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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?

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up vote 9 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.

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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
I wonder if the downvote signifies an objection to the answer or to the answerer. – Asaf Karagila Feb 22 '15 at 17:45

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.

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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 $\langle A \rangle = \langle B \rangle$.

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