Prove or disprove that Q[√2] is a field

Question

Let $R = \mathbb{Q}[\sqrt2]$ = {$\alpha + \beta\sqrt2 \;|\; \alpha, \beta \in \mathbb{Q}$} and $z = a + b\sqrt2$ with $a,b \in \mathbb{Q}$

1. Prove or disprove R is a field.
2. Prove $(a + b\sqrt2)(a - b\sqrt2) \neq0$ where $a,b \in \mathbb{Q}$
3. Prove that $(a+b\sqrt2) \in U(R)$

My attempt: 1. Working on it...

2. $(a + b\sqrt2)(a - b\sqrt2) = a^2 - (\sqrt2)^2b^2$. Since $\sqrt2$ is not a rational number, $a^2 - 2b^2 \neq 0$

3. Confused. I know I need to show $(a + b\sqrt2)(a - b\sqrt2)$ = 1 to be in $U(R)$ but not exactly sure how to proceed.

Answer 1

$$\frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}\sqrt{2}=(a+b\sqrt{2})^{-1}$$ whenever $a$ and $b$ are not simultaneously $0$

Answer 2

Hint: (1) Let $\alpha = \sqrt{2}$ and $J = \{f(x) \in R[x] ; f(\alpha) = 0\}$

$$\frac{ R[x]}{J} \simeq R[\alpha]$$

$J$ is maximal then $ \frac{R[x]}{J}$ is a field.