Factor $3+2\sqrt{3}i$ as a product of irreducible elements in the ring $\mathbb{Z}\left [ \sqrt{-3} \right ]$ The question I am having trouble with is:

Factor $3+2\sqrt{3}i$ as a product of irreducible elements in the ring $\mathbb{Z}\left [ \sqrt{-3} \right ]$

I don't really understand how to go about this besides guessing. Is there any better way to do this?
 A: You want to express $\alpha=a+b\sqrt{3}i=\beta \gamma$, where $\beta, \gamma \in \Bbb{Z[\sqrt{-3}]} $ and they are not units. Using the idea of norms $N(\alpha)=\alpha \bar{\alpha}=a^2+3b^2 \in \Bbb{Z}_{\geq 0}$, we can show that the norm is multiplicative, i.e. $N(\beta \, \gamma)=N(\beta) \, N(\gamma)$. We get the following:
\begin{align*}
\alpha&=\beta \gamma\\
N(\alpha)&=N(\beta \, \gamma)\\
N(\alpha)&=N(\beta) \, N(\gamma).
\end{align*}
In this particular case , where $\alpha=3+2\sqrt{-3}$. We have $N(\alpha)=9+3(4)=21$. So we want $\beta, \gamma$ such that
$$N(\beta) N(\gamma)=21.$$
To make sure that we have a non-trivial factorization, we want to avoid $\beta$ and $\gamma$ being units. It can be shown easily that $\beta$ is a unit $\iff$ $N(\beta)=1$. In this ring, it only happens when $\beta =\pm 1$ (check this!!).
Coming back, we want those $\beta, \gamma$ such that $N(\beta)=7$ AND $N(\gamma)=3$ or vice versa. Now ask yourself, can we have $\beta=c+d\sqrt{-3}$ such that $c^2+3d^2=7$. This give us $c=\pm 2$ and $d=\pm 1$. So $\beta=\pm 2 \pm \sqrt{-3}$. Similarly you can get $\gamma=0\pm \sqrt{-3}$. Now you can verify that $\alpha=\beta \, \gamma$.
But one questions remains, are these $\beta$ and $\gamma$ irreducible? Can (say) $\beta$ be factored non-trivially?
Hint: Look at the norm of $\beta$, is there something special about it?
A: Yes. The absolute value gives you a multiplicative map
$$N \colon ℤ[\sqrt {-3}] → ℕ_0,~z ↦ \lvert z \rvert^2 = z \overline z.$$
So for $z, w ∈ ℤ[\sqrt {-3}]$ with $z \mid w$ in $ℤ[\sqrt {-3}]$, we have $N(z) \mid N(w)$ in $ℕ_0$ and furthermore we may verify
$$z ∈ ℤ[\sqrt {-3}]^× \iff N(z) = 1.$$
Now, $N(3 + 2\sqrt 3 \mathrm i) = 9 + 12 = 21$.
