$\DeclareMathOperator{\Spf}{Spf}\DeclareMathOperator{\Spa}{Spa}$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.


1 Answer 1


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 $\mathscr{O}(\mathfrak{X})=R$.


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