Show that $\text{Frac}(\mathbb{Z}(\sqrt{2}))$ is isomorphic to $\mathbb{Q}(\sqrt{2})$.
Here $\text{Frac}(\mathbb{Z}(\sqrt{2}))$ is the fiel of fractions of $\mathbb{Z}(\sqrt{2})$.
I already proved that $\mathbb{Z}(\sqrt{2})$ is an integral domain. Thus we can construct its field of fracions $F$. Define $\alpha:F\rightarrow \mathbb{Q}(\sqrt{2})$ via $$\alpha([a+b\sqrt{2},c+d\sqrt{2}])=(a+b\sqrt{2})(c+d\sqrt{2})^{-1}=\frac{ac-2bd}{c^2-2d^2}+\frac{bc-ad}{c^2-2d^2}\sqrt{2}$$ It's easy to see check that $\alpha $ is a homomorphism. Now since $\text{Ker}(\alpha)$ is an ideal of $F$, we have that $\text{Ker}(\alpha)=\{0\}$ or $F$. But $\alpha([1,1])=1\neq 0$. Thus $\text{Ker}(\alpha)=\{0\}$, and hence $\alpha$ is one-to-one.
How can I prove that $\alpha$ is onto?
I know that $\alpha(F)\subseteq \mathbb{Q}(\sqrt{2})$, but I don't see clear that $\mathbb{Q}(\sqrt{2})\subseteq \alpha(F) $. In This answer they give the double inclusion, but I can't quite understand why. If you could extend this argument or tell me how with what I have done I can show that $\alpha$ is onto, I would appreciate it