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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?

user134070
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1 Answers1

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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

  • any automorphism $\tau$ must map $\root p\of a\mapsto \zeta_p^k\root p\of a$ and $\root q\of a\mapsto \zeta_q^\ell\root q\of a$ for some $k,\ell, 0\le k<p, 0\le \ell<q$, and
  • specifying the pair $(k,\ell)$ uniquely identifies the automorphism $\tau$, and
  • you know by other means (the extension degree) that the Galois group has order $pq$,
  • you can deduce that all combinations of $(k,\ell)$ must occur.
  • At which point it is safe to conclude that $\sigma$ and $\rho$ are automorphisms (corresponding to $(k,\ell)$ pairs $(1,0)$ and $(0,1)$ respectively). The conclusion is that you, indeed, get $C_p\times C_q$.

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 $\sqrt{5}\in\Bbb{Q}(\zeta_5)\subset F$.

Jyrki Lahtonen
  • 133,153