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let G be the subset of $\operatorname{aut}_{K} K(x)$ consisting of the three automorphisms

$$ x \mapsto x $$ $$ x \mapsto 1/(1-x)$$ $$ x \mapsto (x-1)/x$$

then G is a subgroup of $\operatorname{aut}_K K(x)$. determine the fixed field of $G$

solution:(wrong) let $ f/g \in K(x)$ with $f$ and $g$ relatively prime in K[x], and suppose that $f/g$ is in the fixed field then $ f/g = 1/(1-(f/g))$ which gives $ f^2 -fg=g^2$

$g^2 = f(g-f)$ so we have that $ f \mid g^2$ so we must have that $f$ is a constant since $f$ and $g$ are relatively prime. and by symmetry we have that $g$ must be a constant which is a contradiction so we must have that $f/g$ is not in the fixed field of G, but this is true for every $f/g \in $ K(x) so the fixed field of G must be empty.

is that right im not sure if that is a contradiction or not?

new solution:

let $\dfrac{ax+b}{cx+d} \in Aut_{K}K(x)$ $$ \sigma_{1} :x \mapsto x $$ $$ \sigma_{2} :x \mapsto 1/(1-x)$$ $$ \sigma_{3} :x \mapsto (x-1)/x$$

$\dfrac{ax+b}{cx+d} = x$ gives $ax+b=cx^2+dx$, which gives $c=0$, $b=0$, and $a=d$ or $(a,b,c,d)= (a,0,0,a)$

$\dfrac{ax+b}{cx+d} =1/(1-x)$ gives $ax+b-ax^2-bx =cx+d$ , gives $a=0$, $b=d$, and $c=-b$ or $(a,b,c,d)=(0,b,-b,b)$

$\dfrac{ax+b}{cx+d} = (x-1)/x$ gives $ax^2+bx=cx^2+dx-cx-d$ gives $a=c$, $d=0$, $b=-c$ or $(a,b,c,d)= (a,-a,a,0)$

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If $\sigma \in \mathrm{Aut}(K(x)/K)$, then $\sigma(f(x)/g(x)) = f(1/(1-x))/g(1/(1-x))$, not $1/(1-(f(x)/g(x)))$... –  Brandon Carter Apr 1 '12 at 20:05
well i know that $aut_{K} K(x)$= { $ x \mapsto (ax+b)/(cx+d) : a,b,c,d \in K$ with $ ad-bc\not = 0$} but i dont see how that helps but i do see that k is a subset of the fixed field but im not sure that if there are other elements in the fixed field or not. @Alex Youcis –  Jason Apr 1 '12 at 21:49
The constants are always fixed; if your set was not a group, it might well happen that no more than the constants were fixed. –  Lubin Apr 2 '12 at 0:23
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2 Answers

This is not a complete answer, but it might get you started.

Let's call the three given automorphisms $1,\sigma,\tau$. Convince yourself that $\sigma^2=\tau$ and $\sigma\tau=1$. Now let $f$ be any element of $K(x)$. Can you see that $f+\sigma f+\tau f$ is in the fixed field of $G$?

For example, if we just take $f=x$, then $$f+\sigma f+\tau f=x+{1\over1-x}+{x-1\over x}={x^3-3x+1\over x(x-1)}$$ so $(x^3-3x+1)/(x(x-1))$ is in the fixed field of $G$.

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The answer given by @Gerry is nicest, since it works for any constant field $K$, but here’s an approach strongly dependent on using the complex numbers. I’m going to use a primitive cube root $\omega$ of unity, so satisfying the equation $\omega^2+\omega+1=0$. Think of your transformation $z\mapsto 1/(1-z)$ as defining a transformation of the extended complex plane. You see that $-\omega$ and $-\omega^2$ are fixed points under this transformation. Now use these to see that there is a nice fractional-linear element of ${\mathbb{C}}(z)$ that behaves particularly well under $z\mapsto 1/(1-z)$, namely $g(z)=(z+\omega^2)/(z+\omega)$. This has a simple zero at $-\omega^2$ and a simple pole at the conjugate point $-\omega$. And lo and behold, when you replace $z$ in $g$ by $1/(1-z)$, the result is $\omega g$. This means that $g^3$ is fixed under the action of your group! (Of course this is all linear algebra, and I’ve hidden the diagonalization process that I used for finding $g$.)

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