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This is prop 2.6b on p.28 of Silverman's the Arithmetic of Elliptic curves.

It says that let $\phi: C_1 \rightarrow C_2$ be a non-constant map of projective non-singular irreducible curve. (probably over an algebraically closed field, but I am not too sure) Then for all but finitely many $Q \in C_2$, #$\phi^{-1} (Q) = deg_s (\phi)$, where RHS is the separability degree of the function fields.

I don't understand Silverman's proof.

The proof just says that it is Hartshorne II.6.8, and I don't understand how it is related to this proposition at all. Hartshorne II.6.8 roughly states that if $f: X \rightarrow Y$ is a morphism where $X$ is a complete nonsingular curve over an algebraically closed field $k$, and $Y$ is any curve over $k$, then either $f(X) = pt$ or $Y$, and in the latter case, $f$ is finite morphism and $[K(X):K(Y)] < \infty$.

Can anyone show a proof of the proposition?

I failed to show that the set of all such $Q$ is open myself, can anyone shed some light on this? Thanks!

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You need an algebraically closed base field in the statement. Otherwise there can be points $Q\in C_1$ such that the degree $[k(Q):k(\phi (Q))]>1$ in which case the number of points in the fibre $\phi^{-1}(\phi (Q))$ is less than the degree. –  Hagen Knaf Jan 26 '10 at 10:04
    
Can you give an example? I have very bad intuition about these stuff. –  Ho Chung Siu Jan 26 '10 at 11:15

3 Answers 3

up vote 6 down vote accepted

Here is a complete proof: as remarked in the answer by Norondion, we can reduce to the case when $C\_1 \rightarrow C\_2$ is generically separable, i.e. $k(C\_1)$ is separable over $k(C\_2)$. Let $A \subset k(C\_1)$ be a finite type $k$-algebra consisting of the regular functions on some non-empty affine open subset $U$ of $C\_2$ (it doesn't matter which one you choose), so that $k(C\_2)$ is the fraction field of $A$.

By the primitive element theorem, we may write $k(C\_1) = k(C\_2)[\alpha]$, where $\alpha$ satisfies some polynomial $f(\alpha) = \alpha^n + a_{n-1}\alpha^{n-1} + \cdots + a_1 \alpha + a_0 = 0,$ for some $a_i$ in $K(C\_2)$.

Now the $a_i$ can be written as fractions involving elements of $A$, i.e. each $a_i = b_i/c_i$ for some $b_i,c_i \in A$ (with $c_i$ non-zero). We may replace $A$ by $A[c\_0^{-1},\ldots,c\_{n-1}^{-1}]$ (this corresponds to puncturing $U$ at the zeroes of the $c_i$), and thus assume that in fact the $a_i$ lie in $A$.

The ring $A[\alpha]$ is now integral over $A$, and of course has fraction field equal to $k(C_2)[\alpha] = k(C_1)$. It need not be that $A[\alpha]$ is integrally closed, though. We are going to shrink $U$ further so we can be sure of this.

By separability of $k(C_1)$ over $k(C_2)$, we know that the discriminant $\Delta$ of $f$ is non-zero, and so replacing $A$ by $A[\Delta^{-1}]$ (i.e. shrinking $U$ some more) we may assume that $\Delta$ is invertible in $A$ as well.

It's now not hard to prove that $A[\alpha]$ is integrally closed over $A$. Thus $\text{Spec }A[\alpha]$ is the preimage of $U$ in $C_1$ (in a map of smooth curves, taking preimages of an affine open precisely corresponds to taking the integral closure of the corresponding ring).

In other words, restricted to $U \subset C_2$, the map has the form $\text{Spec }A[\alpha] \rightarrow \text{Spec }A,$ or, what is the same, $\text{Spec }A[x]/(f(x)) \rightarrow \text{Spec A}$.

Now if you fix a closed point $\mathfrak m \in \text{Spec }A,$ the fibre over this point is equal to $\text{Spec }(A/\mathfrak m)[x]/(\overline{f}(x)) = k[x]/(\overline{f}(x)),$ where here $\overline{f}$ denotes the reduction of $f$ mod $\mathfrak m$. (Here is where we use that $k$ is algebraically closed, to deduce that $A/\mathfrak m = k,$ and not some finite extension of $k$.)

Now we arranged for $\Delta$ to be in $A^{\times}$, and so $\bar{\Delta}$ (the reduction of $\Delta$ mod $\mathfrak m$, or equivalently, the discriminant of $\bar{f}$) is non-zero, and so $k[x]/(\bar{f}(x))$ is just a product of copies of $k$, as many as equal to the degree of $f$, which equals the degree of $k(C_1)$ over $k(C_2)$. Thus $\text{Spec }k[x]/(\bar{f}(x))$ is a union of that many points, which is what we wanted to show.

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Do you have questions to Hartshorne's proof or just how to deduce Silverman's result from it?

You can factor the field extension of the function field into a purely separable and purely inseparable extension, so WLOG $\phi$ is separable as a purely inseparable morphism is a universal homeomorphism. As $f$ is finite, it is affine, so it looks locally like $\mathrm{Spec}(B) \to \mathrm{Spec}(A)$. As $\phi$ is separable, the discriminant of $B/A$ is $\neq 0$, which gives us that $f$ is unramified outside a finite set of points (the primes which don't divide the discriminant).

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Thanks for your answer. I don't know how to deduce Silverman's result from the proposition in Hartshorne. In fact I don't even see how they are related. As for your proof, the last line "f is unramified outside a finite set of points ... " is not something I was aware of. I know that for a number field $K$, $p$ ramifies in $K$ iff $p$ divides disc $K$. So are you saying that the relative version is also true? If so, where may I find a reference? Thanks! Also, is "f is finite" what Silverman's referring to when he points to that proposition in Hartshorne? –  Ho Chung Siu Jan 26 '10 at 10:00
    
It is true for Dedekind schemes (which curves are). Check out this script of Szamuely: renyi.hu/~szamuely/gal6-7.pdf Yes, I use that $f$ is finite. –  Timo Keller Jan 26 '10 at 15:23

Example of a finite morphism with inert points:

Define $C_1 := \mathrm{Spec}(\mathbb{R}[x,y])$, where $\mathbb{R}[x,y]:=\mathbb{R}[X,Y]/(X^2+Y^2+1)$ and $C_2 := \\mathbb{A}^1_\mathbb{R}$.

Let the morphism $\phi$ be given by the ring extension $\mathbb{R}[x,y] / \mathbb{R}[x]$.

Then the fibre above every rational point of $C_2$ consists of one element only, because the equation $X^2+Y^2+1=0$ has no real solutions. However $\phi$ has degree $2$.

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