0

Let $\xi$ be a 14-th primitive root of unity. I want to compute its conjugates.

In this question I gave the description of the process that I believed should be followed for computing the conjugates:

Assume $\alpha = \xi + \xi^{-1}$, $\sigma(\xi) = \xi^6$, $\tau(\xi) = \xi^2$

$Gal(\mathbb{Q}(\alpha)/\mathbb{Q}) \cong Gal(\mathbb{Q}(\xi)/\mathbb{Q})/Gal(\mathbb{Q}(\xi)/\mathbb{Q}(\alpha)) \cong Gal(\mathbb{Q}(\xi)/\mathbb{Q})/\langle \sigma \rangle$

I observe that $Gal(\mathbb{Q}(\xi)/\mathbb{Q}(\alpha))$ has degree 2 and by Artin's theorem $|Gal(\mathbb{Q}(\xi)/\mathbb{Q}(\alpha))| = 2$ this is the reason why the quotient is taken with $\langle \sigma \rangle$.

The resulting group has order 3 and therefore is cyclic, its conjugancy classes are $\{1,\sigma\}$, $\{\tau,\sigma\tau \}$, $\{\tau^2,\sigma\tau^2 \}$.

Then I just took one automorphism of each class and computed its image. But the people in the answers claim that the conjugates are not $\xi^4+\xi^{-4},\xi^2+\xi^{-2}$ as I thought but the say they are $\xi^3+\xi^{-3},\xi^5+\xi^{-5}$.

Why is it the case?

user1868607
  • 5,791
  • The conjugates of $\xi$ are the primitive 14-th roots of unity, so the $\xi^a$ with $a$ coprime to $14$ (so including 3 and 5 but not 2 or 4). – Angina Seng Nov 30 '17 at 19:31
  • 1
    As Lord Shark explained there are no automorphisms $\xi\mapsto\xi^2$ or $\xi\mapsto\xi^4$ because $\gcd(2,14)=\gcd(4,14)=2\neq1$. If you are so sold on using exponents $2$ and $4$, then you should apply them to $-\xi$ which is a primitive seventh root of unity. The minimal polynomial of $-\alpha$, a conjugate of $2\cos(2\pi/7)$ has been handled many times on our site, for example here (sorry about promoting a version I answered myself, that was just quicker to locate). If you find the minimal polynomial of $-\alpha$, surely you can manage :-) – Jyrki Lahtonen Nov 30 '17 at 19:57
  • How did you get those formulas for $\sigma$ and $\tau$? – Eric Wofsey Nov 30 '17 at 23:15

1 Answers1

1

In Maple:

P1:= numtheory:-cyclotomic(14,xi);
alpha:= RootOf(P1,xi)+1/RootOf(P1,xi);
evala(Norm(alpha-z));

$$ (z^3 - z^2 - 2z + 1)^2 $$

Thus the minimal polynomial of $\alpha$ is $z^3 - z^2 - 2 z + 1$.

Since $\xi^3$ and $\xi^5$ are also primitive 14'th roots of unity, $\xi^3 + \xi^{-3}$ and $\xi^5 + \xi^{-5}$ will satisfy the same polynomial, and it can be verified that they are different from $\xi + \xi^{-1}$ and from each other. So these are the conjugates.

Robert Israel
  • 448,999