I left out some hypotheses in the title to keep things short, so here is the full form:
Let $r$ be a primitive $m$th root of unity for $m>1$ and let $k$ be a positive integer such that $\gcd(m,k)=1$. Show $\frac{r-1}{r^k- 1}$ is:
- An algebraic integer
- An element of $\mathbb Q(r)$, where $\mathbb Q$ is the set of rationals and $\mathbb Q(r)$ is the algebraic extension with $r$.
If it helps $\dfrac{r-1}{r^k- 1}(1 + r + ... + r^{k-1}) = 1$.
$\zeta := r^k$ is a primitive $m$-th root of unity because of $\mathrm{gcd}(m,k) = 1$. So there is some nonnegative integer $j$ with $r = \zeta^j$. Now $\frac{r-1}{r^k-1} = \frac{\zeta^j-1}{\zeta-1} = 1 + \zeta + \dots + \zeta^{j-1}$ is a sum of roots of unity and hence an algebraic integer. The second item holds because of $\mathbb{Q}(r) = \mathbb{Q}(\zeta)$.