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.
 A: 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.
A: 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.
A: Kind of late to the party but I just finished this question for an assignment. We know that $\mathbb{Q}(\sqrt2)$ is a field if $+,\cdot$ are associative and commutative, $\cdot$ distributes over $+$, $+$ has an identity and an inverse, and $\cdot$ has an identity and an inverse.
For associativity and commutativity, using the correct operations with $(a+b\sqrt2)\ast(c+d\sqrt2), [(a+b\sqrt2)\ast(c+d\sqrt2)]\ast(e+f\sqrt2), etc.$ shows this property holds.
Next, show that $(a+b\sqrt2)[(c+d\sqrt2)+(e+f\sqrt2)]\in\mathbb{Q}(\sqrt2)$.
Third, remember the definitions of additive inverse and additive identity. Can you think of a way to rewrite $0$ and $-a$ (negation of an element of $\mathbb{Q}(\sqrt2)$, not the literal value $a$)in a way such that they are elements of $\mathbb{Q}(\sqrt2)$?
Fourth, rewrite the definitions of multiplicative identity and multiplicative inverse. HINT: For identity, how can you write 1 so that it is in $\mathbb{Q}(\sqrt2)$? For inverse, follow the steps from marwalix's post by rationalising the denominator by using a conjugate.
This way, the proof is a bit more descriptive for readers who may not be familiar with abstract algebra of for professors evaluating work.
Good question!
