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Let $K$ be a finite extension of the $p$-adic numbers with valuation ring $\mathcal{R}$ and uniformizer $\pi$. Consider a smooth and connected rigid $K$-variety $X=Sp(A)$ and assume that the affine formal model $\mathfrak{X}=Spf(A^{\circ})$ is normal (i.e. $A^{\circ}$ is a normal integral domain). My question is whether the fact that $X$ admits a normal affine formal model is stable under finite extensions of the ground field. That is, let $L$ be a finite extension of $K$ (separable as $K$ is of characteristic zero) with valuation ring $\mathcal{R}_{L}$ and uniformizer $\omega$. Is it true that $X_{L}=Sp(A\otimes_{K}L)$ admits a normal formal model? Letting $A_{L}=A\otimes_{K}L$, is $A_{L}^{\circ}$ a normal ring in general? Under which conditions does it hold that $A^{\circ}_{L}= A^{\circ}\otimes_{\mathcal{R}}\mathcal{R}_{L}$?

Context: By the reduced fiber theorem there is a finite extension $L$ of $K$ such that $A^{\circ}_{L}$ has geometrically reduced special fiber. As this process introduces roots of $\pi$, the induced morphism $\mathcal{R}\rightarrow \mathcal{R}_{L}$ will not be étale, just finite flat. Thus, the morphism $A^{\circ}\rightarrow A^{\circ}\otimes_{\mathcal{R}}\mathcal{R}_{L}$ is not neccesarily étale, and $A^{\circ}\otimes_{\mathcal{R}}\mathcal{R}_{L}$ is not necessarily normal. I am interesting in finding out if for every smooth irreducible rigid $K$-variety with a normal affine formal model with geometrically irreducible special fiber there is a finite extension $L$ such that $X_{L}$ admits an affine formal model with integral geometric fiber.

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As for your first question, $X_L$ indeed admits a normal formal model by virtue of normalisation. Whether $A^{\circ}\otimes_R R_L$ is already normal (which is equivalent to $A^{\circ}\otimes_R R_L=(A_L)^{\circ}$, as the latter is normal) is less transparant and is related to issues of wild ramification.

Here is an example to show such base change may fail to be normal. The elliptic curve $y^2=x^3+2$ over $\mathbb{Q}_2$ has bad reduction, but has good reduction over the wild Kummer extension $\mathbb{Q}_2(\sqrt{2})$. Over $\mathbb{Q}_2$ the minimal normal crossings model contains a component of multiplicity 2, above which the base change has a non-normal component. So to obtain a counterexample we can take formal completion and let $\mathrm{Spf}(A)$ be some affine piece containing this multiplicity 2 component.

Using logarithmic geometry one may prove there is no obstruction in certain 'tame' situations: suppose $A^{\circ}/R$ is log regular (for instance coming from a snc degeneration as in the above example), where we equip $A^{\circ}$ and $R$ with the standard log structure of nonzero elements. Then if $A^{\circ}$ is log smooth or $R_L/R$ is log étale (i.e. $L/K$ is tame), we have that $A^{\circ}\otimes_R R_L$ is log regular and hence normal. Section 3.1 of this paper may be relevant. Whenever nice normal crossings models are available, components with multiplicity prime-to-$p$ are log smooth and so above these there is no problem. In particular, if you're only concerned with pieces of semistable formal models of algebraic varieties, all is good.

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  • $\begingroup$ Dear @ArtW , thank you for your answer. Just one question, how do you know that $(A_{L})^{\circ}$ is a normal ring? As far as I know, the fact that this is a ring of power-bounded elements implies that it is integrally closed in its generic fiber $(A_{L})^{\circ}\otimes_{\mathcal{R}_{L}}L=A_{L}$. Does this imply that $(A_{L})^{\circ}$ is integrally closed in its field of fractions? $\endgroup$
    – FPV
    Commented Dec 12, 2022 at 10:42
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    $\begingroup$ that's right, $A_L$ is normal by your smoothness assumptions, so by the fact you mention $(A_L)^{\circ}$ is normal too $\endgroup$
    – ArtW
    Commented Dec 12, 2022 at 16:11

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