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?

Well, the argument advanced by mcmat23 et al is of course the standard method:  $f(x) \in \Bbb Q[x] \;\; \text{irreducible} \; \; \Leftrightarrow \Bbb Q[x]/\langle f(x) \rangle \; \; \text{is a field}$; it is easy to see that $x^2 -2$ is irreducible over $\Bbb Q$ since any factor of degree one would yield a square root of $2$ in $\Bbb Q$:  NOT!!!
You can also proceed a little more intuitively if you realize that elements of $\Bbb Q[x]/\langle x^2 - 2 \rangle$ can be thought of as numbers of the form $a + b\sqrt{2}$ with $a, b \in \Bbb Q$.  Why?  The Euclidean algorithm says that any polynomial $f(x) \in \Bbb Q[x]$ can be uniquely written as $f(x) = q(x)(x^2 - 2) + (a + bx)$; $x^2 - 2 = 0$ in $\Bbb Q[x]/\langle x^2 - 2 \rangle$, so, identifying $x$ with $\sqrt{2}$ makes the correspondence $a + bx \leftrightarrow a + b \sqrt{2}$ legitimate.  This is the essence of the discussion of Kaj Hansen, albeit in a somewhat diluted form.  In any event, we have
$(a + b\sqrt{2})^{-1} = \dfrac{a - b\sqrt{2}}{a^2 - 2b^2}; \tag{1}$
this is always valid since $\sqrt{2} \ne \Bbb Q$, so $a^2 - 2b^2 \ne 0$ for all $a, b \in \Bbb Q$.  Indeed,
$(a + b\sqrt{2})\dfrac{a - b\sqrt{2}}{a^2 - 2b^2} = \dfrac{a^2 - 2b^2}{a^2 - 2b^2} = 1; \tag{2}$
since any $a + b\sqrt{2}$ has inverse, the set of all such $a + b\sqrt{2}$ is in fact a field.  Note the strong analogy with the standard formula for the inverse of a complex $z \in \Bbb C$:
$z^{-1} = \dfrac{\bar z}{\vert \bar z z \vert}; \tag{3}$
with $z = x + iy$ this reads
$(x + iy)^{-1} = \dfrac{x - iy}{x^2 + y^2}; \tag{4}$
each formula involves a sort of "conjugation":  $a + bi \leftrightarrow a - bi$, $a + b\sqrt{2} \leftrightarrow a - b \sqrt{2}$;  this if course is not accidental!
Hope this helps. Cheerio,
and as ever,
Fiat Lux!!!
