Stone–Čech Compactification of $\mathbb{Z}$ with Fürstenberg Topology The Stone–Čech Compactification of $\mathbb{N}$ as a discrete space has been extensively studied and can be represented using ultrafilters.
Consider $X=(\mathbb{Z},\mathcal{T})$, where $\mathcal{T}$ is the Fürstenberg topology generated by arithmetic sequences. Equipped with this exotic topology, $X$ is a topological ring, metrizable, and totally disconnected. 
Since $X$ is metrizable, it is Tychonoff and the map from $X$ to its image in $\beta X$ (its compactification) is a homeomorphism. 
Has $\beta X$ been studied? Is there a straightforward description analogous to the compactification of $\mathbb{N}$ with the discrete topology? 
 A: Topologically speaking $\mathbb{Z}$ with the topology mentioned above is just (homeomorphic to) the space of rational numbers. The space $\beta\mathbb{Q}$ has been studied a lot (not as much as $\beta\mathbb{N}$), for example by Eric van Douwen in Remote Points (MR link), freely available here at DMLPL.
A: We can also describe $\beta(\mathbb{Z},\mathcal{T})$ in terms of ultrafilters on Boolean algebras. I claim that $\beta(\mathbb{Z},\mathcal{T})$ is the space of ultrafilters on the Boolean algebra of clopen sets in $(\mathbb{Z},\mathcal{T})$ where $\mathcal{T}$ is the Fürstenberg topology.
Recall that a space $X$ is zero-dimensional if it has a basis of clopen sets, and recall that a zero set on a space $X$ is a set of the form $f^{-1}(0)$ for some continuous $f:X\rightarrow\mathbb{R}$. A completely regular space $X$ is said to be strongly zero-dimensional if the Stone-Čech compactification $\beta X$ is zero-dimensional. It can be shown that a completely regular space $X$ is strongly zero-dimensional if and only if whenever $Z_{1},Z_{2}\subseteq X$ are disjoint zero sets, there is a clopen set $C\subseteq X$ with $Z_{1}\subseteq C,Z_{2}\subseteq C^{c}$ [1 p. 85]. In other words, a completely regular space is strongly zero-dimensional iff every pair of zero sets is separated by a clopen set. If $X$ is zero-dimensional, then let $\mathfrak{B}(X)$ denote the Boolean algebra of clopen subsets of $X$ and let $\zeta X$ be the space of ultrafilters on $\mathfrak{B}(X)$. Then $\zeta X$ is in a sense the maximal zero-dimensional compactification of $X$ which is called the Banaschewski compactification. If $X$ is strongly zero-dimensional, then the Banaschewski compactification $\zeta X$ is precisely the Stone-Čech compactification. In [1. p. 86] it states that zero-dimensionality and strong zero-dimensionality are equivalent in Lindelöf spaces. Therefore since $(\mathbb{Z},\mathcal{T})$ is zero-dimensional and Lindelöf, the space $(\mathbb{Z},\mathcal{T})$ is strongly zero-dimensional. We conclude that $\beta(\mathbb{Z},\mathcal{T})=\zeta(\mathbb{Z},\mathcal{T})$ is the space of ultrafilters on $\mathfrak{B}(\mathbb{Z},\mathcal{T})$.
In order to clear up some confusion about the space $(\mathbb{Z},\mathcal{T})$ and its Stone-Čech compactification, I will outline some basic facts about $(\mathbb{Z},\mathcal{T})$ and $\beta(\mathbb{Z},\mathcal{T})$.
I claim that the space $(\mathbb{Z},\mathcal{T})$ has an infinite partition into clopen sets. It is not too hard to give an explicit example of such a partition. For a more slick proof, assume that $(\mathbb{Z},\mathcal{T})$ has no partition into countably many clopen sets. If $\mathcal{U}$ is an open cover of $\mathbb{Z}$, then there is a clopen cover $\{C_{n}|n\in\mathbb{N}\}$ that refines $\mathcal{U}$. If we set $D_{n}=C_{n}\setminus(C_{0}\cup...\cup C_{n-1})$ for all $n$, then $\{D_{n}|n\in\mathbb{N}\}$ is a partition of $(\mathbb{Z},\mathcal{T})$ into finitely many clopen sets that refines $\mathcal{U}$, so $(\mathbb{Z},\mathcal{T})$ is compact. This is a contradiction. Therefore $(\mathbb{Z},\mathcal{T})$ has a partition into countably many clopen sets.
In particular, there is a continuous surjective function $f:(\mathbb{Z},\mathcal{T})\rightarrow\mathbb{N}$ where $\mathbb{N}$ has the discrete topology. Therefore the map $f$ extends to a continuous surjective function $\bar{f}:\beta(\mathbb{Z},\mathcal{T})\rightarrow\beta\mathbb{N}$. Since $|\beta\mathbb{N}|=2^{\mathbb{c}}$, we conclude that $|\beta(\mathbb{Z},\mathcal{T})|=2^{\mathbb{c}}$ as well. We conclude that the Stone-Cech compactification $\beta(\mathbb{Z},\mathcal{T})$ is much larger than the pro-finite completion of $\mathbb{Z}$.
[1] The Stone-Čech Compactification, Russell Walker (1970)
A: Edit. The comments and the other answers reveal that my proof has some gap. But I won't delete it, instead I've rewritten it as an attempt to prove $\beta \mathbb{Z} = \hat{\mathbb{Z}}$. I hope that the failure of this naive proof motivates to read the more sophisticated answers.

When $\mathbb{Z}$ is equipped with the Fürstenberg topology, do we have $\beta \mathbb{Z} = \hat{\mathbb{Z}}$?
The Fürstenberg topology is the subspace topology induced by the profinite completion $\hat{\mathbb{Z}} = \lim_{n>0} \mathbb{Z}/n$. The embedding $\mathbb{Z} \to \hat{\mathbb{Z}}$ is dense, hence for every compact Hausdorff space $X$ we get an injective map $\hom(\hat{\mathbb{Z}},X) \to \hom(\mathbb{Z},X)$. The question is whether it is surjective, because this would mean that $ \hat{\mathbb{Z}}$ satisfies the defining universal property of $\beta \mathbb{Z}$.
Let $f : \mathbb{Z} \to X$ be a continuous map. This means that for every $a \in \mathbb{Z}$, every open subset $U \subseteq X$ containing $f(a)$ already contains $f(a+n \mathbb{Z})$ for some $n>0$. Let $a=(a_1,a_2,\dotsc) \in \hat{\mathbb{Z}}$, i.e. $a_n \equiv a_m \bmod n$ for $n|m$. Since $X$ is compact, the net $(f(a_n))_{n>0}$ (using divisibility for the indices) has a convergent subnet, say $(f(a_{n(i)}))_{i \in I} \longrightarrow x$.
Actually any two subnets have the same limit: Assume that $(f(a_{m(j)}))_{j \in J} \longrightarrow y$. Choose open neighborhoods $U,V$ of $x,y$, it is enough to prove $U \cap V \neq \emptyset$ since $X$ is Hausdorff. For large $i$ we have that $f(a_{n(i)}) \in U$, and for large $j$ we have $f(a_{m(j)}) \in V$. Choose $b>0$ with $f(a_{n(i)} + b \mathbb{Z}) \subseteq U$ and $f(a_{m(j)} + b \mathbb{Z}) \subseteq V$. We may assume $n(i),m(j)|b$. For $p=n(i) m(j)$ we have $a_p \equiv a_{n(i)} \bmod n(i)$, hence $a_p \equiv a_{n(i)} \bmod b$. Similarily we get $a_p \equiv a_{m(j)} \bmod b$. Hence $f(a_p) \in U \cap V$.
Hence $\tilde{f}(a) := $(the limit of some subnet of $f(a_n)$) gives a well-defined map $\hat{\mathbb{Z}} \to X$. Clearly it agrees with $f$ on constant sequences. But now the problem seems to be that $\tilde{f}$ is not continuous ...
