I've been stuck on this problem tonight.

Suppose $E=\mathbb{C}(t)$ where $t$ is transcendental over $\mathbb{C}$, and let $\omega$ be a primitive cube root of unity. Let $\sigma$ be the automorphism of $E/\mathbb{C}$ (fixing $\mathbb{C}$) defined by $\sigma(t)=\omega t$, and $\tau$ the automorphism of $\mathbb{C}(t)$ defined by $\tau(t)=t^{-1}$.

I'm trying to show that the group $G=\langle\sigma,\tau\rangle$ has fixed field $\mathbb{C}(u)$ where $u=t^3+t^{-3}$.

I've derived the equalities $\sigma^3=1=\tau^2$ and $\tau\sigma=\sigma^{-1}\tau$, and that $G$ is a group of order $6$, with elements $G=\{1,\sigma,\sigma^2,\tau,\tau\sigma,\tau\sigma^2\}$. I see that an element of $\mathbb{C}(t)$ is fixed by $G$ iff it is fixed by $\sigma$ and $\tau$. It was easy to show that $\mathbb{C}(u)\subset\operatorname{Inv}(G)$, the subfield of elements fixed by $G$. I'm stuck on the other inclusion.

I suppose $g(t)f(t)^{-1}\in\mathbb{C}(t)$ is fixed by $G$. Then I think $$\sigma(g(t)f(t)^{-1})=g(\sigma(t))f(\sigma(t))^{-1}=g(\omega t)f(\omega t)^{-1}$$ and $$\tau(g(t)f(t)^{-1})=g(\tau t)f(\tau t)^{-1}=g(t^{-1})f(t^{-1})^{-1}$$ so altogether $$g(t)f(t)^{-1}=g(\omega t)f(\omega t)^{-1}=g(t^{-1})f(t^{-1})^{-1}$$

So I think $g(t)f(t)^{-1}$ is unchanged by change of variable $t\mapsto \omega t$ or $t\mapsto t^{-1}$. Intuitively I feel like this implies $g(t)f(t)^{-1}$ must actually be a rational function in variable $u=t^{3n}+t^{-3n}$, but I don't know how to rigorously show this is true and that it must by $u=t^3+t^{-3}$.

I got this problem from Section 4.5 of Jacobson's Algebra I. Thanks.


$$[\mathbb{C}(t) : \text{Inv}(G)] = 6$$

$$[\mathbb{C}(t) : \mathbb{C}(u)] \leq 6$$


$$[\text{Inv}(G) : \mathbb{C}(u)] \leq 1$$

  • $\begingroup$ Thanks, but how do you get the first two equalities? It's not clear to me that $E$ is Galois over $\mathbb{C}$, to apply the fact that $|G|=[E:\operatorname{Inv}(G)]$, and how do you get $[\mathbb{C}(t) : \mathbb{C}(u)] = 6$ so quickly? $\endgroup$ – Chelsea Dirks Apr 1 '13 at 5:38
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    $\begingroup$ $\mathbb{C}$ doesn't matter; $E$ is Galois over $\text{Inv}(G)$ because $\text{Inv}(G)$ is the fixed field of $G$. $\endgroup$ – user14972 Apr 1 '13 at 6:02
  • $\begingroup$ As for $[E : \mathbb{C}(u)] \leq 6$, it is because I can write down a polynomial in $\mathbb{C}(u)[X]$ that has $t$ as a root. If you multiply the defining equation for $u$ through by $t^3$, a polynomial immediately suggests itself. $\endgroup$ – user14972 Apr 1 '13 at 6:05

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