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Recall that in EGA $IV_3$ (Théorème (8.12.6)), Grothendieck calls the following theorem ``Zariski's Main Theorem":

Let $Y$ be a quasi-compact separated scheme, and $f : X \to Y$ is a separated quasi-finite finitely presented morphism. Then it factors into $X \to Z \to Y$, where the first is an open immersion and the second is finite.

The way I always remembered this was as a quasi-finite version of a compactification theorem (e.g. Negata), so let's call $Z \to Y$, a compactification of $X \to Y$.

Regarding it, here is my question:

Under the situation of the above ZMT, in addition suppose both $X$, $X$ are smooth integral affine $k$-schemes of finite type over a field $k$, with a sufficiently high dimension $>0$. Suppose we have a given finite set $S \subset Y$ of points. Then can we find a compactification $Z \to Y$ such that the image of the ``bad set" $B:= Z \setminus X$ in $Y$ under the finite morphism does not intersect the given finite set $S$?

Maybe I am making some redundant assumptions here. Note that since $Z \to Y$ is finite, the image of the bad set is a proper closed subset of $Y$. So, if the above holds, then I can actually find an affine open neighborhood $U$ of $S$, over which $X_U \to U$ itself is finite from the beginning.

I hope someone may know an answer, or may have an idea that may become useful in its resolution.

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    $\begingroup$ There are trivial counterexamples to your question. For example, let $Y = \mathbb P^n$, let $X = \mathbb A^n$, and let $S = \{p\}$ be any point at infinity, i.e. $p \not \in X$. Whatever compactification of the inclusion $\mathbb A^n \to \mathbb P^n$ you choose, the boundary locus should always include a point above $p$. $\endgroup$ – R. van Dobben de Bruyn Nov 3 '18 at 17:22
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    $\begingroup$ More generally, if the stalk $X \times_Y \operatorname{Spec} \mathcal O_{Y, p}$ is not finite over $\mathcal O_{Y, p}$ for some $p \in S$, then $Z \setminus X$ needs to contain a point above $p$. This is your comment after the question as well; it should be an assumption in the statement. $\endgroup$ – R. van Dobben de Bruyn Nov 3 '18 at 17:24
  • $\begingroup$ @R.vanDobbendeBruyn Thank you. Let me add the assumption that $X \to Y$ is surjective, and over all points of S, it is already known to be finite. That would remove trivial counterexamples. (That was the case in the problem I was working on... from which this question came out.) $\endgroup$ – Jinhyun Park Nov 3 '18 at 17:25
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There is in fact a 'preferred' choice of Zariski factorisation:

Definition. Let $f \colon X \to Y$ be a quasi-finite morphism of varieties over a field $k$. Then let $Z$ be the normalisation of $Y$ in $X$, i.e. $Z = \operatorname{\underline{Spec}}_Y \mathscr A$ with $\mathscr A$ the normalisation of $\mathcal O_Y$ in $f_* \mathcal O_X$.

Note that $g \colon Z \to Y$ is finite [Tag 0BXS], and the map $\iota \colon X \to Z$ is an open immersion [Tag 02LR].

Lemma. Let $U \subseteq Y$ be an open above which $f$ is finite. Then $\iota$ is an isomorphism above $U$.

Proof. This is immediate from [Tags 035K and 03GP]. $\square$

If $f$ is finite above $\operatorname{Spec} \mathcal O_{Y,y}$ for some $y \in Y$, then it is finite in a neighbourhood of $y$. Thus, there exists a (nonempty) maximal open $U \subseteq Y$ above which $f$ is finite, and above this set the open immersion $\iota \colon X \to Z$ is an isomorphism. We conclude that $$g(Z \setminus X) = Y \setminus U,$$ so $g(Z \setminus X)$ avoids a set $S$ if and only if $S \subseteq U$, i.e. if and only if $f$ is already finite above $S$.

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