Problem on Galois theory and irreducible polynomial Let $p,q$ be primes, estimate the degree $[\Bbb Q(\sqrt[p]{2}\cdot\sqrt[q]{2}):\Bbb Q]$ and prove that the polynomial $X^q-2$ is irreducible in the ring $\Bbb Q(\sqrt[p]{2})[X]$
I found this problem on mathematical competition without any solution and it would be very interesting if we can find a proof for this.
 A: Assuming $p\ne q$... I claim the degree is $pq$ (and thus $X^q-2$ is irreducible over $Q(\sqrt[p]{2})[X]$).
First note $[Q(\sqrt[q]{2}:Q] = q$, since $X^q-2$ is irreducible over $Q$ by Eisenstein. (Same with $X^p-2$)
The field $Q(\sqrt[p]{2},\sqrt[q]{2})$ is the compositum of $Q(\sqrt[p]{2})$ and $Q(\sqrt[q]{2})$. Let $L := Q(\sqrt[p]{2})\cap Q(\sqrt[q]{2})$, then its well known that
$$[Q(\sqrt[p]{2},\sqrt[q]{2}):L] = [Q(\sqrt[p]{2}):L]\cdot[Q(\sqrt[q]{2}):L]$$
Thus it suffices to prove that $L = Q$, but this is clear from the fact that $[L:Q]$ divides both $[Q(\sqrt[q]{2}):Q] = q$ and $[Q(\sqrt[p]{2}):Q] = p$, and $q,p$ are coprime, so $[L:Q] = 1$.
A: Of course $p$ and $q$ are different primes. Then $\sqrt[p]2\cdot\sqrt[q]2=2^{(q+p)/pq}$, and so your big field is certainly contained in $\Bbb Q(2^{1/pq})$. On the other hand, since $p+q$ is relatively prime to $pq$, there are integers $A$ and $B$ with $A(p+q)+Bpq=1$. Now, $2^B\cdot\bigl(2^{(p+q)/pq}\bigr)^A=2^{(Bpq+A(p+q))pq}=2^{1/pq}$. That is, if $\mu=2^{1/pq}$ and $\rho=2^{(q+p)/pq}$, we have $\mu=2^B\rho^A$, so that the opposite inclusion $\Bbb Q(2^{1/pq})\subset\Bbb Q(2^{1/p + 1/q})$ also holds, and the degree over $\Bbb Q$ is exactly $pq$.
For your second question, I believe that the analysis of @oxeimon is just right.
Maybe I could say that a much more advanced approach, embedding $\Bbb Q$ into $\Bbb Q_2$, the $2$-adic numbers, and the very simplest properties of the Newton Polygon, would get your answer much sooner.
