TITLE: Finding a subfield with a desired Galois group
QUESTION [7 upvotes]: I am trying to find a subfield $F$ of the field of complex rational functions $\mathbb{C}(X)$ such that $\mathbb{C}(X)/F$ is galois with galois group $S_3$. 
My approach was the following. I wanted to find a group of automorphisms $G$ that is isomorphic to $S_3$, and by the theorem of Artin obtain that the fixed field of $G$ will be such that $\mathbb{C}(X)/\mathbb{C}(X)^G$ is galois with group $G$. (So in particular this will be my $F$). 
Since $S_3$ is isomorphic to the dihedral group of order $6$ I constructed two automorphisms of $\mathbb{C}$. 
The map $s$ to be defined to fix the variable $X$ and conjugate the complex numbers, and the map $r$ to be defined to do $r(X)=e^{2\pi i/3}X$ and it fixes the complex numbers. 
Then we can check that $r$ has order $3$, $s$ has order $2$, and $rs=sr^{-1}$, so the group generated by these two maps will give us the result. 
My question is how to do it using a hint that was given in the hw which reads: "Consider the automorphisms of $\mathbb{C}(X)$ given by $X\mapsto \frac{aX+b}{cX+d}$ for $a,b,c,d\in \mathbb{Z}$"?
thanks

REPLY [2 votes]: Let $f$ be an automorphism of $\mathbb C(x)$ generated by sending $X$ to $aX+b/(cX+d)$ for $a,b,cd \in \mathbb Z$ such that $ad-bc=1$ and fixing $\mathbb C$. Then we can naturally associate $f$ to an element of $\mathrm{SL}_2(\mathbb Z)$. Furthermore if we take two such automorphisms then the representative of their composition is exactly the product of their reprsentatives in $\mathrm{SL}_2(\mathbb Z)$. So it suffices to embed $S_3$ in $\mathrm{SL}_2(\mathbb Z)$. It's not particularly hard to find a faithful $2$-dimensional reprsentation of $S_3$ with $\mathbb Z$-coefficients. See here if you get stuck.