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!

  • $\begingroup$ 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. $\endgroup$
    – Hagen Knaf
    Jan 26, 2010 at 10:04
  • $\begingroup$ Can you give an example? I have very bad intuition about these stuff. $\endgroup$
    – user709
    Jan 26, 2010 at 11:15

3 Answers 3


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.


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).

  • $\begingroup$ 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? $\endgroup$
    – user709
    Jan 26, 2010 at 10:00
  • $\begingroup$ 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. $\endgroup$
    – user19475
    Jan 26, 2010 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 $2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.