$\mathbb{Q}[\sqrt n]$ is isomorphic to the field of fractions of $\mathbb{Z}[\sqrt n]$

Question

To prove the above statement, I have defined the following map: $(a+b\sqrt n,c+d\sqrt n) \rightarrow (a+b\sqrt n)(c+d\sqrt n)^{-1}$ where $c+d\sqrt n \ne0$ and $(c+d\sqrt n)^{-1} = \frac{c}{c^2-d^2n}-\frac{d}{c^2-d^2n}\sqrt n$

However, I'm stuck at showing that the above homomorphism is surjective. Any suggestions?

Answer 1

You’re making it too complicated. Let $F = \operatorname{Quot} ℤ[\sqrt n]$ be the field of fractions of $ℤ[\sqrt n]$. Then clearly $F ⊆ ℚ(\sqrt n) = ℚ[\sqrt n]$ because $ℤ[\sqrt n] ⊆ ℚ(\sqrt n)$. But also $ℚ ⊆ F$ and $\sqrt n ∈ F$. Hence …