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Let $\mathbb{C}\left(x\right)$ be the field of complex rationals functions.
Find a subfield $\mathbb{K}$ of $\mathbb{C}\left(x\right)$ such that $Gal\left(\mathbb{C}\left(x\right)/\mathbb{K}\right)$ is isomorphic to $S_3$.

Using the Galois Correspondence we can see that $\mathbb{K}$ is the fixed field of $S_3$. But i would like of explicit form for $\mathbb{K}$. My professor gave me a tip:

Prove that given a subgroup ${G}$ of $Gal\left(\mathbb{K}\left(x\right)/\mathbb{K}\right)$ a fixed field of ${G}$ is of form: $\mathbb{K}\left(a_i\right)$ where $a_i$ is a coefficient of polynomial

$$ p(t)= \prod_{\substack{j\\}} \left(t – \sigma_j(x)\right)$$ where $ \sigma_j\ \in {G}$.

I need some help.

Junior Soares
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  • Off the top of my head : I think most ${\mathbb C}(f(x))$, where $f$ is irreducible of degree $6$, will do. The minimal polynomial of $x$ over $\mathbb K$ will then be $f(x)-c$ with $c\in{\mathbb K}$. Do you know about Luroth's theorem ? https://en.wikipedia.org/wiki/L%C3%BCroth%27s_theorem – Ewan Delanoy Jun 20 '15 at 12:55
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    No, @EwanDelanoy, I believe that this will typically give you a nonnormal extension. What you say about the minimal polynomial is right up to a point: I would say that the minimal polynomial over $\Bbb C(f)$ is $f(X)-f$, as confusing as that looks. Considering your field $\Bbb C(f)$ as $\Bbb C(t)$, the minimal becomes $f(X)-t$. – Lubin Jun 22 '15 at 02:52
  • @Lubin you are right. In fact, to get a normal extension ith group $S_n$ one must first take an irreducible polynomial of degree $n$, then take the normal closure. So here I should have started with a polynomial of degree $3$ not $6$ – Ewan Delanoy Jun 22 '15 at 04:53
  • @EwanDelanoy: I'm not sure I understand what you mean by an irreducible cubic here. Surely not you mean that $f$ would be an irreducible cubic in $\Bbb{C}[x]$ because there aren't any. But when you add another variable, say $y$, you go outside of $\Bbb{C}(x)$ and lose Lüroth. For example, if you try the splitting field of $y^3+Ay+B-x^2$ adjoining $y$ gives you a genus $1$ function field. – Jyrki Lahtonen Jun 22 '15 at 13:23
  • @JyrkiLahtonen But the decomposition field of $y^3+Ay+B-(x^3+Ax+B)$ over (${\mathbb C}(x^3+Ax+B)$) does work, doesn't it – Ewan Delanoy Jun 22 '15 at 13:27

1 Answers1

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First you should find a subgroup of automorphisms of $\Bbb{C}(x)$ that is isomorphic to $S_3$. I would use the group generated by the Möbius transformations $s_1:x\mapsto 1/x$ and $s_2:x\mapsto 1-x$. We then have $s_1^2=id=s_2^2$, $s_1s_2:x\mapsto 1/(1-x)$, $s_2s_1:x\mapsto (x-1)/x$, $s_2s_1s_2:x\mapsto x/(x-1)$, $s_1s_2s_1=s_2s_1s_2$. (This is a relatively well known copy of $S_3$ inside the Möbius group)

Then you can start looking for the fixed field. As commented by Ewan Delanoy, the fixed field is of the form $\Bbb{C}(u)$ for some rational function $u$ as a consequence of Lüroth's theorem.

Unless I made a mistake, expanding out the product $m(T)=\prod_{s\in\langle s_1,s_2\rangle}(T-s(x))$ gives $$ m(T)=T^6-3T^5-uT^4+(2u+5)T^3-uT^2-3T+1, $$ where $$ u=\frac{x^6-3x^5+5x^3-3x+1}{x^2(x-1)^2}. $$ So $u$ is invariant under $\langle s_1,s_2\rangle$, and clearly $[\Bbb{C}(x):\Bbb{C}(u)]=6$.

Jyrki Lahtonen
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