Sum of roots of unity an algebraic integer proof

Question

Let S be the sum of a finite number of nth roots of unity (where n is fixed, and the sum is non-zero). How do I go about showing that S is an algebraic integer in the cyclotomic field of order n ?

Answer 1

Let $K= \mathbb{Q}(\zeta)$. We have $$(\zeta^{a} + \zeta^{b})\zeta^{c} = \zeta^{c+a} + \zeta^{c+b},$$ so we can form a matrix of the action of $\zeta^{a} + \zeta^{b}$ on the basis $\{\zeta^{i}\}_{i = 0}^{n - 2}$ (since $(\zeta^{a} + \zeta^{b})\mathbb{Z}[\zeta] \subseteq \mathbb{Z}[\zeta]$ and $\mathbb{Z}[\zeta]$ is a finitely generated additive subgroup of $\mathbb{C}$). The characteristic polynomial then shows that $\zeta^{a} + \zeta^{b}$ is an algebraic integer.

Say $n = 7$ and $a = 2, b = 3$. Then $\zeta^{2} + \zeta^{3}$ acts on $\{\zeta^{i}\}$ as $$M = \begin{bmatrix} 0 & 0 & 1 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 0 & 1 & 1\\ -1 & -1 & -1 & -1 & -1 & 0\\ 0 & -1 & -1 & -1 & -1 & -1\\ -1 & 0 & -1 & -1 & -1 & -1 \end{bmatrix}.$$ For example, the fourth row comes from $$(\zeta^{2} + \zeta^{3})\zeta^{3} = \zeta^{5} + \zeta^{6} = \zeta^{5} + \sum_{i = 0}^{5} -\zeta^{i}.$$ This gives us the equation $$(\zeta^{2} + \zeta^{3})\begin{bmatrix} 1\\ \zeta\\ \zeta^{2}\\ \zeta^{3}\\ \zeta^{4}\\ \zeta^{5} \end{bmatrix} = M\begin{bmatrix} 1\\ \zeta\\ \zeta^{2}\\ \zeta^{3}\\ \zeta^{4}\\ \zeta^{5} \end{bmatrix}.$$ But this says that $\zeta^{2} + \zeta^{3}$ is an eigenvalue of $M$, so is a zero of its characteristic polynomial $\text{det}(xI - M)$.