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$\newcommand{\Gal}{\mathrm{Gal}}$ Suppose I have a field $L$ which is a Galois p-extension over a smaller field, $K$. Suppose further that $K$ is a p-extension over $F$ (a p-extension is a Galois extension with a Galois group of order $p^a$ for some $a \in \Bbb Z_+$. To prove that $L$ is a p-extension over $F$ I am finding it necessary to assume that $K$ is fixed by all the automorphisms $\sigma_i \in \Gal(\overline L/F)$, where $\overline L$ is the Galois closure of $L$ over $F$, i.e., $\sigma_i(K) = K$ for all $\sigma_i \in \Gal(\overline L/F)$. Why is this necessary?

My thanks for any advice you can give me.

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2 Answers 2

up vote 3 down vote accepted

Think about $F$ the rationals, $K$ the extension by $\sqrt2$, $L$ the extension by $\root4\of2$. Then $L$ is not Galois over $F$, so you have to go to the Galois closure.

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The moral here is that you can never count on normality to be transitive. –  JSchlather Mar 7 '13 at 3:05

Why don't you use the tower of fields degree theorem?

$$[L:F]=[L:K][K:F]=p^ap^b=p^{a+b}$$

and we're done...!

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But that doesn't show $L$ is Galois over $K$, which is what OP seems to want (and which need not be true). –  Gerry Myerson Mar 7 '13 at 2:36
    
As far as I can understand the question, he's interested only in the $p-$extension thing –  DonAntonio Mar 7 '13 at 2:42
    
The OP writes "a p-extension is a Galois extension with a Galois group of order $p^a$". It's a bit confusing because earlier he writes Galois $p$-extension, implying your interpretation of $p$-extension. –  JSchlather Mar 7 '13 at 3:00

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