Why $\mathbb Q[\sqrt 2]$ is a field? I was reading this question (that has been changed a bit). By definition $\mathbb Q[\sqrt 2]$ is a ring. It's the ring $\{a+b\sqrt 2\mid a,b\in\mathbb Q\}$.
Q1) Does $\mathbb Q[\sqrt 2]=\{a+b\sqrt 2\mid a,b\in\mathbb Q\}$ by definition or $\mathbb Q[\sqrt 2]\cong \{a+b\sqrt 2\mid a,b\in\mathbb Q\}$ ?
Q2) why does $\mathbb Q[\sqrt 2]=\mathbb Q(\sqrt 2)$ (where $\mathbb Q(\sqrt 2)$ is the fraction field of $\mathbb Q[\sqrt 2]$) ?
 A: This is one point where rationalizing the denominator is important, even though it wasn't in grade school.
$$\frac1{a+b\sqrt 2}=\frac{a-b\sqrt2}{a^2-2b^2}$$
Since $\sqrt2$ is irrational, $a^2-2b^2\neq 0$ as long as either $a$ or $b$ is not zero. 
A: Other than computations to find the inverse of any $a+b\sqrt 2\ne0$, you can observe that it is a finite dimensional $\mathbf Q$-vector space, and multiplication by $a+b\sqrt 2$ is an injective $\mathbf Q$-linear map, hence it is bijective. In particular, $1$ is attained, i.e. there exists $c+d\sqrt 2$ such that
$$(a+b\sqrt 2)(c+d\sqrt 2)=1.$$
A: Q1) The second one ! 
Q2) Because $\mathbb Q[\sqrt 2]$ is a field. If $a+b\sqrt 2\in \mathbb Q[\sqrt 2]$, you can easily find $c+d\sqrt 2$ s.t. $(a+b\sqrt 2)(c+d\sqrt 2)=1$
A: $\mathbb Q[\sqrt 2]$ is a field because $\sqrt{2}$ is an algebraic number i.e. it is the solution of a polynomial with rational coefficients, namely $x^2=2$. This means, for example, that $\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2} \in \mathbb Q[\sqrt 2]$.
On the other hand $\mathbb Q[\pi]$ is not a field because $\pi$ is transcendental, so $\frac{1}{\pi} \notin \mathbb Q[\pi]$.
