Irreducibles polynomials How can I show that the polynomial $f(X)=X^5-2X^3+2X-2$ is irreducible over $\mathbb{Q}[\sqrt[3]{2}]$? Obviously, it is irreducible over $\mathbb{Q}$, by Eisenstein criterion, but how can I consider the extension  $\mathbb{Q}[\sqrt[3]{2}]$? Thanks in advance. 
 A: Here is what I have tried:
Let $\alpha$ be a root of $f(X)$ in $\mathbb{C}$, and denote $\beta = \sqrt[3]{2}$.
Now we consider tower of field extensions
$$
\mathbb{Q} \subset \mathbb{Q}(\alpha) \subset \mathbb{Q}(\alpha, \beta), \Rightarrow [\mathbb{Q}(\alpha, \beta): \mathbb{Q}] = [\mathbb{Q}(\alpha, \beta): \mathbb{Q}(\alpha)][\mathbb{Q(\alpha)}:\mathbb{Q}]
$$
$$
\mathbb{Q} \subset \mathbb{Q}(\beta) \subset \mathbb{Q}(\alpha, \beta), \Rightarrow [\mathbb{Q}(\alpha, \beta): \mathbb{Q}] = [\mathbb{Q}(\alpha, \beta): \mathbb{Q}(\beta)][\mathbb{Q(\beta)}:\mathbb{Q}]
$$
We know that 
$$[\mathbb{Q(\alpha)}:\mathbb{Q}] = 5, [\mathbb{Q(\beta)}:\mathbb{Q}] = 3$$ 
$$[\mathbb{Q}(\alpha, \beta): \mathbb{Q}(\beta)] \leq 5, [\mathbb{Q}(\alpha, \beta): \mathbb{Q}(\alpha)] \leq 3$$
Therefore, we conclude that
$$[\mathbb{Q}(\alpha, \beta): \mathbb{Q}(\beta)] = 5$$
i.e. $f$ is irreducible in $\mathbb{Q}(\beta)$.
If this proof works well, you can generalize to two irreducible polynomials in $\mathbb{Q}[x]$ with their degree coprime.
