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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: $\operatorname{Gal}(K(X)/K)$ contains only finite (proper) subgroups.

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

Now I know that the group $\operatorname{Gal}(K(X)/K)$ 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.

Samuel Adrian Antz
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Dubious
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  • Related: https://math.stackexchange.com/questions/13129/automorphism-of-the-field-of-rational-functions?noredirect=1&lq=1 – Watson Jan 08 '17 at 12:50
  • Possibly related: https://arxiv.org/pdf/1004.5036.pdf – Watson Oct 18 '17 at 20:45
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    Unfortunately I think Lemma 2 is wrong. For instance if $char(K)=0$ then there is an infinite subgroup $H$ (isomorphic to $\mathbb{Z}$) generated by $x\mapsto x+1$. Your proof fails because $Fix(H)=K$ in this case. – Juan Carlos Ortiz Apr 29 '19 at 17:58

1 Answers1

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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 $\mathrm{PGL}_2 (K)$.

Zhen Lin
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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 '16 at 21:39
  • Why isn't sending X to X^2 while fixing K an automorphism? It is linear, multiplicative and injective, no? – Elie Bergman Nov 16 '17 at 00:34