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Let $L/K$ be a Galois extension of number fields with Galois group $G$. Let $O_K$ and $O_L$ be the ring of algebraic integers of $K$ and $L$ respectively. Let $P\subseteq O_K$ be a prime. Let $Q\subseteq O_L$ be a prime lying over $P$.

The decomposition group is defined as $$D(Q|P)=\lbrace \sigma\in G\text{ }|\text{ }\sigma(Q)=Q\rbrace$$

The $n$-th ramification group is defined as $$E_n(Q|P)=\lbrace \sigma\in G:\sigma(a)\equiv a\text{ mod } Q^{n+1}\text{ for all } a\in O_L\rbrace$$

I want to compute the decomposition group and ramification groups of the cyclotomic field $\mathbb{Q}(\zeta)$ over $\mathbb{Q}$ where $\zeta$ is a primitive $p$-th root of unity ($p$ is a prime).

In general, how to calculate it for an arbitrary cyclotomic field ?

learning_math
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    Do you want to compute the decomposition group and ramification groups of a prime ideal of the cyclotomic field $\Bbb Q(\zeta_p)$? For instance a prime ideal over the integer prime $p$? You question isn't clear otherwise (in my opinion). – Watson Nov 07 '16 at 17:05
  • Yes, that is exactly what I want. – learning_math Nov 07 '16 at 17:16
  • Since $p$ is totally ramified in $\Bbb Q(\zeta_p)$, it follows that $D(P|p) = \mathrm{Gal}(K/\Bbb Q) =: G$ where $P$ is a prime ideal of $\mathcal O_K$ above $p$ and $K=\Bbb Q(\zeta_p)$, since the order of $D(P)$ is $e_{P \mid p} \cdot f_{P \mid p}$. – Watson Nov 07 '16 at 17:18
  • What about the ramification groups ? – learning_math Nov 07 '16 at 17:20
  • You know that $\sigma(\zeta_p) = \zeta_p^k$ for some $k$ coprime with $p$. Then try to see when $\zeta_p^k -\zeta_p \in (1-\zeta_p)^{n+1}$. – Watson Nov 07 '16 at 17:26
  • Related: https://math.stackexchange.com/questions/279264/ – Watson Jan 02 '17 at 23:12
  • See also http://wstein.org/wiki/attachments/ant07(2f)projects/ooka.pdf. – Watson Jan 04 '17 at 13:15
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    See Sublemma 1 in An Elementary Proof of the Kronecker-Weber Theorem by M. J. Greenberg : if $K/\Bbb Q$ is abelian of degree $p$ (odd prime number), such that $p$ is the only ramified prime, then the second ramification group $E_2(P/p)$ is trivial (for any $P$ over $p$). – Watson Jan 04 '17 at 14:12
  • Moreover, we can localize at $P$, so that we can assume that $P$ is principal (then you just have to check $\sigma(a) \equiv a \pmod{Q^{n+1}}$ for the uniformizer $a$ (and maybe $\sigma \in E_{n-1}$), to conclude that $\sigma \in E_n$). – Watson Jan 12 '17 at 10:01

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