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Let $\mu \in \mathbb{Q}(\zeta_n)$ lie above the rational prime $p$, and let the prime ideal $\mathscr{P}\subset \mathbb{Z}[\zeta_n]$ have ramification index $a$ over $\mu$.

Why is it then true that $\mathscr{P}$ has ramification index $\frac{n}{(a,n)}$ over $p$?

In "Reciprocity Laws: from Euler to Eisenstein'', it's stated that this is from the decomposition law in Kummer extensions (as $\mathbb{Q}(\zeta_n, \mu^{1/n})/\mathbb{Q}(\zeta_n)$ is Kummer) but I cannot find any reference about this. Does anyone know of one?

I assume it's irrelevant, but just in case: $\mu$ here is the $n$th power of a Gauss sum of a residue character.

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    $\begingroup$ What does it mean for an element of $\mathbb{Q}(\zeta_n)$ to lie above $p$? Do you mean it's an element of $p\mathbb{Z}[\zeta_n]$? Then what does it mean for $\mathcal{P}$ to have ramification index $a$ over an element $\mu$? What if $(\mu)$ isn't prime? Do you mean that all the ramification indices over the prime ideals dividing $(\mu)$ are the same and are equal to $a$? $\endgroup$ – Will Chen Oct 10 '15 at 18:22
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    $\begingroup$ And what is $m$? is $m = \mu$? If $m$ is not an integer, what does $(a,m)$ mean? $\endgroup$ – Will Chen Oct 10 '15 at 18:23
  • $\begingroup$ @oxeimon The latter was a typo, and $\mathscr{P}$ has ramification index $a$ over element $r \in \mathbb{Z}(\zeta_n)$ if it has exponent $a$ in the prime ideal factorisation of $(r)$. $\endgroup$ – Meow Oct 10 '15 at 20:22
  • $\begingroup$ @Alyosha, in the first line shouldn't $\mathscr P$ lie in the integers of $\mathbf Q(\zeta_n,\sqrt[n]{\mu})$, not $\mathbf Q(\zeta_n)$? $\endgroup$ – KConrad Oct 11 '15 at 14:26
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This follows from basic properties of Kummer extensions and the Eisenstein polynomial of your prime ideal $\mathfrak{p}$.

You can find it for example in Koch's book on Algebraic Number Theory.

If you still have problems with it, it might make for a good math.stackexchange question.

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