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 $F$.


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