Show that $\mathbb{Q}(\sqrt{2})$ is a field.

Question

Proof: Since $\mathbb{Q}$ is a field, then $\mathbb{Q}$ is a domain.

(Theorem: if $R$ is a domain, then $R[x]$ is a field.)

By the theorem, $\mathbb{Q}[x]$ is a field.

So, letting $x = \sqrt{2}$, we get $\mathbb{Q}[\sqrt{2}]$ is a field. QED.

I am wondering if this is an acceptable proof? Also, I am wondering if it can be proven using facts of irreducibility? Isn't this what Galois Theory is about? I am curious, thanks.

Answer 1

This doesn't work because $\mathbb{Q}[x]$ is not in fact a field, since for instance $x$ has no inverse.

The conceptual way to prove that $\mathbb{Q}[\sqrt{2}]$ is a field is to use the fact that $(x^2-2)$ is a maximal ideal of $\mathbb{Q}[x]$. Alternately, it's not too hard to show that the inverse of any non-zero element $a+b\sqrt{2}$ is $\frac{a-b\sqrt{2}}{a^2-2b^2}$.