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I'm trying to understand Belyi's theorem, as presented here:

He defines a curve $X$ over a field $C$ to be a smooth projective geometrically connected integral separated scheme of dimension 1 with a morphism $p : X\rightarrow\text{Spec}(C)$ of finite type. (Essentially a variety over $K$)

Further, if $X/C$ is a curve over $C$ and $\sigma\in\text{Aut}(C)$, we define $X^\sigma/C$ to have exactly the same underlying scheme as $X/C$, but whose structure morphism is twisted by $\sigma$. Ie, if $p : X\rightarrow\text{Spec}(C)$ is the structure morphism for $X/C$, then $p^\sigma : X\rightarrow\text{Spec}(C)\stackrel{\text{Spec}(\sigma)}{\longrightarrow}\text{Spec}(C)$ is the structure morphism for $X^\sigma/C$.

In particular, this means that the function fields of $X/C$ and $X^\sigma/C$ are literally the same, so that a morphism between $X/C$ and $X^\sigma/C$ is an $\textit{auto}$morphism of the scheme $X$ (that is compatible with the different structure morphisms), and so induces an $\textit{auto}$morphism of its function field.

Now, let $X$ be a curve over a field $C$, and let $t\in K(X)$ be a morphism to $\mathbb{P}^1_C$. Let $Q$ be a point on $\mathbb{P}^1_C$ above which $t$ is unramified, and let $P\in X$ be any preimage of $Q$ under $t$. Now, in the middle of page 260 in the paper, he defines $U(X,t,P)$ to be the subgroup of $\text{Aut}(C)$ such that there is an isomorphism $f_\sigma : X^\sigma\rightarrow X$ such that $f_\sigma(P^\sigma) = P$ and a certain diagram commutes. (I can only refer you to the diagram in middle of page 260 through the link provided above). Here, he defines $P^\sigma$ as the point on $X^\sigma$ corresponding to $P$, by which he presumably means exactly the same point of $P$, but viewed as a point on $X^\sigma$.

I understand why there can be at most one such isomorphism $f_\sigma$ for each $\sigma$, but what I don't understand is why the automorphism induced on $K(X)$ by $f_\sigma$ fixes $t$. Specifically, when he says:

"Thus, mapping $\sigma$ to the automorphism of the function field $K(X)$ induced by $f_\sigma$ yields an action of $U(X,t,P)$ on $K(X)$ by $C$-semilinear field automorphisms which fix $t\in K(X)$"

As I understand, the automorphism $f_\sigma$ induces is defined by sending any $g\in K(X)$ to $f^*g = g\circ f$, but on $t$, we have $f^*t = t\circ f$, which by the commutative diagram is $\text{Proj}(\sigma)\circ t^\sigma$. However, he never defines $t^\sigma$, and I can only assume that it's the "twin" of $t$ living on $K(X^\sigma) = K(X)$. But since $\text{Proj}(\sigma)$ is not (generally) the identity, there's no way that $f^*t$ can equal $t$.

I apologize for asking such a long question, but I've been stuck on this detail for many hours now, and feel like it's probably a very simple, though subtle fact that I'm missing.


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So apparently the way to see this is to first note that while the function fields of $X$ and $X^\sigma$ have the ``same'' function field, the same element, viewed as a function in one function field, is not the same that element viewed as a function in the other function field.

In short, the best way to note that the automorphism induced by $f_\sigma$ indeed does fix $t$ is to look at the induced diagram on the stalks at the generic point.

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If someone else votes up this answer, it will not be bumped to the front page by MOUser again. – Steven Gubkin May 29 '12 at 20:50

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