$\mathbb{Q}(\sqrt{p},\sqrt[3]{q})= \mathbb{Q}(\sqrt{p}\cdot \sqrt[3]{q})$ ??

Question

Let $p,q$ be primes, $p≠q$, then I have to show that $\mathbb{Q}(\sqrt{p},\sqrt[3]{q})= \mathbb{Q}(\sqrt{p}\cdot \sqrt[3]{q})$

So far I've tried a lot of things with minimal polynomials and bases, but I'm far from any argument that I find sufficient.

This is an example from my algebra course, first chapter about field extension. So far we know about basic definitions, minimal polynomials and their properties, algebraic elements and algebraic fields.

Could someone maybe give me a hint about how to proceed? Thanks a lot!!!

Answer 1

Hint: Let $K = \Bbb Q(\sqrt{p}\sqrt[3]{q})$ and $\alpha = \sqrt{p}\sqrt[3]{q}$. Then $\alpha^3/(pq) = \sqrt{p} \in K$ and $\alpha/\sqrt{p} = \sqrt[3]{q} \in K$.