Is $\sqrt{2}\in\mathbb{Q}(\sqrt[8]{3})$ or not?

Here's one approach to the problem:

If $\sqrt{2} \in \mathbb{Q}(\sqrt[8]{3})$, we have $$8 = [\mathbb{Q}(\sqrt[8]{3}):\mathbb{Q}] = [\mathbb{Q}(\sqrt[8]{3}):\mathbb{Q}(\sqrt{2})]\cdot [\mathbb{Q}(\sqrt{2}):\mathbb{Q}] = 2 \cdot [\mathbb{Q}(\sqrt[8]{3}):\mathbb{Q}(\sqrt{2})],$$
and so $\sqrt[8]{3}$ must satisfy an irreducible polynomial of degree 4 over $\mathbb{Q}(\sqrt{2})$, call it $g(x)$. Then necessarily, $g(x) \mid x^8-3$ over $\mathbb{Q}(\sqrt{2})$. Thus the four roots of $g(x)$ are of the form $\sqrt[8]{3} \zeta_8^i$ for some $i$. In particular, the constant term of $g(x)$ is of the form $\sqrt[2]{3} \zeta_8^j$. But this constant term is in $\mathbb{Q}(\sqrt{2})$, so it must be real, hence $\zeta_8^j = \pm 1$. But then $\sqrt{3} \in \mathbb{Q}(\sqrt{2})$, but I think you can show this is not true pretty easily.

Your hunch can be justified and generalized by using some algebraic number theory, specifically the discriminant $D_K$ of a number field $K$. The discriminant has the following two important properties:

• If $K \subset L$, then $D_K$ divides $D_L$.
• The primes that divide $D_K$ are precisely the primes $p$ which ramify in $K$.

It is not hard to see on the one hand that $3$ ramifies in $\mathbb{Q}(\sqrt[8]{3})$ and on the other hand that $2$ is the only prime that ramifies in $\mathbb{Q}(\sqrt{2})$, so the discriminant of the former cannot divide the discriminant of the latter.