Finding $\mathbb Z[\zeta_p]^∗$, the group of units of $\mathbb Q(\zeta_p$)

Question

Let $p$ be an odd prime and $\zeta_p$ be a primitive $p$-th root of unity. I'm trying to prove that $\mathbb Z[\zeta_p]^∗$, the group of units of $\mathbb Q(\zeta_p$) is $(\zeta_p)\mathbb Z[\zeta_p + \zeta_p^{-1}]^∗$. How does one show this?

Answer 1

Choose an embedding of $\mathbb Q[\zeta_p]$ in $\mathbb C$. If $u \in \mathbb Z[\zeta_p]^\times$, then $u/\overline{u}$ is an algebraic integer all of whose Galois conjugates have absolute value $1$ (because $\text{Gal}(\mathbb Q(\zeta_p)/\mathbb Q)$ is abelian, so the action of any element commutes with complex conjugation). Hence $u/\overline{u}$ is a root of unity $\zeta_p^k$. Write $-k=2h$ in $\mathbb Z/p\mathbb Z$ and put $v = \zeta_p^h u$. Then $\overline{v} = \zeta_p^{-h}\overline{u} = \zeta_p^{-h}\zeta_p^{-k}u = \zeta_p^h u = v$, so $v = \overline{v}$ is a unit in the ring of integers of the maximal real subfield of $\mathbb Q[\zeta_p]$, which is $\mathbb Z[\zeta_p + \overline{\zeta_p}]$. So $u = \zeta_p^{-k} v \in \left<\zeta_p\right> \times \mathbb Z[\zeta_p + \zeta_p^{-1}]^\times$.