# $\mathbb Q(\sqrt{2},i)=\mathbb Q(\sqrt{-2},i)$

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 $\mathbb Q,\sqrt{2},i$.

You already saw that $\mathbb Q(\sqrt{-2},i)\subseteq \mathbb Q(\sqrt{2},i)$, because $\sqrt{-2}\in \mathbb Q(\sqrt{2},i)$. But the other containment is also clear. Hence we have equality. So both fields are the minimal subfield of $\mathbb{C}$ containing $\mathbb{Q}$ and either $\sqrt{2},i$, or $\sqrt{-2},i$.
• So "=" is the mere equality of sets? But it also must be true that $\mathbb Q(\sqrt{-2},i)$ is the minimal subfield of $\mathbb C$ containing $\sqrt{2},i, \mathbb Q$. Why is that true? – user437309 Apr 13 '18 at 19:04
• By definition, $\mathbb Q(\sqrt{-2},i)$ is the minimal subfield of $\mathbb{C}$ containing $\mathbb{Q}$ and $i,\sqrt{-2}$. Because of the above equality, it is also the minimal subfield containing $\mathbb{Q}$ and $i,\sqrt{2}$. – Dietrich Burde Apr 13 '18 at 19:05
To prove the equality in $\mathbb{A}=\mathbb{B}$, you have to show that both $\mathbb{A}\subseteq\mathbb{B}$ and $\mathbb{B}\subseteq\mathbb{A}$.
Note that $$\mathbb{Q}(\sqrt{-2},i)=\mathbb{Q}(i\sqrt2,i)$$ and both $i,\sqrt2\in\mathbb{Q}(\sqrt2, i)$, and also $$\mathbb{Q}(\sqrt2, i)=\mathbb{Q}(-i\sqrt{-2},i)$$ and both $i,\sqrt{-2}\in\mathbb{Q}(\sqrt{-2},i)$.