TITLE: Stone-Čech compactification of completely regular space
QUESTION [7 upvotes]: Suppose $X$ is a completely regular space. Let $M$ be the set of nonzero algebra homomorphisms from $BC(X,\mathbb{R})$ to $\mathbb{R}$, equipped with the topology of pointwise convergence. Show that $M$ is homeomorphic to $\beta X$ (the Stone-Čech compactification of $X$).
I've noted that if $x\in X$, then $\hat{x}(f)=f(x)$ defines $\hat{x}\in M$.
There was also another problem which I've done where $X$ was compact Hausdorff and $M'$ was the set of nonzero algebra homomorphisms from $C(X,\mathbb{R})$ to $\mathbb{R}$. In that case, $M'$ is homeomorphic to $X$ via $x\mapsto\hat{x}$.
My thoughts about this problem are as follows:
(1) Bounded continuous functions from $X$ to $\mathbb{R}$ have unique continuous extensions to $\beta X$. Perhaps this enables identification of $C(\beta X,\mathbb{R})$ with $BC(X,\mathbb{R})$, and thus identification of $M$ with the set of nonzero algebra homomorphisms from $C(\beta X,\mathbb{R})$ to $\mathbb{R}$. 
(2) The set of nonzero algebra homomorphisms from $C(\beta X,\mathbb{R})$ to $\mathbb{R}$ is homeomorphic to $\beta X$.
If (1) is right, then together with (2) it should give the result.

REPLY [2 votes]: It seems you are essentially done. 
$\beta : C_b(X) \rightarrow C(\beta X)$ is an isometry:
$\beta(f + g)$ = $\beta f + \beta g$ since both sides extend $f+g$ to $\beta X$.
Similarly $\alpha \beta f = \beta ( \alpha f)$, $(\beta f) (\beta g) = \beta(fg)$ so $\beta$ is an algebra homomorphism. 
Restriction to $X$ is an algebra homomorphism $C(\beta X) \rightarrow C_b(X)$ which is inverse to $\beta$, so $\beta$ is an isomorphism.
The uniform norm of $\beta f$ is the same as that of $f$ by density of $X$ in $\beta X$, so $\beta$ is an isometry and thus $M$ can be identified with the set of nonzero continuous algebra homomorphisms from $C(\beta X)$ to $\mathbb{R}$  (it should not be difficult to check that the identification is continuous with respect to the topology of pointwise convergence.)
We can now finish with your remark (2).