A lemma in my lecture notes states that if $K$ is a subfield of $\mathbb{C}$ and $\zeta=\exp(2\pi i/p)$ then $K(\zeta)/K$ is Galois. They proved it by arguing that the minimal polynomial of $\zeta$ divides $x^p-1$ which is a separable polynomial, and all its roots are powers of $\zeta$, hence $K(\zeta)/K$ is a separable normal extension, thus Galois.

I understand that to show that an extension is Galois it is sufficient to show that it is separable and normal, for which it suffices to show that the minimal polynomial of any element splits with distinct roots in the extension, but I don't see how we've proved that here. We've proven it for the particular element $\zeta$ but not for any other elements it seems. I'd appreciate it if someone could explain this.

  • $\begingroup$ It is a classical fact that a field extension generated by separable elements is separable. You should find that result easily on the internet or in the classical book "Algebra" by S. Lang $\endgroup$ – AlexL Apr 4 at 15:07
  • $\begingroup$ See here , theorem 3.5 and its associated proof, for instance : kconrad.math.uconn.edu/blurbs/galoistheory/separable1.pdf which is exactly the statement you want. $\endgroup$ – астон вілла олоф мэллбэрг Apr 4 at 15:48
  • $\begingroup$ Oh I see, so it is a nontrivial result that is rather strangely just used implicitly with no justification. Thanks for the reference though. $\endgroup$ – AlephNull Apr 4 at 15:52

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