Automorphisms of the field extension $\Bbb C(x)/\Bbb C$ and their fixed field

Question

Consider automorphisms of the field extension $\mathbb{C}(X)/\mathbb{C}$, where $\sigma$,$\tau$ is defined as:

$$ \sigma:X \rightarrow \frac{X+i}{X-i} $$

$$ \tau :X \rightarrow \frac{iX-i}{X+1} $$

We assume that group $G$ is generated by $\sigma$, $\tau$.

Prove that: $G$ is isomorphic with the alternating group $A_4$. Moreover, compute the fixed field of $G$.

I really have no idea how to solve it, please help me, thanks.

Answer 1

If $G$ is a finite subgroup of $PGL_2(\Bbb{C})$ then let $$\prod_{g\in G} (T-g\cdot x)=\sum_{n=0}^{|G|} a_n T^n \in \Bbb{C}(x)[T]$$

Some $a_n\not\in \Bbb{C}$ because $x$ is algebraic over $\Bbb{C}(a_0,\ldots,a_{|G|})$.

Take one such $a_n$, it will have $\le |G|$ poles whence $[\Bbb{C}(x):\Bbb{C}(a_n)] \le |G|$.

Also $a_n\in \Bbb{C}(x)^G$ and $[\Bbb{C}(x):\Bbb{C}(x)^G] = |G|$ whence $$\Bbb{C}(x)^G=\Bbb{C}(a_n)$$

Answer 2

It helps to know about Mobius transformations to solve this exercise. The field $\mathbb{C}(x)$ is the field of meromorphic functions defined on the Riemann sphere, and automorphisms of this field can be seen as transformations on the sphere.

The group of automorphisms of the sphere is equal to the group of Mobius transformations. I will describe those two:

• $\sigma$ is somewhat well-known. You can see that it transforms the real line into the unit circle (because the modulus of numerator and denominator are the same if $X$ is real). It follows that it transforms one half-plane into the unit disk (the lower half-plane, since for $X = -i$ it is zero), and the other one into the exterior of the disk.

• $\tau$ is equal to the composition of $\frac{X-1}{X+1}$ with multiplication by $i$ (which is a counter-clockwise rotation of angle $\pi/2$). The map $\frac{X-1}{X+1}$ obviously fixes the real line.

This simple description (and some computations) reveals that both $\sigma, \tau$ interchange the real line, the imaginary axis, and the unit circle. These tree lines divide the sphere into 8 octants, and these octants are also interchanged by $\sigma, \tau$. But, each octant $O$ has an opposite octant $O'$, and both $\sigma, \tau$ respect these "pairs" of octants, so, they induce permutations of these $4$ pairs of octants. You can check that these two permutations are even, and that they span the full alternate group $A_4$.

The field of invariants was already computed by @reuns.