$x^2-2$ is irreducible in $\mathbb{Q}[x]$ implies that $\mathbb{Q}[\sqrt{2}]$ is a field

Question

I am having a bit of trouble thinking of where to start. Clearly, $\sqrt{2}$ is a root of $x^2-2$ but I am not really sure how this helps. I would also be interested into how this might be applied more generally.

Thanks!

Answer 1

$\mathbb{Q}$ is a field so that $\mathbb{Q}[x]$ is a PID. In a PID, an element is irreducible if and only if it is prime. In a PID, prime ideals are maximal. Notice that $x^2-2$ has no roots and is quadratic so that it is irreducible. Then $\langle x^2-2\rangle$ is a maximal ideal so that $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}[x]/\langle x^2-2\rangle$ is a field. This holds more generally: if $\alpha$ is a root of an irreducible (monic) polynomial, $p_\alpha(x)$, then by the logic above, $\mathbb{Q}(\alpha) \cong \mathbb{Q}[x]/\langle p_\alpha(x)\rangle$ is a field. The polynomial $p_\alpha(x)$ is called the minimal polynomial for $\alpha$.