# the cyclotomic units are algebraic units

Let $L=\mathbb{Q}(\zeta_{m})$ be a cyclotomic extension of $\mathbb{Q}$ with $m$ chosen to be minimal and let $\mathcal{O}_{L}$ be the ring of integers  ($=\mathbb{Z}(\zeta_{m})$), recall that the cyclotomic units are the elements of the form

 $\displaystyle\eta=\frac{\zeta^{r}-1}{\zeta^{s}-1}$

with $r$ and $s$ relatively prime to $m$ (where $\zeta=\zeta_{m}$). Here we prove that these elements are indeed algebraic units, i.e. $\eta\in\mathcal{O}_{L}^{\times}$.

###### Proof.

In order to prove the lemma, we will check that both $\eta$ and $\eta^{-1}$ are algebraic integers  , thus $\eta$ is a unit. Notice that it suffices to prove that $\eta$ is an algebraic integer, because the rest follows from interchanging the role of $r$ and $s$.

Let $r,s\in\mathbb{Z}$ be relatively prime to $m$, thus $r\mod m,s\mod m$ are units in $\mathbb{Z}/m\mathbb{Z}$ and we can find an integer $a$ such that:

 $a\cdot s\equiv r\mod m$

Note also that it follows that $\zeta^{r}=\zeta^{as}$. Moreover, using the equality of polynomials  :

 $x^{as}-1=(x^{s}-1)\cdot(x^{s(a-1)}+x^{s(a-2)}+\ldots+x^{s}+1)$

we get:

 $\displaystyle\eta$ $\displaystyle=$ $\displaystyle\frac{\zeta^{r}-1}{\zeta^{s}-1}=\frac{\zeta^{as}-1}{\zeta^{s}-1}$ $\displaystyle=$ $\displaystyle\frac{(\zeta^{s}-1)\cdot(\zeta^{s(a-1)}+\zeta^{s(a-2)}+\ldots+% \zeta^{s}+1)}{\zeta^{s}-1}$ $\displaystyle=$ $\displaystyle\zeta^{s(a-1)}+\zeta^{s(a-2)}+\ldots+\zeta^{s}+1\in\mathcal{O}_{L% }=\mathbb{Z}[\zeta]$

Hence the result. ∎

Title the cyclotomic units are algebraic units TheCyclotomicUnitsAreAlgebraicUnits 2013-03-22 14:13:17 2013-03-22 14:13:17 alozano (2414) alozano (2414) 4 alozano (2414) Theorem msc 11R18 AlgebraicInteger