-2

This is a review question, basically the question was written where $\alpha$ was a root of $x^{7}-1$ such that $\alpha \ne 1$, and we were asked to determine, $$[\mathbb{Q}(\alpha):\mathbb{Q}],$$ and I have determined that to be $6$, since we can assume $\alpha^{7}=1$ since the group of unity corresponding to this question is of order $7$. I'm now tasked with finding $[\mathbb{Q}(\alpha + \alpha^{-1}):\mathbb{Q}]$. I know that, $\alpha^{-1} = \alpha^{6}$, so this question can be rewritten as find: $$[\mathbb{Q}(\alpha + \alpha^6):\mathbb{Q}].$$ I know I could let $\omega = \alpha + \alpha^{6}$ and attempt to find an irreducible polynomial by proceeding with: $\omega^2 = \alpha^5 + 2 + \alpha^2$, but I was wondering if there was an easy way to determine this without resorting to that tactic. Any help would appreciated! Thanks.

Greg Martin
  • 92,241
Tobias Rieper
  • 633
  • 3
  • 8
  • For a different approach observe $\alpha^{-1}$ is the complex conjugate of $\alpha$. Show first that $\alpha+\alpha^{-1}$ is stable under complex conjugation. What does Galois theory tell you? – Jyrki Lahtonen Jun 24 '22 at 16:14
  • Search the site for questions about cyclotomic fields. I'm sure we have covered this one (under one of the mentioned tags). – Jyrki Lahtonen Jun 24 '22 at 16:16
  • Hint:- $\alpha\neq 1$ is a root of $1+x+...+x^{6}$. And this polynomial is irreducible(known as the cyclotomic polynomial) . See Dummit and Foote pg 554 . – Mr. Gandalf Sauron Jun 24 '22 at 16:30
  • 1
    "we can assume $\alpha^7=1$": that's literally the given assumption "$\alpha$ is a root of $x^7-1$". – Greg Martin Jun 24 '22 at 16:35
  • 4
    In general, $[\Bbb Q(\zeta_n+\zeta_n^{-1}):\Bbb Q]=\phi(n)/2$, see here. – Dietrich Burde Jun 24 '22 at 16:44

1 Answers1

2

Lets assume that $\alpha\not=1$ (in that case there's no extension) and $\alpha=e^{2\pi i/7}$, all of the roots are primitive so it does not care. Since $e^{2\pi i/7}=\cos(2\pi/7)+i\sin(2\pi/7)$ and $|\alpha|=1$, it implies that $\alpha^{-1}=\bar\alpha=cos(2\pi/7)-i\sin(2\pi/7)$. It follows that $\alpha+\alpha^{-1}=2\cos(2\pi /7)$. Now there are two ways to calculate the degree of the extension:

First one: $$\alpha=cos(2\pi/7)-i\sin(2\pi/7)\implies (\alpha-cos(2\pi/7))^2=-\sin^2(2\pi/7)\implies \alpha^2-2\cos(2\pi/7)\alpha+\cos^2(2\pi/7)=-\sin^2(2\pi/7)\implies \alpha^2-2\cos(2\pi/7)\alpha+1=0$$ So $\alpha$ is root of the irreducible polynomial $f(t)=t^2-2\cos(2\pi/7)t+\cos^2(2\pi/7)\in\mathbb{Q}(\cos(2\pi/7))[t]$. So it implies that $[\mathbb{Q}(\alpha):\mathbb{Q}(\cos(2\pi/7)]=2$, and since $[\mathbb{Q}(\alpha):\mathbb{Q}]=6$, we have that $[\mathbb{Q}(\cos(2\pi/7):\mathbb{Q}]=3$

Second one:

The minimal polynomial of $\alpha$ over the rationals is the cyclotomic polynomial of order $7$, $g(t)=t^6+t^5+t^4+t^3+t^2+t+1$. We can multiplie both sides by $t^{-3}$, getting: $$t^{-3}g(t)=(t^3+t^{-3})+(t^2+t^{-2})+(t+t^{-1})+1$$ It's easy to manipulate it to get the irreducible polynomial $$t^{-3}g(t)=x^3+x^2-2x-1$$ where $x=t+t^{-1}$. So $\alpha+\alpha^{-1}$ is root of $h(t)=t^3+t^2-2t-1$, so it implies that $[\mathbb{Q}(\alpha+\alpha^{-1}):\mathbb{Q}]=3$

Guillermo García Sáez
  • 8,159