and ≤ be the partial ordering induced on X by the topology.Let C = {a i } i∈I be a chain in X and a ∈ X.If {a} = C, then a = inf(C).
Proof.From {a} = C, we get a ≤ a i for every i ∈ I. Now let b ∈ X such that b ≤ a i for each i ∈ I so that C ⊆ {b}.It follows that C = {a} ⊆ {b}, whence b ≤ a.Therefore, we have a = inf(C).
The following result gives an example of topological spaces satisfying ( * ).Proposition 3.13.Let R be a ring.Then the spaces X = Spec(R) verify ( * ).
Proof.Let C = {p i } i∈I be a chain of prime ideals of R, then p = ∩[p i ; i ∈ I] is the infimum of the family {p i } i∈I [12,Theorem 9].For x ∈ A, consider the special open subset D(x) = {q ∈ Spec(R) | x ∉ q} of Spec(R) containing p. Since x ∉ p, there is some j ∈ I such that x ∉ p j so that p j ∈ D(x) ∩ C and p lies in C.

Remark 3.14.
There exists a nonspectral topological space satisfying ( * ) (see Example 6.4).Proposition 3.15.Let (X, T ) be a topological space such that the topology T is compatible with a partial ordering ≤.If X satisfies property ( * ) and Gold(X) = X, then X is partially well ordered.
Proof.Let C be a chain in (X, ≤).Since X satisfies property ( * ), there is some x ∈ X such that x = inf(C) and {x} = C by Proposition 3.12.Then we have x ∈ C, if this does not hold, then for every y ∈ C, we have x < y and C ⊆ {x} .Now x ∈ Gold(X) implies that {x} is closed by Proposition 3.2.This leads to a contradiction: x ∈ C ⊆ {x} .Thus C is well ordered.Proposition 3.16 generalizes a result of Picavet [14, Section II, Proposition 5] (see also Fontana and Maroscia [7], Ramaswamy and Viswanathan [16]).
A topologyon X is called a principal topology [18] or a good topology [1] if arbitrary intersections of open subsets of X are open.We review some properties of principal topological spaces studied in [1].Let X be a set equipped with a binary relation and x ∈ X; we denote by l (x) the subset of all elements y ∈ X such that y = x or there exist finitely many elements x 1 ,x 2 ,...,x n such that x 1 = y, x n = x, and x i x i+1 for i ∈ {1,...,n − 1}.The collection { l (x) | x ∈ X} is a base for a topology on X called the left --topology and is denoted by T l ().We have proved that a topological space (X, T ) is principal if and only if there exists a binary relation on X such that T is the left--topology [1].The following properties hold: Recall that a topology T on X is said to be compatible with the binary relation if {x} = r (x) for each x ∈ X. Proposition 3.16.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 of X has an immediate successor; (3) T is a principal topology on X.
Proof.(1)⇒(2).Let x be a nonmaximal element of X, then {x} is closed by Proposition 3.2, but {x} is a nonempty chain in X.We get {x} = {y} for some element y by property ( * ).Thus y is the smallest element of {x} since {x} is closed.It follows that y is the immediate successor of x.
(2)⇒(3).Let x be an element of X.If x is a maximal element of X, then ] ↓ x] = X is an open subset of X.When x is not maximal, let y be the immediate successor of x.We get {y} = [y ↑ [= X\] ↓ x] since T is compatible with ≤.This proves that ] ↓ x] is an open subset of X.Therefore, T is a principal topology.
(3)⇒( 1).For all x ∈ X, we have Remark 3.17.The total ordering hypothesis is essential in the previous proposition (see Example 6.5).Proposition 3.18.Let (X, ≤) be a partially ordered set such that each element of x ∈ X has a finite height ht(x).Assume that {x ∈ X | ht(x) = n} has finitely many elements for each integer n.Then Gold(X) = X for each topology T on X which is compatible with ≥.
Proof.Denote by X n the set of all x ∈ X such that ht(x) ≤ n.Then X 0 is finite and X n+1 = X n ∪ (X n+1 \X n ).Moreover, X 0 is closed since it is a union of finitely many closed points.Suppose that X i is a closed subset when 1 ≤ i ≤ n and set {x 1 ,...,x p } = X n+1 \X n .Then {x 1 ,...,x p } is a union of closed points in the open set X\X n .We get It follows that X n+1 is a closed subset of X.By induction, X n is closed for each integer n.Let x ∈ X such that ht(x) = n ≥ 1.Then {x} is closed in X\X n−1 .In that case, x belongs to Gold(X).It follows that Gold(X) = X. 4. Goldman topology.Picavet [14] introduced the Goldman topology (Gtopology) on the spectrum of a ring R.This topology is defined as follows.Let Y be a subset of Spec(R).The closure of Y with respect to the G-topology is the set of all prime ideals of R which are an intersection of some elements of Y .Then a prime ideal The family {V (I) ∩ D(x) | I is an ideal of R, x ∈ R} is a base for the Gtopology on Spec(R).We show that this topology can be defined on any topological space.Proposition 4.1.Let X be a topological space, ᏼ(X) the set of all its subsets, and G is a Kuratowski operator closure (thus providing a topology on X, called the Goldman topology or G-topology, for short).
Proof.We show four properties: (1) 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 U be an open subset of X containing x. From x ∈ µ(A) ∩ {x}, we deduce that there is some y ∈ U ∩ µ(A) ∩ {x} so that y ∈ {x} and y ∈ A ∩ {y}.Therefore, U ∩ (A ∩ {y}) is nonempty and so is U ∩ (A ∩ {x}) since {y} ⊆ {x}.It follows that x ∈ A ∩ {x}.Thus we get {x} = A ∩ {x}.This proves that µ(µ(A)) = µ(A).To end, we show (4).It is enough to see that It is easy to see that the G-topology on Spec(R) is the same as the G-topology defined by Picavet [14].A closed (resp., open) set for the G-topology is termed The family of all locally closed subsets of X is a base for the G-topology on X.More precisely, {U ∩ {x} | U is open and x ∈ U } is a base for the G-topology.
(3) The set g(A) is equal to g(A G ) and A G ⊆ g(A) ∩ A for all A ⊆ X.
Proof.We first prove that every locally closed subset if A is open, then A meets {x} and x ∈ A. In any cases, x∈A and A = A G .
Let O be a G-open subset of X; we must prove that O is the union of some locally closed subsets of X.We set F = X\O; let x be an element of O so that Corollary 4.4.Let R be a ring.Then the G-closure of Gold(R) is Spec(R) and every strongly dense subset of Spec(R) contains Gold(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).A G-dense subset Y of Spec(R) is strongly dense.Hence Y contains Gold(R) by Proposition 3.6.
Remark 4.5.The G-topology on a topological space X is finer than the initial topology.In particular, for A ⊆ X, we have the containments where A, A G , and A are, respectively, the interior in the initial topology, the interior in the G-topology, the closure in the G-topology, and the closure in the initial topology.
Proposition 4.6.Let X be a topological space and a ∈ X.
Thus we have {x} = {a}, that is to say, x ∈ Cl(a).The converse is straightforward.
Proposition 4.7.Let (X, T ) be a topological space.Denote by (X, G) the space X equipped with the Goldman topology induced by T .The following statements are equivalent: ( Proof.We first show (1)⇒(2).Let x = y be two elements of X.Since X is a T 0 -space, there exists an open subset U of X such that (x ∈ U and y ∉ U ) or (y ∈ U and x ∉ U ). Set F = X\U .We have U ∩ F = ∅ with (x ∈ U and y ∈ F ) or (x ∈ F and y ∈ U).Therefore, the G-topology is Hausdorff.Now we show (4)⇒ (1).Let a and b be two elements of X such that {a} = {b} so that Cl(a) = Cl(b).In view of Proposition 4.6, we get {a} G = {b} G .Since (X, G) is a T 0 -space, we have a = b.Therefore, (X, T ) is a T 0 -space.
Remark 4.8.It may be checked easily that the following properties hold.
(1) If (X, T ) is a T 1 -space, then (X, G) is a discrete topological space.
(2) The set (X, G) is a discrete topological space if and only if X = Gold(X).In particular, if (X, G) is Hausdorff and nondiscrete, (X, T ) is a T 0 -space and not a T 1 -space.
(3) Every locally closed subset A of X is a clopen set in (X, G) and satisfies Proposition 4.9.Let (X, T ) be a topological space.If the G-topology on X is quasicompact, then the topological space (X, T ) is Noetherian.
Proof.Let U be an open subset of X.Because U is G-closed by Proposition 4.2, U is G-quasicompact.Then U is quasicompact since the G-topology on X is finer than the original topology on X.Therefore, X is a Noetherian space.
Remark 4.10.The converse of the above proposition is not true (see Example 6.3).Nevertheless, things are nicer for a spectral space.
The family of all closed subsets and all quasicompact open subsets of X = Spec(R) is a subbase of closed sets for a topology called the patch topology on X [10].The family of all quasicompact open subsets and their complements is an open subbase for the patch topology.A patch in X is a closed set for the patch topology.It is easily seen that the G-topology is finer than the patch topology.
Proposition 4.11.Let X be a spectral space.Then the following statements are equivalent: (1) the G-topology on X is compact; (2) X is a Noetherian space; (3) the G-topology on X coincides with the patch topology on X.
Proof.Thanks to [14, Section I, Proposition 4], ( 2) and ( 3) are equivalent.Then (1)⇒(2) follows from Proposition 4.9 and (3)⇒( 1) is a consequence of the compactness of the patch topology on X [10].Proposition 4.12.Let (X, T ) be a topological space and Y ⊆ X.Then Y is strongly dense in Y G (for the topologies induced by T ). where Proposition 4.13.Every G-closed subset of a spectral space X has at least a minimal element for the natural order defined by the topology (more precisely, every G-closed subset of X is ≥-inductive).
Proof.Let X = Spec(R), where R is a ring, and Y ⊆ X a G-closed subset.Let {p i | i ∈ I} ⊆ Y be a nonempty chain and p the intersection of all p i .Since Y G is the set of all prime ideals of R which are an intersection of some prime ideals belonging to Y , we see that p ∈ Y .Therefore (Y , ⊇) is inductive.
Remark 4.14.The above proposition does not hold for an arbitrary T 0space (see Example 6.3).Proof.According to [10, Section 7], a patch set is an intersection of constructible subsets.Moreover, a constructible subset is a union of finitely many locally closed quasicompact subsets.Theorem 4.16.Let X be a T 0 -space.
(1) If X is a sober space, every G-closed subset of X is sober.
(2) Every sober subspace of X is a G-closed subset of X.

Proof. We show (1). Let F be an irreducible closed subset of a
Hence F is an irreducible closed subset of X.Therefore, F has a generic point x, that is to say, F = {x} because X is sober.It follows that F = Y ∩ {x} and F = {x} = Y ∩ {x}.This implies that x ∈ Y by G-closeness of Y .Moreover, we have x ∈ Y ∩ F = F , which proves that x is the generic point of F (in the subspace Y ).Now we prove (2) Remark 4.17.In view of Theorem 4.16, the G-closed subsets of X are the sober subspaces of X when X is a sober space.Nevertheless, when X is a T 0space, a G-closed subset of X need not be sober (see Example 6.3).Proposition 4.18.Let X be a T 0 -space and Y a G-dense subset.Then, for each x ∈ X, x = inf({x}∩Y ) (X is equipped with the partial ordering ≤ induced by the topology).

Proof. For each y
such that z ≤ y for any y ∈ Y ∩{x}, and assume that x ∉ {z}.Then {x}∩(X\{z}) is nonempty and locally closed.Since Y is strongly dense, we get {x} ∩ (X\{z}) ∩ Y = ∅.Hence, there exists some y ∈ {x} ∩ Y such that y ∈ X\{z}, that is to say, z y; this is a contradiction.It follows that z ≤ x, proving that x = inf({x}∩Y ).
We recall some topological definitions and link them with the G-topology.Let Min(R) denote the set of all minimal prime ideals of a ring R equipped with the relativization of the Zariski topology.It is well known that Min(R) is Hausdorff.
According to Hochster, a topological space X is said to be minspectral if there is some ring R such that X is homeomorphic to Min(R) [11].As in [11], an m-base Ꮾ for a Hausdorff space X is a base of open sets such that every subset S of Ꮾ which has the finite intersection property has nonempty intersection.
A topology is said to be scattered if every point has a base of clopen neighborhoods.Such a topology is totally disconnected.A Hausdorff space X is said to be completely regular (or Tychonoff ) if for each point x ∈ X and every closed subset V of X such that x ∉ V , there is a continuous map ϕ : X → [0, 1] such that ϕ(x) = 1 and ϕ(a) = 0 for every a ∈ V .Proposition 4.19.Let X be a topological space.
(1) If X is a spectral space, then the G-topology on X is minspectral.
(2) If X is a T 0 -space, then the G-topology on X is scattered.
(3) Every scattered topology is completely regular.In particular the G-topology on a T 0 -space is completely regular.
Proof.With the hypothesis of (1) being granted, the G-topology on X is Hausdorff since X is T 0 .Let Ꮾ be the family of all subsets Y of X such that Y = F ∩ O, where F is closed and O is quasicompact open.Then Ꮾ is a base for the G-topology and its elements are patches.Since the patch topology is compact [10], Ꮾ is an m-base.Thanks to Hochster's result [11,Theorem 1], the G-topology on X is minspectral.We show (2).Every locally closed subset is G-open and G-closed.Let x ∈ X, then the family of locally closed subsets of X containing x is a base of G-clopen neighborhoods of x.Now, (3) is a consequence of the following considerations.Let X be a scattered topological space, x ∈ X, and V a closed subset of X such that x ∉ V .There exists a clopen subset O of X containing x such that O ⊆ X\V .Since {O, X\O} is an open covering of X, we can define a continuous map ϕ : X → [0, 1] such that ϕ(O) = {1} and ϕ(X\O) = {0}.We have thus checked out the three conditions.(1) The canonical injection X → S(X) identifies X with a subspace of S(X), X is strongly dense in S(X) and S(X) is a Noetherian spectral space.In particular, the G-topology on S(X) is the patch topology and induces the G-topology on X.
(2) The set Gold(X) is equal to Gold(S(X)).
Proof.Let X be a Noetherian topological T 0 -space.Then its sobrification S(X) is a Noetherian spectral space.Indeed, S(X) is Noetherian whenever X is Noetherian [9, Chapter 0, Section 3, Proposition 7.6].In this case, S(X) is Noetherian and sober, whence a spectral space.Then X can be identified with a subspace of S(X) and X is strongly dense in S(X) by [9, Chapter 0, Section 3, Proposition 7.1(b)].We get Gold(S(X)) ⊆ X by Proposition 3.6 because X is strongly dense.It follows that Gold(S(X)) ⊆ Gold(X).Now Gold(S(X)) is strongly dense in X so that Gold(X) = Gold(S(X)).

A topological characterization of the Jacobson prime spectrum of a commutative ring.
Recall that a topological space X is said to be a Jacobson space if the set Ꮿ(X) of all closed points of X is strongly dense in X [9, Chapter 0, Section 3, Proposition 8.1].Therefore, X is a Jacobson space if and only if X = Ꮿ(X) G .Obviously, when X is a topological space, Jac(X) = Ꮿ(X) G is a Jacobson space; we call it the Jacobson subspace of X.
Let R be a ring.We denote by Jac(R) the Jacobson subspace of Spec(R).Following Picavet [14], a prime ideal p of R is in Jac(R) if and only if p is the Example 6.5 (a topological space showing that the total ordering hypothesis in Proposition 3.16 is essential).(1) Let R be an integral domain with finite dimension and X = Spec(R).Assume that R is not a G-domain.Then the second statement of Proposition 3.16 holds while the first statement fails.
The family Ꮾ = {] ↓ x]} x∈X∼aω ∪ (∪ n∈Z * {a ω } ∪ {a p | |p| ≥ |n|}) is a base for a topology T compatible with the partial ordering ≤.We have Gold(X) = X, but the topology T is not the left topology.

Proposition 4 . 15 .
Let R be a ring.Every patch of Spec(R) is an intersection of Zariski quasicompact G-open sets of Spec(R).

Theorem 4 .
20.Let X be a sober topological space and Y ⊆ X.Then Y G is homeomorphic to S(Y ).Proof.In view of Theorem 4.16, Y G is a sober space so that Y G S(Y G ). On the other hand, the inclusion Y → Y G is a quasihomeomorphism (see Proposition 4.12).Hence S(Y ) S(Y G ) by Theorem 2.2.Therefore Y G is homeomorphic to S(Y ).Proposition 4.21.Let X be a Noetherian T 0 -topological space.
. Let Y be a sober subspace of X and x ∈ Y G , that is to say, {x} = Y ∩ {x}.Since Y ∩ {x} is irreducible, so is Y ∩ {x}.Hence Y ∩ {x} has a generic point y ∈ Y ∩ {x} so that Y ∩ {x} = Y ∩ {y}.Moreover, x and y belong to Y G .This yields {x} = {y}.It follows that x