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I want to study the idelic Artin map for the, say, 7th cyclotomic field explicitly. In other words I want to see how a particular idele can be constructed which gets sent to the identity element in Galois group,which means, in the idealic language, p=29 splits completely in the cyclotomic extension mapping p=29 to identity Frobenius element. I am not clear how to phrase this in the idelic language.

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If I understand you correctly, what you want to see is how the idelic Artin map evaluated at the idele corresponding to 29 (whatever this is) gives you the Frobenius $\left(\frac{29}{\mathbb Q(\zeta_7)/\mathbb Q}\right) = 1$.

In general, to compute the idelic Artin map $\phi_{L/K}$ of an abelian extension $L/K$ of number fields at an idele ${\bf a} = (a_v)_v \in J_K$ one can proceed as follows. First, find an $\alpha \in K^\times$ such that $\alpha {\bf a} \in J_{K,\mathfrak m}^+$, where $\mathfrak m$ is an admissible modulus for $L/K$. This means that $\alpha a_v >0$ for every real place $v$ and $\alpha a_v \equiv 1 \pmod{\mathfrak p_v^{\text{ord}_v\mathfrak m}}$ for every finite place $v$. Then $$ \phi_{L/K}({\bf a}) = \left(\frac{\prod_{v<\infty} \mathfrak p_v^{\text{ord}_v(\alpha a_v)}}{L/K}\right) \in \text{Gal}(L/K). $$

Now if a prime $\mathfrak p$ of $K$ is unramified in $L$ (so $\mathfrak p\nmid \mathfrak m$), the idelic way of studying its splitting type is by considering the associated "prime idele" ${\bf n}_{\mathfrak p}(\pi) = (1,\ldots,1,\pi,1,\ldots) \in J_K$ where $\pi$ is a prime element of $K_\mathfrak p$: specifically, the order of $\phi_{L/K}({\bf n}_{\mathfrak p}(\pi))$ is the residue degree of $\mathfrak p$. This is immediate from Artin reciprocity and the above description of $\phi_{L/K}$.

For example, if $K=\mathbb Q$, $L=\mathbb Q(\zeta_7)$, $\mathfrak m = 7\mathbb Z$, and $p\neq 7$, then the above recipe (with $\alpha=1$) gives $\phi(n_p(p)) = \left(\frac{p}{\mathbb Q(\zeta_7)/\mathbb Q}\right)$, which is the map $\zeta_7 \mapsto \zeta_7^p$ in $\text{Gal}(L/K)$, as expected.

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  • $\begingroup$ In your example, how do you know that $\mathfrak m = 7$ is admissible? Don't you need $1 + 7 \mathbb{Z}_7$ to be contained in the group of local norms from $\mathbb{Q}_7(\zeta_7)$? I'm not sure if that's necessarily true. $\endgroup$
    – D_S
    Jul 22, 2015 at 14:25

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