TITLE: Finding a fixed subfield of $\mathbb{Q}(t)$
QUESTION [10 upvotes]: I'm looking at automorphisms $t \to 1-t$ and $t \to \frac{1}{t}$ of the field $\mathbb{Q}(t)$. By looking at the relations between these I think I've found the group generated by them to be $S_3$. 
Now I'm trying to find the fixed fields, call these $F$. For the first one I noticed that the polynomial $f = t(t-1)$ is invariant and as we know that $[\mathbb{Q}(t) : F] \ge 2$ and then the polynomial $g = t(t-1)-f$ is a two degree polynomial in $\mathbb{Q}(f)[t]$ then I deduced $F = \mathbb{Q}(f)$.
However I'm not having trouble applying this method to find the fixed field of the second automorphism. I can't actually find any rational functions that are fixed by it.
Thanks for any help

REPLY [4 votes]: Consider the following rational function
$$
w=\frac{(t^3-3t^2+1)(t^3-3t+1)}{t^2(t-1)^2}.
$$
It is relatively easy to check that $w$ is invariant under both $\sigma:t\mapsto 1-t$ and $\tau:t\mapsto 1/t$. The automorphism $\sigma$ maps the factors in the numerator to the negatives of each other (check this), and clearly does the same to the factors of the denominator. Invariance under $\tau$ follows essentially from the observation that the factors in the numerator are reciprocal polynomialss of each other. The degrees match nicely to cancel the effect of $\tau$ to the denominator.
Clearly $t$ is a zero of the sextic polynomial
$$
p(x)=(x^3-3x^2+1)(x^3-3x+1)-x^2(x-1)^2w\in\Bbb{Q}(w)[x],
$$
so $[\Bbb{Q}(t):\Bbb{Q}(w)]\le6$. OTOH, as pointed out by Hurkyl, the fixed field $K$ of
$G=\langle\sigma,\tau\rangle$ ($\subset Aut(\Bbb{Q}(t))$) satisfies $[\Bbb{Q}(t):K]=6$.
We just saw that $\Bbb{Q}(w)\subseteq K$, so we can conclude that
$$
\operatorname{Inv}(G)=\Bbb{Q}(w).
$$

A `funny' thing happend in that $\operatorname{Inv}(G)\simeq\Bbb{Q}(t)$ (send $w$ to $t$). This is not a coincidence. Lüroth's theorem says that all the intermediate fields properly between $\Bbb{Q}$ and $\Bbb{Q}(t)$ are isomorphic to $\Bbb{Q}(t)$.

How did I find $w$? In this earlier question I used the trace to find a non-trivial element 
$$u=t+\frac1{1-t}+\frac{t-1}t=\frac{t^3-3t+1}{t(t-1)}$$
of the fixed field for the index two subgroup $\langle \sigma\tau\rangle\le G$. Here $w=-u\sigma(u)$ and that is clear the invariant of both $\sigma$ and $\sigma\tau$, hence all of $G$.