Let $R$ be a local ring. I wonder if the henselization map $R \to R^h$ or the strict henselization map $R \to R^{sh}$ etale? I manage to prove that it is flat and unramified. But I cannot prove it is finitely presented.
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1Just to note, there are some examples where this is true and $R$ is not already strictly Henselian (e.g. $R=\mathbb{R}$ and $R^\mathrm{sh}=\mathbb{C}$), but they are exceedingly rare. – Alex Youcis Feb 26 '23 at 14:29
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No, the strict Henselization map is not etale in general because it will essentially never be of finite presentation.
For a very concrete, hands-on example let us take $R=k[x]_{(x)}$ where, for simplicitly, let's assume that $k$ is algebraically closed. Then, one can explicitly compute $R^\mathrm{h}=R^{\mathrm{sh}}$. Namely, it is the algebraic closure of $R$ in $\widehat{R}=k[\![x]\!]$ (see this post). This is evidently not of finite presentation because then (as $\mathfrak{m}R^h$ is the maximal ideal of $R^h$) the induced map on fraction fields would be of finite presentation, but this evidently doesn't happen here as the sequence of extensions $\sqrt[n]{x+1}\in R^h$ shows.
Alex Youcis
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