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I am stuck on a step of the proof of Lemma 18.4 of Patrick Morandi, Field and Galois Theory.

Let $p$ be a prime number and let $F$ be a field with $\mbox{char}(F) \neq p$. Morandi defines the $p$-closure $F_p$ of $F$ as the composite of all the Galois extensions of $F$ (in some fixed algebraic closure $F^{\mbox{ac}}$ of $F$) of degree a power of $p$. He claims that if $L$ is a finite extension of $F$ such that $L \subseteq F_p$, then $[L : F]$ is a power of $p$. He proves this is the following way:

"$L = F(a_1, \ldots, a_n)$ for some $a_i \in L$. From the definition of $F_p$, for each $i$ there is a Galois extension $E_i/F$ such that $a_i \in E_i$ and $[E_i : F]$ is a power of $p$."

I do not understand the bold implication. In general, if $K$ is the composite of some family $(K_j)_{j \in J}$ of fields, and $F(a_1, \ldots, a_n)$ is a subfield of $K$, then I see no reason why it should be that for each $i$ we have $a_i \in K_j$ for some $j \in J$. In fact, $K$ is not just the union of the $K_j$'s, but it can contains much more elements.

Thank you in advance for any suggestion.

P.S. Anyway, others ways to prove that "if $L$ is a finite extension of $F$ such that $L \subseteq F_p$, then $[L : F]$ is a power of $p$" are welcome.

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  • $\begingroup$ Has he already shown the composite of two $p$ extensions is a $p$ extension ? $\endgroup$ – Rene Schipperus Nov 7 '14 at 12:41
  • $\begingroup$ @ReneSchipperus I suppose by $p$-extension of $F$ you mean a Galois extension of $F$ of degree a power of $p$. In such a case, the answer is no. However, if the composite of two $p$-extension of $F$ is a $p$-extension of $F$ then what should happen? Thanks. $\endgroup$ – Marvolo Nov 7 '14 at 16:33

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