I am trying to prove the following:
Let $E/F$ be an arbitrary extensions. Show that $E/F$ is normal if and only if $E$ is the union of all those intermediate fields $K$ such that $K$ is the splitting field for some $f(X) \in F[X]$.
My attempt at the solution is the following:
Let $A$ be the union of all those intermediate fields $K$ such that $K$ is the splitting field for some $f(X) \in F[X]$. Assume $E/F$ is normal. Clearly $A \subseteq E$. To prove the opposite containment let $\alpha \in E$ and $f(X)=\text{min}_F(\alpha)$. Since $E/F$ is normal we know that $f(X)$ splits over $E$, and so $E$ contains the splitting field of $f(X)$, call it $K_f$. Since $f(X) \in F[X]$ we know $K_f \in A$ and thus $\alpha \in A$. Therefore $E\subseteq A$.
Assume $E=A$ and let $\alpha \in E$ and $f(X)=\text{min}_F(\alpha)$. The splitting field of $f(X)$ is contained in $E$ by definition, so $f(X)$ splits completely over $E$. Thus $E/F$ is normal.
One immediate problem with my proof is that fact that I implicitly assume $E/F$ is algebraic when I start talking about the minimal polynomial of an element of $E$. My question is:
Is there any way to fix this proof so as to not assume $E/F$ is algebraic?