Welcome to MathOverflow

Question

6

1

Is there a nice algebraic way of determining the ramification of a morphism between curves? Ie, some analog of the discriminant of number fields?

Specifically, I'm trying to prove that if $X$ is a curve over $\mathbb{C}$ that's defined over $\overline{\mathbb{Q}}$, and $t : X\rightarrow\mathbb{P}^1_\mathbb{C}$ is a morphism also defined over $\overline{\mathbb{Q}}$, then the critical values of $t$ (ie, the points of $\mathbb{P}^1_\mathbb{C}$ above which $t$ is ramified) are $\overline{\mathbb{Q}}$-rational.

flag	asked May 1 2012 at 6:08 oxeimon 889●2●7

1

The critical values of $t$ comprise the support of the relative sheaf of differentials for $X$ over $\mathbb{P}^1_{\mathbb{C}}$ This is the (slightly more global) function field analogue of the co-different ideal (that is, the inverse of the different). It's just the co-kernel of the map $t^*:\Omega^1_{\mathbb{P}^1}\to\Omega^1_X$ . If $t$ is defined over $\overline{\mathbb{Q}}$, then so is its relative sheaf of differentials, and so also is its support. – Keerthi Madapusi Pera May 1 2012 at 7:00

I was actually trying to avoid talking about differentials, as defining the sheaf of relative differentials seems unnecessarily complicated, especially when I would end up using them in a relatively trivial way. I also thought about perhaps using a Lefschetz principle-type argument, but then making that precise presents its own challenges. – oxeimon May 1 2012 at 8:11

@oxeimon: Keerthi's suggestion is the correct one. But if you insist on avoiding direct mention of relative differentials, then consider the degeneracy locus of the "trace pairing". As $\overline{\mathbb{Q}}(t) \to \overline{\mathbb{Q}}(X)$ is a finite, separable field extension, the trace map $\text{Tr}:\overline{\mathbb{Q}}(X) \to \overline{\mathbb{Q}}(t)$ defines a bilinear pairing $\overline{\mathbb{Q}}(X)\otimes_{\overline{\mathbb{Q}(t)}\overline{\mathbb{Q}}(X)\to \overline{\mathbb{Q}}(t)$ by $a\otimes b \mapsto \text{Tr}(ab)$. This is "integral" and degenerates at ramification. – Jason Starr May 1 2012 at 8:54

1

An alternative to above suggestions is to view the function field of $X$ as an extension of the rational function field given by an equation $f(t,x)=0$ with algebraic numbers as coefficients. The ramification points are then either infinity or a subset of the points with $\partial f / \partial x = 0$. Taking the resultant in $x$ of $f$ and that partial derivative gives an equation with algebraic coefficients for the branch points. – Felipe Voloch May 1 2012 at 10:31

analog for the discriminant of number fields in the function field case?

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.