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For a normal local ring $(R,m)$ over complex number field C, consider the henselization $R^h$ and the completion $\widehat{R}$.

Question: Is $R^h$ algebraically closed in $\widehat{R}$?

I think this is true generally for normal local ring over Dedekind domain or generally excellent ring $S$ of dimension 1. Please teach me.

Pierre MATSUMI

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  • $\begingroup$ The answer is affirmative for any excellent normal noetherian local domain (no need to bring in $\mathbf{C}$). Excellence is inherited by henselization, so we may also assume $R$ is henselian. Let $K={\rm{Frac}}(R)$, so $\widehat{K}:={\rm{Frac}}(\widehat{R})$ is separable over $K$. Hence, if $K'$ is nontrivial finite over $K$ inside $\widehat{K}$ and $R'$ is the $R$-finite integral closure of $R$ in $K'$ then $R'$ is local and $K'/K$ is separable, so $\widehat{K}\otimes_K K'$ contains the non-domain $K'\otimes_K K'$. But it is a localization of the ring $\widehat{R}'$ that is a domain.QED $\endgroup$
    – user29720
    Jun 12, 2013 at 13:25

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