Let $X$ be an affinoid K-space. That is $X=Spec A$ where $A$ is a quotient of the Tate algebra. In Bosch 'lectures on Formal and Rigid Geometry' we find the curious statement that the canonical map $O_X \to O_{X,x}$ factors as $O_X \to A_m \to O_{X,x}$ where $m$ is the maximal ideal corresponding to $x\in X$, implying that the local ring $O_{X,x}$ can be larger than $A_m$. Is this true? Is there an explicit example?

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    $\begingroup$ This should not be surprising at all. Namely, the stalk is computed as the direct limit of the rings $\mathcal{O}(U)$ as $U$ travels over rational neighborhoods of the point $\mathfrak{m}$. But, note that if one has something like $U=U\left(\frac{p}{T}\right)$ (let's say in the case when $A=k\langle T\rangle$) then the value of $\mathcal{O}$ on $U$ is $k\langle T,Y\rangle/(TY-p)$. That said $A_\mathfrak{m}$ is the direct limit over things like $A_f$ where $D(f)$ travels over the distinguished opens of $\text{Spec}(A)$. In other words, these two constructions essentially $\endgroup$ – Alex Youcis Oct 7 '16 at 15:57
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    $\begingroup$ differ by a completion. Perhaps all of this is more clear in the case of formal schemes where one is completing with respect to more apparent topologies (an adic topology opposed to an '$f$-adic' one). Namely, there the value of $\mathcal{O}$ on a $D(f)$ is not $A_f$ but $\widehat{A_f}$ (the completion of $A_f$) and thus the stalk is surely going to differ from the stalk on $\text{Spec}(A)$. $\endgroup$ – Alex Youcis Oct 7 '16 at 16:01
  • $\begingroup$ Thank you! This cleared things up for me. $\endgroup$ – Lee Wang Sep 5 '17 at 10:44

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