Show that algebraic integers in $\mathbb{Q}[\zeta_3]$ are exactly $\mathbb{Z}[\zeta_3]$.

Question

If $\zeta_3$ is a primitive cube root of unity, then I'm trying to show that $\mathbb{Z}[\zeta_3]$ is the ring of algebraic integers in $\mathbb{Q}[\zeta_3]$. I have shown that $\mathbb{Z}[\zeta_3]$ is contained in the ring of algebraic integers, but I am unsure of how to prove the converse. I'm aware that there are general proofs for the case where $\zeta$ is a primitive $p$-th root of unity where $p$ is any prime, but I'm wondering if there is a more elementary proof that will work for $p = 3$.

Answer 1

One immediate way is to show that $ \mathbf Z[\zeta_3] $ is an Euclidean domain by giving an explicit algorithm for performing Euclidean division there. This, of course, immediately implies that it is integrally closed in its field of fractions, which is $ \mathbf Q(\zeta_3) $; and clearly it is integral over $ \mathbf Z $, so it is the full ring of integers. Another way is to use the well-known result about the classification of quadratic rings of integers: the ring of integers of $ \mathbf Q(\sqrt{d}) $, for squarefree $ d $ is $ \mathbf Z[(1 + \sqrt{d})/2] $ for $ d \equiv 1 \pmod{4} $, and $ \mathbf Z[\sqrt{d}] $ otherwise. Now, recall that $ \zeta_3 = (-1 + \sqrt{-3})/2 $, and you are done.