-3
$\begingroup$

Let $l$ be an odd prime number and $\zeta$ be a primitive $l$-th root of unity in $\mathbb{C}$. Let $K = \mathbb{Q}(\zeta)$. Let $A$ be the ring of algebraic integers in $K$. Let $\epsilon$ be a unit of $A$.

My question: Is the following proposition true? If yes, how would you prove this?

Proposition There exists a real unit $\eta$ such that $\epsilon = \zeta^g\eta$.

Related questions:

On a certain property of units in the cyclotomic number field of an odd prime order

The group of roots of unity in the cyclotomic number field of an odd prime order

Is there a purely imaginary unit in the cyclotomic number field of an odd prime degree?

$\endgroup$
7
$\begingroup$

The proposition is true (it probably goes back to Kummer), and here is a proof:

Let $U$ be the group of units in the cyclotomic field, and $U^+$ denote the group of units in its totally real subfield. By the unit theorem, $U^+$ has finite index in $U$.

Consider $\epsilon/\overline{\epsilon}$. This lies in the group $U^-$ of units such that $\overline{u} = u^{-1}$. Since $U^+ \cap U^- = \{ \pm 1\}$, and $U/U^+$ is finite, we see that $U^-$ is finite. By the answer to another of your questions, $U^-$ thus has order dividing $2l$, and so $U^-/\{\pm 1\}$ has order $l$. In particular, it has odd order, and each of its elements is a square of some other element (a general property of finite abelian groups of odd order). So we may write $\epsilon/\overline{\epsilon} = \pm u^2$ for some element $u \in U^-$. Without loss of generality, we may further assume that $u$ has odd order, so that $u = \zeta^g$ for some $g$.

Then $\overline{\epsilon u^{-1}} = \overline{\epsilon} u = \pm \epsilon u^{-1}$. If the sign is $+1$, then we see that $\epsilon u^{-1}$ is a real unit, and we are done. If the sign is $-1$, then $\epsilon u^{-1}$ is purely imaginary, which is not possible (see this answer).

$\endgroup$

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.