Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Throughout I would like to work over an algebraically closed field of characteristic 0 (so no separability issues), say $k$. My question is the following:

Do there exist two curves $X$ and $Y$ and a necessarily finite morphism $f:X \rightarrow Y$ such that $$[k(X):k(Y)] > \text{max}_{P \in X}\{e_P\}$$ where $k(X)$ denotes the function field of $X$ and $e_P$ denotes the ramification index of the map at a point $P \in X$?

My thoughts/attempts so far: Hurwitz's Theorem tells us that for a finite separable morphism of curves $f:X \rightarrow Y$ we have $$2g(X)-2=(\text{deg }f)(2g(Y)-2)+\text{deg }R$$ where $\text{deg }f=[k(X):k(Y)]$, $g(X)$ is the genus of $X$ and $R$ is the ramification divisor. That is, $\text{deg }R=\sum_{P\in X} (e_P-1)$.

Using this I can note that a simple example may come from a morphism between two genus 0 curves that gives a field extension of degree 3 and ramifies at 4 points. In this case each ramification index would have to be 2 and hence the condition would be satisfied. I thought I had an example of higher genus using the double cover of the projective line over $k$ by an elliptic curve, however at one of the four points of ramification the index would have to be at least equal to the field extension degree.

A related extension to this question is whether it is possible to have such a map that any possible ramification type occurs - that is, given two fixed genus values (necessarily $g(X)\geq g(Y)$ by Hurwitz) and $[k(X):k(Y)]$ fixed, can a morphism be constructed that ramifies at any allowable number of points with indices? Now I do not insist on the ramification indices all being smaller than the degree of the field extension. By allowable here I mean that deg $R$ is controlled by the fixed values by Hurwitz's Theorem, and so the greatest number of ramification points is deg $R$ each with index 2, but other partitions of deg $R$ with integers $\geq 2$ are possible.



share|cite|improve this question
up vote 1 down vote accepted

Take an étale covering $f:X\to Y$ of degree $n\gt 1$ (i.e. with $n$ sheets) between two elliptic curves ( such coverings exist for any $n $ ) .
You have $[k(X):k(Y)]=n$ but all the $e_P=1 $ and thus $$[k(X):k(Y)]=n\gt \text{max}_{P \in X}\{e_P\}=1$$

Edit: an explicit example.
To show how simple it is to construct such étale coverings, consider an arbitrary lattice $\Lambda =\mathbb Z\omega_1\oplus \mathbb Z\omega_2\subset \mathbb C$ and the related lattice $\Lambda' =\mathbb Z\omega_1\oplus \mathbb Z n\omega_2\subset \mathbb C$.
The morphism $\mathbb C/\Lambda \to \mathbb C/\Lambda':[z] \mapsto [nz]' $ is then an étale covering of degree $n$ .
(In the jargon of abelian varieties: an isogeny with kernel $\lbrace [0],[\frac {1}{n}\omega_1],[\frac {2}{n}\omega_1],...,[\frac {n-1}{n}\omega_1]\rbrace $)

share|cite|improve this answer
For the other questions you might take a look here – Georges Elencwajg Feb 9 '12 at 15:53
Thanks, this is great. The link too is very interesting. – Andrew Davies Feb 10 '12 at 12:52

Assume $Y=\mathbb{P}^1$ and consider the cyclic cover $f:X\rightarrow Y$ defined by an equation of the form

$ y^n = (x-a_1)^{v_1}\ldots (x-a_m)^{v_m}, $

that is take the affine curve $C\subset\mathbb{A}^2$ defined by this equation, embed it into $\mathbb{P}^2$ and take the closure, finally normalize it to remove the singularities.

One has to assume that the $a_i\in k$ be pairwise distinct, and that for every prime $p$ dividing $n$, there exists $v_k$ not divisible by $p$. The latter condition ensures the irreducibility of the curve $C$ and thus $X$.

Then a point $x\in X$ is ramified if and only if:

  • $f(x) = [a_k:1]$ for some $k$,
  • $f(x) = [1:0]$.

In the first case the ramification index equals $\frac{n}{\mathrm{gcd} (n,v_k)}$, in the second case $\frac{n}{\mathrm{gcd}(n,m)}$.

Note that $[K(X):K(Y)]=n$. Thus, if $n$ is not prime, one can arrange the $v_k$ and $m$ in such a way, that the requirements are fullfilled.

share|cite|improve this answer
Are there some more conditions that are needed on the numbers you choose? For example if I take $y^6=x^2(x-1)^2(x-2)^3$ then Hurwitz tells me $2g(X)=-10+ramification=-10+(3+3+3+2-4)=-3$ unless I have made a silly error. – Andrew Davies Feb 10 '12 at 12:57
You did not list all ramification points on the right hand side: for example there are 2 ramification points on $X$ lying above the point $[0:1]$. The same holds for $[1:1]$, while there are 3 ramification points lying above $[2:1]$ and $[1:0]$. Altogether this yields $2g(X)=-10 + 2*2 + 2*2 + 3*1 + 3*1 = 4$. By the way: I don't understand the term -4 in your computation. – Hagen Knaf Feb 12 '12 at 0:18
Ah, thank you. I had misread your answer totally, my apologies. The -4 term was since I thought there were 4 points of ramification, so I just collected the $-1$'s from the 4 $e_P-1$ terms. By the way, I would upvote if I could! – Andrew Davies Feb 13 '12 at 9:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.