Irreducibility of $X^5-7$ over $\mathbb{Q}(\sqrt[7]{2})[X]$ and degree of spitting field I have worked through these two questions but am unsure if I got the right idea, please may you help me?

Prove that $X^5-7$ is irreducible over $\mathbb{Q}(\sqrt[7]{2})[X]$

Can we say that $f(X)=X^5-7$ is irreducible over $\mathbb{Q}$ by Eisenstein's criterion?
Or does $f(\sqrt[7]{2}) \neq 0$ imply that $f(X)$ is irreducile over $\mathbb{Q}(\sqrt[7]{2})$? Does that work?

Let $L$ be the splitting field of $X^5-7$. Find $[L : \mathbb{Q}]$

Clearly $\alpha=\sqrt[5]{7}$ is a root of $f(X)$. I think that
$$f(X)=(X-\alpha)(X-\alpha\zeta)(X-\alpha\zeta^2)(X-\alpha\zeta^3)(X-\alpha\zeta^4),$$
is that correct? I am really not sure.
The splitting field is $\mathbb{Q}(\lambda_1, ..., \lambda_n)$ where $\lambda_i$ are the roots of $f(X)$ that are not in $\mathbb{Q}$.
So I think the splitting field may be $L=\mathbb{Q}(\zeta_5)$ and so $[L : \mathbb{Q}]=4$. Is $4$ correct since the cyclotomic polynomial $\Phi_5(X)$ has degree $4$?
 A: *

*Yes, $X^5-7$ is irreducible over $\Bbb{Q}$ by Eisenstein's criterion.

*It is not true that $f(\sqrt[7]{2}) \neq 0$ implies that $f(X)$ is irreducible over $\Bbb{Q}(\sqrt[7]{2})$. Consider $X^4-4$; it has no roots in $\Bbb{Q}$ or $\Bbb{Q}(\sqrt[7]{2})$, but it is still reducible.

*Indeed the splitting field of $X^5-7$ over $\Bbb{Q}$ contains $\Bbb{Q}(\zeta_5)$ as a subfield, but as you noted already it also contains $\sqrt[5]{7}$, so the splitting field is strictly larger than $\Bbb{Q}(\zeta_5)$.


Once you know the degree of the splitting field over $\Bbb{Q}$, it quickly follows that $X^5-7$ is irreducible in $\Bbb{Q}(\sqrt[7]{2})$ by comparing it to the degree of $\Bbb{Q}(\sqrt[7]{2})$ over $\Bbb{Q}$.
A: we use when two field extension are coprime dimension the composite have dimension egaul to the product dimension.
so, $X^5-7$ is irreducible over $\Bbb{Q}(^7\sqrt{2})$ , also $X^7-2$ irreducible over $\Bbb{Q}(^5\sqrt{7})$
for the same raison the composite of   $\Bbb{Q}(\zeta_5)$  and  $\Bbb{Q}(^5\sqrt{7})$ give $[L:\Bbb{Q}]=20$.
