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We have a bijective morphism $f:X\to Y$ of quasi-affine varieties (say over $\Bbb{C}$).

Can $f$ fail to be an isomorphism if $X$ and $Y$ are smooth?

Heitor Fontana
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  • I would think there is a bijective morphism between $\Bbb A^1_{\Bbb C}$ and the variety in $\Bbb A^2_{\Bbb C}$ defined by $y^2 - x^3 + x - 1 = 0$, but they are not isomorphic. I'm posting as a comment because I am not certain. – Arthur Sep 25 '15 at 09:57
  • @arthur: this equation defines a (non-singular) elliptic curve, which surely cannot be isomorphic to $\mathbb{A}^1$ – Dima Sustretov Sep 25 '15 at 10:32
  • @DimaSustretov The non-isomorphism I am certain about. The existence of a bijective morphism, on the other hand... – Arthur Sep 25 '15 at 12:22

2 Answers2

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We might as well think that our morphism is bijective on scheme-theoretic points, i.e. is quasi-finite.

By Zariski's main theorem a quasi-finite map $X \to Y$ factors as a composition of an open immersion and a finite map, $X \to W \to Y$. A degree 1 finite morphism ($W \to Y$) with normal target, which $Y$ is since it is smooth, has to be isomorphism in char 0 (would be radicial in general): the field extension it enduces on the general fibre is trivial, since it is of degree 1.

Since the composition is surjective $X \to W$ is an isomorphism too, and we are done.

You don't actually need quasi-afineness assumption.

KReiser
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Dima Sustretov
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    This is certainly nice. That said, you could remark that in the case that $X$ and $Y$ are projective and we're over $\mathbb{C}$ one can deduce this from GAGA and the relatively simple analogue analytically (that a bijective holomorphic map is an isomorphism) – Alex Youcis Sep 25 '15 at 11:36
  • Also, you don't need quasi-affiness, but you should remark that you are using several things (notably separatedness). – Alex Youcis Sep 25 '15 at 11:36
  • Also, lastly, (more for the OP, since I assume you know this) we can deduce quasi-finiteness from finiteness on closed points, at least if the map is proper. We can then use Chevalley's theorem to say that fiber dimension is upper-semi-continuous which says that ${y:\dim X_y\geqslant 1}$ is a closed subvariety wth no $\mathbb{C}$-points, so it's empty. – Alex Youcis Sep 25 '15 at 11:40
  • Thank you Dima. What are the standard references for ZMT stated in this form? – Heitor Fontana Sep 25 '15 at 13:35
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    Note that you need isomorphism on residue fields, otherwise $\operatorname{Spec}\mathbb{C}\rightarrow\operatorname{Spec}\mathbb{R}$ is bijective on points but not an isomorphism, Arthur's answer is also a counter-example. There is the notion of seminormal scheme : $Y$ is seminormal if every finite morphism $X\rightarrow Y$ bijective on points and inducing isomorphism on the residue fields is an isomorphism. Your answer proves that normal schemes are seminormal. There is also a purely algebraic characterization of seminormality due to Swan. – Roland Sep 25 '15 at 13:44
  • @Heitor: I remember that there is a proof in Raynaud's book "Anneaux locaux henseliennes", but I expect an exposition in English to be in Stacks project by now (haven't checked it though). – Dima Sustretov Sep 25 '15 at 14:00
  • @Roland: yes, the statement is really about a bijection on $k$-points of $k$-varieties – Dima Sustretov Sep 25 '15 at 14:04
  • A finite extension $A \subset B$ of degree zero always induces equality of fraction fields, after all the degree is defined to be the degree of the field extension $[Q(A): Q(B)]$. But why does the morphism $W \to Y$ have degree $0$? – red_trumpet Sep 05 '19 at 16:06
  • Thinking about it, it should be degree $1$, right? – red_trumpet Sep 05 '19 at 16:17
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Hartshorne gives the Froebenius morphism as an example (exercise I.3.2 b). If $k=\bar k$, and $\operatorname{char}(k)=p\neq 0$, then the map $\Bbb A^1_k\to \Bbb A^1_k$ given by $x\mapsto x^p$ is bijective (and even a homeomorphism), but not an isomorphism of varieties.

Arthur
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