Are $A=\Bbb Q(\sqrt{2}),B=\Bbb Q(i),C=\Bbb Q(i\sqrt2),D=\Bbb Q(i,\sqrt2)$ normal over $\Bbb Q$? Are $A=\Bbb Q(\sqrt{2}),B=\Bbb Q(i),C=\Bbb Q(i\sqrt2),D=\Bbb Q(i,\sqrt2)$
Normal over $\Bbb Q$?

My attempt is in the answer below. Is this theorem stated correctly? Are there hidden obstructions for using it, that I haven't considered?
 A: I believe this theorem can be used to solve all of these:

Theorem: If field extension $L/K$ is a splitting field for some $f\in K[x]$, then $L$ is finite and normal.

$A=\Bbb Q(\sqrt2)$. We have $f(x)=x^2-2=(x-\sqrt2)(x+\sqrt2)$ splits over $A$. $[\Bbb Q(\sqrt{2}):\Bbb Q]=2$ and since this polynomial does not split over $\Bbb Q$, $A$ is the splitting field for some polynomial, and hence is finite and normal.

$B=\Bbb Q(i)$. We have $f(x)=x^2+1=(x+i)(x-i)$ splits over $B$. $[\Bbb Q(i):\Bbb Q]=2$ and since this polynomial does not split over $\Bbb Q$, $B$ is the splitting field for some polynomial, and hence is finite and normal.

$C= \Bbb Q(i\sqrt2)$. We have $f(x)= x^2+2=(x-i\sqrt2)(x+i\sqrt2)$ splits over $C$. $[\Bbb Q(i\sqrt{2}):\Bbb Q]=2$ and since this polynomial does not split over $\Bbb Q$, $C$ is the splitting field for some polynomial, and hence is finite and normal.

$D=\Bbb Q(i,\sqrt2)$. We have $f(x)=(x^2+1)(x^2-2)=(x+i)(x-i)(x+\sqrt{2})(x-\sqrt{2})$ splits over $D$.
Now $[\Bbb Q(i):\Bbb Q]=2$ and $[\Bbb Q(\sqrt{2}):\Bbb Q]=2$ so $[\Bbb Q(\sqrt{2},i):\Bbb Q(i)]=1$ or $2$, but $\Bbb Q(i)\ne \Bbb Q(\sqrt{2})$ so $[\Bbb Q(\sqrt{2},i):\Bbb Q(i)] =2$ and the tower law gives:
$$[\Bbb Q(\sqrt{2},i):\Bbb Q]=4$$
With basis $\{1,\sqrt2,i,\sqrt2i\}$. 
This is the splitting field of $(x^2+1)(x^2-2)$ since the splitting field of $x^2+1$ is $\Bbb Q(i)$ and the splitting field of $(x^2-2)$ is $\Bbb Q(\sqrt{2})$, the splitting field of both of them must be $\Bbb Q(\sqrt{2},i)$.
