# Normal Closure of an Algebraic Extension

$$\bf{Q.}$$ 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?

• The extensions are not given to be finite, right? – астон вілла олоф мэллбэрг Mar 20 at 14:59
• It's only given to be algebraic. – Larsson Mar 20 at 15:00
• 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. – астон вілла олоф мэллбэрг Mar 20 at 15:06
• there's some editing error, can you please write down an answer. The lower bound part is not so clear to me. – Larsson Mar 20 at 15:10
• I have done that. – астон вілла олоф мэллбэрг Mar 20 at 15:32

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.