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Problem statement: If $K/F$ is a finite separable extension, and for any field extension $M/F$, $[M:F]$ is divisible by a fixed prime $p$, show that $[K:F]$ is a power of $p$.

Primitive element theorem tells us $K = F[\alpha]$. I am thinking about picking some element and use tower lemma, but I seem to miss pieces?

leoli1
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SmoothKen
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    I assume that $p\mid [M:F]$ is only supposed to hold for extensions $M\ne F$. – leoli1 Sep 02 '21 at 06:17
  • @leoli1 By the way, is field $F$ necessarily perfect? – SmoothKen Sep 02 '21 at 07:05
  • No, consider a non-perfect field $k$ of characteristic $p$ and let $F$ be the separable closure of $k$. Then $F$ is not perfect and all algebraic extensions of $F$ are purely inseparable, in particular have degree divisible by $p$ (if they are not trivial). – leoli1 Sep 02 '21 at 07:36
  • @leoli1 Is it possible to prove that in the non-perfect case, we must have $char = p$ (without using separable closure)? – SmoothKen Sep 03 '21 at 02:05
  • A non-perfect field always has prime characteristic and as such a field admits a purely inseparable extensions this prime has to be the same as $p$ (by the divisibility assumption). – leoli1 Sep 03 '21 at 06:53
  • How to ask a good question. This post misses the mark. Try to read it, or parts of it; surely you'll find a couple suggestions that you can take with you to edit and improve this question. – amWhy Sep 04 '21 at 18:35

1 Answers1

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Hint: By replacing $K$ with the normal hull over $F$ we may assume that $K$ is finite and Galois over $F$. Then consider the fixed field of a Sylow $p$-subgroup of $\operatorname{Gal}(K/F)$.

leoli1
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