Commutative Algebra and Algebraic Geometry (5): Spectrum of Boolean ring and Stone Duality.

Boolean Ring and its Spectrum.Basic property and New definition of Boolean Ring.Equivalence between Boolean Algebra and Boolean Ring.Spectrum of Boolean Ring.Stone Duality.$\mathrm{Stone}\circ \mathrm{Spec}\cong {\mathrm{Id}}_{\mathrm{Bool}}$$\mathrm{Spec}\circ \mathrm{Stone}\cong {\mathrm{Id}}_{\mathrm{BoolSp}}.$Definition of ${\eta }_S$ ${\eta }_S$ is injective${\eta }_S$ is surjective${\eta }_S$ is continuous${\eta }_S^{-1}$ is continuousThe naturalnessCorollary: Finite Boolean Algebra representation theoremA really interesting example.

Some basic Lattice theory.

Let $(L,\vee ,\wedge )$$(L,\vee,\wedge)$ be a lattice, the order will defined as follows

$$\begin{array}{}\text{(1)}& a\wedge b=a⟺a\le b⟺a\vee b=b\end{array}$$

Boolean Ring and its Spectrum.

Basic property and New definition of Boolean Ring.

Classic Definition. Let $R$$R$ be a ring. $R$$R$ is called a Boolean Ring if $\mathrm{\forall }a\in R,a^2=a$$\forall a\in R,a^2=a$​.

Proposition. Let $R$$R$ be a Boolean Ring, $\mathrm{Char}(R)=2$$\mathrm{Char} (R)=2$. i.e. $a+a=0\mathrm{\forall }a\in R$$a+a=0\forall a\in R$.

Proof. $x+x=(x+x{)}^2=x^2+x^2+x^2+x^2=x+x+x+x$$x+x=(x+x)^2=x^2+x^2+x^2+x^2=x+x+x+x$, hence $x+x=0$$x+x=0$. $◻$$\square$

Proposition. Let $R$$R$ be a Boolean Ring, then $R$$R$ is comuutative.

Proof. $x+y=(x+y{)}^2=x^2+xy+yx+y^2=x+xy+yx+y^2$$x+y=(x+y)^2=x^2+xy+yx+y^2=x+xy+yx+y^2$. Hence $xy+yx=0.◻$$xy+yx=0.\quad\square$​​​

I do not like the old definition, so I will give a equivalent definition. We know that every ring $R$$R$ is a subring of a ${\mathrm{End}}_{\mathrm{Ab}}(R,+)$$\mathrm{End}_{\mathrm{Ab}}(R,+)$​, if we define the embedding map as: $\iota :a⟼f_a$$\iota:a\longmapsto f_a$, where $f_a(x)=a\cdot x$$f_a(x)= a\cdot x$ for $a,x\in R$$a,x\in R$. $f_a\in {\mathrm{End}}_{\mathrm{Ab}}(R,+)$$f_a\in \mathrm{End}_{\mathrm{Ab}}(R,+)$ since

$$\begin{array}{}\text{(2)}& f_a(x+y)=a(x+y)=ax+ay=f_a(x)+f_a(y)\end{array}$$

But for a Boolean Ring, $f_a$$f_a$ is not only a group homomorphism, it is a ring homomorphism!

That is the special of Boolean Ring. This lead to the new definition.

My Definition. Let $R$$R$ be a ring. $R$$R$ is a Boolean ring iff R is a subring of ${\mathrm{End}}_{\mathrm{Ring}(R)}\text{ via }\iota :a⟼f_a$$\mathrm{End}_{\mathrm{Ring}(R)}\text{ via }\iota:a\longmapsto f_a$.

Proof.

$⟹$$\Longrightarrow$ If $a^2=a$$a^2=a$, then $f_a(xy)=axy=a^{2}xy=axay=f_a(x)f_a(y)$$f_a(xy)=axy=a^2xy=axay=f_a(x)f_a(y)$.

$⟸$$\Longleftarrow$ If $f_a\in {\mathrm{End}}_{\mathrm{Ring}(R)}$$f_a\in\mathrm{End}_{\mathrm{Ring}(R)}$, then $f_a(xy)=a(x)a(y)=a(xy)$$f_a(xy)=a(x)a(y)=a(xy)$. Let $x=y=1$$x=y=1$ we get $a^2=a◻$$a^2=a\quad\square$​

Proposition. The subobject and quotient object exists in $\mathrm{Bool}$$\mathrm{Bool}$. i.e. A subring of Boolean Ring is Boolean Ring, the quotient ring of Boolean Ring is Boolean Ring.

Proof. It is clear that a subring $S\subseteq R$$S\subseteq R$ is a Boolean ring as well since $s^2=s$$s^2=s$ for all $s\in S$$s\in S$. Let $I$$I$ be an ideal of $R$$R$,

let $\pi :R\to R/I$$\pi:R\to R/I$ be the quotient map, then $\pi (a)\pi (a)=\pi (a^2)=\pi (a)$$\pi(a)\pi (a)=\pi(a^2)=\pi (a)$. Hence Boolean Ring agagin. $◻$$\square$

Proposition. Let $R$$R$ be a Boolean ring, then every prime ideal $\mathfrak{p}$$\mathfrak p$ is maximal and $R/\mathfrak{p}\cong {\mathbb{F}}_2$$R/\mathfrak p\cong\mathbb F_2$.

Proof. Notice that $R/\mathfrak{p}$$R/\mathfrak p$ is Boolean integral domain.

$\mathrm{\forall }x\in R/\mathfrak{p},x(1-x)=x-x^2=x-x=0⟹x=0\vee x=1$$\forall x\in R/\mathfrak p,x(1-x)=x-x^2=x-x=0\implies x=0\vee x=1$. i.e. $R/\mathfrak{p}\cong {\mathbb{F}}_2$$R/\mathfrak p\cong\mathbb F_2$. $◻$$\square$

Equivalence between Boolean Algebra and Boolean Ring.

Let $B$$B$ be a Boolean Ring, then we can define $x\wedge y:=xy,x\vee y:=x+y+xy,\mathrm{\neg }x:=1+x$$x\wedge y:=xy,x\vee y:=x+y+xy,\neg x:=1+x$. Then you get a Boolean Algebra. Conversely, let $\mathfrak{B}$$\mathfrak B$ be a Boolean Algebra, then define $xy:=x\wedge y,x+y:=(x\vee y)\wedge (x\wedge y{)}^c$$xy:=x\wedge y,x+y:=(x\vee y)\wedge (x\wedge y)^c$​. Then you get a Boolean Ring back.

Proposition. Every finite generated ideal of a Boolean Ring is principal ideal.

Proof. Indeed, $(a_1,...,a_n)=(a_1\vee a_2\vee ...\vee a_n)$$(a_1,...,a_n)=(a_1\vee a_2\vee...\vee a_n)$​.

It is clear that $(a_1\vee a_2\vee ...\vee a_n)\subseteq (a_1,...,a_n)$$(a_1\vee a_2\vee...\vee a_n)\subseteq (a_1,...,a_n)$ since $a_1\vee ...\vee a_n$$a_1\vee...\vee a_n$ is a polynomial of $a_1,...,a_n$$a_1,...,a_n$.

Conversely, $a_i\le (a_1\vee ...\vee a_n)⟹a_i\wedge (a_1\vee ...\vee a_n)=a_i$$a_i\le(a_1\vee...\vee a_n)\implies a_i\wedge (a_1\vee...\vee a_n)=a_i$. Hence each $(a_i)\subseteq (a_1\vee ...\vee a_n).◻$$(a_i)\subseteq (a_1\vee...\vee a_n).\quad\square$

Examples of Boolean Algebra/Ring.

${\mathbb{F}}_2$$\mathbb F_2$ is a Boolean Ring.

The functor ${\mathrm{Hom}}_{\mathrm{Set}}(-,2):{\mathrm{Set}}^{\mathrm{op}}\to \mathrm{Bool}$$\mathrm{Hom}_{\mathrm{Set}}(-,2):\mathrm{Set^{op}}\to\mathrm{Bool}$ give us a kind of example of Boolean Ring.

The addition and product are defined pointwise.

Notice that this functor natural isomorphic to $P(-):{\mathrm{Set}}^{\mathrm{op}}\to \mathrm{Bool}$$P(-):\mathrm{Set^{op}}\to\mathrm{Bool}$, whcih maps $S\to (P(S),\mathrm{\Delta },\cap )$$S\to (P(S),\Delta,\cap)$. Here $\mathrm{\Delta }$$\Delta$ is symmetric difference. $A\mathrm{\Delta }B:=A\cup B-A\cap B$$A\Delta B:=A\cup B-A\cap B$​.

The Borel set functor also could gives us a functor from $\mathrm{Top}\to \mathrm{Bool}$$\mathrm{Top}\to\mathrm{Bool}$​​. See Math Essays: Category of Measurable space and relative functor (marco-yuze-zheng.blogspot.com).

Let $\mathcal{B}=\{e_1,...,e_n\}$$\mathcal B=\{e_1,...,e_n\}$ be a orthogonal basis over a inner product space $V$$V$. Then $P(\mathcal{B})$$P(\mathcal B)$ is a Boolean Algebra.

Also, for $U,V\subseteq \mathcal{B}$$U,V\subseteq\mathcal B$, we have

$$\begin{array}{}\text{(3)}& \mathrm{Span}(U\cup V)=\mathrm{Span}(U)+\mathrm{Span}(V),\mathrm{Span}(U\cap V)=\mathrm{Span}(U)\cap \mathrm{Span}(V),\mathrm{Span}(U^c)=\mathrm{Span}(U{)}^{\mathrm{\perp }}\end{array}$$

The span is injective hence the image of span is a Boolean Algebra.

Define the Stone functor $\mathrm{Stone}:{\mathrm{Top}}^{\mathrm{op}}\to \mathrm{Bool}$$\mathrm{Stone}:\mathrm{Top}^{\mathrm{op}}\to\mathrm{Bool}$ as follows.

For a topological space $(X,\tau )$$(X,\tau)$, $\mathrm{Stone}(X,\tau )$$\mathrm{Stone}(X,\tau)$ is the clopen sets in $\tau$$\tau$ with usual complement.

For a continious function $f:(X,\tau )\to (Y,{\tau }^{\prime })$$f:(X,\tau)\to (Y,\tau')$, $\mathrm{Stone}(f)=f^{-1}$$\mathrm{Stone}(f)=f^{-1}$​.

Spectrum of Boolean Ring.

Let $\mathfrak{B}$$\mathfrak B$ be a Boolean Ring, $X=\mathrm{Spec}(\mathfrak{B})$$X=\mathrm{Spec}(\mathfrak B)$.

Proposition. Every principal open set $D(f)$$D(f)$​ is clopen.

Proof. By previous proposition, we know that $R/\mathfrak{p}\cong {\mathbb{F}}_2$$R/\mathfrak p\cong \mathbb F_2$. Hence $f(x)=0⟺(1+f)(x)\ne 0$$f(x)=0\iff (1+f)(x)\ne 0$.

Hence we get that $V(f)=D(1+f)$$V(f)=D(1+f)$ and $D(f)=V(1+f)$$D(f)=V(1+f)$. In other word, $D(f)=X-D(1+f)$$D(f)=X-D(1+f)$ $◻$$\square$

Proposition. Finite union of principal open set in $X$$X$​ is still a principal open set.

$$\begin{array}{}\text{(4)}& \bigcup _{i=1}^{n}D(f_i)=\bigcup _{i=1}^n(X-D(1+f_i))=X-\bigcap _{i=1}^{n}D(1+f_i)=X-D(\prod _{i=1}^n(1+f_i))=D(1+\prod _{i=1}^n(1+f_i))◻\end{array}$$

Remark. There exists a connection with inclusion-exclusion theory when we consider the sigma algebra. We do something similar when we prove inclusion-exclusion theory. But the difference is the value each characteristic function is in $\mathbb{R}$$\mathbb R$, not ${\mathbb{F}}_2$$\mathbb F_2$. Hence $\pm$$\pm$ matter.

Math Essays: Measure space and inclusion-exclusion Theorem (marco-yuze-zheng.blogspot.com)

Proposition. The only clopen sets in $X$$X$​ is those principal open set.

You need to farmilar with point set topology.

Proof. We already know that for any ring $R$$R$, $\mathrm{Spec}(R)$$\mathrm{Spec}(R)$ is quasi compact, hence the closed subset is quasi compact as well.

Since each clopen set is open as well, so it could be write as finite union of $D(f_i)$$D(f_i)$. But finite union of principal open set is still principal open set. $◻$$\square$

This proposition tells us that if we take the Stone functor at a Spectrum of Boolean Ring, then it will give us those principal open sets.

Proposition. $X$$X$​ is a Compact Hausdorff space.

Proof. We already know that spectrum of a ring is compact. Now let us prove Hausdorff property.

For $x\ne y$$x\ne y$, pick $f\in {\mathfrak{p}}_x-{\mathfrak{p}}_y$$f\in\mathfrak p_x-\mathfrak p_y$, then $D(f)$$D(f)$ and $D(1+f)$$D(1+f)$ separate $x,y$$x,y$. $◻$$\square$

Stone Duality.

Definition. Let $X$$X$ be a Compact Hausdorff space. We call $X$$X$ Boolean space if the clopen set form a topological basis.

Denote the Category of Boolean Space as $\mathrm{BoolSp}$$\mathrm{BoolSp}$. It is clear that $\mathrm{Spec}:{\mathrm{Bool}}^{\mathrm{op}}\to \mathrm{BoolSp}$$\mathrm{Spec}:\mathrm{Bool}^{\mathrm{op}}\to\mathrm{BoolSp}$.

Proposition. ${\mathrm{Bool}}^{\mathrm{op}}\cong \mathrm{BoolSp}.$$\mathrm{Bool^{op}}\cong\mathrm{BoolSp}.$

Proof. Let us prove $\mathrm{Stone}\circ \mathrm{Spec}\cong {\mathrm{Id}}_{\mathrm{Bool}}$$\mathrm{Stone}\circ\mathrm{Spec}\cong\mathrm{Id_{Bool}}$​ first.

$\mathrm{Stone}\circ \mathrm{Spec}\cong {\mathrm{Id}}_{\mathrm{Bool}}$$\mathrm{Stone}\circ\mathrm{Spec}\cong\mathrm{Id_{Bool}}$

Let $h:A\to B$$h:A\to B$ be a Boolean Algebra homomorphism.

$$\begin{array}{}\text{(5)}& \begin{array}{c}\begin{array}{ccc}A& \stackrel{{\eta }_A}{\to }& \mathrm{Stone}\circ \mathrm{Spec}(A)\\ h\downarrow & & \downarrow \mathrm{Stone}\circ \mathrm{Spec}(h)& \\ B& \stackrel{{\eta }_B}{\to }& \mathrm{Stone}\circ \mathrm{Spec}(B)\end{array}\end{array}\end{array}$$

Let us check each $\eta$$\eta$ are isomorphism.

Consider the corresponding Boolean Ring of $A$$A$ and its spectrum, we know the only clopen set is those principal open set. Hence $\mathrm{Stone}\circ \mathrm{Spec}$$\mathrm{Stone\circ\mathrm{Spec}}$ gives us each $D(f)$$D(f)$. Notice that $D(f)\subseteq D(g)⟺V(f)\supseteq V(g)⟺\sqrt{(f)}\subseteq \sqrt{(g)}$$D(f)\subseteq D(g)\iff V(f)\supseteq V(g)\iff \sqrt{(f)}\subseteq\sqrt{(g)}$. But this is Boolean Ring, hence $\mathrm{\forall }a\in A,a^2=a⟹(f)=\sqrt{(f)},(g)=\sqrt{(g)}.$$\forall a\in A,a^2=a\implies (f)=\sqrt{(f)},(g)=\sqrt{(g)}.$ i.e.

$$\begin{array}{}\text{(6)}& D(f)\subseteq D(g)⟺(f)\subseteq (g)⟺f\in (g)⟺\mathrm{\exists }a\in A,f=ag,fg=agg=ag=f⟺f\wedge g=f\end{array}$$

Hence

$$\begin{array}{}\text{(7)}& D(f)\subseteq D(g)⟺f\le g\end{array}$$

i.e.

$$\begin{array}{}\text{(8)}& {\eta }_X:X\cong \mathrm{Stone}\circ \mathrm{Spec}(\mathrm{X}),{\eta }_X(f)=D(f)\end{array}$$

For the morphism $h$$h$, we can see that $\mathrm{Stone}\circ \mathrm{Spec}(h)={(}^{a}h{)}^{-1}$$\mathrm{Stone}\circ\mathrm{Spec}(h)=(^ah)^{-1}$, we know that ${(}^{a}h{)}^{-1}(V(f))=V(h(f))$$(^ah)^{-1}(V(f))=V(h(f))$​.

Hence it maps $V(f)$$V(f)$ to $V(h(f))$$V(h(f))$, i.e. $D(f)$$D(f)$ to $D(h(f))$$D(h(f))$.

$$\begin{array}{}\text{(9)}& {\eta }_B^{-1}\circ \mathrm{Stone}\circ \mathrm{Spec}(h)\circ {\eta }_A(f)=h(f)\end{array}$$

Hence we proved that $\mathrm{Stone}\circ \mathrm{Spec}\cong {\mathrm{Id}}_{\mathrm{Bool}}$$\mathrm{Stone}\circ\mathrm{Spec}\cong\mathrm{Id_{Bool}}$​.

$\mathrm{Spec}\circ \mathrm{Stone}\cong {\mathrm{Id}}_{\mathrm{BoolSp}}.$$\mathrm{Spec}\circ\mathrm{Stone}\cong\mathrm{Id}_{\mathrm{BoolSp}}.$

Now let us prove $\mathrm{Spec}\circ \mathrm{Stone}\cong {\mathrm{Id}}_{\mathrm{BoolSp}}.$$\mathrm{Spec}\circ\mathrm{Stone}\cong\mathrm{Id}_{\mathrm{BoolSp}}.$​

Let $f:X\to Y$$f:X\to Y$​ be a continuous function between two Boolean Spaces.

Then we need to prove the following natural isomorphism.

$$\begin{array}{}\text{(10)}& \begin{array}{c}\begin{array}{ccc}X& \stackrel{{\eta }_X}{\to }& \mathrm{Spec}\circ \mathrm{Stone}(X)\\ f\downarrow & & \downarrow \mathrm{Spec}\circ \mathrm{Stone}(f)& \\ Y& \stackrel{{\eta }_Y}{\to }& \mathrm{Spec}\circ \mathrm{Stone}(Y)\end{array}\end{array}\end{array}$$

Let us prove ${\eta }_S:S\cong \mathrm{Spec}\circ \mathrm{Stone}(S)$$\eta_{S}:S\cong \mathrm{Spec}\circ\mathrm{Stone}(S)$ first.

Take the clopen sets in $S$$S$ and we get a Boolean Algebra $\mathrm{Stone}(S)$$\mathrm{Stone}(S)$, then consider the spectrum of the Boolean Ring respect to $\mathrm{Stone}(S)$$\mathrm{Stone}(S)$. We need to prove that ${\eta }_S$$\eta_S$​ is a homeomorphism.

Definition of ${\eta }_S$$\eta_S$

Define ${\eta }_S$$\eta_S$ as follows, for $s\in S,{\eta }_X(s)=\{U\in \mathrm{Stone}(S):s\notin U\}$$s\in S,\eta_X(s)=\{U\in\mathrm{Stone}(S):s\notin U\}$. If you view $\mathrm{Stone}(S)$$\mathrm{Stone}(S)$ as a Boolean Ring (each elements as $f:S\to {\mathbb{F}}_2$$f:S\to\mathbb F_2$). Then $\eta (s)$$\eta(s)$ is the kernel of the evaluation map at $s$$s$, hence a prime ideal, or a point in $\mathrm{Spec}\circ \mathrm{Stone}(S)$$\mathrm{Spec}\circ\mathrm{Stone}(S)$.

We will check that ${\eta }_S$$\eta_S$ is injective and surjective, continuous, and finally, ${\eta }_S$$\eta_S$ is continuous as well.

${\eta }_S$$\eta_S$ is injective

Firstly, let us check that ${\eta }_S$$\eta_S$ is injective. Let $s\ne t$$s\ne t$, then there exists a clopen set $U_t$$U_t$ such that $s\notin U,t\in U$$s\notin U,t\in U$.

Hence $U\in {\eta }_S(s)$$U\in\eta_S(s)$ but $U\notin {\eta }_S(t)$$U\notin\eta_S(t)$​. This assert that ${\eta }_S(s)\ne {\eta }_S(t)$$\eta_S(s)\ne\eta_S(t)$, hence ${\eta }_S$$\eta_S$​ is injective.

${\eta }_S$$\eta_S$ is surjective

Now let us prove that ${\eta }_S$$\eta_S$ is surjective. i.e. For all ${\mathfrak{p}}_x\in \mathrm{Spec}\circ \mathrm{Stone}(S)$$\mathfrak p_x\in\mathrm{Spec\circ\mathrm{Stone}}(S)$, there exists a $x\in S$$x\in S$ such that ${\eta }_S(x)={\mathfrak{p}}_x$$\eta_S(x)=\mathfrak p_x$.

Define ${\displaystyle K=\bigcap \{U\in \mathrm{Stone}(\mathrm{S}):U\notin {\mathfrak{p}}_x\}}$$\displaystyle K=\bigcap{\{U\in\mathrm{Stone (S)}:U\notin\mathfrak p_x\}}$. Hence each $U({\mathfrak{p}}_x)=1$$U(\mathfrak p_x)=1$. If ${\mathfrak{p}}_x={\eta }_S(x)$$\mathfrak p_x=\eta_S(x)$, then $K$$K$ is the intercetion of all the clopen set that contain $x$$x$.

We need to prove that $K\ne \mathrm{\varnothing }$$K\ne\empty$, and indeed, $K=\{\ast \}$$K=\{*\}$​.

To prove $K\ne \mathrm{\varnothing }$$K\ne\empty$​​, we need a definition and a lemma.

Definition. Finite intercetion property.

Let $X$$X$ be a set and $\mathcal{A}$$\mathcal A$ a nonempty family of $X$$X$. Then $\mathcal{A}$$\mathcal A$ is said to have finite intercetion property if for any non empty finite subfamily has non empty intercetion.

Lemma.

Let $X$$X$ be a topological space and $F$$F$ be a family of closed subsets of $X$$X$.

$X$$X$ is compact $⟺$$\iff$ For any closed set family $\mathcal{F}$$\mathcal F$ with finite intercetion property(FIP), $\bigcap _{V\in \mathcal{F}}V\ne \mathrm{\varnothing }$$\bigcap_{V\in\mathcal F} V\ne\empty$.

Proof.

Suppose $X$$X$ is compact, and closed set family $\mathcal{F}$$\mathcal F$ with finite intercetion property, we want to prove $\bigcap _{V\in \mathcal{F}}V\ne \mathrm{\varnothing }$$\bigcap_{V\in\mathcal F} V\ne\empty$.

Assume ${\displaystyle \bigcap _{V\in \mathcal{F}}V=\mathrm{\varnothing }}$$\displaystyle \bigcap_{V\in\mathcal F} V=\empty$, then $\{V^c:V\in \mathcal{F}\}$$\{V^c:V\in\mathcal F\}$ is a open covering of $X$$X$. Hence there exists a finite subcovering $V_1^c,...,V_n^c$$V_1^c,...,V^c_n$.

Hence ${\displaystyle \bigcap _{i=1}^n{V}_i=\mathrm{\varnothing }}$$\displaystyle\bigcap_{i=1}^n V_i=\empty$, that is a contradiction with $\mathcal{F}$$\mathcal F$ have FIP property.

Conversely, let assume for any closed set family $\mathcal{F}$$\mathcal F$ with finite intercetion property(FIP), $\bigcap _{V\in \mathcal{F}}V\ne \mathrm{\varnothing }$$\bigcap_{V\in\mathcal F} V\ne\empty$, and $U_{\alpha }$$U_{\alpha}$ is an open covering of $X$$X$. Let us consider the family of closed set $U_{\alpha }^c$$U_{\alpha}^c$. If any finite intercetion of elements in $U_{\alpha }^c$$U_{\alpha}^c$ is not empty, then by FIP, ${\displaystyle \bigcap U_{\alpha }^c\ne \mathrm{\varnothing }}$$\displaystyle\bigcap U_{\alpha}^c\ne \empty$, i.e. ${\displaystyle \bigcup U_{\alpha }\ne X}$$\displaystyle\bigcup U_{\alpha}\ne X$. That is a contradcition.

Hence there exists $U_1^c,...,U_n^c$$U_1^c,...,U_n^c$ such that ${\displaystyle \bigcap _{i=1}^n{U}_i^c=\mathrm{\varnothing }}$$\displaystyle \bigcap_{i=1}^n U_{i}^c=\empty$, i.e. ${\displaystyle \bigcup _{i=1}^n{U}_i=X}$$\displaystyle \bigcup_{i=1}^n U_i=X$, $X$$X$ is compact. $◻$$\square$

Now let us prove ${\displaystyle K=\bigcap \{U\in \mathrm{Stone}(\mathrm{S}):U\notin {\mathfrak{p}}_x\}\ne \mathrm{\varnothing }}$$\displaystyle K=\bigcap{\{U\in\mathrm{Stone (S)}:U\notin\mathfrak p_x\}} \ne\empty$.

We only need to prove that finite intercetion of $\{U\in \mathrm{Stone}(S):U\notin {\mathfrak{p}}_x\}$$\{U\in\mathrm{Stone}(S):U\notin\mathfrak p_x\}$ is not empty.

Suppose that ${\displaystyle \bigcap _{i=1}^n{U}_i=\mathrm{\varnothing }\in {\mathfrak{p}}_x}$$\displaystyle \bigcap_{i=1}^n U_i=\empty\in\mathfrak p_x$, then there exists $U_i\in {\mathfrak{p}}_x$$U_i\in\mathfrak p_x$ since $\cap$$\cap$ is product, $\mathrm{\varnothing }=0$$\empty=0$ and ${\mathfrak{p}}_x$$\mathfrak p_x$ is prime ideal.

That is a contradiction with $U_i\notin {\mathfrak{p}}_x$$U_i\notin\mathfrak p_x$. Hence $K\ne \mathrm{\varnothing }$$K\ne\empty$.

Let us prove that $K=\{\ast \}$$K=\{*\}$. Suppose that $x,y\in K$$x,y\in K$ with $x\ne y$$x\ne y$, then we could find $U,V$$U,V$ separate $x,y$$x,y$ since $X$$X$ is Hausdorff space. Since $X$$X$ is Boolean Space, we could find a clopen set $T$$T$ such that $x\in T\subseteq U$$x\in T\subseteq U$, hence $y\notin T$$y\notin T$. Hence $T\in {\mathfrak{p}}_x$$T\in\mathfrak p_x$ and $y\in T^c$$y\in T^c$. That is, $T^c\notin {\mathfrak{p}}_x$$T^c\notin\mathfrak p_x$ but $x\notin T^c$$x\notin T^c$. Hence $x\notin K$$x\notin K$, that is a contradiction. Therefore, $K=\{\ast \}$$K=\{*\}$.

Now let us prove ${\eta }_S(\ast )=\{U\in \mathrm{Stone}(X):\ast \notin U\}={\mathfrak{p}}_x$$\eta_S(*)=\{U\in\mathrm{Stone}(X):*\notin U\}=\mathfrak p_x$. That follows from $U\in {\eta }_S(\ast )⟺\ast \notin U⟺U\in {\mathfrak{p}}_x$$U\in\eta_S(*)\iff *\notin U\iff U\in\mathfrak p_x$

Hence we get that ${\eta }_S$$\eta_S$​ is surjective.

${\eta }_S$$\eta_S$ is continuous

Now let us prove that ${\eta }_S$$\eta_S$ is continuous. Since the principal open set $D(U)$$D(U)$ form a basis, we only need to check that the preimage of ${\eta }_S$$\eta_S$ of $D(U)$$D(U)$ is still open. Recall the definition: ${\eta }_S(s)=\{U\in \mathrm{Stone}(S):s\notin U\}$$\eta_S(s)=\{U\in\mathrm{Stone}(S):s\notin U\}$

$$\begin{array}{}\text{(11)}& {\eta }_S^{-1}(D(U))=\{x\in S|{\eta }_S(x)\in D(U)\}=\{x\in S|U\notin {\eta }_S(x)\}=\{x\in S|x\in U\}=U\end{array}$$

Since $U$$U$ is clopen set, hence ${\eta }_S$$\eta_S$ is continuous.

${\eta }_S^{-1}$$\eta_S^{-1}$ is continuous

We already know that ${\eta }_S$$\eta_S$ is a bijection and ${\eta }_S^{-1}(D(U))=U$$\eta_S^{-1}(D(U))=U$, hence ${\eta }_S^{-1}$$\eta_S^{-1}$ is a Boolean Algebra isomorphism. Hence ${\eta }_S$$\eta_S$ is a Boolean Algebra isomorphism, ${\eta }_S(U)=D(U)$$\eta_S(U)=D(U)$. Hence finaly, we prove that ${\eta }_S$$\eta_S$ is a homeomorphism. $◻$$\square$​

The naturalness

Now we need to check the naturalness.

$$\begin{array}{}\text{(12)}& \begin{array}{c}\begin{array}{ccc}X& \stackrel{{\eta }_X}{\to }& \mathrm{Spec}\circ \mathrm{Stone}(X)\\ f\downarrow & & \downarrow \mathrm{Spec}\circ \mathrm{Stone}(f)& \\ Y& \stackrel{{\eta }_Y}{\to }& \mathrm{Spec}\circ \mathrm{Stone}(Y)\end{array}\end{array}\end{array}$$

$$\begin{array}{}\text{(13)}& {\eta }_Y^{-1}\circ \mathrm{Spec}\circ \mathrm{Stone}(f)\circ {\eta }_X(x)={\eta }_Y^{-1}\circ \mathrm{Spec}\circ \mathrm{Stone}(f)({\mathfrak{p}}_x)={\eta }_Y^{-1}{\circ }^a(f^{-1})({\mathfrak{p}}_x)\end{array}$$

but

$$\begin{array}{rl}{}^a(f^{-1})({\mathfrak{p}}_x)& =\{V\in \mathrm{Stone}(Y)\mid f^{-1}(V)\in {\mathfrak{p}}_x\}\\ & =\{V\in \mathrm{Stone}(Y)\mid x\notin f^{-1}(V)\}\\ & =\{V\in \mathrm{Stone}(Y)\mid f(x)\notin V\}\\ & ={\mathfrak{p}}_{f(x)}\end{array}$$

and ${\eta }_Y^{-1}({\mathfrak{p}}_{f(x)})=f(x)$$\eta_Y^{-1}(\mathfrak p_{f(x)})=f(x)$

Hence

$$\begin{array}{}\text{(14)}& {\eta }_Y^{-1}\circ \mathrm{Spec}\circ \mathrm{Stone}(f)\circ {\eta }_X(x)={\eta }_Y^{-1}\circ \mathrm{Spec}\circ \mathrm{Stone}(f)({\mathfrak{p}}_x)=f(x)◻\end{array}$$

Corollary: Finite Boolean Algebra representation theorem

Lemma. Let $X$$X$ be a Hausdorff space, then singleton sets are closed.

Proof.

Let $x\in X$$x\in X$ and consider $X-\{x\}$$X-\{x\}$. Then for any $x^{\prime }\in X-\{x\}$$x'\in X-\{x\}$, we could find a open set $U_{x^{\prime }}$$U_{x'}$ such that $x^{\prime }\in U_{x^{\prime }},x\notin U_{x^{\prime }}$$x'\in U_{x'},x\notin U_{x'}$​.

Hence $\bigcup _{x^{\prime }\in X-\{x\}}{U}_{x^{\prime }}=X-\{x\}$$\bigcup_{x'\in X-\{x\}} U_{x'}=X-\{x\}$ is a open set, $\{x\}$$\{x\}$ is closed. $◻$$\square$

Lemma. Let $X$$X$ be a Compact Hausdorff space, then the following conditions are equivalent.

$†.X$$\dagger. X$ is finite.

$†.X$$\dagger.X$ is discrete, i.e. every subset of $X$$X$ is clopen.

Proof. If $X$$X$ is finite, since $X$$X$ is Hausdorff, each $\{x\}$$\{x\}$ is closed set, hence for any $U\subseteq X,U$$U\subseteq X,U$ is closed.

Let $X$$X$ be a discrete space, then the open covering ${\displaystyle \bigcup _{x\in X}\{x\}=X}$$\displaystyle \bigcup_{x\in X}\{x\}=X$ have a finite sub covering, hence $X$$X$ is finite.

Corollary. Every finite Boolean Algebra $\mathcal{B}$$\mathcal B$ isomorphic to $P(S)$$P(S)$ for some finite set $S$$S$.

Proof. By Stone duality, we know every $\mathcal{B}\cong \mathrm{Stone}\circ \mathrm{Spec}(\mathcal{B})$$\mathcal B\cong\mathrm{Stone\circ\mathrm{Spec}}(\mathcal B)$. Since $\mathrm{Spec}(\mathcal{B})$$\mathrm{Spec}(\mathcal B)$ is a finite Compact Hausdorff space, hence it is discrete. Therefore $\mathcal{B}\cong \mathrm{Stone}\circ \mathrm{Spec}(\mathcal{B})=P(\mathrm{Spec}(\mathcal{B}))$$\mathcal B\cong \mathrm{Stone\circ\mathrm{Spec(\mathcal B)}}=P(\mathrm{Spec}{(\mathcal B)})$. $◻$$\square$

Remark. The last result looks like every finite dimension vector space over $\mathbb{F}$$\mathbb F$ is isomorphic to ${\mathbb{F}}^n$$\mathbb F^n$ for some $n$$n$.

The connection is here: in fact,for every $|S|=n,P(S)\cong {\mathbb{F}}_2^n$$|S|=n,P(S)\cong\mathbb F_2^n$.

A really interesting example.

Let $m=p_1...p_n$$m=p_1...p_n$ be a square free number. In other word, $(m)$$(m)$ is a radical ideal.

The divisors of $m$$m$, denote as $M$$M$ could form a Boolean Algebra with the following operator:

$$\begin{array}{}\text{(15)}& a\vee b:=\mathrm{lcm}(a,b),a\wedge b:=gcd(a,b),a^{\prime }:=\frac{m}{a}.\end{array}$$

Whcih is isormorhic to $P(S),S=\{p_1,...,p_n\}$$P(S),S=\{p_1,...,p_n\}$, the isomorphism is given by $\mathrm{factor}:a⟼\{p_i\in S:p_i|a\}$$\mathrm{factor}:a\longmapsto\{p_i\in S:p_i|a\}$.

I already talking about something about it last year on my blog. Let us have a new point of view about that.

Notice that $(m)$$(m)$ is a radical ideal, hence $V(m)\subset \mathrm{Spec}(\mathbb{Z})$$V(m)\subset\mathrm{Spec}(\mathbb Z)$ is isomorphic to $\mathrm{Spec}(\mathbb{Z}/m\mathbb{Z})$$\mathrm{Spec}(\mathbb Z/m\mathbb Z)$.

Also, the Zariski topology on $V(m)$$V(m)$ or $\mathrm{Spec}(\mathbb{Z}/m\mathbb{Z})$$\mathrm{Spec}(\mathbb Z/m\mathbb Z)$​ is discrete topology, ${\displaystyle \mathrm{Spec}(\mathbb{Z}/m\mathbb{Z})\cong \coprod _{i=1}^n\mathrm{Spec}\mathbb{Z}/p_i\mathbb{Z}}$$\displaystyle\mathrm{Spec}(\mathbb Z/m\mathbb Z)\cong\coprod_{i=1}^n\mathrm{Spec}\mathbb Z/p_i\mathbb Z$​.

In this case, the prime ideal of the Boolean Ring coincide with the prime ideal of $\mathbb{Z}/m\mathbb{Z}$$\mathbb Z/m\mathbb Z$.

Hence again, you get a Boolean Algebra, which is isomorphic to $P(S)$$P(S)$.

We know that there is a one-one corresponding between radical ideal and closed set and

$$\begin{array}{}\text{(16)}& V(I+J)=V(I)\cap V(J),V(I\cap J)=V(I\cdot J)=V(I)\cup V(J)\end{array}$$

We get a Boolean Algebra on $(\sqrt{I},\cap ,+)$$(\sqrt I,\cap,+)$. The complement is given by $(m):(a)$$(m):(a)$, i.e. the quotient ideal.

Since let $Y,Z$$Y,Z$ be two closed set.

Then $I(Y-Z)=\{f\in R|f(x)=0\mathrm{\forall }x\in Y-Z\}=\{f\in R|f(x)g(x)=fg(x)=0\mathrm{\forall }x\in Y,g\in I(Z)\}$$I(Y-Z)=\{f\in R|f(x)=0\forall x\in Y-Z\}=\{f\in R|f(x)g(x)=fg(x)=0\forall x\in Y,g\in I(Z)\}$​.

But $\{f\in R|f(x)g(x)=fg(x)=0\mathrm{\forall }x\in Y,g\in I(Z)\}=\{f\in R|f\cdot I(Z)\subseteq I(Y)\}=I(Y):I(Z)$$\{f\in R|f(x)g(x)=fg(x)=0\forall x\in Y,g\in I(Z)\}= \{f\in R| f\cdot I(Z)\subseteq I(Y)\}=I(Y):I(Z)$.