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


*

*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 $\mathbb Q[\sqrt2+1]\subseteq\mathbb Q[\sqrt2]$.

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