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Let $R$ be a complete valuation ring of rank $1$ (e.g., a complete discrete valuation ring) and let $K$ be its field of fractions. Consider a proper $R$-scheme $X$ that is, say, normal (if needed). There are two ways to attach a rigid analytic space over $K$ to $X$:

  1. The rigid analytic space associated to the finite type $K$-scheme $X_K$;
  2. The rigid analytic generic fiber of the formal $\varpi$-adic completion $\hat{X}$ of $X$, where $\varpi \in R$ is some nonzero nonunit element (e.g., a uniformizer in the DVR case).

Are these two rigid analytic spaces canonically isomorphic over $K$? And if so, then why (a reference would suffice and would be much appreciated)?

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    $\begingroup$ Yes. More generally, if you relax "proper" to "separated of finite type" then there is a canonical quasi-compact open immersion $i_X:{\widehat{X}}^{\rm{rig}} \hookrightarrow (X_K)^{\rm{an}}$ (seen most concretely for $X = \mathbf{A}^n_R$, in which case $i_X$ is the open immersion of the closed unit $n$-disc into rigid-analytic affine $n$-space over $K$), and it is an isomorphism if $X$ is proper. For a proof, see Theorem A.3.1 in math.stanford.edu/~conrad/papers/irredpaper.pdf $\endgroup$
    – nfdc23
    Commented Jul 21, 2016 at 4:18
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    $\begingroup$ Wonderful. Thank you very much! In the published version it's Theorem 5.3.1. $\endgroup$
    – O-Ren Ishii
    Commented Jul 21, 2016 at 15:09
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    $\begingroup$ One needs flatness and finitely presentedness when R is not DVR... $\endgroup$
    – S. Li
    Commented Feb 14, 2017 at 19:17

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