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.

Question

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 : https://math.stackexchange.com/questions/324186/f-ab-sqrt7-mid-a-b-in-mathbbq-closed-under-addition-subtracti (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
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. — Angina Seng, 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

score 9 · Accepted Answer · edited Jul 08 '20 at 20:35

9

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.

edited Jul 08 '20 at 20:35

J. W. Tanner

60,406

answered Apr 26 '18 at 15:47

marwalix

16,773

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
3

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. https://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 00:45
There is a factor of $\sqrt 2$ missing from $b /(a^2 - 2b^2)$. Minor oversight, *+1, endorsed!!!* – Robert Lewis Jul 27 '19 at 18:22
Edited thanks a lot – marwalix Aug 03 '19 at 10:15

score 2 · Answer 2 · answered Apr 26 '18 at 16:35

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.

answered Apr 26 '18 at 16:35

David C. Ullrich

89,985

score 0 · Answer 3 · answered Jun 23 '20 at 23:46

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!

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.

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