Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is $\mathbb{Q}[\sqrt2]$ = $\mathbb{Q}[\sqrt2+1]$?

I think so because

$$\mathbb{Q}[\sqrt{2}+1] = \{\sum_{i=0}^{n}c_i(\sqrt{2}+1)^i\mid n\in\mathbb{N}, c_i\in\mathbb{Q}\}$$

$$= \{\sum_{i=0}^{n}c_i(\sqrt{2})^i\mid n\in\mathbb{N}, c_i\in\mathbb{Q}\} = \mathbb{Q}[\sqrt{2}].$$

This could be worked out with Binomial theorem right?

share|improve this question

3 Answers 3

up vote 10 down vote accepted

Actually show that: $$\mathbb Q[\sqrt2]=\left\{a+b\sqrt2\mid a,b\in\mathbb Q\right\}\\ \mathbb Q[\sqrt2+1]=\left\{a+b(\sqrt2+1)\mid a,b\in\mathbb Q\right\}\\$$

So you do not need the "entire" binomial argument, not to take arbitrarily long combinations. Two is enough.

Now you can either use the argument Alex gave, or we can show two-sided inclusions directly:

  1. Suppose that $a+b(\sqrt2+1)$ is in $\mathbb Q[\sqrt2+1]$, take $c=a+b$ (which is a rational number) and then $a+b(\sqrt2+1)=a+b\sqrt2+b=c+b\sqrt2\in\mathbb Q[\sqrt2]$. Therefore $\mathbb Q[\sqrt2+1]\subseteq\mathbb Q[\sqrt2]$.

  2. Take now $a+b\sqrt2\in\mathbb Q[\sqrt2]$ then $a+b\sqrt2=a+(-b+b)+b\sqrt2=(a-b)+b(\sqrt2+1)\in\mathbb Q[\sqrt2+1]$. Therefore $\mathbb Q[\sqrt2]\subseteq\mathbb Q[\sqrt2+1]$.

Therefore we have equality.

share|improve this answer
    
I think you are missing $b$ in number two :) –  Theorem Jun 12 '12 at 18:27
    
@Theorem: Thanks. I corrected that. –  Asaf Karagila Jun 12 '12 at 18:29

Yes. Note that $\sqrt{2}=(\sqrt{2}+1)-1\in \mathbb Q[\sqrt{2}+1]$ so $\mathbb Q[\sqrt{2}]\subseteq \mathbb Q[\sqrt{2}+1]$ and $\sqrt{2}+1\in \mathbb Q[\sqrt{2}]$ so $\mathbb Q[\sqrt{2}]\subseteq \mathbb Q[\sqrt{2}+1]$. Thus the two are equal.

share|improve this answer
3  
Really? 12 upvotes? I'll never understand you, math.SE. –  Alex Becker Jun 12 '12 at 20:20
    
It's because this simple observation is the simple answer. –  Hagen von Eitzen Sep 15 '12 at 16:45
    
Easily the best, least complicated and most straightforward answer. –  Ryker Nov 6 '13 at 23:35

A basis for $\mathbb{Q}[\sqrt2]$ is $\mathcal{A}=\{1,\sqrt2\}$. A basis for $\mathbb{Q}[\sqrt2+1]$ is $\mathcal{B}=\{1,\sqrt2+1\}$. You can write $$ \mathcal{B}= \pmatrix{ 1 & 0 \\ 1 & 1} \mathcal{A} $$ Since this matrix is invertible, we have $\langle A \rangle = \langle B \rangle$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.