# Proof of Kummer's lemma

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!

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 $$\alpha$$.