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Let $X$ be an affine variety and $Y$ a projective variety, both integral (reduced and irreducible). Assume that $\phi:X\to Y$ is a birational morphism. I would venture to say that $\phi$ can not be surjective, but I have no proof for the statement. So, my question: Can $\phi$ be surjective? If no, why?

Jesko Hüttenhain
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  • How about the identity map of the empty scheme? (One may be tempted to disregard it as a variety, though.) – Ben Mar 03 '15 at 17:06
  • @Ben: I am not so much interested in pathologic cases, you may safely assume that all the objects are actually interesting. – Jesko Hüttenhain Mar 03 '15 at 18:11
  • The only thing that comes to my mind immediately is that if $Y$ is factorial, then $\phi^{-1}$ can't be defined in codimension one, because otherwise $\phi$ would have to be an isomorphism, leaving stupid pathological cases only. This, however, doesn't seem very helpful. – Ben Mar 03 '15 at 18:28
  • Do you require $X,Y$ intergal? You could maybe play some sort of game with $\mathbb{A}^1\cup{pt}\to \mathbb{P}^1$, but birationality doesn't really work for non-integral schemes... – KReiser Mar 04 '15 at 18:03
  • @KReiser: Yea, I edited my question, this only makes sense for irreducible varieties. – Jesko Hüttenhain Mar 04 '15 at 18:08
  • Such a map would give you an isomorphism of an open dense subset $U$ of $X$ with $Y$. Thus $U$ is projective. Now isn't any projective subvariety closed (e.g. by the the fact that it is proper over the base)? So $U$ has to be closed, thus $X = U$ and for this case we know, that no such thing exists. – karl_christ Mar 04 '15 at 18:15
  • @karl_christ: Such a map would only give me an isomorphism of a dense open subset $U\subseteq X$ with another dense open subset $V\subseteq Y$ of $Y$. This proves nothing. – Jesko Hüttenhain Mar 04 '15 at 18:18
  • Ah, but then what do you mean by surjective? Dominant? – karl_christ Mar 04 '15 at 18:18
  • @karl_christ: No, I mean surjective. Every fiber is nonempty. – Jesko Hüttenhain Mar 04 '15 at 18:19

1 Answers1

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Here's a thought to show that this cannot happen if $\dim Y>0$. It's 5:30 AM though, and so I can't in good conscience post this not as community wiki. :) Anyone should feel free to point out a silly mistake, or finish the proof in some way. It proceeds as classic dévissage.

Indeed, suppose first that $X$ and $Y$ are regular curves. Then, this certainly cannot happen. Indeed, open embed $X$ into a regular projective curve $C$. Suppose that the birational map gives an isomorphism $X\supseteq U\xrightarrow{\approx} V\subseteq Y$ and let $f:V\to U$ be the inverse isomorphism. Then, consider the composition $V\to U\to C$. By the curve-to-projective theorem this extends to a map $X\to C$ which must be an isomorphism since it's degree $1$. But, in particular, by the surjectivity of $X\to Y$, this implies that $C=X$ which contradicts that $X$ is affine.

Suppose now that $X\to Y$ is as in the problem statement, but now with no assumptions on $\dim Y$. Take a curve $C\subseteq Y$. Then, consider the fiber product

$$\begin{matrix}X_C & \to & X\\ \downarrow & & \downarrow\\ C & \to & Y\end{matrix}$$

By assumption, the map $X_C\to C$ is surjective and birational (edit: this may need to be tweaked, but I think it's fixably tweakable. see comments below edit v 2.0: this is possible by a simple application of Bertini's lemma, when $Y$ is normal. To show that it suffices that $Y$ is normal, take its normalization and show, similarly to below, that everything works out OK) . Moreover, since $C$ closed embeds into $Y$ it's projective, and since $X_C$ closed embeds into $X$ it's affine.

Now, by checking component by component we may (hopefully I'm not making a silly error here--I haven't checked through all the details) assume that $C$ and $X_C$ are irreducible. Moreover, by passing to their reduced subschemes, we may assume they are reduced, and thus integral.

Now, consider the normalization $C'\to C$ and form, once again, the fibered diagram

$$\begin{matrix}X_{C'} & \to & X_C\\ \downarrow & & \downarrow\\ C' & \to & C\end{matrix}$$

Using similar ideas to above, we see that $X_{C'}$ is affine, $C'$ is projective, and we may assume that $X_{C'}$ is integral. Then, finally, by passing to the normalization of $X_{C'}$, call it $X'_{C'}$ and looking at the composition

$$X'_{C'}\to X_{C'}\to C'$$

we obtain a surjective birational map with $X'_{C'}$ affine, and $C'$ projective. But, since both of these objects are a curve, this contradicts our original case.

Alex Youcis
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  • @GeorgesElencwajg Will probably be able to spot any error in the above, if he'd be so kind as to read it. – Alex Youcis Mar 10 '15 at 12:45
  • Dear Alex, I have not read your answer yet but I don't think I'll find an error since you are one of the most reliable, competent and ingenious users on this site. What is puzzling me is that I came to this page on my own, without being advised that you had mentioned me in your comment (is "pinged me" the correct term ?) Is that because there is no interval between Georges and Elencwajg or because you wrote only fifteen minutes ago or because of the answer being community wiki or for some other reason? Might I ask you to make a new experiment below this comment ? – Georges Elencwajg Mar 10 '15 at 13:06
  • I might have missed something, but I'm not so certain that $X_C\to C$ is birational. Even if $C$ is not entirely contained in the fundamental locus of the inverse of $X\to Y$, the fibre product $X_C$ might be too big. This can be fixed by passing to a suitable irreducible component of $X_C$, right? – Ben Mar 10 '15 at 13:58
  • @GeorgesElencwajg Thanks for the kind words! It likely didn't ping you since you weren't in the thread. :) Let me know if you find any issues with the proof. – Alex Youcis Mar 10 '15 at 17:11
  • @Ben Yeah, this is a good point. I think choosing our $C$ correctly (densely intersecting fundamental locus, which we should be able to do), and as you mention, passing to a suitable component of $X_C$ should work--namely any component interesecting with the preimage of the fundamental locus. I'll have to ruminate more to make sure that this is entirely possible. Thanks for pointing it out! – Alex Youcis Mar 10 '15 at 17:14
  • For any readers, the correct statement, which eluded me last night, is that the base change of birational by dominant is birational. I still think this issue might be fixable in this special case, but I just wanted to point out how to fix the incorrect statement in general. – Alex Youcis Mar 10 '15 at 17:32
  • You're welcome! To me, the proof seems fine, except precisely this point, where, starting with irreducible $C$ right away, say, we need to make sure that there is an irreducible(!) component of $X_C$ which surjects onto $C$. It sounds plausible, but I think it's non-trivial. – Ben Mar 10 '15 at 17:33
  • @Ben Yeah, I agree. When I have time tonight/tomorrow I'll think of how this can, if it can be, fixed. It's also worth mentioning, in case there is some obvious counterexample to the general case ($C$ intersecting good locus densely but no component of pullback intersecting good locus densely surjects) that we can always just pick a different $C$. So, we could just not worry about generalities by thinking if we can pick such a $C$ that does our job. PS, as you noted, and as is clear in the above proof, we may as well assume that $C$ is irreducible to begin with. – Alex Youcis Mar 10 '15 at 17:41
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    @Ben This is figured out. I'll write it up later. By normalizing Y you may assume it's normal then use Bertini to find the desired curve. – Alex Youcis Mar 10 '15 at 19:42
  • @JeskoHüttenhain No problem. I think there is probably an easier way with messing around with normalization maps, but this is intuitively nice for me (even though it requires a bit more in the way of justification): such a map clearly doesn't exist for curves, and any higher-dimensional example would produce an example with curves. – Alex Youcis Mar 11 '15 at 10:09