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

Question

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

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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?

Answer 1

Let $g\in\Bbb{Z}[\sqrt{-2}]$ be a generator of the ideal generated by $3$ and $2-2\sqrt{-2}$ in $\Bbb{Z}[\sqrt{-2}]$. Then $$3=gx\qquad\text{ and }\qquad 2-2\sqrt{-2}=gy,$$ for some $x,y\in\Bbb{Z}[\sqrt{-2}]$. It follows that the norm of $g$ divides $3$ because \begin{eqnarray*} \mathcal{N}(g)\mathcal{N}(x)&=&\mathcal{N}(gx)=\mathcal{N}(3)&=&9,\\ \mathcal{N}(g)\mathcal{N}(y)&=&\mathcal{N}(gy)=\mathcal{N}(2-2\sqrt{-2})&=&12, \end{eqnarray*} and $\gcd(9,12)=3$. Because the norm is positive, either $\mathcal{N}(g)=1$ or $\mathcal{N}(g)=3$. As $g$ generates the smallest ideal containing $3$ and $2-2\sqrt{-2}$, we first check whether $\mathcal{N}(g)=3$ is possible; we find that if $g=a+b\sqrt{-2}$ then $$3=\mathcal{N}(g)=\mathcal{N}(a+b\sqrt{-2})=a^2+2b^2,$$ from which it easily follows that $a=\pm1$ and $b=\pm1$. Multiplying by $-1$, a unit, if necessary we find that $g=1\pm\sqrt{-2}$. And indeed $$3=(1+\sqrt{-2})(1-\sqrt{-2})\qquad\text{ and }\qquad 2-2\sqrt{-2}=2(1-\sqrt{-2}),$$ which shows that $g=1-\sqrt{-2}$ is a generator. And for good measure we can verify that $$(1-\sqrt{-2})\cdot3+(-1)\cdot(2-2\sqrt{-2})=1-\sqrt{-2}=g,$$ which shows that indeed $g$ is contained in the ideal generated by $3$ and $2-2\sqrt{-2}$.

Answer 2

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 $\alpha = \beta_{1}\times 3 + \beta_{2} \times (2 - 2\sqrt{-2})$.