Show that $E$ is the splitting field of some polynomial in $F[x].$ Let $K$ be the splitting field of some polynomial over $F$. If $E$ is a field extension contained in $K$ and $[E:F]=2,$ I want to show that $E$ is the splitting field of some polynomial in $F[x].$ I believe I can use something related to Galois theory to deal with this but I am confused here.
 A: If $[E:F]=2$, then there is an element $\alpha\in E$ that’s not in $F$. Dimension $2$ says that $\{1,\alpha,\alpha^2\}$ is an $F$-linearly dependent set, and $\alpha\notin F$ says that the scalar multiplying $\alpha^2$ in your relation of linear dependence is nonzero. So you may get from your relation of dependence a monic $F$-polynomial, say $g$, of degree $2$ of which $\alpha$ is a root. Since $F\ne F(\alpha)$ and $F(\alpha)\subset E$, multiplicativity of dimension says that $F(\alpha)=E$
and is the splitting field of $g$.
No Galois theory, no question about separability.
A: Remember that a field extension of degree two is always normal. Since $\;E/F\;$ is clearly separable and algebraic as a subextension of a Galois extension, it is Galois and thus $\;E\;$ is the splitting field of some set of polynomials over $\;F\;$
A: Since $K$ is the splitting field of a polymomial over $F$, it is Galois over $F$. Thus there is a correspondence between the intermediate fields of $K/F$ and the subgroups of $Gal(K/F)$. Let $N \subseteq Gal(K/F)$ be the subgroup corresponding to $E$ under the Galois correspondence. 
Note that $|Gal(K/F)| = [K:F] = [K:E][E:F] = 2[K:E]$.
But $[K:E] = |N|$. So
\begin{align*}
[Gal(K/F):N] &= \frac{|Gal(K/F)|}{|N|}\\
&= \frac{|Gal(K/F)|}{[K:E]}\\
&=2.
\end{align*}
So $N$ is a normal subgroup of $Gal(K/F)$. But then this implies that $E$ is Galois over $F$. And so $E$ is the splitting field of a polynomial in $F[x]$.
