1
$\begingroup$

Let $\zeta$ be the $p$th root of unity, with $p$ an odd prime number.

Let $\mathbb{Q}(\zeta)$ be the $p$th cyclotomic field and let $\mathcal{O}=\mathbb{Z}(\zeta)$ the ring of integers of $\mathbb{Q}(\zeta)$. Let $u$ be a unit in $\mathcal{O}$, with $u \ne \zeta$.

I'm struggling to find a lower bound for the algebraic integer $\alpha = \zeta - 2 u$. In particular it should be easy to show that $N(\alpha) > 1$, thus $\alpha$ is not a unit.

Letting $p(x)$ the the minimal polynomial for $u$, there should be some way to bound the discriminant of the minimal polynomial for $2 u - \zeta$.

Any hints?

Thanks in advance

Improve this question
$\endgroup$
4
  • 3
    $\begingroup$ This is obviously wrong if $u = \zeta$. $\endgroup$
    – Franz Lemmermeyer
    Commented Oct 22, 2021 at 18:51
  • $\begingroup$ Of course. I'll edit the question specifying "excluding trivial cases". $\endgroup$
    – ptass
    Commented Oct 24, 2021 at 8:55
  • $\begingroup$ Is there any reason why your claim should be true? Or interesting? $\endgroup$
    – Franz Lemmermeyer
    Commented Oct 24, 2021 at 9:36
  • 1
    $\begingroup$ $p=5$ and $u=(\zeta+1)^2$ is the easiest example (of many) that $u$ can be a unit while $\zeta-2u$ is a unit, too. $\endgroup$
    – Chris Wuthrich
    Commented Oct 24, 2021 at 10:11

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .