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$\bf{Q.}$ Let $K/F$ be an algebraic extension. Show that there is an algebraic extension $L/K$ such that $L/F$ is normal and if $M$ is another normal extension of $F$ such that $F\subseteq K\subseteq M\subseteq L $, then $K=M$

$\bf{My Attempt:}$

$\because$ $K/F$ is algebraic, so for any $\alpha\in K \hspace{2ex} \exists$ a minimal polynomial $f_\alpha\in F[X]$ such that $f_\alpha(\alpha)=0$, then define $K:=$ splitting field of all these polynomials $f_\alpha$. Then $K/F$ is normal, Now suppose $M$ is another normal extension of $F$ such that $F\subseteq K\subseteq M\subseteq L $, if possible say $\exists \theta \in L-M$. Now $K$ is generated by roots of $f_\alpha$ s. So, $\theta$ is a finite linear combination of some finite products of some roots of $f_\alpha$s, e.g. say $\theta =abc+d$, Let $a_1, b_1,c_1,d_1$ be corresponding conjugates of $a,b,c,d$ in $K$. Then I want to say that $a_1 b_1 c_1+d_1$ should be a conjugate of $\theta=abc+d$ in $K$. Is that true? If so, how do I show that?

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  • $\begingroup$ The extensions are not given to be finite, right? $\endgroup$ – астон вілла олоф мэллбэрг Mar 20 at 14:59
  • $\begingroup$ It's only given to be algebraic. $\endgroup$ – Larsson Mar 20 at 15:00
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    $\begingroup$ Why don't you use Zorn's lemma then? Consider an algebraic closure $\bar F$ of $F$ which contains $K$, then it is a normal extension of $F$. Now, the set of extensions of $M$ over $F$ contained in $\bar F$ which are normal, is a collection which can be ordered by reverse inclusion and is non-empty containing $\bar F$. If you have a chain, then show that the intersection is the lower bound. $\endgroup$ – астон вілла олоф мэллбэрг Mar 20 at 15:06
  • $\begingroup$ there's some editing error, can you please write down an answer. The lower bound part is not so clear to me. $\endgroup$ – Larsson Mar 20 at 15:10
  • $\begingroup$ I have done that. $\endgroup$ – астон вілла олоф мэллбэрг Mar 20 at 15:32
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Zorn's lemma will help you out if you have an infinite extension, in showing existence.

For example, consider an algebraic closure of $F$ containing $K$, say $\bar F$. Now, $\bar F$ over $F$ is certainly normal. What we do, is consider the following collection : $S = \{L : F \subset M \subset L \subset \bar F , L / F \text{ is normal}\}$.

The collection $S$ is non-empty, since $\bar F \in S$. Now, order $S$ be reverse inclusion i.e. for $A,B \in S$ , we let $A \geq B$ if and only if $A \subset B$ i.e. the smaller the extension the "greater" the extension.

Now a maximal element of $S$ would be one such that no element of $S$ is greater than it i.e. no subextension of it is normal. Thus, we are looking for a maximal element of $S$, and now we just need to verify the chain condition.

If we have a chain $A_1 \leq A_2 \leq A_3 \leq ...$ of elements of $S$, then consider the intersection of all the extensions, $\cap A_i$. Clearly, since each $A_i$ contains $M$, we have that $\cap A_i$ contains $M$. However, if $\alpha \in \cap A_i$, then $\alpha \in A_j$ for all $j$, so every $A_j$ contains all the conjugates of $\alpha$, therefore $\cap A_i$ contains all the conjugates of $A$. Therefore, $\cap A_i$ is normal over $M$ and therefore is an upper bound of the chain.

Applying Zorn's lemma provides us with a maximal element of $S$ which will be a minimal normal extension of $M$ over $F$. Note that such an extension must be unique, for if there were two such extensions we could take their intersection and obtain a smaller normal extension.

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