Jtdylktuy

Let $X$ be a compact, totally disconnected space.

It must be shown that $X$ is homeomorphic to $mathrm{Spec}(A)$, where
$$ A:=C(X, mathbb{Z}_2), $$
and $mathbb{Z}_2$ is endowed with the discrete topology.

At first, it is easy to see that $A={ chi_S: S; mathrm{is,open,and,closed,in;}X }$ (no matter how ugly or cool $X$ is)[further, such $A$ is a boolean ring]

And, naturally, there is a correspondence $${ chi_S: S; mathrm{is,open,and,closed,in;}X } rightleftarrows { Ssubseteq X: S; mathrm{is,open,and,closed,in;}X }=:Gamma_X $$

I think I've read somewhere (sorry, I forgot where), that if $B$ is a boolean ring, then $B$ is isomorphic to $Gamma_{mathrm{Spec}(B)}$. Let us denote this result by $(*)$.

I think that, under the assumption of $(*)$, $Acong Gamma_{mathrm{Spec}(A)}$, but if $X$ is totally disconnected, then $Gamma_{mathrm{Spec}(A)}=mathrm{Spec}(A)$, is this right ?

Also, $X$ is in a one-to-one correspondence with ${ chi_{{x } }: xin X }$, which gives us
$$ Xcong AcongGamma_{mathrm{Spec}(A)}=mathrm{Spec}(A). $$

I've not checked continuity yet, which shouldn't be too complicated. I'm stuck at this point because $Acong mathrm{Spec}(A)$ shocks me.

Can you help me, please?
Is my reasoning right?

I consider directly $Y=operatorname{Spec}(A)$: because $A$ is a Boolean $mathbb{Z}_2$-algebra, the unique clopen subsets of $Y$ are
begin{equation*}
Y_f={mathfrak{p}in Ymid fnotinmathfrak{p}},,fin A
end{equation*}
(cfr. [AM] chapter 1, exercises 23.(i) and 23.(iii)); every prime ideal $mathfrak{p}$ of $A$ is maximal and $A_{displaystyle/mathfrak{p}}congmathbb{Z}_2$ (cfr. [AM] chapter 1, exercise 11.(ii)).

Because $A$ is the ring of regular morphisms on $Y$, that is
begin{gather*}
forall fin A,,f^{*}:Ytocoprod_{mathfrak{q}in Y} A_{displaystyle/mathfrak{q}},\
f^{*}(mathfrak{p})=[f]_{mathfrak{p}}in A_{displaystyle/mathfrak{p}}congmathbb{Z}_2,
end{gather*}
then
begin{equation*}
Y_f={mathfrak{p}in Ymid f^{*}(mathfrak{p})=1inmathbb{Z}_2}.
end{equation*}
Analogously, one can define
begin{equation*}
forall fin A,,X_f={xin Xmid f(x)=1inmathbb{Z}_2};
end{equation*}
easily one can to show that ${X_fsubseteqq X}_{fin A}$ is an open covering of $X$ and
begin{equation*}
forall f,gin A,,X_fcap X_g=X_{fg},
end{equation*}
that is ${X_fsubseteqq X}_{fin A}$ is a topological base of $X$.

It is an exercise to show that
begin{equation*}
forall xin X,,mathfrak{p}_x={fin Amid f(x)=0inmathbb{Z}_2}
end{equation*}
is a prime ideal of $A$; because $X$ is totally disconnected, it is a Fréchet space and
begin{equation*}
forall xneq yin X,,exists fneq gin Amid xin X_f,yin X_g,xnotin X_g,ynotin X_fRightarrowmathfrak{p}_xneqmathfrak{p}_yin Y.
end{equation*}
By previous reasoning, the map
begin{equation*}
varphi:xin Xtomathfrak{p}_xin Y
end{equation*}
is injective and continuous because:
begin{equation*}
forall fin A,,varphi^{-1}(Y_f)=X_f.
end{equation*}
Because $Y$ with Zariski topology is a compact Hausdorff space (cfr. [AM], chapter 1, exercise 23.(iv)) and, in general, $X$ is not Hausdorff: one can not prove that $varphi$ is a homeomorphism.

[AM] M. F. Atiyah, I. G. MacDonald - Introduction to Commutative Algebra, Addison-Wesley Publishing Company (1969)