Show that the set $\mathbb{Q}[\sqrt{2}] = \{a + b \sqrt{2} \mid a, b \in \mathbb{Q}\}$ is a field with the usual multiplication and addition.

Show that the set $\mathbb{Q}[\sqrt{2}] = \{a + b \sqrt{2} \mid a, b \in \mathbb{Q}\}$ is a field with the usual multiplication and addition.

It is easy enough to show that it is closed under addition and multiplication ( $\forall a,b \space)$ we have $a+b \in \mathbb{Q}\sqrt{2}$ and $a * b = ab \sqrt{2} \in \mathbb{Q}\sqrt{2}$.

However, I had trouble proving the other axioms (associativity, commutativity, unique neutral element, unique inverse, and distributivity of multiplication over addition). I would really appreciate it if someone could help me with those.

Solution 1:

Associativity, commutativity and distributivity of multiplication over addition come from the fact that $\Bbb{Q}[\sqrt{2}]\subset\Bbb{R}$, which is a field. The neutral element for addition is $0$ and for multiplication is $1$. We just need to prove that the inverse in $\Bbb{R}$ of $a+b\sqrt{2}\neq 0$ belongs in fact to $\Bbb{Q}[\sqrt{2}]$. One has

$$\begin{align}\left(a+b\sqrt{2}\right)^{-1}=&{1\over a+b\sqrt{2}}\\=&{a\over a^2-2b^2}-{b\over a^2-2b^2}\sqrt{2}\in \Bbb{Q}[\sqrt{2}]\end{align}$$

where the denominator $a^2-2b^2\neq 0$ because $\sqrt2$ is irrational.

Solution 2:

Or show that $\Bbb Q[\sqrt 2]$ is isomorphic to $\Bbb Q[x]/<x^2-2>$, which is field because $x^2-2$ is irreducible.