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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.

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  • $\begingroup$ Related : math.stackexchange.com/questions/324186/… (It has $7$ instead of $2$, but that shouldn't change much). $\endgroup$ – Arnaud D. Apr 26 '18 at 15:40
  • $\begingroup$ 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. $\endgroup$ – David Hill Apr 26 '18 at 15:41
  • $\begingroup$ How so? Could you elaborate on that? $\endgroup$ – Matthijs Bjornlund Apr 26 '18 at 15:41
  • $\begingroup$ 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. $\endgroup$ – Lord Shark the Unknown Apr 26 '18 at 15:42
  • $\begingroup$ As in, how would you prove that it is a sub-field of $\mathbb{R}$? $\endgroup$ – Matthijs Bjornlund Apr 26 '18 at 15:43
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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

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  • $\begingroup$ 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? $\endgroup$ – Matthijs Bjornlund Apr 26 '18 at 15:51
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    $\begingroup$ 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 $\endgroup$ – giobrach Apr 26 '18 at 15:55
  • $\begingroup$ Thank you. That was very helpful. $\endgroup$ – Matthijs Bjornlund Apr 27 '18 at 0:45
  • $\begingroup$ There is a factor of $\sqrt 2$ missing from $b /(a^2 - 2b^2)$. Minor oversight, +1, endorsed!!! $\endgroup$ – Robert Lewis Jul 27 at 18:22
  • $\begingroup$ Edited thanks a lot $\endgroup$ – marwalix Aug 3 at 10:15
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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.

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