I'm a bit at a loss about what I can say in this situation. Do I have to show that $\zeta_{p} + \zeta_{p}^{-1}$ form an integral basis ? If I do, I have no idea how to do it.

If not, can I use the fact that $\mathbb{Q}(\zeta_{p} + \zeta_{p}^{-1}) \subseteq \mathbb{R}$ at my advantage ?

Thanks in advance for any hint or answer,


  • $\begingroup$ Let $\tau_p=\zeta_p+\zeta_p^{-1}$, then you have to prove that $\tau_p$ forms a power basis, this means that the $1,\tau,\tau^2,\ldots$ form an integral basis. $\endgroup$ – Marc Bogaerts Sep 30 '14 at 13:06
  • $\begingroup$ Thanks, I tried it but it seems I lack some results which could help me proving that $\tau_{p}$ is an algebraic integer. $\endgroup$ – Jérôme Sep 30 '14 at 14:03
  • $\begingroup$ It is not an easy problem. Some things that could be useful are lacking like a nice form of the minimal polynomial of $\tau_p$ and the lack of knowledge of the discriminant of the ring of algebraics. Where does this question come from? $\endgroup$ – Marc Bogaerts Sep 30 '14 at 14:34
  • $\begingroup$ I don't have a book reference. It comes from my teacher as a homework. Could it help to know that $p$ is an odd prime ? I just noticed I forgot to mention that. $\endgroup$ – Jérôme Sep 30 '14 at 14:37
  • $\begingroup$ Have you seen cyclotomic extensions? $\endgroup$ – Marc Bogaerts Sep 30 '14 at 14:38

Let $\tau_p=\zeta_{p} + \zeta_{p}^{-1}$ then since $\bar{\zeta_p}=\zeta_{p}^{-1}$ we have that $\tau_p \in \mathbb{R}$ so that $\mathbb{Q}(\tau_p) \subset \mathbb{R}$. We know that $\{\zeta,\zeta^2,\ldots,\zeta^{p-1}\}$ form an integral basis for the cyclotomic integers. Let $a=\sum_{i=1}^{p-1}b_i\zeta_p^i$ be a real cyclotomic integer, then $a=\bar{a}$. Since $\bar{\zeta_p^i}=\zeta_p^{p-i}$ we must have $b_i=b_{p-1}$ but then $a=\sum_{i=1}^{(p-1)/2}\tau_p^i$ which shows that the $\{\tau,\tau^2,\ldots,\tau^{(p-1)/2}\}$ form an integer basis of the real quaternion integers.

  • $\begingroup$ Ah, I didn't think of writing the general form of a real cyclotomic integer. Thank you very much. I read it carefully. I still have some questions, unfortunately. The deduction that $a = \overline{a}$, do we use it here ? Is it that evident ? $\endgroup$ – Jérôme Sep 30 '14 at 15:32
  • $\begingroup$ If you think this is the answer to this question then please check my answer as "accepted". $\endgroup$ – Marc Bogaerts Sep 30 '14 at 15:39

You show that every element of $\mathbb{Q}(\zeta_{p} + \zeta_{p}^{-1})$ which is a root of a monic integer polynomial is actually in $\mathbb{Z}[\zeta_{p} + \zeta_{p}^{-1}]$.

  • $\begingroup$ Thanks for answering this quickly. Do I use the contradiction : Let $\alpha \in \mathbb{Q}(\zeta_{p} + \zeta_{p}^{-1}) $ a root of a monic integer polynomial. We suppose that $\alpha \notin \mathbb{Z}[\zeta_{p} + \zeta_{p}^{-1}]$. And we get a contradiction ? Or is there something more direct ? $\endgroup$ – Jérôme Sep 30 '14 at 13:35

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