Prove that $x^4 + \alpha x^3 + {\alpha}^2 x^2 + {\alpha}^3 x + {\alpha}^4$ is irreducible over $\Bbb{Q}(\alpha)$ where $\alpha ^5 = 2$. My initial idea was to follow the same idea as in the case for a quadratic (as in here: Prove that $x^2$ + $\alpha$ x + ${\alpha}^2$ is irreducible over $\Bbb{Q}(\alpha)$ where $\alpha ^3 = 2$.), but it could factor into two squares and even the case of having a linear factor is quite complex. Can someone give me an idea that would work?

First, if $p(x) = x^4 + \alpha x^3 + \alpha^2 x^2 + \alpha^3 x + \alpha^4$ and $q(t) = t^4 + t^3 + t^2 + t + 1$, note that $p(x) = \alpha^4 q(x / \alpha)$; from this, it follows that $p$ is irreducible over $\mathbb{Q}(\alpha)$ if and only if $q$ is.
On the other hand, every monic factor of $q$ must have coefficients in the splitting field of $q$ over $\mathbb{Q}$.  This splitting field is the cyclotomic field $\mathbb{Q}(e^{2\pi i/5})$, which has degree 4 over $\mathbb{Q}$.  (So for example, the monic quadratic factors must be of the form $(t - e^{2\pi i k/5}) (t - e^{2\pi i \ell/5})$ for some $k, \ell$.)  Therefore, if we have a factorization $q(t) = q_1(t) q_2(t)$ where $q_1$ and $q_2$ are monic, then $q_1, q_2 \in (\mathbb{Q}(\alpha) \cap \mathbb{Q}(e^{2\pi i/5})[t]$.  However, $[\mathbb{Q}(\alpha) \cap \mathbb{Q}(e^{2\pi i/5}) : \mathbb{Q}] \mid [\mathbb{Q}(\alpha) : \mathbb{Q}] = 5$, and also $[\mathbb{Q}(\alpha) \cap \mathbb{Q}(e^{2\pi i/5}) : \mathbb{Q}] \mid [\mathbb{Q}(e^{2\pi i/5}) : \mathbb{Q}] = 4$.  Thus, $[\mathbb{Q}(\alpha) \cap \mathbb{Q}(e^{2\pi i/5}) : \mathbb{Q}] = 1$, so $\mathbb{Q}(\alpha) \cap \mathbb{Q}(e^{2\pi i/5}) = \mathbb{Q}$.  Now, we have $q_1, q_2 \in \mathbb{Q}[t]$, and since $q$ is irreducible over $\mathbb{Q}$, we see that either $q_1$ or $q_2$ is a unit.  This shows that $q$ is also irreducible over $\mathbb{Q}(\alpha)$, implying that $p$ is irreducible over $\mathbb{Q}(\alpha)$ as well.
