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I would like to show the following (note: it is not an assigned problem, so it may be false) (EDIT: Indeed it is false, see end of post):

Suppose $f:X \rightarrow Y$ is a finite, surjective morphism of integral nonsingular curves over $k$ algebraically closed. If the fiber over some point $p$ of $Y$ contains exactly one point of $X$, then $f$ is birational. (Since a finite map is affine, we can reduce to the case $X=SpecA$, $Y=SpecB$)

In browsing around trying to prove this, I came upon Vakil's definition of the degree of a finite morphism at a point (Section 13.7 of his online notes). He defines the degree of $f$ at $p$ to be the dimension of $(f_*O_X)_p \otimes _{O_{Y,p}} k(p)$ as a $k(p)$ vector space, where $k(p)$ denotes the residue field of $Y$ at $p$. In an exercise, he asserts that degree is an upper semicontinuous function of the point on $Y$.

My feeling is that this solves my problem. Degree as defined above is the dimension of the vector space of functions on the fiber over the field $k(p)$ (=$k$, since I assumed algebraically closed). Morally, this should be the number of points in the fiber, with a basis of functions being those which send one point in the fiber to 1 and the others to 0. Then, since degree counts the fiber, is upper semicontinuous, and equal to 1 at some point, then $f$ must be injective in a neighborhood, hence birational.

My problem is that "morally" is not a proof and weird commutative algebra things can happen. So if my thought process is correct, can someone prove it to me? Also, if I am indeed on the right track, then in what kind of generality can one say that "degree (as defined above) counts the fiber"? Thanks.

EDIT: Li Yutong gives the example of projection of the curve $xy = z^2$ in $\mathbb{P}^2$ to $\mathbb{P}^1$. It is a 2:1 map except over the points (0:1) and (1:0) where it is 1:1. Since my proposition fails in such a nice case, my thinking about Vakil's "degree" must be way wrong. Can somebody point out where I'm jumping the shark?

Cass
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  • I happen to ask a very similar question, and the following link might be helpful:http://math.stackexchange.com/questions/341281/number-of-points-in-the-fibre-and-the-degree-of-field-extension – Li Yutong Aug 31 '13 at 23:58
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    Also the characteristic of $k$ should be $0$, otherwise the Frobenius morphism will give you a counter example. – Li Yutong Sep 01 '13 at 00:00
  • Thanks Li Yutong. I had skimmed through that thread earlier, but I don't see how the posted answer would address my question. I don't know anything about the fiber over the generic point in this case. – Cass Sep 01 '13 at 00:23
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    If $p$ is arbitrary point, then I doubt the claim might be fails. Consider the example : $xz=y^2 \subset \mathbb{P}^2$ maps to $\mathbb{P}^1$ by $[x,y,z] \mapsto [x,z]$, this is a $2-1$ map but over $[0,1]$, the fibre is a single point. – Li Yutong Sep 01 '13 at 00:42
  • Ah wow, thanks. So number of points in the fiber is not upper semicontinuous in even the nicest of cases. Something must be dramatically wrong with my interpretation of degree. Can somebody see what it is? – Cass Sep 01 '13 at 00:50
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    Dear Cass, You are not taking into account ramification, or (equivalently) multiplicity in the fibres. Another way to describe @LiYutong's example is that it is the map $z \mapsto z^2$ from $\mathbb A^1$ to itself, extended (in the unique manner) to a map from $\mathbb P^1$ to itself. The fibre over a point $z$ is given by solving $w^2 = z$. At most points (namely $z \neq 0, \infty$) this gives two distinct values of $w$. Over $0$ and $\infty$, you get just one point ($w = 0$ or $\infty$) but with multiplicity two; concretely, $0$ is a double root of the equation $w^2 = 0$. ... – Matt E Sep 01 '13 at 01:56
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    ... The scheme theoretic fibre also knows this; the scheme-theoretic fibre over $z = 0$ is Spec $k[w]/(w^2)$, which is a single point, but the ring $k[w]/(w^2)$ is two-dimensional as a $k$-algebra, so the degree is two, not one. Regards, – Matt E Sep 01 '13 at 01:57
  • Ok cool, so we can say that degree counts the fiber with multiplicity? I guess I was being very thick, because I did happen to know that the fiber over my point p in the proposition was a single point WITH multiplicity 1. So if I modify the hypothesis to say that the degree over p is 1, the conclusion birational follows from semicontinuity of degree, yes? – Cass Sep 01 '13 at 02:18

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