TITLE: Is $\mathbb{Q}[\sqrt2]$ = $\mathbb{Q}[\sqrt2+1]$?
QUESTION [6 upvotes]: 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?

REPLY [7 votes]: 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 $\langle A \rangle = \langle B \rangle$.