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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.

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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 '12 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 '12 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 '12 at 8:54
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 '12 at 10:31

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