This is part of a question from a book, I'll post the rest if anyone would like me to, but the bit I'm stuck with is:

Suppose $\deg(L/K) = p$, a prime not equal to $\operatorname{char}K$, and $L$ is algebraically closed. Suppose (if possible) that $p \gt 2$. Show that the cyclotomic polynomials $\Phi_p $ and $\Phi_{p^2}$ both split over $K$.

It eventually leads to a contradiction from assuming $p \gt 2$. But the splitting of $\Phi_{p^2}$ is the only bit I can't get: I can see that $\Phi_p$ splits ...

  • 2
    $\begingroup$ Are you reading a proof of the Artin--Schreier theorem? The premise that an algebraically closed field $L$ contains a proper subfield $K$ in which $[L:K]$ is finite forces $[L:K] = 2$ by the Artin--Schreier theorem, so your whole premise is far more restrictive than you suggest unless your intended goal is to prove $p = 2$ as part of the proof of the Artin--Schreier theorem. $\endgroup$ – KCd Apr 26 '15 at 21:29
  • $\begingroup$ It's an exercise from Garling's 'Course in Galois Theory', and it pretty much guides you to a proof of what I now know is called the Artin-Schreier theorem, thank you! if $deg(L/K) \lt$ infinity, then the Galois group contains a subgroup of order p for all p dividing that degree: and for p not equal to 2, this gives a contradiction. Then, there's a bit about how $deg(L/K) = 4$ is also impossible, and it $L/K$ is a splitting field for $x^2 + 1$ $\endgroup$ – Latimer Leviosa Apr 27 '15 at 7:50
  • $\begingroup$ The original proof by Artin and Schreier uses cyclotomic polynomials, but that can be avoided. See math.uconn.edu/~kconrad/blurbs/galoistheory/artinschreier.pdf. $\endgroup$ – KCd Apr 27 '15 at 12:22
  • $\begingroup$ Thanks, I'll have a look :) $\endgroup$ – Latimer Leviosa Apr 27 '15 at 14:30

The fact that $\Phi_p$ splits follows already from $\deg \Phi_p=p-1$, which is not a divisor of $p$ (for $p>2$). So $K$ contains all primitive $p$th roots of unity. In order to show that $\Phi_{p^2}$ splits it suffices to find just one primitive $p^2$th root of unity in $K$, for all other roots are powers of this.

Let $\xi\in L$ be a primitive $p^3$th root of unity and let $\zeta:=\xi ^p$. An element $\sigma$ of the Galois group must act by mapping $\xi\mapsto \xi^m$. As $\sigma^p=1$, we have $m^p\equiv 1\pmod {p^3}$. We conclude $m\equiv 1\pmod p$. Write $m=1+ap$. Then (using $p>2$) we have $m^p\equiv 1+ap^2\pmod{p^3}$ so that $a\equiv 0\pmod p$. Hence $m=1+bp^2$ for some $b$. Then $\sigma(\zeta)=\zeta^m=\zeta\cdot \zeta^{bp^2}=\zeta$ because $\zeta$ is a $p^2$th root of unity. Since $\zeta$ is fixed under the Galois group, we conclude $\zeta\in K$.

  • $\begingroup$ Thank you very much. Don't feel too ashamed at not getting it, if that's the simplest way.... :O $\endgroup$ – Latimer Leviosa Apr 27 '15 at 7:46

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