$\bf{Q.}$ 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?

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.