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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.

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Makoto Kato
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    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
  • For the argument for (3) in the general case, see https://mathoverflow.net/questions/17289/quick-proof-of-the-fact-that-the-ring-of-integers-of-mathbb-q-zeta-n-is-m – 19021605 Nov 18 '25 at 04:00

1 Answers1

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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
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  • By u bar do you mean the conjugate of u? If yes then could you please explain why u and u conjugate are equal? Thanks! – Ayush Jul 06 '21 at 19:46
  • and wont there only be l-1 elements in the integral basis of Q[zeta] ? – Ayush Jul 07 '21 at 05:56
  • Ah since u is an element of a real field its conjugate must be equal to itself! – Ayush Jul 07 '21 at 06:22
  • +1 for the nice argument. I think the argument holds even if $\ell$ is not prime. – Paramanand Singh Aug 12 '23 at 06:05
  • Shouldn't $u$ be $u = \sum_{i = 1}^{\ell - 1} u_i \zeta^i$ and $u=\sum_{i=1}^{(l-1)/2}u_i(\zeta^i+\zeta^{-i})$ since (I think) we're using the integral basis ${\zeta,\zeta^2,\dots,\zeta^{l-1}}$ for $\Bbb Z[\zeta]$? – 19021605 Nov 18 '25 at 03:32
  • @Ayush Yes, there should be $l-1=[\Bbb Q[\zeta]:\Bbb Q]$ elements in an integral basis of $\Bbb Q[\zeta]$. – 19021605 Nov 18 '25 at 03:33
  • @ParamanandSingh If $l$ is not odd prime, $l$ may be even so $(l−1)/2$ in (2) is not an integer. Even if we replace $l-1$ by $\phi(l-1)$, a basis is given by $1,\zeta,\zeta^2,...,\zeta^{\phi(l)-1}$, to which I'm not sure the same argument applies since $\zeta^{l-i}$ may mot be on the list for $0\leq i<\phi(l)$. (1) and (2) are true after modifications according to wiki. Dylan Moreland's arguments still work. – 19021605 Nov 18 '25 at 03:57