Let $f:X\to Y$ be a morphism of algebraic varieties over $\mathbb C$. Assume that

a) $f$ is bijective on $\mathbb C$-points

b) $X$ is connected

c) $Y$ is normal.

Does it imply that $f$ is an isomorphism? If not, are there stronger reasonable conditions under which it is true?

  • 6
    $\begingroup$ That is true and follows from Grothendieck's formulation of Zariski's Main Theorem. Since $f$ is finite type, separated and has finite fibers, there exists a factorization $i:X\hookrightarrow \overline{X}$, $\overline{f}:\overline{X}\to Y$ with $i$ a dense open immersion and $\overline{f}$ a finite morphism. By Zariski's Main Theorem, $\overline{f}$ is an isomorphism. Thus, $f$ is an open immersion. Since $f$ is surjective, $f$ is an isomorphism. $\endgroup$ Mar 9, 2017 at 15:03
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    $\begingroup$ @JasonStarr: Of course, ZMT is the part about the existence of $i$ (for which we can arrange that $\overline{X}$ is also a variety; e.g., reduced, etc.), not the fact that $\overline{f}$ is an isomorphism. :) $\endgroup$
    – nfdc23
    Mar 9, 2017 at 16:26
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    $\begingroup$ This seems to be a duplicate of several MO questions, e.g. mathoverflow.net/questions/73321/… or mathoverflow.net/questions/113858/… $\endgroup$ Mar 9, 2017 at 16:28
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    $\begingroup$ @nfdc23 I should have been more precise. First I use Grothendieck's formulation for existence of $i$, but then I use the original formulation to know that $\overline{X}\to Y$ is an isomorphism. I expanded the comment in the answer below. $\endgroup$ Mar 9, 2017 at 17:19
  • $\begingroup$ what is an "algebraic variety"? $\endgroup$
    – user138661
    Apr 21, 2019 at 18:09

1 Answer 1


I am just expanding the comment above. The statement is local on $Y$, so assume that $Y$ is connected. Since $Y$ is normal, this is equivalent to assuming that $Y$ is irreducible. Grothendieck's formulation of Zariski's Main Theorem is EGA $\textrm{III}_2$, Théorème 4.4.3, p. 136. According to that theorem, there exists a factorization of $f$, $f=\overline{f}\circ i$ where $i:X\hookrightarrow\overline{X}$ is a dense open immersion, and where $\overline{f}:\overline{X}\to Y$ is a finite morphism. Identify $X$ with the image of $i$.

Since $f$ is bijective on points, there exists a unique irreducible component $\overline{X}_0$ of $\overline{X}$ that dominates $Y$. The restriction of $\overline{f}$ to $\overline{X}_0$ is a finite, surjective morphism that is generically one-to-one. Since the characteristic is $0$, this morphism is birational (in positive characteristic, it might be purely inseparable but not birational). Thus, by the classical form of Zariski's Main Theorem, $\overline{f}$ restricts to an isomorphism from $\overline{X}_0$ to $Y$, cf. Mumford's Red Book of Varieties and Schemes, p. 288.

By way of contradiction, assume that $\overline{X}_0$ is a proper subset of $\overline{X}$. Denote by $Z \subset \overline{X}$ the union of all irreducible components different from $\overline{X}_0$. Since $X$ is a dense open in $\overline{X}$, also $X\cap Z$ is a dense open in $Z$. Thus, also $(X\cap Z) \setminus (\overline{X}_0\cap Z)$ is a dense open subset of $Z$. Denote by $W$ the image in $Y$ of this open subset. Since $f$ is injective, the constructible subset $W$ is disjoint from the dense open $V=f(X\cap \overline{X}_0)$ of $Y$. Thus, the closure of $W$ is disjoint from $V$. Since $\overline{f}$ is finite, this closure equals the image of $Z$. Thus, $X\cap \overline{X}_0$ is disjoint from $\overline{X}_0\cap Z$, i.e., $X\cap \overline{X}_0\cap Z$ is empty. Thus, the open subsets $\overline{X}_0\setminus (\overline{X}_0\cap Z)$ and $Z\setminus (Z\cap \overline{X}_0)$ of $\overline{X}$ pullback to disjoint open subsets of $X$ that cover $X$. Since $X$ is connected, this is a contradiction.

Therefore, $\overline{X}_0$ equals all of $\overline{X}$. Since $\overline{X}_0\to Y$ is an isomorphism, it follows that $f$ is an open immersion. Since $f$ is also surjective, $f$ is an isomorphism.


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