Extension of a field I am reviewing for my exam and i need your help to answer the following problem:
Let $F$ be a field, $E$ an extension of $F$ of degree 2, and $L$ an extension of $E$ of degree 2 . Let $\alpha \in L$ and $f(x) \in F[x]$ be the minimal polynomial of $\alpha$ over $F$.

Show that the degree of $f$ divides 4 .
Show that there is a extension $K$ of $F$ whose degree over $F$ divides 8 such that $f$ splits in $K$.

Thanks
 A: For the the first part:
$[L:E]=2$ and $[E:F]=2$. So $$[L:F]=[L:E][E:F]=4.$$
The mininmal polynomial of $\alpha\in L$ over $F$ is $f$. Then 
$$[F(\alpha):F]=\deg (f)$$
But we know that $$[L:F]=[L:F(\alpha)][F(\alpha):F],$$
From the second and the third equtaion it follows that $\deg(f)$ divides $[L:F]=4$. 
Fact 1:

Let $[L(\theta):L]=2$. Then the minimal polynomial of $\theta$ over $L$ splits over $L(\theta)$.  

Proof:
Let $p(x)=x^2+bx+c$ be the minimalpolynomial of $\theta$ over $F$. Let $\theta'$ be the other root $p$. Then $\theta+\theta'=-b$, so $\theta'=-b-\theta\in F(\theta)$.
For the second part:
There are two possibilities: $\deg (f)=2$ or $\deg (f)=4$. If $\deg(f)=2$, then (by Fact 1) $f$ splits over $F(\alpha)$  and we are done. 
Assume that $\deg (f)=4$. Let $g$ be the minimal polynomial  of $\alpha$ over $E$ (so that $\deg (g)=2$). Then $f=gh$ for some $h\in E[x]$  with $\deg(h)=2$. Then (by fact 1) $g$ splits over $K$. If $h$ splits over $K$ then, we are done. Otherwise, define $L=K(\beta)$ where $\beta$ is a root of $g$. Then $[L:K]=2$ and $h$ splits in $L$ (again by Fact 1). So $f=gh$ splits in $L$ and  $[L:F]=[L:K][K:F]=8$; and we are done.
A: You can do this by hand. Without loss
$$E=F(\sqrt{a})$$ where $a\in F$
and 
$$L=E(\sqrt{p+q\sqrt{a}})$$
its now easy to see that 
$$L(\sqrt{p-q\sqrt{a}})$$ is a splitting field.
