How can I prove that the cyclotomic integers $\frac{\zeta_p^r - 1}{\zeta_p^s - 1}$, with $\ p\nmid rs$, are units?

Question

I am reading a paper in which the cyclotomic integers $$\frac{\zeta_p^r - 1}{\zeta_p^s - 1},\ p\nmid rs$$ are claimed to be units, but I'm not sure how to show that this is the case.

Taking the norm, we have

$$N\left( \frac{\zeta_p^r - 1}{\zeta_p^s - 1} \right) = \prod_{\sigma \in \operatorname{Gal(\Bbb Q(\zeta_p)/\Bbb Q)}} \sigma\left( \frac{\zeta_p^r - 1}{\zeta_p^s - 1}\right) = \prod_{\sigma \in \operatorname{Gal(\Bbb Q(\zeta_p)/\Bbb Q)}} \frac{\sigma^r(\zeta_p) - 1}{\sigma^s(\zeta_p) - 1}$$

This should simplify to be $1$ if the element is a unit, but I can't see how this simplifies in this way.

Is there a simpler way to see that these are units explicitly? (The paper states that there is an obvious inverse, but multiplying the two together doesn't make it obvious that this is the inverse.)

Answer 1

Let $\zeta = \zeta_p$, we can choose positive integer $u$ such that $s|u$ and $u\equiv r \pmod{p}$.

Then $$\frac{\zeta^r - 1}{\zeta^s -1} = \frac{\zeta^u - 1}{\zeta^s -1} \in \mathbb{Z}[\zeta]$$

From the symmetry of $r$ and $s$ shows its reciprocal $(\zeta_s -1)/(\zeta_r - 1)$ is also in $\mathbb{Z}[\zeta]$.

Answer 2

Here’s another argument, from local principles:

At every prime $q\ne p$, $\zeta_p\not\equiv1\pmod q$, so that the numerator and denominator both are units with respect to $q$. At $p$, however, the fact that $v_p(\zeta_p-1)=1/(p-1)$ implies that $v_p(\zeta_p^u-1)=1/(p-1)$ as well. So quotient has $v_p$ equaling $1$. Unit everywhere, so a unit.

Answer 3

Let $\zeta$ be a primitive $p$-th root of $1$. Since $p\nmid rs, \zeta^r$ and $\zeta^s$ are also primitive $p$-th roots of $1$. Then $\zeta^r$ will be a power of $\zeta^s$ and your fraction will belong to $\mathbf Z[\zeta]$. By permuting $r$ and $s$, the inverse fraction also belongs to $\mathbf Z[\zeta]$. QED