Take the integers $\mathbb{Z}$ and the addition
\begin{align\*}
 +: \mathbb{Z} \times \mathbb{Z} &\to \mathbb{Z}
 \\
 (a,b) &\mapsto a+b.
\end{align\*}
Using the Stone-Čech compactification $\beta\mathbb{Z}$ in two steps, $A$ can be "continuously" extended to
\begin{equation\*}
 +: \beta\mathbb{Z} \times \mathbb{Z} \to \beta\mathbb{Z}
\end{equation\*}
and then to
\begin{equation\*}
 +: \beta\mathbb{Z} \times \beta\mathbb{Z} \to \beta\mathbb{Z}.
\end{equation\*}
The resulting operation $+$ is continuous in its right argument, and, when the right argument is an integer, it is continuous on its left argument.


Call $\mathbb{Z}^\* := \beta\mathbb{Z} \setminus \mathbb{Z}$.
For any $p \in \mathbb{Z}^\*$, we have the right ideal $I\_p := p + \beta\mathbb{Z}$.


Is it possible to partition $\mathbb{Z}^\*$ in two (non trivial) pieces $A$ and $B$ such that
\begin{equation\*}
 p \in A \text{ and } q \in B
 \Rightarrow
 I\_p \cap I\_q = \emptyset?
\end{equation\*}


In other words, if I generate an equivalence relation based on $p \sim q$ whenever $I\_p \cap I\_q \neq \emptyset$, will this equivalence relation partition $\mathbb{Z}^\*$ in more then one piece?



Partitioning $\beta \mathbb{Z} \setminus \mathbb{Z}$
Yes, this is possible.


First of all, let me suggest a way of thinking about the $+$ operation. If you're familiar with the idea of taking limits along an ultrafilter, then given $p,q \in \mathbb Z^\*$, you can think of $p+q$ as 
$$p+q = q\text{-lim}\_{n \in \mathbb Z} \ (p+n).$$
I tend to look at this dynamically: we have a map (the shift map) that sends an ultrafilter $p$ to $p+1$, and $p+q$ just represents one limit point of the set $\{p+n \,:\, n \in \mathbb Z\}$, the (full) orbit of $p$ under the shift map. Also, this makes it clear that $I\_p = p+\beta \mathbb Z = \overline{\{p+n \,:\, n \in \mathbb Z\}}$ for any $p \in \mathbb Z^\*$.


To answer your question, one partition of $\mathbb Z^\*$ that does what you want is the partition into the two clopen sets $\mathbb N^\*$ and $(-\mathbb N)^\*$ (the ultrafilters concentrating on the positive and negative integers, respectively). If $p$ concentrates on the positive integers, then so does $p+n$ for any $n$. In other words, $p \in \mathbb N^\*$ implies $p+n \in \mathbb N^\*$ for all $n$. This means that $p+\beta \mathbb Z = \overline{\{p+n \,:\, n \in \mathbb Z\}} \subseteq \mathbb N^\*$. Similarly, if $p \in (-\mathbb N)^\*$ then $p+\beta \mathbb Z = \overline{\{p+n \,:\, n \in \mathbb Z\}} \subseteq (-\mathbb N)^\*$.


No other partition of $\mathbb Z^\*$ into clopen sets has the property you want. (I'll sketch a proof of this below.) 


But there are other, more interesting partitions of $\mathbb Z^\*$ that do what you want. Recall that $X \subseteq \mathbb Z^\*$ is a weak $P$-set if $X \cap \overline{D} = \emptyset$ for any countable $D \subseteq \mathbb Z^\* \setminus X$. In [this paper](https://www.impan.pl/en/publishing-house/journals-and-series/fundamenta-mathematicae/all/235/2/91586/g-delta-and-co-meager-semifilters), Jonathan Verner and I show that there is a minimal right ideal $R$ of $\mathbb Z^\*$ that is also a weak $P$-set. This means that if $p \in R$, then $p + \beta \mathbb Z = R$ (because it's a minimal right ideal), and if $p \notin R$ then $p+\beta \mathbb Z = \overline{\{p+n \,:\, n \in \mathbb Z\}} \subseteq \mathbb Z^\* \setminus R$ (because $R$ and its complement are both shift-map-invariant, and because $R$ is a weak $P$-set). Therefore the partition of $\mathbb Z^\*$ into $R$ and $\mathbb Z^\* \setminus R$ has the property you want. In fact, the proof in my paper with Jonathan shows that there are $2^{\mathfrak{c}}$ distinct right ideals of this kind; because each minimal right ideal that is a weak $P$-set constitutes an equivalence class, it follows that the equivalence relation you describe has the maximum possible number of equivalence classes, namely $2^{\mathfrak{c}}$.


(Suppose $\mathcal P$ is a partition of $\mathbb Z^\*$ into clopen sets other than the one described above. Some $A^\* \in \mathcal P$ has the property that either (1) $\mathbb N^\* \cap A^\* \neq \emptyset$ and $\mathbb N^\* \setminus A^\* \neq \emptyset$, or else (2) $(-\mathbb N^\*) \cap A^\* \neq \emptyset$ and $(-\mathbb N^\*) \setminus A^\* \neq \emptyset$. Without loss of generality, let us say $\mathbb N^\* \cap A^\* \neq \emptyset$ and $\mathbb N^\* \setminus A^\* \neq \emptyset$. This means that the set $E$ of $n \in \mathbb N$ such that $n \in A$ but $n+1 \notin A$ must be infinite. If $p$ is an ultrafilter that concentrates on $E$, then we have $p \in A^\*$ while $p+1 \notin A^\*$. Thus $p$ and $p+1$ are in different partition elements, but clearly $p + \beta \mathbb Z = \overline{\{p+n \,:\, n \in \mathbb Z\}} = \overline{\{(p+1)+n \,:\, n \in \mathbb Z\}} = (p+1)+\mathbb Z^\*$. Therefore this partition does not have the property you want.)


