Let $f=x^p-\theta \in \mathbb{Q} [x]$ be irreducible for some $\theta \in \mathbb{Q}$, $p$ prime. Show that $\Gamma_{\mathbb{Q}} (f)$ is generated by two elements $\sigma$ and $\tau$ st.
$\sigma ^p=\tau ^{p-1}=1$ and $\sigma^k \tau=\tau \sigma$, where $\bar{k}$ is a generator for $\mathbb{Z}^*_p$.
My thoughts go like this:
Let $\omega$ be a primitive $p^{th}$ root of unity. We can factor $f$ as $f=(x-\sqrt[p]{\theta})(x-\omega \sqrt[p]{\theta})...(x-\omega^{p-1}\sqrt[p]{\theta})$ so the splitting field for $f$ over $\mathbb{Q}$ is $\mathbb{Q}(\omega, \sqrt[p]{\theta})=L$.
So the elements of $\Gamma_{\mathbb{Q}} (f)$ will permute roots of $f$ and it is clear that this group will be generated by two elements.
Observe first extension $\mathbb{Q}\subset \mathbb{Q}(\omega)$.
The minimal polynomial for $\omega$ over $\mathbb{Q}$ will be $p^{th}$ cyclotomic polynomial $x^{p-1}+ ...+x+1$ and extension $\mathbb{Q}\subset \mathbb{Q}(\omega)$ is Galois.
Further, $Gal(\mathbb{Q}(\omega)/\mathbb{Q})\cong (\mathbb{Z}/p\mathbb{Z})^*$ which is cyclic, so there is an element $\tau$ such that $\tau^{p-1}=1$.
Observe now extension $\mathbb{Q}(\omega)\subset L$.
The minimal polynomial for $\sqrt[p]{\theta}$ over $\mathbb{Q}(\omega)$ is $f$ ($f$ is irreducible over $\mathbb{Q}(\omega)$ since any factor would contain $[\sqrt[p]{\theta}]^i$ in its coefficients for some $i <p$, but $[\sqrt[p]{\theta}]^i \notin \mathbb{Q}(\omega)$ for $i<p$) and $\mathbb{Q}(\omega)\subset L$ is Galois extension $\Rightarrow |Gal(L/\mathbb{Q}(\omega))|=p$ .
There is a theorem which tells that $Gal(L/\mathbb{Q}(\omega))$ is cyclic so there is $\sigma$ such that $\sigma^p=1$.
Are these observations okay and how to proceed with last part?
