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My exercise:

Let $f:X\rightarrow Y$ be a dominant morphism of curves. For any dominant morphism, the degree of it is defined to be $[K(X):K(Y)]$ with $K(Y)$ identified with $f^*(K(Y))$.
Prove that the fibres of $f$ have at most $deg(f)$ points if $Y$ is non-singular.

The notes we are working out skim over the whole degree/ramification story. Other sources(i.e. Shafarevich Book 1) treat more material, but only for $f$ finite.

Could anyone elaborate(e.g. give some concrete examples) on what the connection is between the algebraic definition of degree and the behaviour of $f$ on the varieties and/or give a hint for the exercise so I can figure this connection out for myself?

bbnkttp
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  • Mohamed, if the below doesn't answer your question, please let me know :) – Alex Youcis May 04 '14 at 11:56
  • I think http://math.stackexchange.com/questions/341281/number-of-points-in-the-fibre-and-the-degree-of-field-extension?rq=1 might work for me.. – bbnkttp May 04 '14 at 12:03

1 Answers1

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I assume by curves you mean integral projective curves over some field $k$. Then, maybe the question will be answered by the following fact:

Theorem: If $f:X\to Y$ is a map of curves over $k$. Then, either $f$ is constant, or it is surjective. If it is the latter, then it is finite.

The proof of this is not too difficult. Since $X$ and $Y$ are projective, by the Cancellation Theorem (see Vakil), $f$ is projective, and so closed. Thus, $f(X)$ is an irreducible closed subset of $Y$. Since $Y$ is a curve, this forces to $f(X)$ to be a point, or all of $Y$.

Now, if $f$ is surjective, then it is necessarily quasi-finite. Indeed, it's clear that the only point which maps to the generic point is the generic point (because $f$ is closed). If $x\in Y$ is a closed point, then $f^{-1}(y)\subseteq X$ is closed. If it's not finite, then since $X$ carries the cofinite topology, it must be dense, which contradicts that $f$ isn't constant. But, $f$ is then quasifinite and projective, and so finite (by ZMT).

[There is an easier proof of this theorem, I believe in Hartshorne, under some regularity assumptions. I don't recall it though].

Alex Youcis
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  • The domain is allowed to be singular and both curves are allowed to be non-complete, so We do have dominant iff quasi-finite, but ZMT does not give finite. – bbnkttp May 04 '14 at 12:02
  • @MohamedHashi You can always lift this to a map between projective curves by embedding the domain in a smooth projective model, similarly for the codomain, and lifting to a morphism by the valuative criterion for properness. – Alex Youcis May 04 '14 at 12:05
  • @MohamedHashi If this doesn't help you though, let me know, and I can think some more. Or, if you're happy with the algebra result you linked to, that's fine too :) – Alex Youcis May 04 '14 at 12:06
  • I prefer your method, but where do you get the embedding from? My notes only state that there is a complete smooth curve birationally equivalent to it. I will think about the other parts. – bbnkttp May 04 '14 at 12:11
  • @MohamedHashi Ah, yes, you're right. You don't need any regularity conditions to find a birational model, but you do need regularity to get an open embedding (see 17.4.2 of Vakil). So, while the proof I wrote works for any two projective curves (both of which can be singular), to show we can always reduce to the projective case, it looks like we might need to assume that both are regular. – Alex Youcis May 04 '14 at 12:13
  • If you assume regularity and get an open immersion, it follows that $\hat{f}$, the extension of $f$, is finite. Note that the restriction of a proper map to a open set does not have to be proper. A map is finite iff it is proper and quasi-finite. The properness of $f$ is not guaranteed by properness of $\hat{f}$, so the same holds for finiteness. If this can be resolved, I get my result, since it is clear that the degree of the extension of $f$ is $deg(f)$. – bbnkttp May 04 '14 at 13:17
  • @MohamedHashi I don't quite follow. Morphisms between projective things are always proper. This follows from the Cancellation Theorem, as in Vakil. I think that's what you're asking? – Alex Youcis May 04 '14 at 13:21
  • let us continue this discussion in chat – bbnkttp May 04 '14 at 13:36
  • @MohamedHashi I'm not sure if you were fully happy with our chat. If so, you might want to accept this to let people know it's been solved. – Alex Youcis May 04 '14 at 22:31
  • This is not an answer since regularity is not assumed. – bbnkttp May 06 '14 at 09:28
  • @MohamedHashi Take the normalization, and consider the composition. Then, apply our chat. This seems to never end. – Alex Youcis May 06 '14 at 09:38