Is $\sqrt[6]{3} \in \mathbb{Q}(\sqrt[8]{21})$ and/or $\sqrt[4]{5} \in \mathbb{Q}(e^{\frac{2 \pi i}{25}})$?

Question

I need to check if $\sqrt[6]{3} \in \mathbb{Q}(\sqrt[8]{21})$ and/or $\sqrt[4]{5} \in \mathbb{Q}(e^{\frac{2 \pi i}{25}})$.

For the first one:

Assume $\sqrt[6]{3} \in \mathbb{Q}(\sqrt[8]{21})$, then $\mathbb{Q}(\sqrt[6]{3}) \subseteq \mathbb{Q}(\sqrt[8]{21})$. So $[\mathbb{Q}(\sqrt[8]{21}): \mathbb{Q}] = [\mathbb{Q}(\sqrt[8]{21}): \mathbb{Q}(\sqrt[6]{3})] \cdot [\mathbb{Q}(\sqrt[6]{3}) : \mathbb{Q}]$.

Since $[\mathbb{Q}(\sqrt[8]{21}): \mathbb{Q}] = 8$ and $[\mathbb{Q}(\sqrt[6]{3}) : \mathbb{Q}] = 6$, it follows that $6 \mid 8$, which is not true. Therefore $\sqrt[6]{3} \notin \mathbb{Q}(\sqrt[8]{21})$. Is this correct?

For the second one:

First let's denote $\omega := e^{\frac{2 \pi i}{25}}$ as the $25$-th root of unity.

I tried the same technique, but ended up with $[\mathbb{Q}(\omega) : \mathbb{Q}(\sqrt[4]{5})] = 5$, since $[\mathbb{Q}(\omega) : \mathbb{Q}] = \varphi(25) = 20$ and $[\mathbb{Q}(\sqrt[4]{5})] : \mathbb{Q}] = 4$, where $\varphi$ is Euler's totient function. So the assumption $\sqrt[4]{5} \in \mathbb{Q}(\omega)$ might be true, if i manage to find some irreducible $f \in \mathbb{Q}(\sqrt[4]{5})[X]$ with $f(\omega) = 0$ and $\deg(f) = 5$. All this seems far to complicated for this kind of question. (?)

It might be easier to express $\sqrt[4]{5}$ in terms of $\omega$, but i can't see how.

Answer 1

Your proof for the first question is good, and with a little more work the same idea can be used for the second one.

The maximal real subfield of $\newcommand{\Q}{\mathbb{Q}} \Q(\omega)$ is $\Q(\omega + \omega^{-1})$ and $[\Q(\omega) : \Q(\omega + \omega^{-1})] = 2$ so $[\Q(\omega + \omega^{-1}) : \mathbb{Q}] = 10$. Since we can choose $\sqrt[4]{5} \in \mathbb{R}$, if $\sqrt[4]{5} \in \Q(\omega)$, then $\sqrt[4]{5} \in \Q(\omega + \omega^{-1})$. But $[\Q(\sqrt[4]{5}) : \Q] = 4 \nmid 10 = [\Q(\omega + \omega^{-1}) : \mathbb{Q}]$, so this is impossible.

Answer 2

$\sqrt[4]5$ has a minimal polynomial $f(x) = x^4 - 5$. This polynomial does not split over $K = \mathbb Q(\sqrt[4]5)$. Consider its Galois closure $L$. Then $[L:K] = 2$ since $L = K(i)$. So $[L:\mathbb Q] = 8$. But $\mathbb Q(e^{2\pi i/25})$ is also a Galois extension over $\mathbb Q$, so if it contains $\sqrt[4]5$ it also contains $L$ but it cannot.