Galois group of order 6

\mathbf{Problem:} Let E/\mathbb{C}(t) where t is transcedental over \mathbb{C} and let w\in \mathbb{C} satisfy \displaystyle w^3=1, w\neq 1. Let \sigma be the automorphism of E/\mathbb{C} such that \tau(t)=t^{-1}. Show that

\displaystyle \sigma^3=1=\tau^2,\quad \tau\sigma=\sigma^{-1}\tau.

Show that the group of automorphism generated by \sigma and \tau has order 6 and the subfield F=\text{Inv}G=\mathbb{C}(u) where u=t^3+t^{-3}.

\mathbf{Proof:} It is easy to verify the structure of group G. And G is the dihedral group D_6

Let H_1=\langle\sigma\rangle\leq G, H_2=\langle\tau\rangle\leq G. Then \displaystyle \text{Inv}H_1=\left\{\frac{f(t)}{g(t)}|g(t)\neq 0, \frac{f(t)}{g(t)}=\frac{f(\omega t)}{g(\omega t)}\right\}=\left\{\frac{f(t^3)}{g(t^3)}|g(t) \neq 0\right\}.

Consider \displaystyle \text{Inv}H_2=\left\{\frac{f(t)}{g(t)}|g(t)\neq 0, \frac{f(t)}{g(t)}=\frac{f(t^{-1})}{g(t^{-1})}\right\}, any \displaystyle \frac{f(t)}{g(t)}\in \text{Inv}H_2\exists\, h, l\in \mathbb{C}(t) such that

 \displaystyle 2\frac{f(t)}{g(t)}=\frac{f(t)}{g(t)}+\frac{f(t^{-1})}{g(t^{-1})}=\frac{f(t)g(t^{-1})+g(t)f(t^{-1})}{g(t)g(t^{-1})}=\frac{h(t+t^{-1})}{l(t+t^{-1})}

So \displaystyle \text{Inv}H_2=\left\{\frac{f(t+t^{-1})}{g(t+t^{-1})}\lvert g\neq 0\right\}.

So \text{Inv}G=\text{Inv}H_1\cap\text{Inv}H_2=\mathbb{C}(t^3+t^{-3})

\text{Q.E.D}\hfill \square

\mathbf{Remark:}Jacobson p243

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