Why $[\mathbb{Q}(\sqrt[n]{p}, \sqrt[m]{q}):\mathbb{Q}] = nm$?

Question

As title says, I want to prove that $\mathbb{Q}(\sqrt[n]{p}, \sqrt[m]{q})$ is degree $nm$ extension of $\mathbb{Q}$ when $p \neq q$ are distinct primes. By Eisenstein's criterion, $x^{n} - p$ is irreducible over $\mathbb{Q}$ and so $[\mathbb{Q}(\sqrt[n]{p}):\mathbb{Q}] = n$. However, I wonder if it is trivial that $x^{m} - q$ is irreducible over $\mathbb{Q}(\sqrt[n]{p})$, assuming the general form of Eisenstein's criterion over integral domain (see Wikipedia for the statement). I'm not sure if the polynomial satisfies assumptions of the Eisenstein's criterion.

Also, I tried some $p$-adic ($q$-adic will be more appropriate) strategy. If we assume that $\mathbb{Q}(\sqrt[n]{p})$ can be embedded in $\mathbb{Q}_{q}$, then the irreducibility of $x^{m} - q$ over $\mathbb{Q}(\sqrt[n]{p})$ follows from it over $\mathbb{Q}_{q}$, which is a result of Newton polygon. However, the assumption fails in general, and when we enlarge $\mathbb{Q}_q$ to its finite extension that contains $\sqrt[n]{p}$, the situation gets worse.

Another direction (which I believe that shouldn't be easy) is to prove that the polynomial $$ f(x) = \prod_{0\leq i < n, 0 \leq j < m} (x - \sqrt[n]{p}\zeta_{n}^{i} - \sqrt[m]{q}\zeta_{m}^{j}) $$ is irreducible over $\mathbb{Q}$. Then the claim follows from the another proposition $$\mathbb{Q}(\sqrt[n]{p} + \sqrt[m]{q}) = \mathbb{Q}(\sqrt[n]{p}, \sqrt[m]{q})$$ (this is not trivial, but I have a proof by the light use of Galois theory.) According to this, this proposition implies our main claim when $n, m$ are coprime.

Can we use Eisenstein's criterion in this case? If not, is there a simple way to prove this?

Maybe related: Why $\sqrt[3]{3}\not\in \mathbb{Q}(\sqrt[3]{2})$?

Answer 1

An elementary proof produced with the assistance of @KCd.

Let $\alpha = \sqrt[n]{p}$ and $\beta = \sqrt[m]{q}$ and let $L=\mathbb{Q}(\alpha,\beta)$.

If $i\not\equiv 0\pmod n$ and $j\not\equiv 0\pmod m$, then some minimal power $k>1$ of $\gamma=\alpha ^i\beta ^j$ is a rational. Now let $\omega$ be a primitive $k$th root of unity and consider the polynomial $x^k-\gamma ^k$ with roots $$\gamma, \gamma \omega ,\gamma \omega ^2, ...,\gamma \omega ^{k-1}.$$ If $x^k-\gamma ^k$ is not a minimal polynomial for $\gamma $ over $\mathbb{Q}$, there would be a polynomial over $\mathbb{Q}$ with roots a non-trivial subset of $\gamma, \gamma \omega ,\gamma \omega ^2, ...,\gamma \omega ^{k-1}$ and therefore with a product $\gamma ^x\omega ^y\in \mathbb{Q}$ where $0<x<k$. Then $\omega ^y$ is real and is therefore $\pm 1$ and so $\gamma ^x\in \mathbb{Q}$, a contradiction of minimality.

We conclude that $x^k-\gamma ^k$ is a minimal polynomial. Let $K=\mathbb{Q}(\gamma )$ and then

$$\mathrm{Tr}_{L/\mathbb{Q}}(\gamma )=[L:K]\mathrm{Tr}_{K/\mathbb{Q}}(\gamma )=0.$$ Now suppose that there are rationals $a_{ij}$ such that $$\sum_{i=0}^{n-1} \sum_{j=0}^{m-1}a_{ij}\alpha ^i\beta ^j=0.$$ Then for any integers $u$ and $v$, $$\sum_{i=0}^{n-1} \sum_{j=0}^{m-1}a_{ij}\alpha ^{i-u}\beta ^{j-v}=0.$$

Taking $\mathrm{Tr}_{L/\mathbb{Q}}$ we see that $a_{ij}=0$ if $i\equiv u\pmod n$ and $j\equiv u\pmod m$ and therefore all $a_{ij}=0$. Then the $\alpha ^i\beta ^j$ are linearly independent over $\mathbb{Q}$ and $[L:\mathbb{Q}]=mn$.

Answer 2

See the earlier question $\operatorname{dim} \mathbb{Q}[2^{1/n},3^{1/m}] / \mathbb{Q} = nm$, where my answer explains why this problem over all $m$ and $n$ is equivalent to the case $m=n$ (for all $n$) and you can replace 2 and 3 with a more general pair of numbers in the base field.