Find a generator of the ideal generated by $3$ and $2-2\sqrt{-2}$ in ${\mathbb{Z}\left[\sqrt{-2}\right]}$ Exercise: Find a generator of the ideal generated by $3$ and $2-2\sqrt{-2}$ in ${\mathbb{Z}\left[\sqrt{-2}\right]}$.

I know that such a generator exists since $\mathbb{Z}\left[\sqrt{-2}\right]$ is a Euclidean domain and thus a PID, with a generator being given by an arbitrary element of $I$ with minimal norm, where $I$ denotes the ideal generated by $3$ and $2-2\sqrt{-2}$. Furthermore,  $3$ has norm $9$ and $2-2\sqrt{-2}$ has norm $12$.
How do I continue?

We want to find $\alpha\in \mathbb{Z}[\sqrt{-2}]$ such that $I = (3, 2 - 2\sqrt{-2}) = (\alpha)$. For such $\alpha$, there exists $\beta, \gamma \in \mathbb{Z}[\sqrt{-2}]$ such that $3 = \alpha\beta, 2 - 2\sqrt{-2} = \alpha \gamma$. If we observe norms of them, we have $9 = N(\alpha)N(\beta)$ and $12 = N(\alpha)N(\gamma)$, so $N(\alpha) = 1$ or $3$. When $N(\alpha) = 1$, $\alpha = \pm 1$ and one can show that $1 \not \in I$ by several ways (brute force actually works) Since there are only finitely many $\alpha \in \mathbb{Z}[\sqrt{-2}]$ with $N(\alpha)$ (and there are only two actually), and only one of them satisfies $\alpha | 3$ and $\alpha = 2 -2\sqrt{-2}$. Now we have $(3, 2 - 2\sqrt{-2}) \subseteq (\alpha)$, and you also need to show the opposite inclusion by seeking at $\beta_1, \beta_2$ with $\alpha = \beta_{1}\times 3 + \beta_{2} \times (2 - 2\sqrt{-2})$.
