TITLE: The Stone-Čech compactification of a space by the maximal ideals of the ring of bounded continuous functions from the space to $\mathbb{R}$
QUESTION [7 upvotes]: There is a claim that for any completely regular space, the maximal ideals of the ring of bounded continuous functions from $X$ to $\mathbb{R}$ forms the Stone-Čech compactification of $X$. How is the topological structure of the compactification constructed in this case?

REPLY [5 votes]: Here's the outline of the proof that was taught to me “a few” years ago. Note that much more was proved: for a completely regular Hausdorff space $X$, one can characterize all the compactifications of $X$ by means of subalgebras of $C_b(X)$.

The Zariski topology on the set $\operatorname{Spec}(A)$ of the prime ideals of the commutative ring $A$ has as closed sets those of the form
$$
V(E)=\{P\in\operatorname{Spec}(A): P\supseteq E\}
$$
where $E\subseteq A$. With $\operatorname{Max}(A)$ we denote the set of maximal ideals of $A$. On $\operatorname{Spec}(A)$ and $\operatorname{Max}(A)$ we'll always consider the Zariski topology.
With $C(X)$ and $C_b(X)$ we denote the $\mathbb{R}$-algebras of the continuous (respectively bounded continuous) real functions on $X$.
Fact 1 The condition $\operatorname{Spec}(A)=\operatorname{Max}(A)$ is equivalent to $\operatorname{Spec}(A)$ being Hausdorff.
Fact 2 If $X$ is a topological space and $A$ is a subring of $C(X)$ (continuous functions from $X$ to $\mathbb{R}$) such that $A$ contains the constant functions and is closed with respect to “bounded inversion”, then $\operatorname{Spec}(A)$ is Hausdorff.
Closure with respect to bounded inversion means that if $f\in A$ and $|f(x)|\ge k$, for some $k>0$ and all $x\in X$, then $1/f\in A$.
Fact 3 If $X$ is a topological space, then $\operatorname{Max}(C(X))$ is Hausdorff and compact.
Fact 4 Let $A$ be a subring of $C(X)$ containing the constants and closed with respect to bounded inversion. Then the map $\vartheta^A\colon X\to\operatorname{Max}(A)$ defined by $\vartheta^A(x)=\{f\in A:f(x)=0\}$ is continuous with dense image; moreover it is injective if and only if $A$ separates points of $X$, that is, for $x,y\in X$, $x\ne y$, there is $f\in A$ such that $f(x)\ne f(y)$.
Then we can prove
Stone's theorem Let $X$ and $Y$ be compact Hausdorff spaces; then $X$ is homeomorphic to $Y$ if and only if $C(X)$ is isomorphic as a ring to $C(Y)$.
Compactifications
If $X$ is a completely regular Hausdorff space and $T$ is a compactification of $X$, then the induced map $C(T)\to C(X)$ (restriction to $X$) is injective. Let $A(T)$ be the image of the map; then two compactification $T_1$ and $T_2$ of $X$ are equivalent (that is, homeomorphic via a map that fixes $X$) if and only if $A(T_1)=A(T_2)$.
The compactifications of $X$ are in one to one correspondence with the subalgebras of $C_b(X)$ that contain the constants and separate points and closed sets (if $x\in X$ and $C\subseteq X$ is closed, with $x\notin C$, then there is $f$ in the subalgebra such that $f(C)=0$ and $f(x)\ne0$). The inverse map is just $A\mapsto\operatorname{Max}(A)$.
The order relation on (equivalence classes of) compactifications corresponds to inclusion of the associated subalgebras.
In particular $\operatorname{Max}(C_b(X))$ is the maximum compactification of $X$, that is, the Stone-Čech compactification $\beta X$. One can show also that $\beta X$ is homeomorphic to $\operatorname{Max}(C(X))$.