Does the polynomial $x^7-2$ reduce over the field $\mathbb{Q}(e^{\frac{2 \pi i}{7}})$?

Question

So my question is

Suppose $\xi =\exp{\frac{2 \pi i}{7}}$. Is $x^7-2$ reducible over $\mathbb{Q}(\xi)$?

Answer 1

The splitting field of $X^7-{2}$ is $\mathbb{Q}[\zeta, \sqrt[7]{2}]$ where $\zeta$ is a $7$-th root of unity.

In general, we have that for a prime $p\geq 2$, and $K$ a subfield of $\mathbb{C}$, if $K$ contains $\zeta$ then $X^p-a$ is either irreducible in $K[X]$ or factors as a product of linear factors. In this case, we know that this polynomial factors as $$\prod_{0\leq k \leq 6}{(X-\sqrt[7]{2}\zeta^k)}.$$ Since our extension does not contain $\sqrt[7]{2}$, this polynomial is irreducible in $\mathbb{Q}[\zeta]$.

Answer 2

Other answers raise good points. FWIW I want to give an elementary self-contained argument.

For starters, $p(x)=x^7-2$ is irreducible in $\Bbb{Q}[x]$ by Eisenstein. From the theory of cyclotomic fields we know that $[\Bbb{Q}(\xi):\Bbb{Q}]=6$. Using these pieces of information and basic facts we can deduce that $p(x)$ remains irreducible over $\Bbb{Q}(\xi)$.

Assume contrariwise that $p(x)=p_1(x)p_2(x)$ non-trivially with some $p_1,p_2\in\Bbb{Q}(\xi)[x]$. W.l.o.g. we can assume that $p_1(x)$ is irreducible over $\Bbb{Q}(\xi)$. Let $\alpha$ be a zero of $p_1(x)$. Consider the field $F=\Bbb{Q}(\xi,\alpha)$. It is a degree $m=\deg p_1$ extension of $\Bbb{Q}(\xi)$, so $[F:\Bbb{Q}]=6m$. But it contains $K=\Bbb{Q}(\alpha)$ as an intermediate field. Irreducibility of $p(x)$ over $\Bbb{Q}$ implies that $[K:\Bbb{Q}]=7$. So $$ 6m=[K:\Bbb{Q}]=[K:F]\cdot [F:\Bbb{Q}]=7\cdot [K:F]. $$ Because $\gcd(6,7)=1$, we can conclude that $7\mid m$. This is in violation of the assumption that $p_1(x)$ was to be a proper factor of $p(x)$.

Answer 3

Assume that the polynomial $\prod_{k=0}^6 (X- \omega^ k \sqrt[7]{2})$ is reducible over $\mathbb{Q}(\omega)$. Then for some proper subset $K$ of $\{0, \ldots,6\}$ we have $$\prod_{k\in K} (X- \omega^ k \sqrt[7]{2}) \in \mathbb{Q}(\omega)[X]$$ and in particular, the coefficient of $X^{|K|-1}$, which is $$-\sqrt[7]{2} \sum_{k \in K} \omega^k \in \mathbb{Q}(\omega)$$ Now, $\sum_{k \in K} \omega^k \ne 0$, so we conclude that $$\sqrt[7]{2} \in \mathbb{Q}(\omega),$$ contradiction.

${\bf Added:}$ In fact, for a Galois extension $K$ to split at all an irreducible polynomial $P$ of prime degree, $K$ has to contain the splitting field of $P$.