Proving that the extension field $\mathbb{Q}(\sqrt{2})$ is indeed a field

Question

Every element in the extension $\mathbb{Q}(\sqrt{2})$ of the rational numbers can be written in the form $\alpha+\beta\sqrt{2}$ with $\alpha,\beta\in\mathbb{Q}$. The book states that this, like $\mathbb{Q}$, is a field, but that if $\alpha,\beta\in\mathbb{Z}$ then this is no longer true. I guess the last statement can be proven by considering that the multiplicative inverse of $\alpha+\beta\sqrt{2}$ is not an integer, but how can it be proven that $\mathbb{Q}(\sqrt{2})$ is a field when $\alpha,\beta$ are rationals?

Answer 1

Hint:

You only have to prove that $(\alpha +\beta\sqrt 2)^{-1},\enspace\alpha, \beta\in\mathbf Q\,$ can be written as $(\gamma +\delta\sqrt 2),\enspace\gamma, \delta\in\mathbf Q$. Use the conjugate: $$\frac1{\alpha +\beta\sqrt 2}=\frac{\alpha -\beta\sqrt 2}{(\alpha +\beta\sqrt 2)(\alpha -\beta\sqrt 2)}.$$