I'm interested in the splitting field $\mathbb{Q}(\sqrt[20]{2},\zeta_{20})$ of polynomial $f(x):=x^{20}-2\in\mathbb{Q}[x]$ over $\mathbb{Q}$, where $\zeta_{20}=e^{\tfrac{2\pi i}{20}}$.
Clearly, $f(x)$ is irreducible over $\mathbb{Q}$ by Eisenstein at $p=2$.
But, I'm not sure whether $f(x)$ is irreducible over $\mathbb{Q}(\zeta_{20})$ or not.
In order to show the irreduciblity of $f(x)$ over $\mathbb{Q}(\zeta_{20})$, is it enough to show that $\mathbb{Q}(\sqrt{2})\cap\mathbb{Q}(\zeta_{20})=\mathbb{Q}$?
The reason I thought it was enough to see that is:
Because there are only three quadratic extensions over $\mathbb{Q}$ in $\mathbb{Q}(\zeta_{20})$ as $\mathbb{Q}(\sqrt{5}),\mathbb{Q}(i),\mathbb{Q}(\sqrt{5}i)$, and every nontrivial intermediate field between $\mathbb{Q}$ and $\mathbb{Q}(\zeta_{20})$ is normal over $\mathbb{Q}$ since $\text{Gal}(\mathbb{Q}(\zeta_{20})/\mathbb{Q})$ is abelian.
Thus, $f(x)$ is irreducible over $\mathbb{Q}(\zeta_{20})$ since $\mathbb{Q}(\sqrt{2})\cap\mathbb{Q}(\zeta_{20})=\mathbb{Q}$.
I would be grateful if someone could confirm that my claim is correct or incorrect. Thank you.
Let $K=\mathbf{Q}(\zeta_{20})$ and $L=\mathbf{Q}(\zeta_{20},\sqrt[20]{2})$. We wish to compute $[L:\mathbf{Q}]$.
The intermediate field $K$ has degree $\phi(20)=\phi(5)\phi(4)=4\cdot 2=8$ over $\mathbf{Q}$.
The extension $L/K$ is a so called radical extension since $K$ already contains a primitive $20$th root of unity. From this theorem in field theory, it is a cyclic extension of degree equal to $20/d$, where $d$ is the greatest divisor of $20$ for which $x^d=2$ for some $x\in K$.
We ask ourselves since $20=2^2\cdot 5$, is $\sqrt[10]{2}\in K$? Is $\sqrt[5]{2}\in K?$ Is $\sqrt[4]{2}\in K$? Is $\sqrt{2}\in K$?
Now clearly $\sqrt[10]{2},\sqrt[5]{2}\not\in K$, since those have degree $10$ and $5$ over $\mathbf{Q}$; $10>8$ and $5$ is coprime to $8$..
Furthermore, $\sqrt[4]{2},\sqrt{2}\not\in K$, since as you noticed, $\mathbf{Q}(\sqrt{2})$ is not an intermediate field of $K/\mathbf{Q}$.
Hence $L/K$ is a (cyclic) extension of degree $20$ and we conclude $\boxed{[L:\mathbf{Q}]=8\cdot 20=160}$.
To answer your question: $f$ is indeed irreducible over $K[X]$, but it requires some more work.. :)