10

Let $l$ be an odd prime number and $\zeta$ be a primitive $l$-th root of unity in $\mathbb{C}$. Let $K = \mathbb{Q}(\zeta)$.

My question: Is the following proposition true? If yes, how would you prove this?

Proposition

(1) $K_0 = \mathbb{Q}(\zeta + \zeta^{-1})$ is the maximal real subfield of $K$.

(2) $[K_0 : \mathbb{Q}] = (l - 1)/2$

(3) The ring of algebraic integers $A_0$ in $K_0$ is $\mathbb{Z}[\zeta + \zeta^{-1}]$.

(4) $\zeta + \zeta^{-1}$ and its conjugates constitute an integral basis of $A_0$.

Related questions

This and this.

Community
  • 1
Makoto Kato
  • 42,602
  • 2
    You can get some of this cheaply, I think. (1) and (2) should follow from the fact that $\zeta + \zeta^{-1}$ is fixed by conjugation and the fact that $\mathbb Q(\zeta)$ will split the quadratic $(X - \zeta)(X - \zeta^{-1}) = X^2 - (\zeta + \zeta^{-1})X + 1$. – Dylan Moreland Jul 25 '12 at 05:54
  • Ah, right, sorry for botching the title. Just wanted to be a bit more descriptive. – Dylan Moreland Jul 25 '12 at 06:31
  • No problem, Dylan. Thanks. – Makoto Kato Jul 25 '12 at 06:50

1 Answers1

9

As Dylan points out, parts (1) and (2) are clear. Moreover, $\mathbb{Z}[\zeta + \zeta^{-1}]$ contains $\zeta^j + \zeta^{-j}$ for all $j \ge 1$ (by induction using the binomial theorem); these include all the conjugates of $\zeta + \zeta^{-1}$, so (4) implies (3). Thus it suffices to prove (4), which follows from the corresponding fact for the full cyclotomic field $\mathbb{Q}(\mu_\ell)$ (which is well known), as follows:

Let $u \in A_0$. Because $u$ is an algebraic integer in $\mathbb{Q}(\zeta)$, we can write $u = \sum_{i = 0}^{\ell - 1} u_i \zeta^i$ for some $u_i \in \mathbb{Z}$. But since $u = \overline{u}$, we have $u = \sum u_i \zeta^{-i} = \sum u_i \zeta^{\ell - i}$. Hence $u_i = u_{\ell - i}$. Thus we have $u = \sum_{i = 0}^{(\ell - 1)/2} u_i (\zeta_i + \zeta^{-i})$, and (4) is proved.

David Loeffler
  • 9,584