# On the local properties of rigid analytic varieties

Let $$K$$ be a non-archimedean 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 power-bounded elements. Let $$f\in A^{\circ}$$ be an element such that its clas in $$A^{\circ}/\pi$$ is non-zero. 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 blow-up $$\mathfrak{X}'\rightarrow \mathfrak{X}$$ such that there is an open $$Z\subset \mathfrak{X}'$$ such that its generic fiber is $$Y$$. However, admissible formal blow-up 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 $$X$$?

• 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.
– Z. M
Dec 16, 2022 at 15:17
• 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. Dec 16, 2022 at 16:12
• 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. Dec 16, 2022 at 16:20

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.