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Assuem we have a finite surjective map between two irreducible, separated schemes, $f:X \rightarrow Y$, and for a dense open $U \subset Y$ and for any $y \in U$, $|X_y| =1$, then can we say $f$ is injective everywhere?

We can also assume that $Y$ is normal.

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  • $\begingroup$ I suggest you make the map flat and separated. I don't have a proof, but I don't see a counterexample of the top of my head. It also excludes blow-ups, and such. $\endgroup$
    – jmc
    Dec 19, 2014 at 10:13
  • $\begingroup$ flat for finite maps would be equivalent to have fibers with the same cardinality everywhere, which would give the answer right away. $\endgroup$ Dec 19, 2014 at 10:17
  • $\begingroup$ Wait, I missed that the map is finite. But then the blow-up example didn't work anyway. $\endgroup$
    – jmc
    Dec 19, 2014 at 10:19
  • $\begingroup$ Of course my counterexample also fails, since $f$ finite implies $f$ affine, and that clearly fails in my answer. $\endgroup$
    – jmc
    Dec 19, 2014 at 10:26

3 Answers 3

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With some additional hypotheses this follows from Zariski's Main theorem.

If $f \colon X \to Y$ is a birational projective morphism between noetherian integral schemes, then the inverse image of every normal point of $Y$ is connected. [Hartshorne (1977, Corollary III.11.4)]

See also: https://en.wikipedia.org/wiki/Zariski%27s_main_theorem


If you make the assumption that $X$ and $Y$ are noetherian and integral, I guess you will be there.

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The answer is "no" without some further hypothesis on $X$ and $Y$. For instance, let $Y$ be $\text{Spec}(\mathbb{Z}_{5\mathbb{Z}})$, the localization of $\mathbb{Z}$ at the prime ideal $5\mathbb{Z}$, let $X$ be $\text{Spec}(\mathbb{Z}_{5\mathbb{Z}}[x]/\langle x^2 + 1 \rangle)$, and let $U$ be the distinguished open where $5$ is invertible.

However, for finite type schemes over a field, this does follow from Zariski's Main Theorem.

Edit. I did not see jmc's answer below when I posted this answer. This answer is basically the same as jmc's answer. It is important to note, even for finite type schemes over a field, being injective on points does not necessarily imply the morphism is birational, e.g., Frobenius morphisms.

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  • $\begingroup$ I think I made the edit to my answer at the same time you posted yours. However, you give a counterexample, and an important remark. So I think your answer is "better". $\endgroup$
    – jmc
    Dec 19, 2014 at 12:04
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The situation is much worse; you don't need pathological examples. Just take any blow-up: the blow-down map is surjective and injective almost everywhere. I think your claim is true if both $X$ and $Y$ are projective curves.

Added in proof: sorry, I meant by "finite" generically finite-to-one. Otherwise, "A variety X over a field is normal if and only if every finite birational morphism from any variety Y to X is an isomorphism", see Normal scheme

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  • $\begingroup$ Of course. Good answer! So some kind of "minimal"-condition is needed. $\endgroup$
    – jmc
    Dec 19, 2014 at 10:11
  • $\begingroup$ I am not sure when the blow up would be finite, one may check easily, it is proper and binational. But finite, I don't know! $\endgroup$ Dec 19, 2014 at 10:24
  • $\begingroup$ @SaraCastillejo My answer edited. $\endgroup$ Dec 19, 2014 at 10:38

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