# 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.

• Related : math.stackexchange.com/questions/324186/… (It has $7$ instead of $2$, but that shouldn't change much). – Arnaud D. Apr 26 '18 at 15:40
• If it were me, I would prove that $\mathbb{Q}(\sqrt{2})$ is a sub-field of $\mathbb{R}$. That solves most of your problems. – David Hill Apr 26 '18 at 15:41
• How so? Could you elaborate on that? – Matthijs Bjornlund Apr 26 '18 at 15:41
• It's contained in $\Bbb R$ which has a lot of these properties, e.g., associativity, distributivity, etc., and so will inherit many of them. – Lord Shark the Unknown Apr 26 '18 at 15:42
• As in, how would you prove that it is a sub-field of $\mathbb{R}$? – Matthijs Bjornlund Apr 26 '18 at 15:43

Associativity, commutativity and distributivity of multiplication over additioncome from the fact that $$\Bbb{Q}[\sqrt{2}]\subset\Bbb{R}$$ that 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 $$2$$ is irrational

• I see your point, but how can we assume that $\mathbb{Q}[\sqrt{2}] \in \mathbb{R}$ is a field when that is what we need to prove? – Matthijs Bjornlund Apr 26 '18 at 15:51
• You showed it is closed under addition and multiplication, and @marwalix showed that the multiplicative inverse is in it. This is enough to show that it is a subfield of $\mathbb R$, and it entails that it is a field in its own right. en.wikipedia.org/wiki/Field_extension#Subfield – giobrach Apr 26 '18 at 15:55
• Thank you. That was very helpful. – Matthijs Bjornlund Apr 27 '18 at 0:45
• There is a factor of $\sqrt 2$ missing from $b /(a^2 - 2b^2)$. Minor oversight, +1, endorsed!!! – Robert Lewis Jul 27 at 18:22
• Edited thanks a lot – marwalix Aug 3 at 10:15

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.