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 $a,b,c,d\in \mathbb{Z}$"?



1 Answer 1


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.

  • $\begingroup$ For the faithful 2 - dimensional representation can't we just take the 2 - dimensional irreducible copy sitting inside of the standard representation of $S_3$ on $\Bbb{C}^3$? By the 2 - dimensional copy I mean the orthogonal complement of the line spanned by $(1,1,1)$. $\endgroup$
    – user38268
    Feb 24, 2013 at 2:55
  • $\begingroup$ @BenjaLim Yeah, that works fine. Hence why I said it's not particularly hard ;). $\endgroup$
    – JSchlather
    Feb 24, 2013 at 3:51

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