Stone's Representation Theorem

Introduction

The following result is well-known and is usually called Stone's representation theorem:

Every Boolean algebra is isomorphic to that of simultaneously open and closed sets of some suitable topological space. Moreover the latter can be chosen to be compact Haussdorf and totally discontinuous, and in this case it is unique up to homeomorphism.

However Stone's statement is more general. It deals with any distributive lattice and illustrates the truly innovative ideas that we owe to Stone. Among them one can cite the fundamental idea of considering (before Zariski)See: P. Johnstone, Stone Spaces, Cambridge University Press (1986), for historical details. prime ideals as points, as well as a genuine result of duality several years before the formulation of categorical equivalence by Eilenberg and Mac Lane.

We give below both statement and proof of Stone's complete theorem, in a more modern language than the original article (a bit difficult to read). M. H. Stone, Topological representations of distributive lattices and Brouwerian logics. Časopis Pěst. Mat. Fys. 67 (1937), p. 1-25.

Distributive lattices

Throughout this article a lattice denotes an ordered set \(T\) in which every finite subset has a greatest lower bound and a least upper bound. Hence, by considering the empty set, \(T\) admits a least and a greatest elements, which we denote by \(0\) and \(1.\) If \(x,y\in T\) the lower and upper bounds of the set \(\{x,y\}\) will be denoted by \(x\wedge y\) and \(x\vee y.\) A sublattice of a lattice \(T\) is a subset of \(T\) containing \(0\) and \(1\) and closed by \(\wedge\) and \(\vee.\) A lattice morphism is a map compatible with \(\wedge\) (and therefore non-decreasing) and \(\vee,\) preserving \(0\) et \(1.\) A lattice is said to be distributive if each of the operations \(x\wedge y\) and \(x\vee y\) is distributive with respect to the other.

An ideal of a lattice \(T\) is a subset \(\ga≠\Vide\) of \(T\) satisfying the conditions

• \(({\rm I}_1)\)\(\forall x\in\ga\,,\>\forall y\in T\,,\>(y≤x\implique y\in \ga)\)
• \(({\rm I}_2)\)\(\forall x,y\in\ga\,,\>x\vee y\in\ga.\)

The property \(({\rm I}_1)\) can be replaced by

• \(({\rm I'}_2)\)\(\forall x\in\ga\,,\>\forall y\in T\,,\>x\wedge y\in \ga.\)

Any ideal \(\ga\) of a lattice \(T\) contains \(0,\) et \(\ga=T\) if, and only if, \(1\in\ga.\)
A filter is an ideal for the opposite order.

The ideal generated by a subset \(\Phi\) of a lattice \(T\) is the intersection of all ideals of \(T\) which contain \(\Phi,\) that is the smallest (for the inclusion ordering) ideal of \(T\) which contains \(\Phi.\) If \(T\) is distributive the ideal generated by \(\Phi\) is the set of elements of \(T\) of the form \[\bigvee_{i\in I}a_i\wedge x_i\,,\] where \(I\) is a finite set and \(a_i\in T,\) \(x_i\in \Phi.\)

An ideal \(\ga≠T\) of a lattice \(T\) is a prime ideal if it satisfies the condition \[\forall x,y\in T\,,\>x\wedge y\in\ga\implique (x\in\ga\hbox{ ou }y\in\ga)\,.\] A maximal ideal of a lattice \(T\) is an ideal \(\ga≠T\) which is maximal among the proper ideals of \(T\) for the inclusion ordering between subsets of \(T.\) If the lattice \(T\) is distributive, every maximal ideal is prime. This is a special case of the following fundamental result (take \(\gf=\{1\}):\)

First, since \(\gf≠\Vide,\) \(\gi\) is indeed a proper ideal. Then, by contradiction suppose there exists \(x,y\in T\) such that \(x,y\notin\gi\) and \(x\wedge y\in\gi.\) It is easy to verify that the set \(\gj=\{(a\wedge x)\vee \lambda\,,\>a\in T\,,\>\lambda\in\gi\}\) is an ideal which contains both \(\gi\) and \(x.\) By maximality of \(\gi\) it follows that \(\gj\inter\gf≠\Vide,\) i.e. \[\exists a\in T\,,\>\exists\lambda\in\gi\,,\>\exists f\in\gf\,,\>(a\wedge x)\vee \lambda=f\,.\] We then have \[f\wedge y=(a\wedge x\wedge y)\vee(\lambda\wedge y)\in \gi\,.\] As \(\gi\inter\gf=\Vide,\) we have \(f\notin\gi,\) and we can apply the previous result by making \((y,f)\) play the role of \((x,y).\) We obtain: \(\exists f'\in\gf\,,\>f\wedge f'\in\gi\,.\) But the set \(\gf\) is closed by \(\wedge,\) hence we have \(f\wedge f'\in\gi\inter\gf,\) which contradicts \(\gi\inter\gf=\Vide.\)

Indeed, the set of ideals containing \(\ga\) and not meeting \(\gf\) is clearly inductive for inclusion ordering. By Zorn's lemma it has a maximal element \(\gi,\) which is a prime ideal according to the previous result.

An inclusion is obvious; for the other, let \(\ga\) be an ideal of \(T\) and \(x\in T\setminus \ga\,.\) According to the previous result applied to \(\gf=\{x\},\) there exists a prime ideal containing \(\ga\) but not \(x.\)

Spectrum

In this section \(T\) denotes a distributive lattice. We are going to construct a topological space whose points are the prime ideals of \(T,\) analogous to the spectral space (Zariski topology) of a commutative ring.

The set \(\Spec(T)\) of prime ideals of \(T\) is called the spectrum of \(T.\) If \(A\subset T\) we put \[V_T(A)=\{\ga\in\Spec(T)\tq A\subset \ga\}\,.\] If \(x\in T,\) \(V_T(\{x\})\) is noted \(V_T(x).\)
If \(\ga\) is the ideal generated by \(A\) we obviously have \(V_T(A)=V_T(\ga).\)

• \(\Vide=V_T(T)\) (every prime ideal is a proper subset).
• For every family \((A_i)_{i\in I}\) of subsets of \(T,\) one has \[\Inter_{i\in I}V_T(A_i)=V_T\Big(\Union_{i\in I}A_i\Big)\,.\]
• By definition of a prime ideal, for every \(x,y\in T,\) \[V_T(x)\union V_T(y)=V_T(x\wedge y)\,,\] hence, for all subsets \(A\) and \(B\) of \(E,\) \begin{eqnarray*} V_T(A)\union V_T(B)&=&\Inter_{x\in A}V_T(x)\union \Inter_{y\in B}V_T(y)\\ &=& \Inter_{x\in A}\Inter_{y\in B}V_T(x)\union V_T(y)\\ &=& \Inter_{x\in A}\Inter_{y\in B}V_T(x\wedge y)\,. \end{eqnarray*}

Complements on topological spaces

We recall that a topological space \(X\) is compact if any open covering admits a finite subcovering, and compact Haussdorf if it is both compact and Haussdorf. Any closed subset of a compact space is compact, but when the ambiant space \(X\) is not Haussdorf the compact subsets of \(X\) are not necessarily closed.

We recall that a topological space \(X\) is a Kolmogorov spaceWe use Bourbaki's terminology here. Usually called a \(T_0\)‑space in english. if, for any set of two distinct points of \(X,\) there exists a neighborhood of one of them which does not contain the other. This condition is weaker than for all \(x,y\in X\) s.t. \(x≠y\) there exists a neighborhood of \(x\) that does not contain \(y\),\(T_1\) property, which is equivalent to the one-point sets being closed. itself weaker than \(X\) to be Haussdorf (that is \(T_2\) property).

In any topological space \(X,\) the relation \(≤\) defined by \(\overline{A}\) denotes the topogical closure of a set \(A.\) \[x≤y\ssi y\in\overline{\{x\}}\ssi \overline{\{y\}}\subset\overline{\{x\}}\] is a preorder, which we will call topological preorder.

We say that \(x\) is a closed point of a topological space \(X\) if \(\{x\}\) is closed, in other words if \(x\) is maximal for the topological preorder.

We recall that a topological space \(X\) is irreducible if it satisfies one of the following equivalent conditions:

1. Any intersection of a finite number of nonempty open sets is nonempty;
2. \(X\) is non-empty and the intersection of two non-empty open sets is non-empty;
3. \(X\) is not the union of a finite number of proper closed sets;
4. \(X\) is non-empty and is not the union of two proper closed sets;
5. \(X\) is non-empty and every nonempty open set of \(X\) is dense in \(X\);
6. \(X\) is non-empty and every open set of \(X\) is connected.

Every one-point set is irreducible. Any irreducible Haussdorf space is a one-point set. Any non-empty open set of an irreducible space is irreducible. A subset of a topological space is irreducible if, and only if, its closure is (so that the study of the irreducible subsets of \(X\) comes down to that of irreducible closed sets).

Let \(X\) be a topological space, and \(F\subset X.\) We say that a point \(x\in F\) is a generic point of \(F\) if \(F=\overline{\{x\}}.\) The point \(x\) is then a smallest element of \(F\) for the topological preorder. Any subset of \(X\) admitting a generic point is closed (obvious) and irreducible as the closure of a one-point set.

Let \(X\) be a topological space. The following conditions are equivalent:

1. \(X\) is a Kolmogorov space;
2. The topological preorder is an ordering (called in this case the topological ordering);
3. The map \(x\mapsto\overline{\{x\}}\) is one-to-one;
4. Every closed set of \(X\) has at most one generic point.

A sober space is a topological space \(X\) such that every irreducible closed set of has a unique generic point, or, equivalently, a Kolmogorov space in which every irreducible closed set has a generic point.

Theorem statement

We say that a topological space \(X\) is a spectral space if it satisfies the following conditions:

• \(({\rm S}_1)\)\(X\) is sober;
• \(({\rm S}_2)\)\(X\) is compact;
• \(({\rm S}_3)\)The compact open sets form a basis for the topology;
• \(({\rm S}_4)\)The intersection of two compact open sets is compact.

Given a topological space \(X,\) we denote by \(\oqc(X)\) the set of compact open subsets of \(X,\) equipped with the inclusion ordering.

This is almost a paraphrase of the previous theorem. The uniqueness up to homeomorphism comes from point d), since for any spectral space \(X\) such that \(T\simeq\oqc(X)\) we will then have \[X\simeq\Spec(\oqc(X))\simeq\Spec(T)\,.\]

The theorem can therefore be read in both directions. In one direction, it gives a topological representation of distributive lattices. In the other, it gives a purely topological characterization of spectral topologies of distributive lattices. We will later see that it is more a theorem of duality than of representation.

Proof of the theorem

Proof of point a)

Let \(T\) be a distributive lattice.

For all \(x\in T\) we set \(D_T(x)=\Spec(T)\setminus V_T(x)\) (set of prime ideals not containing \(x).\) The sets \(D_T(x),\) where \(x\in T,\) are open sets, called elementary open sets, and they form a base of open sets of the spectral topology. Indeed, these sets are open sets as complementaries of closed sets, and for any open set \(U\) of \(\Spec(T),\) there exists a set \(A\subset E\) such that \(U=\Spec(T)\setminus V_T(A).\) Therefore \[U=\Spec(T)\setminus\Inter_{x\in A}V_T(x)=\Union_{x\in A}D_T(x)\] is a union of elementary open sets.

The topological space \(\,\Spec(T)\) is Kolmogorov.

Indeed, if \(\ga\) and \(\gb\) are two distinct prime ideals, one of them, let us say \(\ga,\) does not contain the other, and taking \(x\in\gb\setminus\ga,\) \(U=D_T(x)\) is a neighbourhood of \(\ga,\) and \(\gb\notin U.\)

For \(Y\subset\Spec(T),\) let's note \(I_T(Y)\) the ideal \(\Inter_{\ga\in Y}\ga.\) According to for any set \(A\subset T,\) \(I_T(V_T(A))\) is the ideal generated by \(A.\)

The closure of a set \(Y\subset\Spec(T)\) for the spectral topology is \(V_T(I_T(Y)).\)

Indeed, \(V_T(I_T(Y))\) is a closed set containing \(Y,\) and it is the smallest because every closed set \(F\) is of the form \(V_T(A),\) and in this case \(\displaystyle Y\subset F\implique \forall \ga\in Y\,,\>A\subset \ga\) \(\displaystyle\implique A\subset I_T(Y) \implique V_T(I_T(Y))\subset F\,.\) As a special case, taking \(\ga\in\Spec(T),\) one has: \(\overline{\{\ga\}}=V_T(\ga).\)

Let \(\ga\) be an ideal of \(T.\) The closed set \(V_T(\ga)\) is irreducible if, and only if, the ideal \(\ga\) is prime.

Indeed, if \(\ga\) is prime then \(V_T(\ga)=\overline{\{\ga\}}\) is irreducible as a closed set having a generic point. Conversely, suppose \(V_T(\ga)\) is irreducible. The ideal \(\ga\) is proper because otherwise \(V_T(\ga)\) would be empty. Let \(x,y\in T\) such that \(x\wedge y\in\ga.\) By definition of a prime ideal we then have: \(\forall \gb \in V_T(\ga)\,,\>x\in \gb\) or \(y\in \gb,\) in other words: \(V_T(\ga)\) is the union of the two closed sets \(V_T(x)\) and \(V_T(y).\) But \(V_T(\ga)\) is supposed to be irreducible. Hence one of these two closed sets is the whole \(V_T(\ga),\) hence it contains \(\ga,\) i.e. \(x\in \ga\) or \(y\in\ga.\) Therefore \(\ga\) is a prime ideal.

Every irreducible closed set of \(\,\Spec(T)\) has a generic point.

Indeed, the closed sets of \(\Spec(T)\) are the sets of the form \(V_T(\ga)\) where \(\ga\) is an ideal, and according to such a closed set is irreducible if, and only if, \(\ga\) is a prime ideal, i.e. \(\ga\in\Spec(T).\) It then follows from that we have \(V_T(\ga)=\overline{\{\ga\}}.\)

The elementary open sets \(D_T(x)\) of the spectral topology are compact.

Let \(a\in T,\) and suppose we are given a covering \(\{D_T(x)\,,\>x\in \Phi\}\) (\(\Phi\subset T)\) of \(D_T(a)\) by elementary open sets. For any ideal \(\ga\in D_T(a),\) there exists \(x\in\Phi\) such that \(\ga\in D_T(x),\) i.e. \(x\notin \ga\) ; by contraposition: any prime ideal \(\ga\) containing \(x\) contains \(a,\) in other words, according to : the ideal generated by \(\Phi\) contains \(a.\) According to there exists a finite set \(\Psi\subset\Phi\) such that \(a\) belongs to the ideal generated by \(\Psi.\) Then any ideal, and consequently any prime ideal, containing \(\Psi,\) contains \(a,\) that is, by contraposition: \(\{D_T(x)\,,\>x\in \Psi\}\) is a covering of \(D_T(a).\)

Conversely, every compact open set of \(\,\Spec(T)\) is of the form \(D_T(x).\)

Indeed, any open set \(U\) is the union of sets of the form \(D_T(x),\) so if \(U\) is compact it is the union of a finite number of them. We then just have to notice that if \(I\) is a finite set, we have \[\Union_{i\in I}D_T(x_i)=D_T\Big(\bigvee_{i\in I}x_i\Big).\]

Finally all this proves point a) : \(\Spec(T)\) is sober as a Kolmogorov space in which every closed irreducible set has a generic point. The sets \(D_T(x)\) form a basis of compact open sets. Finally \(\Spec(T)=D_T(1)\) is compact, and the intersection of two compact open sets is of the form \(D_T(x)\inter D_T(y)=D_T(x\wedge y)\) hence compact.

We deduce from that the topological ordering in \(\Spec(T)\) (see and ) is the inclusion ordering between prime ideals, hence the closed points of \(\Spec(T)\) are the maximal ideals of \(T.\)

Proof of point b)

It's enough to show that \(\oqc(X)\) is a sublattice of \(\mathcal{P}(X).\) But in every topological space, any finite intersection and any union of open sets is an open set, any finite union of compact subsets is compact, and by definition in a spectral space any finite intersection of compact open sets is compact (the case of the intersection of an empty family comes from the compactness of the whole space).

Proof of point c)

It follows easily from the previous sections that if \(T\) is any distributive lattice, the map \(x\mapsto D_T(x)\) is a lattice morphism from \(T\) onto \(\oqc(\Spec(T)).\) Moreover it is one-to-one, since for any \(x\in T,\) \(I_T(V_T(x))\) is the ideal generated by \(x\) that is \(\{z\in T\tq z≤x\},\) hence \(x=\max(I_T(V_T(x)))\) is uniquely determined by \(V_T(x),\) and consequently by \(D_T(x).\)

Proof of point d)

In the following \(X\) denotes a spectral space.

If \(F\) is a closed set in \(X,\) we set \(\Phi(F)=\{U\in\oqc(X)\tq F\inter U=\Vide\}.\) For any ideal \(\ga\) of the lattice \(\oqc(X)\) we set \(\Psi(\ga)=X\setminus\Union_{U\in \ga}U.\)

\(\Phi\) is a decreasing map from the set of closed sets of \(X\) into the set of ideals of \(\oqc(X),\) and \(\Psi\) is a decreasing map from the set of ideals of \(\oqc(X)\) into the set of closed sets of \(X.\)

This assertion is obvious.

The maps \(\Phi\) et \(\Psi\) are inverse bijections of each other.

For any closed set \(F\subset X,\) we first have \(\Psi(\Phi(F))=F.\) Indeed, the space \(X\) has a basis of compacts open sets. Consequently any open set is the union of all compacts open sets it contains. Conversely, we show that for any ideal \(\ga\) of \(\oqc(X),\) we have \(\Phi(\Psi(\ga))=\ga:\) we clearly have \(\ga\subset\Phi(\Psi(\ga)).\) For the other inclusion one has to show that any compact open set \(V\) that is included in the union of the elements of \(\ga\) lies itself in \(\ga.\) \(V\) is compact so there exists a finite number of elements of \(\ga\) whose union covers \(V.\) But the set \(\ga\) is an ideal (of \(\oqc(X)\) ordered by inclusion), therefore \(V\in \ga.\)

An ideal \(\ga\) of \(\oqc(X)\) is a prime ideal if, and only if, the closed set \(F=\Psi(\ga)\) is irreducible.

Suppose \(F\) is irreducible. On one hand \(F\) is non-empty hence \(\ga\) is a proper ideal (the whole space is compact). On the other hand \(F\) admits a generic point \(x\) (the space \(X\) is sober), that is: \(F=\overline{\{x\}}.\) Let \(U\) and \(V\) be two open compacts sets such that \(U\inter V\in\ga.\) Then \(U\inter V\in\Phi(F),\) hence \(x\notin U\inter V,\) in other words \(x\notin U\) or \(x\notin V,\) so \(F=\overline{\{x\}}\) is disjoint from \(U\) or from \(V,\) or equivalently \(U\in\Phi(F)\) or \(V\in\Phi(F)\) that is \(U\in\ga\) or \(V\in\ga.\) Hence the ideal \(\ga\) is a prime ideal.

Conversely, assume the ideal \(\ga\) is prime. Then \(F≠\Vide\) since \(\ga=\Phi(F)\) is a proper ideal. Suppose \(F=G\union H\) where \(G\) and \(H\) are closed sets distincts from \(F.\) Since every closed set of \(X\) is the intersection of closed sets with compact complements that contain it, there exists two closed sets with compact complements \(G'\) et \(H'\) that do not contain \(F\) but whose union contains \(F.\) Then \(X\setminus(G'\union H')\in \Phi(F)=\ga.\) But the ideal \(\ga\) is prime; hence we deduce \(X\setminus G'\in \ga\) or \(X\setminus H'\in \ga,\) that is a contradiction since \(\ga=\Phi(F)\) and no one of the closed sets \(G'\) or \(H'\) contains \(F.\) Hence \(F\) is irreducible.

Let \(x\in X.\) We put \(\ph(x)=\Phi(\overline{\{x\}}).\) Since \(\overline{\{x\}}\) is irreducible, \(\ph(x)\) is a prime ideal of \(\oqc(X),\) or equivalently a point of the spectral space \(\Spec(\oqc(X)).\) Conversely, let \(\ga\) be a prime ideal of \(\oqc(X),\) and \(x\) the generic point of \(\Psi(\ga).\) The point \(x\) is uniquely determined because \(X\) is sober. We put \(x=\psi(\ga).\) According to et \(\ph\) is one-to-one from \(X\) onto \(\Spec(\oqc(X))\) and \(\psi\) is its inverse map.

The map \(\ph\) is continuous.

Let \(G\) be a closed set of \(\Spec(\oqc(X)).\) By definition of the spectral topology there exists an ideal \(\ga\) of \(\oqc(X)\) such that \(G\) is the set of prime ideals that contain \(\ga.\) Put \(F=\Psi(\ga).\) For every point \(x\in X\) we have \(\displaystyle \ph(x)\in G\ssi \ga\subset \ph(x)\ssi \ga\subset \Phi(\overline{\{x\}})\) \(\displaystyle\ssi\overline{\{x\}}\subset \Psi(\ga)\ssi x\in F\,.\) Hence the inverse image of any closed set under \(\ph\) is closed.

The map \(\psi\) is continuous.

Let \(F\) be a closed set in \(X,\) \(\ga=\Phi(F),\) and \(G\) the (closed) set of \(\Spec(\oqc(X))\) whose elements are the prime ideals that contain \(\ga.\) For any point \(\gb\in\Spec(\oqc(X))\) one has \(\displaystyle \psi(\gb)\in F\ssi \overline{\{\psi(\gb)\}}\subset F\ssi\Psi(\gb)\subset F\) \(\displaystyle\ssi \Phi(F)\subset \gb\ssi \gb\in G\,.\) Hence the inverse image of any closed set under \(\psi\) is closed.

Finally the map \(\ph\) is an homeomorphism from \(X\) onto \(\Spec(\oqc(X)),\) which completes the proof.

The special case of boolean algebras

A Boolean algebra is a distributive lattice \(T\) such that for any \(x\in T,\) there exists \(x'\in T\) such that \(x\wedge x'=0\) and \(x\vee x'=1.\) Then it can easily be shown that \(x'\) is unique; it is denoted \(\neg x.\)

It is well known and easy to prove that a compact Haussdorf topological space is totally discontinuous if, and only if, it admits a basis of clopen (both closed and open) sets. It easily follows that the Haussdorf spectral spaces are the totally discontinuous compact Haussdorf spaces. All together with the following result we deduce that the statement mentioned in the introduction is indeed a special case of Stone's theorem.

• Assume (i). Then every one-point set of \(\Spec(T)\) is closed (see ) but this still does not mean that the space is Haussdorf. But if \(\ga\) and \(\gb\) are two distinct points in \(\Spec(T),\) that is two distinct prime ideals, one of them, for instance \(\ga,\) does not contain the other one. Pick \(x\in\gb\setminus\ga\) and let's consider the set \[\gf=\{s\wedge x\,,\>s\in T\setminus\gb\}\,.\] \(\gf\) is non-empty since it contains \(1\wedge x=x,\) and closed by \(\wedge\) (since \(\gb\) is prime hence \(s,s'\notin \gb\implique s\wedge s'\notin\gb).\) We necessarily have \(0\in\gf:\) otherwise, according to applied to the ideal \(\{0\},\) there would exist a prime ideal \(\gi\) disjoint from \(\gf.\) Then for every \(s\in T\setminus \gb\) one would have \(s\wedge x\notin \gi\) hence \(s\notin\gi,\) which would imply \(\gi\subset\gb,\) hence \(\gi=\gb\) since by assumption every prime ideal is maximal, and that would contradict \(x=1\wedge x\notin\gi.\)
  Therefore there exists \(s\notin\gb\) such that \(s\wedge x=0.\) But then \[\Vide=D_T(0)=D_T(s\wedge x)=D_T(s)\inter D_T(x)\,.\] Hence \(D_T(x)\) et \(D_T(s)\) are two open sets that separate \(\ga\) and \(\gb:\) the space \(\Spec(T)\) is therefore Haussdorf.
• Assume (ii). Then \(\Spec(T)\) is compact hence \(\oqc(\Spec(T))\) is the set of clopen sets of \(\Spec(T).\) Therefore it is a Boolean subalgebra of \(\mathcal{P}(\Spec(T)).\) So the lattice \(T\) is itself a Boolean algebra, since \(T\simeq \oqc(\Spec(T).\)
• Assume (iii). Let \(\ga\) be a prime ideal of \(T,\) and \(x\in T\setminus\ga.\) One has \(x\wedge \neg x=0\in \ga,\) and since the ideal \(\ga\) is prime we deduce that \(\neg x\in \ga.\) But then every ideal that contains \(\ga\) and \(x\) contains \(x\vee\neg x=1\) hence equals \(T.\) In other words: \(\ga\) is a maximal ideal.

Stone duality

Stone's theorem is a lot more than a representation result: it provides a theorem of duality. A duality result asserts the anti-equivalence of two categories \(C_1\) and \(C_2,\) that is an equivalence of categories between \(C_1\) and the opposite category \(C_2^{{\rm op}}.\)

Let us note:

• \(\td\) the category whose objects are the (small) distributive lattices and whose arrows are the lattice morphisms;
• \(\sp\) the category whose objects are the (small) spectral spaces and arrows the spectral maps, i.e. such that the inverse image of any compact open set is a compact open set (for general spectral spaces this condition is stronger than the continuity).

First of all, let's verify the (contravariant) functoriality of \(\Spec:\) given a distributive lattice morphism \(f:T\to T',\) denote \(\Spec(f)\) the map from \(\Spec(T')\) into \(\Spec(T)\) defined for any prime ideal \(\ga\) of \(T'\) by \[\Spec(f)(\ga)=f^{-1}(\ga).\] The inverse image of a prime ideal by a lattice morphism is a prime ideal, so that \(\Spec(f)\) has values in \(\Spec(T).\) Let us then verify that the map \(\Spec(f)\) is indeed spectral: every compact open set of \(\Spec(T)\) has the form \(U=D_T(x),\) \(x\in T.\) We then have, for any prime ideal \(\ga\) of \(T',\) \[\ga\in\Spec(f)^{-1}(U)\ssi \Spec(f)(\ga)\in D_T(x)\ssi x\notin f^{-1}(\ga)\ssi f(x)\notin\ga\] in other words the inverse image of \(U\) under \(\Spec(f)\) is the compact open set \(D_T(f(x)).\) Finally the functoriality follows from \(\Spec(\id_T)=\id_{\Spec(T)}\) and from the relation \[(g\circ f)^{-1}(\ga)=f^{-1}(g^{-1}(\ga)).\]

In the same way we show that we define a contravariant functor \(\oqc\) from \(\sp\) into \(\td\) by setting, for each spectral map \(f:X\to Y\) between spectral spaces and \(U\in\oqc(Y),\) \[\oqc(f)(U)=f^{-1}(U).\]

Let us then check that the isomorphism \(\theta_T:x\mapsto D_T(x)\) from \(T\) onto \(\oqc(\Spec(T))\) is natural in \(T.\) Let \(f:T\to T'\) be a morphism between distributive lattices, and \(x\in T.\) We have: \begin{eqnarray*} (\theta_{T'}\circ f) (x)&=&D_{T'}(f(x))\\ &=&\{\ga\in\Spec(T')\tq f(x)\notin\ga\}\\ &=&\{\ga\in\Spec(T')\tq x\notin f^{-1}(\ga)\}\\ &=&\{\ga\in\Spec(T')\tq f^{-1}(\ga)\in D_T(x)\}\\ &=&\{\ga\in\Spec(T')\tq \Spec(f)(\ga)\in D_T(x)\}\\ &=&\Spec(f)^{-1}(D_T(x))\cr &=&\oqc(\Spec(f))(D_T(x))\\ &=&\oqc(\Spec(f))\circ \theta_T(x)\,, \end{eqnarray*} hence \[\theta_{T'}\circ f=\oqc(\Spec(f))\circ \theta_T\,.\]

In the same way we show that the isomorphism \(\theta'_X:x\mapsto\ph(x)\) from \(X\) onto \(\Spec(\oqc(X))\) defined in the previous section is natural in \(X.\)

Epilogue

The reader will have noticed the analogy between the notions we just studied here and the classical tools of commutative algebra (ideals, prime ideals, spectrum of a ring and Zariski topology). Note that the spectrum notion works in an even simpler way for lattices than for rings: in a commutative ring \(A,\) the intersection of all prime ideals containing some set \(B\subset A\) is the radical ideal generated by \(B.\) Then the map \(x\mapsto D_A(x)\) (set of prime ideals of \(A\) containing \(x)\) is no longer one-to-one: \(D_A(x)=D_A(y)\) means that \(x\) and \(y\) generate the same radical ideal, which is no longer equivalent to \(x=y\) but can be written as \(\exists p,q>0\,,\>y^p\in Ax\) et \(x^q\in Ay.\) In a distributive lattice, where the operation \(\wedge\) plays the role of multiplication, there is no radical, due to the idempotence of any element.

That being said, the Zariski topology of a commutative ring also provides a spectral space as defined at . This can be interpreted by noticing that it is closely related to the lattice structure: it can be shown that radicals of finitely generated ideals of a commutative ring \(A\) form a distributive lattice \(\zar(A)\) (the Zariski lattice of \(A)\) whose spectral topology is homeomorphic to the Zariski topology. In other words the spaces \(\Spec(A)\) (ring spectral space) and \(\Spec(\zar(A))\) (lattice spectral space) are homeomorphic, hence, by Stone's theorem: \(\zar(A)\) is isomorphic to \(\oqc(\Spec(A)),\) which illustrates the fact that from the open compact \(D_A(x)\) of \(\Spec(A)\) one cannot recover \(x\) but only the radical ideal that it generates.

Conversely, it can be shown that any spectral space is homeomorphic to that of a commutative ring (Hochster's theorem).See: M. Hochster, Prime ideal structure in commutative rings. Transactions of the American Mathematical Society. 142. (1969), 43-60. By applying Stone's theorem we deduce that every distributive lattice is isomorphic to the Zariski lattice of a commutative ring.

When \(A\) is a Boolean ring, the ring \(A\) itself becomes a distributive lattice (Boolean algebra) whose prime ideals obviously coincide with those of \(A\) as a ring, from which we deduce that \(\Spec(A)\) is Haussdorf. More generally, commutative rings \(A\) with Haussdorf spectral space are those whose quotient by the nilradical is what Bourbaki calls an absolutely flat ring, which amounts to saying that for all \(x\in A\) there exists \(a\in A\) such that \(x-ax^2\) is nilpotent.See: N. Bourbaki, Éléments de mathématique : Algèbre commutative. Chap. 2. § 3. ex. 16. Hermann, Paris (1961). This condition is therefore equivalent to the lattice \(\zar(A)\) being a Boolean algebra.

Instead of defining \(\Spec(T)\) using the prime ideals of the distributive lattice \(T,\) we could have proceeded in a dual way (by considering the opposite order) from the filters: we then obtain an isomorphism of \(T\) on the lattice of closed sets of compact complement of a spectral space.

If the point of view of ideals illustrates the analogy with commutative algebra, that of filters makes it possible to make the link with logic: filters appear naturally in a minimal logical context based on the only connector of implication. To see it while remaining within a very elementary framework, let us consider a logical system, made up of a set \(E\) (of propositions, or formulas) equipped with a composition law \(\f\) (the implication), and a set \(A\) (of axioms). A formula is said to be provable if it admits a proof using as axioms the elements of \(A\) and as deduction rule the modus ponens rule: from \(p\) and \(p\f q\) one can deduce \(q.\) In other words, the provable formulas are the elements of the theory generated by \(A,\) a theory being defined as a subset of \(E\) closed under modus ponens. We can then define irreducible theories, and by suitably choosing the set of axioms we get a Krull-like theorem (any theory that contains \(A\) is the intersection of the irreducible theories which contain it) as well as a spectral topology on the set \(\Spec_A(E)\) of irreducible theories containing the set of axioms \(A.\)

The condition required on the set \(A\) is that it contains at least Hilbert's axioms of implication in intuitionistic logic, i.e. formulas of the form \[p\f (q\f p)\] as well as those of the form \[(p\f(q\f r))\f((p\f q)\f(p\f r))\,.\] We then move on to classical logic by adding Pierce's law to the set of axioms, that is the set of formulas of the form \[((p\f q)\f p)\f p\,,\] which allows all non-constructive reasoning (reductio ad absurdum, excluded middle, etc.) If the set of axioms \(A\) contains Pierce's law, the topological space \(\Spec_A(E)\) becomes Haussdorf (and vice versa) and the irreducible theories containing \(A\) become maximal (which actually illustrates in an algebraic way the fact that classical logic only needs two truth values: according to Krull's theorem the class of logical equivalence of a formula then only depends on all maximal theories that contain it, and each quotient by a maximal theory has two elements). The moral of this story is that passing from intuitionist to classical logic is characterized by the spectral topology becoming Haussdorf.

In the presence of enough other connectives and adequate axioms, \(\Spec_A(E)\) becomes a genuine spectral space and the set \(E\) quotiented by logical (intuitionistic) equivalence, becomes a distributive lattice \(H,\) and even a Heyting algebra A Heyting algebra is a lattice \(H\) equipped with a composition law \(\f\) satifying, for any \(x,y,z\in H,\) the condition \[x\wedge y≤z\ssi x≤y\f z.\] (intermediate structure between distributive lattices and Boolean algebras). In the quotient structure the theories become the filters of the distributive lattice \(H,\) and \(\Spec_A(E)\) is homeomorphic to \(\Spec(H^{op}).\) In the presence of Pierce's law, the Heyting's algebra \(H\) becomes a Boolean algebra and we recover \(\Spec_A(E)\) being Haussdorf.

As already mentioned, reading P. Johnstone's book shows how far Stone was ahead of his time: Stone's article was published in 1937, Zariski's in 1950, and even then Zariski only considered the case of algebraic varieties.According to Johnstone the idea of a topology over the prime ideals of any commutative ring is due independently to Jacobson (1945) and Grothendieck (1960) to whom we owe the name Zariski Topology.

The reader wishing to go further may consult:

• Johnstone's essential work already mentioned.P. Johnstone, Stone Spaces, Cambridge University Press (1986).
• An excellent paper to make the link with abstract Galois theory and Galois connections. M. Erné, J. Koslowski, A. Melton, G. E. Strecker A primer on Galois connections, Ann. N.Y. Acad. Sci 704, p. 103-125 (1993).
• Stone Duality on Wikipedia.
• Stone Duality on \(n\)Lab.