Prove galois conjugates of primitive $p$-th root of unity $\zeta$ are associate?

Question

Let $p$ be a prime number, let $ζ$ be a primitive $p$-th root of unity, and let $K = \mathbb{Q}(ζ)$. Prove that all the Galois conjugates of $ζ$ are associates in $O_K$ and that all the Galois conjugates of $1 − ζ$ are associates in $O_K$.

Thoughts so far: I know another way to say this for every $\sigma ∈ Gal(K/\mathbb{Q})$ it holds that $\sigma(ζ) = \muζ$ and $\sigma(1−ζ) = \mu'(1−ζ)$ for some invertible elements $\mu, \mu' \in O^{X}_{K}$, I'm not sure if it is easier to prove through this definition?

Answer 1

First, if $\zeta$ is a root of unity, then any Galois conjugate of it is another root of unity. Since roots of unity are always units, and all units are associates, we are done (if $u,v$ are units, look at $u^{-1}v$).

The second statement is trickier, since we know that $1-\zeta$ is not a unit (look at its norm). There are two approaches:

1. Use some algebraic number theory. $K$ is totally ramified at $p$ (the defining polynomial is eisenstein) and its ring of integers is $\mathbb Z[\zeta]$. By Dedekind's theorem, the primes lying over $p$ are of the form $(p,1-\zeta)$. Since $N(1-\zeta) = p$ for any primitive root of unity, the ideal $(1-\zeta)$ already contains $p$, and so the prime lying over $p$ is principal and generated by $1-\zeta$ for any primitive root of unity $\zeta$. In particular $(1-\zeta) = (1-\sigma(\zeta))$, which (because you are in an integral domain) means that the two are associates.
2. If you are very careful, you might worry that the above is circular (as I recall, some proofs that $O_K = \mathbb Z[\zeta]$ pass through this claim about associates, although I am pretty sure some of them don't). So you could also prove it directly. Recall that the conjugates of $\zeta$ are all of the form $\zeta^i$ for $1\leq i < p$. To say that $1-\zeta$ and $1-\zeta^i$ are conjugates is equivalent to their quotient being a unit in $O_K$. It's easy to expand $$\frac{1-\zeta^i}{1-\zeta} = \zeta^{i-1} + ... + \zeta + 1$$ so it is clearly integral, while its norm is obviously $1$ (norm is Galois invariant so the numerator and denominator have the same norm) which makes it a unit of $O_K$.