# Galois actions on extensions of algebraic function fields

Let $k$ be an algebraically closed field, $C/k$ and $C'/k$ be smooth projective curves, and $C'/k \rightarrow C/k$ be a $k$-morphism which is corresponding to the field extension $k(C) \hookrightarrow k(C')$. We assume $C'/k \rightarrow C/k$ is a Galois cover, i.e $k(C')/k(C)$ is Galois.

Given any point $P$ on $C$, let $x \in k(C)$ be a uniformizer at $P$. Let $P'$ be a point on $C'$ which is unramified over $P$. So $x \in k(C')$ is still a uniformizer of $C'$ at $P'$. We know that every "function $f$ on $C'$" , i.e every $f \in k(C')$ can be written as a laurent series of $x$ with coefficients in $k$, i.e $f = \sum_{i=-n}^{\infty} a_i g^i$ with $a_i \in k$ and $n \in \mathbb{Z}$.

Now given any $\sigma \in \mathrm{Gal}(k(C') / k(C))$. We want to calculate $\sigma(f)$. Last night when I was drunk, I just did this as following:

$$\sigma(f) = \color{red}{\sigma(\sum_{i=-n}^{\infty} a_i g^i) = \sum_{i=-n}^{\infty} \sigma(a_i g^i)} = \sum_{i=-n}^{\infty} \sigma(a_i) \sigma(g)^i = \sum_{i=-n}^{\infty} a_i g^i = f$$,

since $a_i$ and $g$ are all in $k(C)$. However this is not true, otherwise it implies that $f \in k(C)$ for all $f \in K(C')$.

But I can't give myself a good reason that why $\sigma$ can't commute with "infinity sum", even when I sober up completely this afternoon. So what is the obstruction for this "commutativity"?

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Presumably $g$ is $x$? Also, notice that it is not true that $f$ can be written as a Laurent series: Laurent series simply do not make sense in $k(C')$... you have some other ring in mind (the completion of something, &c, and why does $\sigma$ act on it?) Finding a real interpretation for what you wrote will surely help. – Mariano Suárez-Alvarez Jul 12 '13 at 7:40