TITLE: Berkovich stalk versus rigid analytic stalk
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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".