Exercise 7.5.13

Proof. Let $\sigma$ the automorphism of $L=ℂ(t)$ defined by $\alpha (t)\mapsto \alpha ({\zeta }_{n}t)$.

For all $\alpha \in ℂ(t)$, for all $k\in \mathbb{N}$, ${\sigma }^k(\alpha (t))=\alpha ({\zeta }_n^{k}t)$. Then ${\sigma }^n=e$, and for $\alpha (t)=t,1\le k\le n-1$, ${\sigma }^k(t)={\zeta }_n^{k}t\ne t$, so ${\sigma }^k\ne e$. Therefore the order of $\sigma$ is $n$, and $G=⟨\sigma ⟩$ is a cyclic group of order $n$.

By Theorem 7.5.3, the extension $L_G\subset ℂ(t)$ is a Galois extension of degree $n$, with Galois group $G=⟨\sigma ⟩$.

We want to specify the field $L_G$.

$\sigma (t^n)={({\zeta }_{n}t)}^n=t^n$, thus $t^n\in L_G$ and so $ℂ(t^n)\subset L_G\subset ℂ(t)$.

By Theorem 7.5.5(c), the extension $ℂ(t^n)\subset ℂ(t)$ has degree $n$, so

therefore $[L_G:ℂ(t^n)]=1$ :

Conclusion:

$ℂ(t^n)\subset ℂ(t)$ is a Galois extension whose Galois group $G$ is cyclic of order $n$. □