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Let $\zeta=e^{\frac{2\pi i}{p}}$ where $p\geq3$ is a prime. Consider the algebraic number field $K=\mathbb{Q}(\zeta)$. Let $k$ be a positive integer such that $(k,p)=1$ i.e., $k$ is co-prime with $p$. Show $\alpha=1+\zeta+\dots+\zeta^{k-1}$ is a unit in $O_K$.

$O_K$ denotes the set of algebraic integers in $K$.

To solve this problem I tried to show $\alpha\in O_K$ (that is easy) and $N(\alpha)=1$, or find $\beta$ such that $\alpha\beta=1$. I have not been successful in doing either.

Many thanks to all the help. One way to show this is as Jyrki pointed out, $$\alpha=\frac{1-\zeta^k}{1-\zeta}$$ and, $$\left(\frac{1-\zeta^k}{1-\zeta}\right)^{-1}=\frac{1-\zeta^{km}}{1-\zeta^k}$$ where $m$ is some integer such that $km\equiv 1$ (mod p). Thus $\alpha$ has inverse and is hence a unit.

A boogie wit a hoodie
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    If I understand your question, these are "cyclotomic units", and take a little work to understand, though it's standard. What's your context? – paul garrett Apr 24 '22 at 05:02
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    There exists an integer $m$ such that $km\equiv1\pmod p$. And $$\left(\frac{1-\zeta^k}{1-\zeta}\right)^{-1}=\frac{1-\zeta^{km}}{1-\zeta^k}.$$ – Jyrki Lahtonen Apr 24 '22 at 05:11
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    ... that is, consider $(1-\zeta^k)/(1-\zeta)$ with $k$ prime to $p$... and look at what its "ord" is at various places. Turns out only $p$ matters, etc. – paul garrett Apr 24 '22 at 05:11

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Almost same as @Jyrki's approach, but here's another approach. Let $r$ be an order of $k$ mod $p$, i.e. smallest $0 < r < p$ with $k^{r} \equiv 1 \,(\mathrm{mod}\,p)$, which always exists by Fermat's little theorem. Then $$ \frac{1 - \zeta^{k}}{1-\zeta} \frac{1 - \zeta^{k^{2}}}{1 - \zeta^{k}} \frac{1 - \zeta^{k^{3}}}{1 - \zeta^{k^{2}}} \cdots \frac{1 -\zeta^{k^{r-1}}}{1 - \zeta^{k^{r-2}}} = 1 $$ so $\alpha = (1 - \zeta^{k}) / (1- \zeta)$ is a unit.

Seewoo Lee
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When $(k,p) = 1$, both $\zeta$ and $\zeta^k$ are powers of each other, so the ratio $(\zeta^k-1)/(\zeta-1)$ lies in $\mathbf Z[\zeta]$ and by the same reasoning $(\zeta-1)/(\zeta^k-1)$ lies in $\mathbf Z[\zeta^k] \subset \mathbf Z[\zeta]$. If a number and its inverse are in $\mathbf Z[\zeta]$, that's exactly what it means to be a unit in $\mathbf Z[\zeta]$.

There is nothing special about roots of unity of prime order here. If $\zeta$ is a root of unity of order $n > 1$ and $(k,n) = 1$ then $(\zeta^k-1)/(\zeta-1)$ is a unit in $\mathbf Z[\zeta]$.

KCd
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