Order of the zero of a meromorphic function under the action of $Gal(\mathbb{C},\mathbb{Q})$

Question

Let's take $X$ a Riemann surface as an algebraic curve in $\mathbb{P}^n$. The group $Gal(\mathbb{C},\mathbb{Q})$ (automorphisms of $\mathbb{C}$ which act as the identity on $\mathbb{Q}$) acts on $X$ through the action on the coefficients of the equations defining $X$.

So, if for example $X$ is defined in $\mathbb{P}^2$ by the equation $\{F=0\}$ then $\sigma\in Gal(\mathbb{C},\mathbb{Q})$ sends $\{F=0\}$ to $\{F^\sigma=0\}$ and this action corresponds to the action of $\sigma$ on the coordinates of the points ($P\mapsto P^\sigma$).

Now let's take $f\in \mathcal{M}(X)$ i.e. a meromorphic function on $X$. If $P\in X$ is a zero for $f$, i want to prove that $ord_P(f)=ord_{P^\sigma}(f^\sigma)$.

If $X=\mathbb{P}^1$ this is straightforward, but i don't know how to deal with two ore more homogeneous variables (i think i'm forced to see $X$ as an algebraic curve in $\mathbb{P}^n$ to see the action of $Gal(\mathbb{C},\mathbb{Q})$, so $f$ is a rational function in the homogeneous coordinates of $\mathbb{P}^n$).

Answer 1

This question amounts to tracing the ideal-theoretic interpretations of all of the objects in question. You have a coordinate ring $S$ for projective space, and a homogeneous ideal $I$ defining $X$. Complex conjugation takes $I$ to $I^\sigma$ (i.e., the coefficients of all functions are conjugated), and the homogeneous coordinate ring $S/I$ of $X$ is taken to the homogeneous coordinate ring $S^\sigma/I^\sigma$ of $X^\sigma$. Points in $X$ are in bijection with maximal relevant ideals $J$ containing $I$, and complex conjugation respects containment.

Any point $P \in X$ has a local ring $R$ composed of the meromorphic functions on $X$ that don't have a pole at $P$ (it is made by localizing $S/I$), and the maximal ideal $m$ is made from those functions that vanish at $P$. The order of vanishing of a meromorphic function $f$ at $P$ is the supremum of integers $k$ such that $f \in m^k$. For your question, it suffices to note that complex conjugation takes $R \to R^\sigma$ and hence $m$ to $m^\sigma$. In particular, $f \in m^k$ if and only if $f^\sigma \in (m^\sigma)^k$.