For the field $F=\{a+b\sqrt{3}\}$, find $d$ so that $b\cdot d=1$, where $b,d\in F$

Question

Let $b=a_1+b_1\sqrt{3}$

Let $d=a_2+b_2\sqrt{3}$

We need to find $a_2,b_2$ such that $b\cdot d=1$

Doing the math, we get that $b\cdot d=(a_1a_2+3b_1b_2)+(a_1b_2+a_2b_1)\sqrt{3}$

But what from here? How do I decide that I want $(a_1a_2+3b_1b_2)$ to equal $0$, and so I want $(a_1b_2+a_2b_1)$ to be $1$. Or I want the first part of the expression to be $1$, and the next part to be $0$. How do I decide?

Your $b\cdot d$ seems to lack some $\sqrt 3$. Correcting this should facilitate your decision — Hagen von Eitzen, Sep 09 '17 at 20:23
Also, just for fun multiply $b$ with its conjugate $a_1-b_2\sqrt 3$ and see if you can find suitable $d$ from that result. — Hagen von Eitzen, Sep 09 '17 at 20:24
What are $a,b$? You haven't quite specifed $F$ unless you tell us where $a,b$ come from. Hint: $(a_1+b_1\sqrt3)(a_1-b_1\sqrt3)$ is an element of the same field $a_1,b_1$ are from. — Jyrki Lahtonen, Sep 09 '17 at 20:25
Okay sorry I added in the $\sqrt{3}$, but still, how will I know what to make $0$ and what to make $1$ — K Split X, Sep 09 '17 at 20:27
@fleablood yes but i realise that i didnt quite understand how we chose the first equation to be set to $0$ and the other one to $1$ — K Split X, Sep 09 '17 at 20:36

fleablood · Accepted Answer · 2017-09-09T21:45:00.623

Just like in you other exact same question that YOU posted Show that the set $\mathbb{Q}(\sqrt{p})=\{a+b\sqrt{p}; a,b,p\in\mathbb{Q},\sqrt{p}\notin \mathbb{Q}\}$ is a field

You note that if $\gamma \in \mathbb Q \subset \mathbb Q[\sqrt{3}]$ and $\gamma$ is of the form $a + b \sqrt{3}; a,b$ then $b = 0$.

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Note the "definition" $F= \{a + b\sqrt{3}\}$ isn't very well defined unless you specify what values $a,b$ might be. I am assuming you are trying to say $F = \mathbb Q[\sqrt{3}]$; a field extension of $\mathbb Q$. As $(\sqrt 3)^2 = 3 \in \mathbb Q$ we can conclude that or any $n \in \mathbb Z$ that $(\sqrt{3})^n= c+d\sqrt{3}$ for some $c,d \in \mathbb Q$ and that $ \mathbb Q[\sqrt{3}] $ which is consists of all linear combinations of rationals and the powers of $\sqrt{3}$ is $\{a+b\sqrt{3}|a,b \in \mathbb Q\}$.

(Thanks to Lykri Lahtonen, for not letting this slide in the comments. I was going to let it pass, even though I shouldn't.)

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You note that because the $\sqrt{3}$ is irrational, and $a_1, b_1,a_2,b_2$ are all rational that $b\cdot d=(a_1a_2+3b_1b_2)+(a_1b_2+a_2b_1)\sqrt{3}=1$ means that

$(a_1b_2+a_2b_1)\sqrt{3} = 1 - (a_1a_2+3b_1b_2)$.

If $a_1b_2+a_2b_1 \ne 0$ then

$\sqrt 3 = \frac {1 - (a_1a_2+3b_1b_2)}{a_1b_2+a_2b_1}\in \mathbb Q$.

This is impossible so $a_1b_2+a_2b_1 = 0$.

So $a_1b_2+a_2b_1 = 0$ and $a_1a_2+3b_1b_2=1$.

So solve those.

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One thing you should realize and should be intuively obvious is if you have

a field $F$ and you extend it with $\omega \not \in F$

$\gamma \in F[\omega]$ so that $\gamma = a + b\omega; a,b \in F$

And you find that $\gamma = c \in F$ then it has to be that $\gamma \in F$ and so for $\gamma = a + b\omega$ that $b = 0$.

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So $(a_1a_2 + 3b_1b_2) + (a_1b_2+a_2b_1)\sqrt 3 = 1 \in \mathbb Q$.

The only elements $a+b\sqrt{3} \in \mathbb Q$ are of the form $a + 0*\sqrt{3}$.

So $a_1a_2+3b_2b_2 =1$ and $a_1b_2 + a_2b_2 = 0$.

For the field $F=\{a+b\sqrt{3}\}$, find $d$ so that $b\cdot d=1$, where $b,d\in F$

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