$\alpha \in O_K$ with all $K$-conjugates $\sigma_i(\alpha)$ , $|\sigma_i(\alpha)| = 1 \implies \alpha$ is a root of unity

Question

I am working on this problem:

(i) Let $m, M \in \mathbb{N}$. Show that the set

$$S_{m,M} := \left\{ \alpha \in \mathbb{C} \mid \alpha \text{ is a root of a monic polynomial } f \in \mathbb{Z}[t] \text{ with } \deg(f) \leq m \text{ and } |\alpha'| \leq M \text{ for all roots } \alpha' \right\}$$

is finite.

(ii) Let $\alpha \in O_K$ such that all $K$-conjugates $\sigma_i(\alpha)$ satisfy $|\sigma_i(\alpha)| = 1$. Show that $\alpha$ is a root of unity.

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This is my solution, I have already proven (i) ,but I am stuck at (ii)

(i) Let $f(t) \in \mathbb{Z}[t]$ be a monic polynomial of degree $n \leq m$ with roots $\alpha_1, \dots, \alpha_n \in \mathbb{C}$. By Vieta’s formulas, the coefficients $a_k$ of $f(t)$ are given by the elementary symmetric polynomials:
$$ a_{n-k} = (-1)^k \sigma_k(\alpha_1, \dots, \alpha_n), $$
where $\sigma_k(\alpha_1, \dots, \alpha_n)$ is the sum of all products of $k$-distinct roots. Since $|\alpha_i| \leq M$ for all $i$, we have:
$$ |\sigma_k(\alpha_1, \dots, \alpha_n)| \leq \binom{n}{k} M^k. $$
Thus, each coefficient $a_{n-k}$ satisfies $|a_{n-k}| \leq \max_{k} \binom{m}{k} M^k$, where $n \leq m$. Since $a_k \in \mathbb{Z}$ and the bound depends only on $m$ and $M$, there are finitely many possible monic polynomials $f(t)$ with these conditions. Each such $f$ has at most $m$ roots, so $S_{m,M}$ is finite.

(ii) Since $\alpha \in O_K$, it is a root of a monic polynomial $f(t) \in \mathbb{Z}[t]$. By assumption, all $K$-conjugates $\sigma_i(\alpha)$ satisfy $|\sigma_i(\alpha)| = 1$. Therefore, $\alpha \in S_{n,1}$, where $n = [K : \mathbb{Q}]$. By part (i), $S_{n,1}$ is finite.

How should I complete this argument?

Answer 1

If $|\sigma_i (\alpha)| \leq 1 \forall i$ then $|\sigma_i (\alpha^k)| \leq 1 \forall i$ for $k=0,1,2,...$ as well. Also the minimal polynomial of $\alpha^k$ has as roots precisely all the conjugates of $\alpha^k$. So similarly to your argument for (i), all the $\alpha^k$ are elements of $S_{n,1}$ (where $n=[K:\Bbb{Q}]$).

Since $1,\alpha, \alpha^2, \alpha^3...$ are all elements of the same finite set there must at some point be a repeat, ie $\alpha^s=\alpha^t$ for some $s < t$. But then since $\alpha \neq 0$ we have $\alpha^{t-s}=1$, ie $\alpha$ is a root of unity.