Show that $\mathbb{Q}[x]/(x^2−2)$ is a field. 
Show that $\mathbb{Q}[x]/(x^2−2)$ is a field.

I have an idea to solve it, but I'm not sure. I thought of using the root of two since it is isomorphic. also to be a field this should be an ideal maximum. Correct me if I am wrong if you can help me would be very helpful.
 A: You idea is about right. In more detail, you can use the following facts, which you likely know.


*

*The quotient of a ring by an ideal is a field if and only of the ideal is a maximal. 

*A polynomial generates a maximal ideal in the polynomial ring over a field if and only if the polynomial is irreducible.

*A degree two polynomial over a field is irreducible if and only if it has no root (in that field). 
Then it remains to show that your polynomial has no rational root. But that  is not hard. 
A: This follows from the quadratic formula: Since the roots of $x^2-2=0$ are given by 
$$
x=\pm \sqrt{2}
$$
are irrational, the polynomial doesn't factor i.e. it is irreducible, over $\mathbb{Q}$. As ideals generated by irreducible elements are maximal in PID's, 
$$
\mathbb{Q}[x
]/(x^2-2)$$
is a field.
Even more directly, we may invert nonzero elements in $\mathbb{Q}[\sqrt{2}]\cong \mathbb{Q}[x
]/(x^2-2)$ 
by taking 
$$
(a+b\sqrt{2})^{-1}=\frac{a-b\sqrt{2}}{a^2-2b^2}
$$
A: This is more general than the mere case of $\sqrt 2$: $x^2-2$ is an irreducible polynomial, hence it generates a non-zero prime ideal in the P.I.D $\;\mathbf Q[x]$.
Then, either you know a P.I.D. has Krull dimension $1$, i.e. non-zero prime ideals are maximal, which settles the question. 
Or you can make a direct reasoning: $\mathbf Q[x]/(x^2-2)$ is an integral domain, which is a finite  dimensional $\mathbf Q$-vector space. Hence multiplication by a non-zero element $p$ is an injective endomorphism of this vector space. As we're in finite dimension, injective  $\iff$ surjective, so $1$ is attained, which means there exists a $q\in\mathbf Q[x]/(x^2-2)$ such that $pq=1$, and we've proved any $p\ne 0$ is invertible in $\mathbf Q[x]/(x^2-2)$.
