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A theorem is stated as follows. For a field $F$ of characteristic $p$, $F$ has a $p$-cyclic extension if and only if for every positive integer $m$, $F$ has a $p^m$-cyclic extension. I wonder if there is an elementary proof of it, without too much discussion on Abelian extension theories. All I am acquainted with is basic Galois group and group theory.

Jyrki Lahtonen
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user11546831
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  • You do know that when the extension degree equals the characteristic, Kummer-extensions need to be replaced with Artin-Schreier? This standard argument can be extended to show that $F$ has a cyclic extension of degree $p$ if and only if the mapping $T:F\to F, z\mapsto z^p-z$ is not surjective. Lifting that to cyclic extensions of degree $p^m$ may be a bit more difficult in general. The route I'm aware of would go via Witt vectors, but it sounds like that is one of the pieces you would rather avoid... – Jyrki Lahtonen May 26 '19 at 05:10
  • @JyrkiLahtonen I appreciate it. So if I have to prove this as quickly as possible, what’s your advice? Do I need to finish the whole chapter 3 of Jacobson III？ – user11546831 May 26 '19 at 05:16
  • Anyway, here is my take on the "easiest" inductive step of going from a cyclic extension of degree two to one of degree four. All using Witt-vector arithmetic. IIRC the arithmetic details become progressively hairier as $p^m$ grows. OTOH you only want existence. May be that is easier, but I am too ignorant. – Jyrki Lahtonen May 26 '19 at 05:18
  • @JyrkiLahtonen THX very much – user11546831 May 26 '19 at 05:23
  • I'm afraid I don't know if there is an easier way. Jacobson is good. If you already worked your way through the corresponding pieces of Kummer theory, then your background is strong enough. – Jyrki Lahtonen May 26 '19 at 05:24

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