Determine the field of fractions of $\mathbb{Z}[\sqrt 2]$

Question

The title says it all. Here the field of fractions of a ring $R$ is: $$ Q=\Big\{\frac{a}{s}:a\in R, s\in R\setminus\{0\}\Big\} $$

Answer 1

Hint:

Elements of $\mathbb{Z}[\sqrt{2}]$ look like $a + b\sqrt{2}$ with $a,b \in \mathbb{Z}$. So elements of the field of fractions look like $\frac{a + b\sqrt{2}}{c + d\sqrt{2}}$ with $a,b,c,d \in \mathbb{Z}$.

Can you rewrite that element so that the denominator is free of radicals?

Answer 2

every element of $\Bbb Z [\sqrt2]$ is the form $a+b\sqrt2$, then every element in field of fraction is $$\frac{a+b\sqrt2}{c+d\sqrt2}=\frac{(a+b\sqrt2)(c-d\sqrt2)}{c^2-2d^2}=\frac{ac-2bd+(bc-ad)\sqrt2}{c^2-2d^2}=\frac{ac-2bd}{c^2-2d^2}+\frac{bc-ad}{c^2-2d^2}\sqrt2 $$

where $a,b,c,d \in \Bbb Z$

it is an element of $\Bbb Q[\sqrt2]$