TITLE: Prove that $\mathbb Q(\sqrt 2, i) = \mathbb Q(\sqrt 2 + i)$
QUESTION [8 upvotes]: I probably don't understand the definition of each properly. What I am thinking is that 
$\mathbb Q(\sqrt 2, i)$ has elements of form $a_1(\sqrt2) ^1 + a_2(\sqrt2) ^2 + ... + a_n(\sqrt2) ^n + a_1i^1 + a_2i^2 + ... + a_ni^n$
$\mathbb Q(\sqrt 2 +i)$ has elements of form $a_1(\sqrt 2 + i)^1 + a_2(\sqrt 2 + i)^2 + ... + a_n(\sqrt 2 + i)^n$
I'm having trouble jumping from one to the other, as by this definition $\mathbb Q(\sqrt 2 +i)$ has elements of form $\sqrt 2 ^ni^m$ and it seems as though $\mathbb Q(\sqrt 2, i)$ doesn't. Can anyone help?

REPLY [4 votes]: You are mistaken in what $\mathbb Q(\sqrt 2,i)$ is. It not only contains linear combinations of all powers of $\sqrt 2$ and of $i$, but also of elements of the form $\sqrt 2^mi^n$.