Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $(a,b,c,d)= (a,-a,a,0)$

share|improve this question
    
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

2 Answers 2

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 $G$.

share|improve this answer

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 $g$.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.