I would like to know if it is true that $\mathbb{Q}(\sqrt{2}-i, \sqrt{3}+i) = \mathbb{Q}(\sqrt{2}-i+2(\sqrt{3}+i))$.
I can prove, that $\mathbb{Q}(\sqrt{2}-i, \sqrt{3}+i) = \mathbb{Q}(\sqrt{2},\sqrt{3},i)$, so the degree of this extension is 8. Would it be enough to show that the minimal polynomial of $\sqrt{2}-i+2(\sqrt{3}+i)$ has also degree 8?
It follows from the proof of the primitive element theorem that only finitely many numbers $\mu$ have the property that $\mathbb{Q}(\sqrt{2}-i, \sqrt{3}+i)\neq \mathbb{Q}(\sqrt{2}-i+\mu(\sqrt{3}+i))$. Obviously $\mu=1$ is one of them, but how to check, whether 2 also has this property?
Thanks in advance,