2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.