Is Stone-Čech compactification of 0-dimensional space also 0-dimensional? What is an example of a 0-dimensional locally compact Hausdorff space $X$ for which the Stone-Čech compactification $\beta(X)$ is not 0-dimensional?
It is known that if $X$ is a 0-dimensional locally compact Hausdorff space which is also paracompact, then $\beta(X)$ is 0-dimensional. (Engelking 1989, Th. 6.2.9). I would expect a counterexample in the non-paracompact case.
Another way of asking the question is to look at a Boolean ring L (without assuming a unit).  If X is the Stone space of L then X is a 0-dimensional locally compact Hausdorff space and L is isomorphic with the ring of compact open sets in X.  The Boolean algebra $\mathrm{Cl}(X)$ of closed-open subsets is clearly a Boolean algebra extension of L, and the Stone space of $\mathrm{Cl}(X)$ is a certain compactification of $X$.  I believe it is easy to see that this compactification of $X$ is just $\beta(X)$ if and only if $\beta(X)$ is 0-dimensional.
Thus the question can be expressed in the algebraic side of the duality.  Find conditions on the Boolean ring $L$ so that $\beta(X)$ is 0-dimensional, where $X$ is the Stone space of $L$.
I think this may be true if, for example, $L$ is a Boolean $\sigma$-ring.  My original question asks for an example showing this is not true without some additional conditions on $L$.
 A: Thanks to KP.  This is the class $\Psi$ of Gillman-Jerison 5I, but apparently Terasawa explained more some 16 years later. 
There is a little history here. As pointed out by Figa-Talamanca and Franklin, "Multipliers of Distributive Lattices", Indian J. Math. 12 (1970) p. 159, John Kelley asserted in his topology book (p. 169) that the Cech-Stone compactification is always zero-dimensional.  They attempt to correct this error by citing an unspecified example in the 1955 paper of Dowker, "Local Dimension of Normal Spaces", Quart. J. Math. 2 (1955) 101-120, and they say this example was pointed out to them by J. Isbell and P. Dwinger.  However, I have not yet been able to verify that any of the non-strongly zero-dimensional examples in Dowker's paper are locally compact.  I wonder which one they had in mind?  If they were right, this would be a much earlier example than Terasawa's $\Psi$.  Could all four of them been mis-reading Dowker?
Here is a link to my copy of the Dowker paper. Does it contain a counterexample?
A: In this paper (Spaces $N\cup\mathscr{R}$ and their dimensions, Topol. Appl.
11(1) 1980 93-102) — I hope the PDF is freely available) Jun Terasawa constructs maximal almost disjoint families on $\mathbb{N}$ whose associated spaces can have any dimension you want. Given an almost disjoint family $\mathcal{A}$ on $\mathbb{N}$ one defines a topology on the union $\mathbb{N}\cup\mathcal{A}$ by declaring each natural number to be isolated and giving each $A\in\mathcal{A}$ a countable local base by putting $$U(A,n)=\lbrace A\rbrace \cup \lbrace i\in A:i\ge n\rbrace$$  for each $n$. This space is locally compact and zero-dimensional but Terasawa could arrange it so that its Čech-Stone Compactification would contain the Hilbert cube, for example.
Dowker's example does have a locally compact version: 
Let us retain the notation of Dowker's paper and use $T$ to denote the
set of countable ordinals and $Q_\alpha$ to denote the $\alpha$th
congruence class as chosen by Dowker.
In addition let $\mathbb{A}$ denote Alexandroff's double arrow space.
This is the product $[0,1]\times\lbrace0,1\rbrace$, ordered lexicographically
and endowed with its order topology and with the two isolated points
$\langle 0,0\rangle$ and $\langle 1,1\rangle$ deleted.
Now consider the product $T\times \mathbb{A}$ and define a quotient
space $X$ by identifying the points 
$\langle \alpha,x,0\rangle$ and $\langle \alpha,x,1\rangle$ 
whenever $x\notin\bigcup_{\beta\ge\alpha}Q_\beta$.
It is elementary to verify that this is an upper semicontinuous decomposition
and that the resulting space is locally compact and zero-dimensional.
The key observation is that for every $\alpha$ the product 
$T_\alpha\times\mathbb{A}$ is compact and open in the domain and its
image is compact and open in $X$.
Furthermore, arguments similar to those given by Dowker will show that
each finite open cover of $T\times\mathbb{A}$ has a refinement of the
form $\mathcal{U}\cup\mathcal{V}$, where $\mathcal{U}$ is a disjoint
open cover of $T_\alpha\times\mathbb{A}$ for some $\alpha$
and $\mathcal{V}$ consists of finitely many sets of the 
form $T\times C$, where $C$ is a clopen interval in $\mathbb{A}$.
The latter can then be used, just as for Dowker's $M$, to show that the bottom
and top lines in $X$ cannot be separated by clopen sets and hence
that $\beta X$ is not zero-dimensional. 
