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Suppose that $K$ is a field, $V$ and $W$ are $K$-varieties, and $f : V \to W$ is a quasi-finite morphism. One form of Zariski's main theorem is that $f$ factors as $V \to V' \to W$ where $V'$ is a $K$-variety, $V \to V'$ is an open immersion, and $V' \to W$ is finite.

Suppose further that $V$ is smooth. Can we take $V'$ to also be smooth? In my naive opinion this seems unlikely, but I would like to know for sure.

If this was true then I think I can produce an open immersion $V \to V'$ where $V'$ is a smooth proper $K$-variety, so we run into issues with resolution of singularities. So I am willing to restrict to characteristic zero.

I would also be interested in a weaker statement. For example, could there be an open subvariety $V_0$ of $V$ such that the restriction of $f$ to $V_0$ factors as $V_0 \to V'_0 \to W$ where $V'_0$ is smooth, $V_0 \to V'_0$ is an open immersion, and $V'_0 \to W$ finite? Or could we at least take $V'$ to not be too "badly singular" in some sense?

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    $\begingroup$ This is not true. Take a finite covering $V_0\rightarrow W$, with $W$ smooth, $V_0$ normal but with a singular point $p$ (that is easy to construct), and $V=V_0\smallsetminus \{p\} $. Then necessarily $V'=V_0$ which is not smooth. $\endgroup$
    – abx
    Commented Oct 9, 2020 at 4:16
  • $\begingroup$ alright, thanks! $\endgroup$
    – Erik Walsberg
    Commented Oct 9, 2020 at 23:34
  • $\begingroup$ @abx could you take $W$ to be $\mathbb{A}^m$ ? I think this is actually the only case that matters for what I am thinking about. $\endgroup$
    – Erik Walsberg
    Commented Oct 10, 2020 at 0:56
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    $\begingroup$ Yes — just take a double covering of $\mathbb{A}^m$ branched along a hypersurface with one singular point. $\endgroup$
    – abx
    Commented Oct 10, 2020 at 5:02

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