An algebraic-integer average of roots of unity is either $0$ or a common root of unity

If ${\zeta }_1=\cdots ={\zeta }_n$, then $\alpha ={\zeta }_1$ and the conclusion holds.

Assume now that the ${\zeta }_i$ are not all equal. By [A2], $\mathrm{\mid }\alpha \mathrm{\mid }<1$. Every algebraic conjugate ${\alpha }^{\mathrm{\prime }}$ of $\alpha$ has the form $({\zeta }_1^{\mathrm{\prime }}+\cdots +{\zeta }_n^{\mathrm{\prime }})\mathrm{/}n$ with each ${\zeta }_i^{\mathrm{\prime }}$ a root of unity by [A1], so $\mathrm{\mid }{\alpha }^{\mathrm{\prime }}\mathrm{\mid }\le 1$ by [A2].

Suppose also that $\alpha \mathrm{\ne }0$. Let $m(X)=X^d+a_{d-1}{X}^{d-1}+\cdots +a_0$ be the monic minimal polynomial of $\alpha$ over $\mathbb{Q}$; then $(-1{)}^d{a}_0$ is the product of the algebraic conjugates of $\alpha$. Because $\alpha$ is an algebraic integer by [F3], that product is a rational algebraic integer, hence an integer by [F1]. But step 2.1 gives its modulus strictly between $0$ and $1$, impossible. So $\alpha =0$.

Under the assumption that the roots are not all equal, step 3.1 forces $\alpha =0$. Together with step 1.1, this proves that $\alpha$ is either $0$ or a common root of unity.