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Let $\mathbb F(t)$ be the rational functions over the field $\mathbb F$ with $t$ as an argument, and Let $\alpha,\beta \in Aut(\mathbb F(t))$ such that $\alpha(t)=\frac 1 t$, $\beta(t)=1-t$.

(a) Define $H=\langle \alpha,\beta \rangle \le Aut(\mathbb F(t))$. Show that $H\cong S_3$.

(b) Find $y\in \mathbb F(t)$ such that $\mathbb F (x)^H=\mathbb F(y)$.

For (a), It's not that hard to show directly that $|\alpha \beta \alpha|=|\alpha|=|\beta|=2$, and $|\alpha \beta|=|\beta \alpha |=3$, and so on, but I couldn't find something more elegant.

For (b) I have no Idea.

S. R
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1 Answers1

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Look at the six rational functions $t, 1-t, 1/t, (t-1)/t, 1/(1-t)\text{ and }t/(t-1)$. The set of these is stable under the action of $H$, so you should try to come up with some expression in those elements that is invariant under the action of $H$. The product of the 6 turns out to equal 1, so that won't work. Similarly their sum equals 3, so again doesn't work. The sum of the squares of the six rational functions should work and isn't too difficult to compute by hand.

sharding4
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