1

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?

John Doe
  • 3,233
  • 5
  • 43
  • 88
  • 2
    The irreducible polynomial of $\omega$ over $\mathbb{Q}$ is $x^4+x^3+x^2+x+1$ so $[\mathbb{Q}(\omega) : \mathbb{Q}] = 4$. – Daniel Schepler Feb 19 '19 at 20:20
  • 1
    @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.) – John Doe Feb 19 '19 at 20:25
  • 1
    Well, my comment doesn't answer the overall question yet, just points out a mistake in the reasoning in the second paragraph. – Daniel Schepler Feb 19 '19 at 20:36

2 Answers2

2

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.

Lubin
  • 62,818
2

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 $x^n-a \in \mathbb{Q}[x]$

Dietrich Burde
  • 130,978