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I am trying to prove the following:

Let $E/F$ be an arbitrary extensions. Show that $E/F$ is normal if and only if $E$ is the union of all those intermediate fields $K$ such that $K$ is the splitting field for some $f(X) \in F[X]$.

My attempt at the solution is the following:

Let $A$ be the union of all those intermediate fields $K$ such that $K$ is the splitting field for some $f(X) \in F[X]$. Assume $E/F$ is normal. Clearly $A \subseteq E$. To prove the opposite containment let $\alpha \in E$ and $f(X)=\text{min}_F(\alpha)$. Since $E/F$ is normal we know that $f(X)$ splits over $E$, and so $E$ contains the splitting field of $f(X)$, call it $K_f$. Since $f(X) \in F[X]$ we know $K_f \in A$ and thus $\alpha \in A$. Therefore $E\subseteq A$.

Assume $E=A$ and let $\alpha \in E$ and $f(X)=\text{min}_F(\alpha)$. The splitting field of $f(X)$ is contained in $E$ by definition, so $f(X)$ splits completely over $E$. Thus $E/F$ is normal.

One immediate problem with my proof is that fact that I implicitly assume $E/F$ is algebraic when I start talking about the minimal polynomial of an element of $E$. My question is:

Is there any way to fix this proof so as to not assume $E/F$ is algebraic?

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3  
In both directions it seems to me that the algebraicity is automatic. Normal extensions are algebraic and so are splitting fields. –  Dylan Moreland Jul 13 '12 at 15:55
2  
+1 Very nicely posted question, with explanation and development of what's been attained so far. –  DonAntonio Jul 13 '12 at 16:03

1 Answer 1

up vote 2 down vote accepted

What is a normal extension? Lang, for example, in his "Algebra", makes it crystal clear there are (at least) three equivalent conditions for this, one of which is "$\,E/F\,$ is a normal extension iff $\,E\,$ is the splitting field of a family of polynomials $\,\mathcal P\,$ in$\,F[x]\,$"

Now, since "the splitting field" carries a meaning of minimality (i.e., the minimal extension containing all the roots of all the polynomials in some such family $\,\mathcal P\,$) , it seems clear that we can "assume" or "agree" on the extension being algebraic, otherwise we could simply take $\,F(\{\alpha\; :\;\alpha\,\,\text{is a root of some element in}\,\,\mathcal P\}) $ , which would be strictly contained in $\,E\,$ if $\,E/F\,$ weren't algebraic, thus contradicting the above mentioned minimality...

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