1
$\begingroup$

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,

Jérôme

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $\mathbb{Z}[\zeta_{p} + \zeta_{p}^{-1}]$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.