# Irreducible polynomial over $\mathbb{Q}(\zeta)$

Show that the polynomial $f(x)=x^5-2$ is irreducible over $\mathbb{Q}(\zeta)$, where $\zeta=e^{2\pi i/5}$.

I tried show that the roots of polynomial $f(x)=x^5-2$, $$\sqrt[5]{2}, \zeta\sqrt[5]{2}, \zeta^2\sqrt[5]{2},\zeta^3\sqrt[5]{2}, \zeta^4\sqrt[5]{2}$$ with $\zeta^5=1$ e $\zeta\neq 1$ no belong in $\mathbb{Q}(\zeta)$ and $f(x)$ can not fatorized in two polynomials, one the degree $2$ and outher the degree $3$. I think the idea is good, but I cannot finish it.

Someone could help me?

First of all consider the splitting field $\mathbb{E}$ of $x^5-2$: It's easy to see that $\mathbb{E}=\mathbb{Q}(\zeta_5, \sqrt[5]{2})$ and has degree 20.

Let's note that $\mathbb{E}$ has degree $5$ over $\mathbb{Q}(\zeta_5)$.

Now suppose that $x^5-2$ splits on $\mathbb{Q}(\zeta_5)$. Then esist an irreducible polynomial $p$ of degree smaller than $5$ and greatest of 1 (because $\mathbb{Q}(\zeta_5) \subsetneq \mathbb{E}$) that divide $x^5-2$ and then, if $\mathbb{F}$ is the splitting field of $p$, we have the containment $$\mathbb{Q}(\zeta_5) \subsetneq \mathbb{F} \subseteq \mathbb{E}$$ Now you have an absurd by the multiplicativity of the degree of extension of fields and observing that if $\deg p <5$ then $5$ doesn't divide the degree of the splitting field of $p$.

Hint: Check that $x^4 + x^3 + x^2 + 1$ is irreducible over $\mathbb Q$, of degree $4$ which is coprime to $5$, and $\zeta$ is a root of this polynomial.

And show this result

Claim:

Let $[K:\mathbb Q] = m$ and $p(x) \in \mathbb Q[x]$ is an irreducible polynomial over $\mathbb Q$ of degree $n$. If $\mathrm {G.C.D}\, \{m,n\} = 1$ then $p(x)$ is an irreducible polynomial over $K$.

Proof: (Sketch) Consider $\alpha$ a root of $p(x)$ and notice that $\mathbb Q[\alpha] \subset K[\alpha]$, suppose $[K[\alpha]: K] = r$ and $[K[\alpha] : \mathbb Q[\alpha]] = s$. Look at

$$n\cdot s = m\cdot r$$

Draw a diagram, it really helps.

The idea is very good indeed. You have to show that a product of $k$ factors from $(x- \zeta^l \sqrt[5]{2}$) is not a polynomial with integer coefficients if $1\le k \le 4$. Indeed, such a product would have the constant term a product of $k$ factors of $(- \zeta^l \sqrt[5]{2})$, whose absolute value is $\sqrt[5]{2}^k$, not an integer.