Prove that $x^4 + \alpha x^3 + {\alpha}^2 x^2 + {\alpha}^3 x + {\alpha}^4$ is irreducible over $\Bbb{Q}(\alpha)$ where $\alpha ^5 = 2$.

Question

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?

Answer 1

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.

Answer 2

Let $f(x) =x^5-2\in\mathbb{Q}[x]$ then $f(x) $ is irreducible over $\mathbb{Q} $. Let $a$ be a root of $f(x) $ and let $K=\mathbb{Q} (a) $ so that $[K:\mathbb{Q}] =5$.

Let $b$ be a root of the polynomial $$g(x) =\frac{f(x)} {x-a} =x^4+ax^3+a^2x^2+a^3x+a^4\in K[x] $$ Then we show that $b\notin K$. We have $$g(b) = b^4p(a/b)$$ where $$p(x) =x^4+x^3+x^2+x+1\in\mathbb{Q}[x]$$ is irreducible over $\mathbb{Q} $ (apply Eisenstein criterion on $p(x+1)$). If $b\in K$ then $c=a/b\in K$ is a root of $p(x) $ and $\mathbb{Q} (c) $ is a subfield of $K$ of degree $4$ over $\mathbb{Q} $. This is absurd as $K$ is of degree $5$ over $\mathbb{Q} $ and $4\nmid 5$.

Let $L=K(b) $ then we have proved above that $1<[L:K]\leq 4$ so that $[L:\mathbb{Q}] \leq 20$. Our job is done if we prove that $[L:K] = 4$ or equivalently $[L:\mathbb{Q}] =20$. This is where we need $c=a/b$ which is a root of $p(x) $. We have $K(b) =K(c) $ and then $L=\mathbb{Q} (a, c) $ and $[L:\mathbb{Q}] $ is divisible by both $[\mathbb{Q} (a) :\mathbb{Q}] =5$ and $[\mathbb{Q} (c) :\mathbb{Q}] =4$ so that $[L:\mathbb{Q}] \geq 20$. It follows that $[L:\mathbb{Q}] =20$ and $[L:K] =4$ and therefore $g(x) $ is irreducible over $K$.