2
$\begingroup$

Given $\zeta_{p}$ is a $p^{th}$ root of unity for some odd prime $p$ I was able to show for any unit $u$ that $\frac{u}{\overline{u}}=\pm \zeta_{p}^{k} $. Thus we have $u=\pm \zeta_{p}^{k} \overline{u}$. The problem is that I can't seem to prove $\overline{u}$ is a real unit (which I suppose would mean $u$ is a unit in $\mathbb{R} \cap \mathbb{Z}[\zeta_{p}]$). I tried by contradiction using the fact $\omega^{k}+\overline{\omega^{k}}$ is a real number and thought when substituted with $\frac{u}{\overline{u}}$ and $\frac{\overline{u}}{u}$ respectfully I'd see their sum equal a complex number. If anything it showed $u$ should be a complex number. Any insight or hints would be greatly appreciated!

$\endgroup$
  • 1
    $\begingroup$ If $\overline{u}$ is real then so is $u$. What you need to show is that if you {\em change} $u$ by a suitable root of unity then the new unit is real. $\endgroup$ – franz lemmermeyer Dec 3 '17 at 18:12
  • 1
    $\begingroup$ See here. Unfortunately I'm not 100% that Bruno's otherwise great argument is watertight towards the end. See S.Gau's comment. $\endgroup$ – Jyrki Lahtonen Dec 3 '17 at 18:12
1
$\begingroup$

In your formula for $u/\bar u$, you must show that only the sign + can occur. At least two kinds of proofs can be found in textbooks, but they both use a zest of arithmetic in a hidden way:

1) Directly for $\mathbf Q(\zeta_p)$, on using divisibility by $(1-\zeta_p)$ in the ring $\mathbf Z[\zeta_p]$. See Washington, prop. 1.5, or Marcus, ex. 12 of chapter 2.

2) Via the algebraic theory of CM-fields which provides, with obvious notations, the formula $Q_K:= [E:WE^+]= 1$ or $2$ for any CM-field $K$ (Wash., thm. 4.12). But for $K=\mathbf Q(\zeta_n)$, the choice between the values 1 and 2 needs arithmetic arguments: the same proof as in 1) gives $Q=1$when $n$ is a prime power; when $n$ is not a prime power, an additional argument modulo some power of $2$ gives $Q=2$.

$\endgroup$

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.