Let $K$ be a nonarchimedean field complete with respect to a discrete valuation with ring of integers $\mathcal{R}$, uniformizer $\pi$ and residue field $k$. Consider an affinoid analytic $K$variety $X=Sp(A)$ with an affine formal model $\mathfrak{X}=Spf(A^{\circ})$ where $A^{\circ}\subset A$ is the set of powerbounded elements. Let $f\in A^{\circ}$ be an element such that its clas in $A^{\circ}/\pi$ is nonzero. As the generic fiber functor sends complete localizations on $\mathfrak{X}$ to Laurent subdomains in $X$, it follows that the affinoid subdomain $X(\frac{1}{f})=\{ x\in X \text{ such that } \vert f(x)\vert \geq 1 \}$ has an affine formal model given by the formal spectrum of $A^{\circ}\langle f^{1}\rangle$. Let $\mathfrak{X}_{k}$ denote the special fiber of $\mathfrak{X}$. By definition, the special fiber associated to the formal scheme $\mathfrak{X}(\frac{1}{f})=Spf(A^{\circ}\langle f^{1}\rangle)$ is a Zariski open of $\mathfrak{X}_{k}$. Hence, all properties of the special fiber $\mathfrak{X}_{k}$ which are local in the Zariski topology (irreducibility, smoothness etc) will also hold in $\mathfrak{X}(\frac{1}{f})_{k}$. My question is as follows: does this generalize to more general Laurent subdomains of $X$? For example, consider $f\in A^{\circ}$ as before, and consider subdomains of the form $Y=X(\frac{\pi^{n}}{f},\frac{f}{\pi^{n}})=\{ x\in X \text{ such that } \vert f(x)\vert = n \}$. As far as I know, these are not necessarily the generic fiber of an affine open subspace of $\mathfrak{X}$, so we cannot argue as above to construct affine formal models of $Y$ with topological properties similar to those of $\mathfrak{X}$. By Raynaud's theory we know there is an admissible formal blowup $\mathfrak{X}'\rightarrow \mathfrak{X}$ such that there is an open $Z\subset \mathfrak{X}'$ such that its generic fiber is $Y$. However, admissible formal blowup seems to have a weird behaviour at the topological level, so I don't know which properties of the topological space assocaited to $\mathfrak{X}$ are preserved under these kinds of maps. I would like to know if the fact that $X$ admits an affine formal model such that its special fiber has some topological property (mainly interested in irreducibility) implies that an affinoid subdomain of the form $X(\frac{\pi^{n}}{f},\frac{f}{\pi^{n}})$ also admits such an affine formal model. What about subspaces of the form $X(\frac{\pi^{n}}{f})$? Would something like this hold for Weierstrass subdomains of $X$?

$\begingroup$ I am confused by your terminology. For me, a formal model for a rigid analytic $K$variety $X$ is a $p$adic formal $\mathcal O_K$scheme $\mathfrak X$ whose adic generic fiber is isomorphic to $X$, and this choice is not a priori necessarily unique. I am not sure how it is related to your concept. $\endgroup$– Z. MDec 16, 2022 at 15:17

$\begingroup$ Dear @Z.M , yes this choice is not unique. I would like to know if in the case above there exist any admissible formal scheme over $Spf(\mathcal{R})$ such that its special fiber is integral and its generic fiber is $Y$. I do not claim that such model is unique, I just would like to know if there is one. $\endgroup$– Fernando Peña VázquezDec 16, 2022 at 16:12

$\begingroup$ In fact, if there is a formal model $\mathfrak{Y}$ of $Y$ which has integral special fiber but It is not affine It would also be enough. $\endgroup$– Fernando Peña VázquezDec 16, 2022 at 16:20
1 Answer
For an affinoid $X=\mathrm{Sp}A$, the number of Shilov points of $X$ is a lower bound for the number of irreducible components of the special fiber of any formal model of $X$. This follows, for instance, from Proposition 2.2 of this paper.
So then it's easy to make concrete examples: $X=\mathrm{Sp}K \langle T \rangle$ has an obvious formal model with irreducible special fiber, but any formal model of the Laurent domain $U=\mathrm{Sp}K \langle T,\pi/T \rangle \subset X$ will have at least two irreducible components in its special fiber, since $U$ has two Shilov points.
Hope this helps!