# Adic generic fiber of a small formal scheme in the sense of Faltings

$$\DeclareMathOperator{\Spf}{Spf}\DeclareMathOperator{\Spa}{Spa}$$In the Definition 8.5 of the paper "integral $$p$$ adic Hodge theory" by Bhatt-Morrow-Scholze, they define the adic generic fiber of a small affinoid smooth formal scheme $$\Spf(R)$$ as $$\Spa(R[\frac{1}{p}],R)$$, here $$R$$ is a smooth $$\mathscr{O}_{\mathbb{C}_{p}}$$-algebra which admits an 'etale map to the formal torus. However, the adic generic fiber defined by Huber should send $$\Spf(R)$$ to $$\Spa(R[\frac{1}{p}],R[\frac{1}{p}]^{\circ})$$, here $$R[\frac{1}{p}]^{\circ}$$ means the subring of all the power bounded elements of $$R[\frac{1}{p}]$$.

My question is that why $$R=R[\frac{1}{p}]^{\circ}$$ ? If $$R$$ is not smooth, a counterexample is $$R=\mathscr{O}_{\mathbb{C}_{p}}\left \langle x,y \right \rangle/(xy-p)=\mathscr{O}_{\mathbb{C}_{p}}\left \langle x,\frac{p}{x} \right \rangle$$(the semistable curve). In this example, $$R[\frac{1}{p}]^{\circ}=\mathscr{O}_{\mathbb{C}_{p}}\left \langle x,y \right \rangle/(xy-1)=\mathscr{O}_{\mathbb{C}_{p}}\left \langle x,\frac{1}{x} \right \rangle$$. Besides, if we consider the admissible blow up of the formal scheme $$\Spf(R)$$, then the ring $$R$$ is changed, however, it does not change the generic fiber. Or my understanding about the adic generic fiber functor is wrong?

Any idea will be appreciated.

• It seems that the command textup isn't working here - hopefully the result after my edit is approximately what you wanted to achieve. Commented Oct 9, 2023 at 11:39

In general, $$R[1/p]^\circ$$ is the integral closure of $$R$$ in $$R[1/p]$$ (see BGR or Bosch's lecture notes). So it is enough to see that since the formal scheme is smooth, it is normal (though nonnoetherian).
By the way, I think you got your example wrong, as $$R=\mathscr{O}\langle x,y\rangle/(xy-p)$$ is likewise normal. This is obvious over $$\mathbf{Z}_p$$ as then $$R$$ is even regular. Over $$\mathscr{O}_{\mathbb{C}_p}$$ this follows since the special fiber is reduced.
To address the last question. If you blow up $$\operatorname{Spf}(R)$$, the resulting formal scheme will usually not be affine. In fact, $$R$$ integrally closed in $$R[1/p]$$ is equivalent to saying that for every admissible blowup $$\mathfrak{X}$$ of $$\operatorname{Spf}(R)$$ we have $$\mathscr{O}(\mathfrak{X})=R$$.