# Berkovich stalk versus rigid analytic stalk

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?

• Hi Helene! The rings $\mathcal{O}_{X^{Ber},x}$ and $\mathcal{O}_{X^{Tate},x}$ are the same, so the answer is yes. Jun 22, 2015 at 10:54
• Hi Jérôme! How do we know that they are the same? I couldn't find it in Berkovich's 1993 IHES paper. Jun 22, 2015 at 11:09
• 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.) Jun 22, 2015 at 11:23
• 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? Jun 22, 2015 at 11:32
• Do as you wish. Jun 22, 2015 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".