Consider $\mathbb Q({\sqrt 2},i)$, which is the minimal subfield of $\mathbb C$ containing $\mathbb Q,i,\sqrt{2}$. How do I show that $\sqrt{-2},i$ generate the same field over $\mathbb Q$?
I believe the question asks to show $\mathbb Q(\sqrt{2},i)=\mathbb Q(\sqrt{-2},i)$. First of all, what does "$=$" mean? Is this "equal" or "isomorphic as $\mathbb Q$-vector spaces"?
I can see that $\sqrt{-2}\in \mathbb Q(\sqrt{2},i)$, so $\mathbb Q(\sqrt{2},i)$ is a subfield of $\mathbb C$ containing $\mathbb Q, \sqrt{-2},i$. I believe what remains to show that it is the minimal among such fields (i.e., intersection of all such fields). Why is that true?
I can also see that $\sqrt 2\in \mathbb Q(\sqrt{-2},i)$ but again cannot see that $\mathbb Q(\sqrt{-2},i)$ is the minimal subfield of $\mathbb C$ containing $\mathbb Q,\sqrt{2},i$.