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Let $K$ be an infinite field, if $K(X)$ is the field of rational function I want to find the Galois group of the extension $K(X)/K$.

Lemma 1: If $L$ is a field such that $K\subsetneq L\subseteq K(X)$ then $[K(X):L]$ is finite.

Proof. It is easy to show that $X$ is algebraic over $L$, so $K(X)/L$ is a finite extension.

Lemma 2: $Gal(K(X)/K)$ contains only finite (proper) subgroups.

proof. Suppose that $H<G$ is infinite; for the lemma 1 we have that $[K(X): Fix(H)]=n$, and so $|Gal(K(X)/Fix(H))|\le n$. But $Gal(K(X)/Fix(H))\supseteq H$ and this is a contraddiction.

Now I know that the group $Gal (K(X)/K)$ is a group with only finite subgroups, but I can't find other informations about its structure. Maybe this group depends strongly from the field $K$.

Thanks in advance.

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You are right, but I don't know were is the mistake. $X$ is algebraic over $L$, infact if $\frac{f}{g}\in L$, then $\frac{f(X)}{g(X)}g(X)-f(X)=0$. – Dubious Jun 9 '13 at 20:56
sorry that comment was wrong. – Ted Jun 9 '13 at 20:58
up vote 10 down vote accepted

The methods of Galois theory are not so good for studying transcendental extensions – instead you should consider the methods of birational geometry. For simplicity, assume $K$ is algebraically closed. Then $K (X)$ is the function field of the projective curve $\mathbb{P}^1_K$, and the group $K$-automorphisms of $K (X)$ is canonically isomorphic to (the opposite of) the group of birational automorphisms of $\mathbb{P}^1_K$. This is called the Cremona group of order 1.

Now, notice that $K (X)$ is generated over $K$ by $X$, so $K$-automorphisms of $K (X)$ are uniquely determined by the image of $X$. By geometrical considerations, we see that the image of $X$ must be of the form $$\frac{a X + b}{c X + d}$$ because otherwise the induced rational map $\mathbb{P}^1_K \to \mathbb{P}^1_K$ would send more than one point to $0$ or more than one point to $\infty$. Moreover, we must have $a d - b c \ne 0$, so that $$\frac{1}{a d - b c} \frac{d X - b}{- c X + a}$$ corresponds to the inverse automorphism. Thus the automorphism group of $K (X)$ over $K$ is none other than the Möbius group $\mathrm{PGL}_2 (K)$.

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Would you mind correcting my understanding. If we map a matrix $$ M = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $$ to the automorphism which maps $X$ to $$ \frac{aX + b}{cX + d} $$ isn't this technically an antihomomorphism (which can of course be corrected to a homomorphism by first inverting $M$)? – Jacob Denson Apr 16 at 21:39

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