Proof of Kummer's lemma

Question

Context: Let $\omega = e^{\frac{2\pi i}{p}}$ for some prime $p$. Kummer’s Lemma states that if $u$ is a unit in $\mathbb{Z}[\omega]$ and $\overline{u}$ is its complex conjugate, then $\frac{u}{\overline{u}}$ is a power of $\omega$.

I'm looking at the following proof. The author argues that there are only finitely many algebraic integers $\alpha$ of degree $\leq n$ such that $\alpha$ and all of the algebraic conjugates have absolute value $1$. Then, he makes a claim that if $\alpha$ is such an algebraic integer, then so is $\alpha ^k$ for any $k \in \mathbb{N}$, eventually yielding the desired result.

My question is why? Why is it true that if all the algebraic conjugates of $\alpha$ have absolute value $1$, then all of the algebraic conjugates of $\alpha^k$ also have absolute value $1$? The explanation on the linked site makes zero sense to me.

Any help is appreciated. Thanks!

Answer 1

Let the distinct conjugates be $\alpha = \alpha_1, \ldots, \alpha_n$, and note the minimal polynomial of $\alpha$ is $$ f(x) = (x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_n). $$ We can see this since $\mathrm{Gal}(\mathbb{C} / \mathbb{Q})$ permutes the roots, the polynomial is fixed; hence, it's in $\mathbb{Q}[x]$. The Galois group also acts transitively on the roots, so the polynomial is irreducible. i.e., it's the minimal polynomial.

Now consider $$ g(x) = (x-\alpha_1^k)(x-\alpha_2^k) \cdots (x-\alpha_n^k). $$ As before, the Galois group permutes the roots, so it must be in $\mathbb{Q}[x]$, and the minimal polynomial of $a^k = a_1^k$ must divide $g(x)$. So any conjugate of $\alpha^k$ must be a power of the conjugate of $\alpha$.