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...
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!!!
$\mathbb Q[x]$ is an euclidean domain and $x^2-2$ is irriducible then the ideal $(x^2-2)$ is prime then the ideal is maximal because $\mathbb Q[x]$ in particular is a PID, then the quotient is a field.
Hopefully you can justify that every element really is of that form using, say, polynomial division. It's not enough to consider elements with $a = 0$ or $b = 0$ since the sum of two units need not be a unit. (Can you think of an example? Maybe something more interesting than $u + (-u)$.)
But otherwise I think your idea can work. Take another element $cx + d$, expand out $(ax + b)(cx + d)$ and see what you need to do in order to make that equal to $1$. The idea of "rationalizing the denominator" from high (?) school could also help. You should probably also justify that this quotient isn't the zero ring, which is not considered to be a field.
