Let $A$ be a strictly affinoid algebra. Let $X^{Ber}$ bet its Berkovich spectrum and $X^{Tate} = \operatorname{Sp} A$ its affinoid variety in the sense of rigid analytic geometry. Let $\mathfrak{m} \subset A$ be a maximal ideal and $x$ the corresponding point in $X^{Ber}$, respectively in $X^{Tate}$. We know that for all $n \in \mathbb{N}$ there are isomorphisms $$A / \mathfrak{m}^n \cong A_\mathfrak{m} / A_\mathfrak{m} \mathfrak{m}^n \cong \mathcal{O}_{X^{Tate},x} / \mathcal{O}_{X^{Tate},x} \mathfrak{m}^n.$$ Does $\mathcal{O}_{X^{Ber},x}$ also fit in? That is, is $\mathcal{O}_{X^{Ber},x} / \mathcal{O}_{X^{Ber},x} \mathfrak{m}^n$ also isomorphic to these rings?

  • $\begingroup$ Hi Helene! The rings $\mathcal{O}_{X^{Ber},x}$ and $\mathcal{O}_{X^{Tate},x}$ are the same, so the answer is yes. $\endgroup$ – Jérôme Poineau Jun 22 '15 at 10:54
  • $\begingroup$ Hi Jérôme! How do we know that they are the same? I couldn't find it in Berkovich's 1993 IHES paper. $\endgroup$ – Helene Sigloch Jun 22 '15 at 11:09
  • 1
    $\begingroup$ In both cases, it is the inductive limit of the rings of functions of the affinoid domains containing the point. (In the Berkovich setting, a rigid point is always in the interior of an affinoid domain that contains it.) $\endgroup$ – Jérôme Poineau Jun 22 '15 at 11:23
  • $\begingroup$ Of course. Thank you! As you answer is only a comment, I can't accept it. Should I delete the question instead, because it was silly? $\endgroup$ – Helene Sigloch Jun 22 '15 at 11:32
  • $\begingroup$ Do as you wish. $\endgroup$ – Jérôme Poineau Jun 22 '15 at 12:19

As Jérôme points out, the rings $\mathcal{O}_{X^{Ber},x}$ and $\mathcal{O}_{X^{Tate},x}$ are the same, thus the answer is "yes".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.