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Let $K/F$ be a finite field extension such that $F$ contains a primitive root of unity of order the degree of $K/F$. A theorem states that if the characteristic of $F$ is $0$ and $Gal(K/F)$ is a solvable group, then $K/F$ is a radical extension (not necessarily simple).

Question: what happens if we don't assume that the characteristic of $F$ is $0$? i.e if $K/F$ is a finite field extension, with $char(F)=p$, such that $F$ contains a primitive root of unity of order $[K:F]$ (which cannot be prime though). If $Gal(K/F)$ is solvable, does it follow that $K/F$ is radical?

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  • $\begingroup$ This is false as stated. Did you omit a condition asserting that the Galois group of the Galois closure is solvable? $\endgroup$ – Starfall Oct 14 '16 at 11:31
  • $\begingroup$ I've edited, thanks! $\endgroup$ – user10 Oct 14 '16 at 11:59
  • $\begingroup$ Now also posted at mathoverflow.net/questions/252202/… (but gathering close votes there). $\endgroup$ – Gerry Myerson Oct 15 '16 at 4:02
  • $\begingroup$ Now deleted from MO by OP, after it was closed. $\endgroup$ – Gerry Myerson Oct 17 '16 at 6:03

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