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 ${𝒪}_L$ be the ring of integers ($=\mathbb{Z}({\zeta }_m)$), recall that the cyclotomic units are the elements of the form

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

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

Lemma 1.

The cyclotomic units are algebraic units.

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 $rmodm,smodm$ are units in $\mathbb{Z}/m\mathbb{Z}$ and we can find an integer $a$ such that:

$$a\cdot s\equiv rmodm$$

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

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

we get:

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

$=$${\displaystyle \frac{({\zeta }^s-1)\cdot ({\zeta }^{s(a-1)}+{\zeta }^{s(a-2)}+\mathrm{\dots }+{\zeta }^s+1)}{{\zeta }^s-1}}$

$=$${\zeta }^{s(a-1)}+{\zeta }^{s(a-2)}+\mathrm{\dots }+{\zeta }^s+1\in {𝒪}_L=\mathbb{Z}[\zeta ]$

Hence the result.∎