It is known that the splitting field of $x^p-a$ over $\mathbb{Q}$ has no $p^2$ roots of unity. We can assume $a\in \mathbb{Q}$ is not a pth power in $\mathbb{Q}$. I came up with the following proof of this statement, which I believe is correct:

The splitting field of $x^p-a$ over $\mathbb{Q}$ is $\mathbb{Q}(\zeta_p, a^{1/p})$. If $\zeta_{p^2} \in \mathbb{Q}(\zeta_p, a^{1/p})$, then by degree considerations $\mathbb{Q}(\zeta_p, a^{1/p}) = \mathbb{Q}(\zeta_{p^2})$. Then $\mathbb{Q}(\zeta_{p^2})/\mathbb{Q}(\zeta_{p})$ and $\mathbb{Q}(\zeta_{p}, a^{1/p})/\mathbb{Q}(\zeta_{p})$ are equal field extensions and so by Kummer theory, exists $\alpha \in \mathbb{Q}(\zeta_{p})$ and $k \in \mathbb{Z}, (k,p)=1,$ such that $a^{1/p} = \zeta_{p^2}^k \alpha^p$. That is, $x^p- \frac{a}{\zeta_p^k}$ has a root $\alpha$ in $\mathbb{Q}(\zeta_{p})$. Now we consider the norm of $\alpha$ in the extension $\mathbb{Q}(\zeta_p)/\mathbb{Q}$: $N(\alpha)^{p^2} = N(\alpha^{p^2})= N(a^p) = a^{p(p-1)}$ since $|\mathbb{Q}(\zeta_p):\mathbb{Q}| = p-1$. Then $N(\alpha) = a^{(p-1)/p} \in \mathbb{Q}$ and so $a^{1/p} \in \mathbb{Q}$, which is a contradiction.

Does someone know of a simpler proof of this? My proof seems too complicated. In particular, is there a proof that avoids Kummer theory? Ideally, I would like a proof that uses only norms...

Thank you.

  • $\begingroup$ The splitting field of $x^2-(-1)$ has roots of unity of order $2^2$. $\endgroup$ – Gerry Myerson Mar 8 '13 at 4:39
  • $\begingroup$ I guess I mean for $p\ne 2$ since the 2nd roots of unity are rationals... $\endgroup$ – sam Mar 8 '13 at 4:54
  • $\begingroup$ @GerryMyerson Why is it so that the splitting field of x^p-a over $\mathbb Q$ is $\mathbb{Q}(\zeta_p, a^{1/p})$ $\endgroup$ – P-S.D Jan 7 '18 at 10:09
  • $\begingroup$ Because all the zeros of that polynomial are in that field, and any other field containing all the zeros of that polynomial must contain that field. $\endgroup$ – Gerry Myerson Jan 7 '18 at 15:43
  • $\begingroup$ Oops, forgot to ping with @P-S.D $\endgroup$ – Gerry Myerson Jan 8 '18 at 22:29

Hint: For $p>2$, $\text{Gal}(\mathbb{Q}(\zeta_p,a^{\frac{1}{p}})/\mathbb{Q})$ is not abelian. The easy way to see this is note that $\mathbb{Q}(a^{\frac{1}{p}})/\mathbb{Q}$ isn't Galois (consider it' automorphism group).

  • $\begingroup$ Thank you! My mistake was thinking that $Gal(\mathbb{Q}(\zeta_p,a^{1/p}))$ is cyclic and that $Gal(\mathbb{Q}(\zeta_p,a^{1/p}))$ and $Gal(\mathbb{Q}(\zeta_{p^2})$ are isomorphic.. $\endgroup$ – sam Mar 8 '13 at 5:32
  • 1
    $\begingroup$ @sam Yeah, that's an easy mistake to make. One is tempted to use the fact that $\def\Q{\mathbb{Q}}$ $\Q(\zeta_p)\cap\Q(a^{\frac{1}{p}})=\Q$, to try and conclude that $\text{Gal}(\Q(\zeta_p,a^{\frac{1}{p}})/\Q)=\text{Gal}(\Q(\zeta_p),\Q)\times \text{Gal}(\Q(a^{\frac{1}{p}})/\Q)$. But, of course this only works if both the extensions $\Q(\zeta_p)/\Q$ and $\Q(a^{\frac{1}{p}})/\Q$ are Galois! $\endgroup$ – Alex Youcis Mar 8 '13 at 5:34

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