# Constructing a minimal extension containing a finite separable extension

Given a finite separable extension $$E \subset F \subset \bar E$$, I am asked to construct

1. A normal extension $$M \subset \bar E$$ such that $$[M:E] \leq [F:E]!$$
2. a unique minimal normal extension $$M \subset \bar E$$ containing $$E$$.

$$E \subset F$$ finite means that it is finitely generated, then $$F = E(a_1, \ldots, a_n)$$ and $$n$$ can be taken to be $$[F：E]$$. Then let $$m_i$$ be the minimal polynomial of $$a_i$$. $$\prod m_i$$ is separable by the uniqueness of minimal polynomials. Then take $$M$$ to be the splitting field of $$\prod m_i$$, we get a normal extension over $$F$$. However how do I make it satisfy each case listed above?

EDIT: should be the minimal normal extension containing $$F$$.

First of all, are you sure the question doesn't ask you to find "2. a unique minimal normal extension $$M \subset \bar{E}$$ containing $$\color{red} F$$."? What I mean is that if this extension should contain $$E$$, then you can take $$M=E$$ and the problem is trivially true. Hence I will prove my version of the problem.
By the Primitive Element Theorem we have that $$F = E(\alpha)$$ for some element $$\alpha \in F$$. Then we have that the minimal polynomial of $$\alpha$$ over $$E$$, call it $$f$$ has degree $$[F:E]$$. For the first case take $$M$$ to be the splitting field of $$f$$ over $$E$$. Then it's not hard to conclude that $$[M:E] \le (\deg f)! = [F:E]!$$. These follows since if you adjoin the roots of $$f$$ one by one the degree of the extensions drop by at least $$1$$. Then use Tower's Theorem to finish off the proof.
To see that this $$M$$ satisfies the second condition use the fact that the splitting field of $$f$$ is the smallest normal field extension of $$E$$ containing $$\alpha$$. Indeed if another normal extension $$M'$$ contains $$F$$ we have that $$E \subset M'$$ and $$\alpha \in M'$$. Since $$M'$$ is normal we have that $$f$$ splits comletely. Thus we must have $$M \subset M'$$, since $$M$$ is the splitting field of $$f$$. Hence we conclude that $$M \subset \bar{E}$$ is the unique minimal normal extension containing $$F$$.