*This is related to a couple recent MO/MSE questions of mine, namely [1](https://mathoverflow.net/questions/427135/what-algebraic-properties-are-preserved-by-mathbbn-leadsto-beta-mathbbn),[2](https://math.stackexchange.com/questions/4498457/which-ultrafilters-are-sort-of-commutators). Belatedly, I've tweaked this post to remove an overly-ambitious secondary question; see the edit history if interested.*


Let $\beta\mathbb{Z}$ be the set of all ultrafilters on $\mathbb{Z}$, and as usual conflate $n$ and $\{A\subseteq\mathbb{Z}:n\in A\}$ for $n\in\mathbb{Z}$. We can extend any binary operation on $\mathbb{Z}$ to a semicontinuous analogue on $\beta\mathbb{Z}$, at the cost of many (most?) algebraic properties. Ultrafilter *addition* is quite well studied (see e.g. [Hindman/Strauss](https://people.dm.unipi.it/dinasso/ULTRA2015/BUMN/Hindman_Strauss%20-%20Algebra%20in%20the%20Stone-Cech%20compactification%20(2012%20-%202nd%20ed).pdf)), but I've been able to find much less about ultrafilter *subtraction*: $$\mathcal{U}\widehat{-}\mathcal{W}=\{A\subseteq\mathbb{Z}: \{k:\{a: a-k\in A\}\in\mathcal{U}\}\in\mathcal{W}\}.$$


Say that an ultrafilter $\mathcal{W}$ is **zeroid** iff $\mathcal{W}=\mathcal{U}\widehat{-}\mathcal{U}$ for some $\mathcal{U}$. My question is:



> 
> Which ultrafilters are zeroid?
> 
> 
> 


To make this actually answerable, I tentatively guess that $(i)$ $0$ is the only zeroid principle ultrafilter but not the only zeroid ultrafilter and $(ii)$ $p$-points are not zeroid; are these guesses true?



Ultrafilter subtraction and "zero"
Both your guesses are correct. To see this, it's helpful to reformulate the way you're thinking about the subtraction operator on $\beta \mathbb Z$. Beginning with subtraction on $\mathbb Z$, you can first extend this to an operator $\beta \mathbb Z \times \mathbb Z \rightarrow \beta \mathbb Z$ by setting $\mathcal U - n = \{B-n :\, B \in \mathcal U\} = \{A \subseteq \mathbb Z :\, A+n \in \mathcal U\}$. Notice that this agrees with your definition of subtraction on $\beta \mathbb Z$ when we identify $n$ with the principle ultrafilter at $n$. So this is the "right" way to think of subtracting an integer from an ultrafilter. But then there is only one way to extend this to a semi-continuous operation $\beta \mathbb Z \times \beta \mathbb Z$: we must define $$\mathcal U - \mathcal W = \textstyle {\mathcal W}\text{-}\!\lim\_n (\mathcal U-n).$$ This again agrees with your definition (it must!), but personally I find it much more intuitive to think of $\mathcal U - \mathcal W$ as a topological limit of the sequence $\mathcal U, \mathcal U - 1, \mathcal U - 2, \dots$.


This description of what's going on makes your two guesses plainly true. For $(i)$, note that if $\mathcal U$ is nonprincipal then so is $\mathcal U - n$ for all $n$, which means (because $\mathbb Z$ is open in $\beta \mathbb Z$) that the limit ${\mathcal W}\text{-}\!\lim\_n (\mathcal U-n)$ is in $\beta Z \setminus \mathbb Z$ for any $\mathcal W$ (including $\mathcal W = \mathcal U$). For $(ii)$, note that ${\mathcal W}\text{-}\!\lim\_n (\mathcal U-n)$ is a limit of a countable sequence of points in $\beta \mathbb Z \setminus \mathbb Z$, hence not a P-point. In fact, let me observe that $\mathcal U - \mathcal W$ is never a weak P-point (for the same reason), which shows that not every ultrafilter is representable as the difference of two others.


