1

Prove that $\mathbb{Q}(\sqrt3) = \{a + b\sqrt{3} | a, b \in \mathbb{Q}\}$ is a field assuming addition and multiplication is commutative since $\mathbb{Q} \subset \mathbb{R}$.

In my book, I have a list of field axioms, but I am not so sure on how to prove each one.

Any help is much appreciated, but I would like it if someone could please explain rather than just posting a solution.

Thank you

SchrodingersCat
  • 24,472
  • 4
    The only point that's hard to prove is that every element of $\mathbb{Q}[\sqrt{3}]$ has a multiplicative inverse in $\mathbb{Q}[\sqrt{3}]$. And even that's not hard, because of a high school trick called rationalizing the denominator. – Matthew Leingang Oct 25 '17 at 15:09
  • The hardest part is showing inverses, which is contained within this question. A similar question that addresses some definition-understanding issues is this question. And this question is also essentially related. – davidlowryduda Oct 25 '17 at 15:13

4 Answers4

2

Since:

  • $\mathbb{Q}[\sqrt3]\subset\mathbb R$;
  • $0,1\in\mathbb{Q}[\sqrt3]$;
  • $x,y\in\mathbb{Q}[\sqrt3]\implies x+y\in\mathbb{Q}[\sqrt3]$;
  • $x,y\in\mathbb{Q}[\sqrt3]\implies x\times y\in\mathbb{Q}[\sqrt3]$,

all that remains to be proved is:

  • $x\in\mathbb{Q}[\sqrt3]\implies-x\in\mathbb{Q}[\sqrt3]$;
  • $x\in\mathbb{Q}[\sqrt3]\setminus\{0\}\implies\frac1x\in\mathbb{Q}[\sqrt3]$.

This is easy: $-(a+b\sqrt3)=-a-b\sqrt3$ and$$\frac1{a+b\sqrt3}=\frac{a-b\sqrt3}{a^2-3b^2}.$$

José Carlos Santos
  • 427,504
2

It's easy to see $\Bbb Q[\sqrt 3]$ is closed under addition, for if

$a_1 + b_1\sqrt 3, a_2 + b_2\sqrt 3 \in \Bbb Q[\sqrt 3], \tag 1$

then clearly

$(a_1 + a_2) + (b_1 + b_2)\sqrt 3 \in \Bbb Q[\sqrt 3]; \tag 2$

similarly we have

$(a_1 + b_1\sqrt 3)(a_2 + b_2\sqrt 3) = (a_1 a_2 + 3b_1 b_2) + (a_1b_2 + a_2 b_1)\sqrt 3 \in \Bbb Q[\sqrt 3]; \tag 3$

all the rules about associativity, commutativity, and distributivity are inherited from $\Bbb R$, in which $\Bbb Q[\sqrt 3]$ may be considered to "live". Also, if $a + b \sqrt 3 \in \Bbb Q[\sqrt 3]$, $-a - b\sqrt 3 \in \Bbb Q[\sqrt 3]$, and clearly $1 \in \Bbb Q[\sqrt 3]$, so all we really need to do is show that $(a + b\sqrt 3)^{-1} \in \Bbb Q[\sqrt 3]$. But

$\dfrac{1}{a + b\sqrt 3} = \dfrac{a - b\sqrt 3}{(a - b\sqrt 3)(a + b\sqrt 3)} = \dfrac{a - b\sqrt 3}{a^2 - 3b^2} = \in \Bbb Q[\sqrt 3], \tag 4$

which works since $a^2 - 3b^2 \ne 0 $ for $a, b \in \Bbb Q$, since $\sqrt 3$ is irrational.

Robert Lewis
  • 71,180
1

Hint: $$ \frac{1}{a+b\sqrt{3}}=\frac{a}{a^2-3 b^2}-\frac{b}{a^2-3 b^2} \sqrt{3}. $$

Leox
  • 8,120
0

With respect to addition, verifying the axioms is straightforward. (I hope)

You would also have to verify group axioms with respect to multiplication.

  1. Multiplicative identity. Is $1\in Q(\sqrt{3})$ true? (Yes, it is)
  2. Inverses - Consider $a+b\sqrt{3}$, $a,b\in\mathbb Q$. Do there exist $c,d\in\mathbb Q$ such that $$(a+b\sqrt{3})(c+d\sqrt{3}) = 1? $$
  3. Associativity of multiplication (should be obvious)

One way of showing existence of inverses is as follows: $$(a+b\sqrt{3})(c+d\sqrt{3}) = (ac +3bd) + (cb+ad)\sqrt{3} = 1 + 0\sqrt{3} $$ We want to find $c,d$ such that $$ac+3bd = 1\quad\mbox{and}\quad bc + ad = 0 $$ what do you do?

AlvinL
  • 8,664