# Degree of $\mathbb{Q}(e^{2i\pi / 5}, \sqrt[5]{2}) / \mathbb{Q}$

I have the field extension $$E = \mathbb{Q}(z,w)$$ over $$F = \mathbb{Q}$$ where $$z = \sqrt[5]{2}$$ and $$w = e^{2i\pi / 5}$$. I want to find the degree of the extension $$[E:F]$$.

I see that $$E$$ is the splitting field of $$x^5 - 2$$. And since this has degree $$5$$ I was initially thinking that the field extension has degree $$5$$. But then I was considering the tower $$F \subseteq \mathbb{Q}(w) \subseteq \mathbb{Q}(z,w)$$. I think each step has degree $$5$$, so by the tower law $$[E:F] = 5^2 = 25$$.

Is that correct?

Just out of curiosity, what does a specific basis look like?

• The irreducible polynomial of $\omega$ over $\mathbb{Q}$ is $x^4+x^3+x^2+x+1$ so $[\mathbb{Q}(\omega) : \mathbb{Q}] = 4$. Feb 19, 2019 at 20:20
• @DanielSchepler Ah, that actually makes sense. Please write this as an answer so that I can accept. (Also, I am still wondering about a specific basis.) Feb 19, 2019 at 20:25
• Well, my comment doesn't answer the overall question yet, just points out a mistake in the reasoning in the second paragraph. Feb 19, 2019 at 20:36

It’s “well known” that $$\Bbb Q(w)$$ has degree $$4$$ over $$\Bbb Q$$, in fact the minimal polynomial for $$w$$ is $$X^4+X^3+X^2+X+1$$. Since $$\sqrt[5]2$$ is quintic over $$\Bbb Q$$, you expect that your larger field will have degree $$20$$ over $$\Bbb Q$$.
Let $$K$$ be the compositum of the fields $$\Bbb Q(w)$$ and $$\Bbb Q(\sqrt[5]2\,)$$. This is, by definition, the smallest field containing the two named fields, and its degree over $$\Bbb Q$$ must be divisible by both $$4$$ and $$5$$, thus by $$20$$. In particular, $$[K:\Bbb Q]\ge20$$, while on the other hand, you easily see that $$\bigl[(\Bbb Q(w))(\sqrt[5]2\,):\Bbb Q(w)\bigr]\le 5$$, with $$K=(\Bbb Q(w))(\sqrt[5]2\,)$$ and by the multiplicativity of field extension degree, $$[K:\Bbb Q]\le20$$.
An explicit basis? The collection $$\lbrace \sqrt[5]2\,^iw^j\rbrace_{0\le i\le4,0\le j\le3}$$ will do.
The field is the splitting field of the polynomial $$x^5-2$$, which has degree $$\phi(5)\cdot 5=20$$ over $$\Bbb Q$$, see this question, for $$x^n-a$$ in general:
Computing the Galois group of polynomials $x^n-a \in \mathbb{Q}[x]$