I have a Galois field extension $E/F$ of degree $p$ and $F$ has characteristic $p$ and contains all the roots of unity. I've trying to show that $E/F$ is not a solvable extension.

My main issue is that as $E/F$ is a prime degree extension then $Gal(E/F) = C_p$ and is therefore solvable? I must be missing something obvious or I have an incorrect definition of solvable.

Failing that, to show $E/F$ is not solvable I'd like to find the corresponding Galois field extensions of $F$ that are contained in $E$ (to find the normal subgroups of $Gal(E/F)$) but again as $p$ is a prime there are none?


  • $\begingroup$ Is it clear that $E/F$ is separable? $\endgroup$ – Hagen von Eitzen Jan 12 '15 at 16:50
  • $\begingroup$ Are you sure it doesn't say "separable" extension? $\endgroup$ – Timbuc Jan 12 '15 at 16:51
  • $\begingroup$ No it says $E/F$ is Galois $\endgroup$ – Wooster Jan 12 '15 at 16:58
  • $\begingroup$ A degree $p$ Galois extension in characteristic $p$ is always of the Artin-Schreier type. In other words $E=F(\alpha)$, where the minimal polynomial $m(x)$ of $\alpha$ over $F$ is of the form $$m(x)=x^p-x+a$$ for some $a\in F$. The Galois group is generated by the automorphism $\sigma$ determined by $\sigma(\alpha)=\alpha+1$. For a proof of this fact see my answer here. This fact does not depend on the presence/absence of roots of unity. $\endgroup$ – Jyrki Lahtonen Jan 12 '15 at 17:38
  • $\begingroup$ But I don't know what is your definition of a solvable extension. As you observed yourself, the Galois group here is certainly solvable, being cyclic of order $p$. $\endgroup$ – Jyrki Lahtonen Jan 12 '15 at 17:41

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