$\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.
 A: 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.
A: Hint : You can write $S_1=\{c+d(\sqrt{2}+2): c,d \in \mathbb{Q}\}$ as  $S_2=\{(c+2d)+d(\sqrt{2}): c,d \in \mathbb{Q}\}$. 
Logically you don't need to find any isomorphism. All you need to know is that as sets $S_1=S_2$. Then since they are both subsets of $\mathbb{R}$ they obviously have the same inherent field structure. 
A: You should have seen that there is an equivalent characterization of $k(S)$ where $k$ is a field, and $S$ is a subset (or element) of a larger field $\Omega$. The definition you’ve apparently seen is “the set of all rational expressions in $S$, with coefficients in $k$”. But you can also define $k(S)$ as the intersection of all subfields of $\Omega$ that contain $k$ and $S$, or in other words, the smallest subfield of $\Omega$ that contains $k$ and $S$. The equivalence of these two definitions is not hard to prove.
With this latter definition it needs no argument whatever to see that if $\lambda\in\Bbb Q$, then $\Bbb Q(s)=\Bbb Q(s+\lambda)$.
