Proof: $\mathbb{Z}[\zeta_6]$ is a PID. I am reading through A First Course in Modular Forms. In Proposition 2.2.3 they claim that $\mathbb{Z}[\zeta_6]$ is known to be a principal ideal domain.
Does anyone have a reference for the proof of this fact? If the proof is simple enough, perhaps you could sketch it below.
 A: Let $\zeta_6 = e^{\frac{\pi i}{3}}$ be a primitive sixth root of unity. We compute
\begin{align}
x^6 - 1 &= (x^3 - 1) (x^3 + 1) \\
&= (x - 1) (x^2 + x + 1) (x + 1) (x^2 - x + 1) \\
&= (x^2 - 1) (x^2 + x + 1) (x^2 - x + 1).
\end{align}
So, $\zeta_6$ is a zero of at least one of the factors in the product above. As we have assumed $\zeta_6$ is a primitive sixth root of unity, it cannot be a zero of $x^2 - 1$. Because $\Im(\zeta_6) > 0$ and $\Im(\zeta_6^2) > 0$, we see that $\Im(\zeta_6^2 + \zeta_6 + 1) > 0$, so $\zeta_6$ cannot be a zero of $x^2 + x + 1$. So, as we have eliminated the possibility that $\zeta_6$ is a zero of the first two factors, we must have
\begin{align}
\zeta_6^2 - \zeta_6 + 1 &= 0.
\end{align}
But $\zeta_6^2 = e^{\frac{2 \pi i}{3}}$ is a primitive third root of unity. Let us denote this primtive third root of unity as $\zeta_3 = \zeta_6^2$. So we have
\begin{align}
\zeta_3 = \zeta_6 - 1 \\
\zeta_6 = \zeta_3 + 1.
\end{align}
The relations above are quite easy to see geometrically. Look at the lattice formed by the sixth roots of unity

The roots $\zeta_6$ and $\zeta_3 = \zeta_6^2$ are separated by a unit distance, which is what our formulas tell us. Continuing, we now see that
\begin{align}
\mathbb{Z}[\zeta_6] = \mathbb{Z}[\zeta_3].
\end{align}
Note that $\mathbb{Z}[\zeta_3]$ is known as the ring of Eisenstein integers. 
We will now show that the $\mathbb{Z}[\zeta_3]$ is a principal ideal domain. We first have to define a norm on $\mathbb{Z}[\zeta_3]$. For any $a + b \zeta_3 \in \mathbb{Z}[\zeta_3]$ we compute
\begin{align}
|a + b \zeta_3|^2 &= (a + b \zeta_3) (a + b \overline{\zeta_3}) \\
&= a^2 + ab (\zeta_3 + \overline{\zeta_3}) + b^2 \\
&= a^2 - ab + b^2.
\end{align}
One easily checks that
\begin{align}
N(a + b \zeta_3) &= a^2 - ab + b^2
\end{align}
defines a norm on $\mathbb{Z}[\zeta_3]$. Now, let $I \subset \mathbb{Z}[\zeta_3]$ be a nonzero ideal. Note that
\begin{align}
N (a + b \zeta_3) &= |a + b \zeta_3|^2 \\
&\geq 0,
\end{align}
with equality if and only if $a = b = 0$. We may select an element $v \in I$ of minimum positive norm. Let $w$ be any element in $I$. 
Now, consider the quotient $\frac{w}{v} \in \mathbb{C}$. Geometrically, $\mathbb{Z}[\zeta_3]$ forms a triangular lattice in the complex plane. So, there is exists $q \in \mathbb{Z}[\zeta_3]$ such that
\begin{align}
\left|q - \frac{w}{v}\right| &\leq \frac{\sqrt{3}}{4}.
\end{align}
Based on the way we defined the norm, this implies
\begin{align}
N \left(q - \frac{w}{v}\right) &= \left|q - \frac{w}{v}\right|^2 \\
&\leq \frac{3}{16}.
\end{align}
Now we note
\begin{align}
N(v q - w) &= N(v) N (q - \frac{w}{v}) \\
&\leq \frac{3}{16} N(v) \\
& < N(v).
\end{align}
But $v q - w \in I$ has norm strictly less than $v$. We assumed that $v$ had the minimum positive norm among all elements in $I$. So, we must have that $w = v q$. Thus $I \subseteq (v)$. Obviously $(v) \subseteq I$. Thus $I = (v)$; that is, $I$ is a principal ideal.
Thus, $\mathbb{Z}[\zeta_6] = \mathbb{Z}[\zeta_3]$ is a principal ideal domain.
