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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.

Thanks,

Andrew

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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 $)

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  • $\begingroup$ For the other questions you might take a look here $\endgroup$ Feb 9, 2012 at 15:53
  • $\begingroup$ Thanks, this is great. The link too is very interesting. $\endgroup$ Feb 10, 2012 at 12:52
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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.

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  • $\begingroup$ 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. $\endgroup$ Feb 10, 2012 at 12:57
  • $\begingroup$ 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. $\endgroup$
    – Hagen Knaf
    Feb 12, 2012 at 0:18
  • $\begingroup$ 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! $\endgroup$ Feb 13, 2012 at 9:21

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