Let $F$ be a field obtained by adjoining all the roots of unity to $\mathbb{Q}$. Given $p<q$ primes and $a\in \mathbb{C}F\setminus$, I am trying to figure out the Galois group of the polynomial $(x^p-a)(x^q-a)$ over $F$. (Both $x^p-a$ and $x^q-a$ are assumed to be irreducible!)
My attempt: Let $L$ be the splitting field of $(x^p-a)(x^q-a)$ over $F$. Since $F$ contains all the roots of unity, $L=F(\sqrt[p]{a})(\sqrt[q]{a})=F(\sqrt[p]{a},\sqrt[q]{a})$. The Galois group is generated by two automorphisms $\sigma:\sqrt[p]{a}\mapsto \sqrt[p]{a}\zeta_p$ and $\rho:\sqrt[q]{a}\mapsto \sqrt[q]{a}\zeta_q$, where $\zeta_p,\zeta_q$ are $p$-th and $q$-th roots of unity, respectively. Therefore $\mathrm{Gal}(L/F)\cong C_p\times C_q$.
Is my attempt correct? If not, how may I fix it?
Best Answer
It follows from this result that unless $a$ happens to be either a $p$th or a $q$th power of an element of $F$ then the two factors $x^p-a$ and $x^q-a$ are both irreducible in $F[x]$.
Assuming that you can easily deduce that $[L:F]=pq$. Therefore the Galois group must have order $pq$ as well, and your argument largely works. There is a shortcoming that to specify an automorphism, like $\sigma$, you need to tell how it maps both $\root p\of a$ and $\root q\of a$. So I would describe $\sigma$ as the automorphism determined by $\root p\of a \mapsto \zeta_p\root p\of a$ and $\root q\of a\mapsto \root q\of a$. The obvious modifications then give you $\rho$.
Because
Don't take my criticism too seriously. You may have gone through these steps many times during the course, and your teacher may give you some slack in the exam (ask them to be sure!)
Otherwise there may be pitfalls. For example $x^2-5$ is not irreducible over $F$ because $\sqrt{5}\in\Bbb{Q}(\zeta_5)\subset F$.