how do I prove that $\mathbb{Q} [x]/\langle x^2 – 2 \rangle$ is a field how do I prove that $\mathbb{Q} [x]/\langle x^2 – 2\rangle$ is a field?  Is it enough to show that each since each class can be written as $[ax+b]$, then if $a=0$, $b$ is a constant which is an element of $\mathbb{Q}$. and if $b = 0$, then there exists $(ax+b)^{-1}$ such that $(ax+b)(ax+b)^{-1}=1\cdot F$ so it is invertible.  And if it is invertible, then it is a field
Is this sufficient?

It is a theorem that, in general, if $f \in F[x]$ is irreducible, then $F[x]/\langle f(x) \rangle \cong F[\alpha]$ where $\alpha$ is a root of $f$.
We can prove this with the isomorphism theorems.  Consider the evaluation homomorphism $ev_{\alpha}:F[x] \rightarrow F[\alpha]$ defined as follows: $g(x) \mapsto g(\alpha)$.  You'll want to prove that this is indeed a homomorphism.  
Now certainly $\ker(\phi) = \langle f(x) \rangle$.  This is because $F[x]$ is a PID and $f(x)$ is irreducible over $F$ with $\alpha$ as a root.  Finally, we know that $ev_\alpha$ maps surjectively to $F[\alpha]$.  
And so...
