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I was going through the proof of Kummer's Lemma (stated below) as done in Serge Langs Cyclotomic fields on page 312.

Now the author states that by class field theory it suffices to show that $\mathbf{Q}(\xi_p, u^{1 / p})$ for a unit $u$ and a primitive $p$-th root of unity $\xi_p$ is unramified over $\mathbf{Q}(\xi_p)$. I do not see how this happens (or: this seems like magic to me) - could anyone explain this in a bit more detail?

Thanks for any help!

Kummer's Lemma: Let $p$ be a regular prime (i.e. an odd prime which does not divide the class number of the $p$-th cyclotomic field) and $\xi$ a primitive $p$-th root of unity; if a unit $u \in \mathbf{Q}(\xi)$ is congruent to an integer modulo $p$, then $u$ is a $p$-th power in $\mathbf{Q}(\xi)$.

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Since $K = \mathbf{Q}(\zeta_p)$ contains the $p$th roots of unity, $K(u^{1/p})$ is the splitting field of $x^p - u \in K[x]$. But $x^p - u$ is either irreducible or splits into linear factors depending on whether $u^{1/p} \in K$, hence $K(u^{1/p})/K$ is Galois of degree dividing $p$. Thus the extension is necessarily abelian, so if it's unramified then it sits inside the Hilbert class field of $K$. But, by class field theory, the Hilbert class field of $K$ has degree over $K$ equal to $\# \mathrm{Cl}(K)$. So if $p$ doesn't divide $\# \mathrm{Cl}(K)$ then this would force $K(u^{1/p}) = K$.

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  • $\begingroup$ Where does the hypothesis that u is congruent to an integer mod p intervene ? $\endgroup$ – nguyen quang do Mar 30 '16 at 6:01
  • $\begingroup$ Where does the hypothesis that u is congruent to an integer mod p intervene ? Note that thm.5.36 in Washington's "Introduction to Cyclotomic Fields" is given a more detailed proof. Two proofs actually, one using class field theory, the other the p-adic regulator. Note also that Kummer's lemma is nowadays considered as an embryo of the celebrated Leopldt conjecture. $\endgroup$ – nguyen quang do Mar 30 '16 at 6:14
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    $\begingroup$ @nguyenquangdo The question asked why showing $\mathbf{Q}(\zeta_p, u^{1/p})/\mathbf{Q}(\zeta_p)$ is unramified is enough. You need the condition that $u$ is congruent to an integer mod $p$ to actually show that it's unramified. $\endgroup$ – Brandon Carter Mar 30 '16 at 11:54
  • $\begingroup$ Actually, the condition that $u^{p-1}$ is congruent to 1 mod (p-1) is enough. See coroll. 2.3 in my paper "Sur l'arithmétique des corps de nombres p-rationnels ", Sém. Théorie des Nombres, Paris 1987-88, pp. 155-200, in Progress in Math., vol.81, Birkhaäuser 1990, which also contains a vast generalization of Kummer's lemma . $\endgroup$ – nguyen quang do Apr 7 '16 at 14:44

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