$\mathbb{Q}(\sqrt2) $ is isomorphic to $\mathbb{Q}(\sqrt2 +2 )$ Show that $\mathbb{Q}(\sqrt2) $ is isomorphic to $\mathbb{Q}(\sqrt2 +2 )$, but is it more? Are these fields equal?
$\mathbb{Q}(\sqrt2)=\{a+b\sqrt2 |a,b \in \mathbb{Q}\}$
$\mathbb{Q}(\sqrt2)=\{c+d(\sqrt2+2) |c,d \in \mathbb{Q}\}$
so we can find homomorphism which send $a +b \rightarrow c+2d +d$ where $a=c+2d$ and $b=d$. Is that correct thinking ? When they are equal they are also isomorphic.

The easiest way here is to show that $\sqrt 2 = (\sqrt 2+2)-2 \in \mathbb Q(\sqrt 2+2)$ and $\sqrt 2+2 \in \mathbb Q(\sqrt 2)$ - which is trivial.
From the first we get $\mathbb Q(\sqrt 2) \subseteq \mathbb Q(2+\sqrt 2)$ and from the second we get the reverse inclusion.
The two fields are therefore equal and therefore isomorphic.
