Find degree of $\mathbb{Q}(i, \sqrt{-2})$ over $\mathbb{Q}$ and over $\mathbb{Q}(i)$

Question

I'm studying for a qualifier and came upon this problem. Now my reasoning is that $\sqrt{-2} = i\sqrt{2}$ does not live in $\mathbb{Q}(i)$ and so $[\mathbb{Q}(i,\sqrt{-2}):\mathbb{Q}]=[\mathbb{Q}(i,\sqrt{-2}):\mathbb{Q}(i)][\mathbb{Q}(i):\mathbb{Q}]=2\cdot2=4$. Does this make sense? Further it seems that $\mathbb{Q}(i)=\mathbb{Q}[i]$ so that $[\mathbb{Q}(i,\sqrt{-2}):\mathbb{Q}(i)]=2$ as well. Any help would be appreciated. Thanks.

Answer 1

You've certainly got everything right. Here's another way of thinking about it:

1) Is the minimal polynomial for $\sqrt{-2}$ still irreducible over $\mathbb{Q}(i)$? If so, what is it's degree?

That should answer both your questions, I think.