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The ideles of $\mathbb{Q}$, lets denote them by $\mathbb{I}$, satisfy the following decomposition:

$\mathbb{Q}^\times\times \mathbb{R}_{>0}\times\prod_p \mathbb{Z}_p^\times \to \mathbb{I}$

The map is given by $(r,t,(u_p))\mapsto (rt,ru_2,ru_3,ru_5,\ldots)$

and it's an isomorphism of topological groups if $\mathbb{Q}^\times$ is given the discrete topology. Now let $K/\mathbb{Q}$ be finite abelian, so that we have the composition of maps given by the reciprocity theorem:

$\mathbb{Q}^\times\times \mathbb{R}_{>0}\times\prod_p \mathbb{Z}_p^\times \to \mathbb{I}\to \textrm{Gal}(K/\mathbb{Q})$

My questions are the following:

  1. How can we identify the kernel of this map on the left side? My guess based on some trivial examples is that if $p$ is unramified, then $\mathbb{Z}_p^\times$ is contained in the kernel. Is this true? If so could anyone show how its done? If $p$ ramfies, can we say anything?

  2. If $K/\mathbb{Q}$ is cyclotomic, then is there a nicer description of the kernel? This would be nice to know, since it would help with identifying a cyclotomic extension that contains a general abelian extension.

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  • $\begingroup$ I can't restrain myself from noting that posing the question in this fashion makes needless use of the class number and units simplicity for the rationals, thus, the possibility of choosing "sections" of natural quotient maps. If, instead, one forces oneself to take a somewhat more "natural" viewpoint that avoids taking advantage of any such coincidences, one is led to more robust versions of such questions. But, yes, as a transitional stage, cyclotomic extensions of the rationals did historically provide motivations, and are (by various such accidents) simpler than the general case. $\endgroup$ – paul garrett Jan 31 '17 at 23:38
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You can find the answers in Sections XIII-9 and XIII-11 of Weil: Basic Number Theory. See in particular Theorem 7 on Page 275, and the last paragraph on Page 287.

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    $\begingroup$ Is Weil's book readable ? :) $\endgroup$ – Alexander Chervov Jan 16 '12 at 18:56
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    $\begingroup$ Well, I learned class field theory from it, but it took me almost a year to cover it. One full day was enough to read 8 pages, if I recall. Later I read Cassels-Fröhlich, but I still tend to look up things in Weil. $\endgroup$ – GH from MO Jan 16 '12 at 19:30
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    $\begingroup$ Once I've started to read Serre's Lie groups and algebras the speed was the same 1 page per hour. I was reading it for several weeks and stopped. Actually strange thing happens - several years later it seems to me that Serre's book is more or less obvious... I more or less know all ideas... Despite I have not read any book, just visited some seminars, discussed with people, thought by myself. The same happened with many other books. My conclusion is that reading books is not the good way to learn (at least for me)... My uncle said books are of two sorts: "good" and "thick" :) $\endgroup$ – Alexander Chervov Jan 16 '12 at 19:40
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    $\begingroup$ @AlexanderChervov, I tend to agree that many "classics" are much less useful for learning than as diagnostics, so that if one can understand what's going on, then one is certified as knowing something about the subject. But one does not (perhaps cannot) reach that state by directly studying that source. Many of the "classics" concerning more esoteric general notions about automorphic forms certainly fall into this class. $\endgroup$ – paul garrett Jan 31 '17 at 23:33
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Class field theory gives a surjective map

$\mathbb{Q}^\times \times \mathbb{R}_{>0}\times$ $\prod_p \mathbb{Z}_p^\times$ $\rightarrow \text{Gal}(\mathbb{Q}^{ab}/\mathbb{Q})$.

The kernel of this map is $\mathbb{Q}^\times \times \mathbb{R}_{>0}\times \{1\}$. The isomorphism

$\prod_p \mathbb{Z}_p^\times\cong\text{Gal}(\mathbb{Q}^{ab}/\mathbb{Q})$

comes from the identification

$\text{Gal}(\mathbb{Q}^{ab}/\mathbb{Q})=\text{Gal}(\mathbb{Q}(\mu_\infty)/\mathbb{Q})\cong\prod_p\text{Gal}(\mathbb{Q}(\mu_{p^\infty})/\mathbb{Q})\cong \prod_p \mathbb{Z}_p^\times$.

The image of inertia at $p$ is the embedded $\mathbb{Z}_p^\times$, since $p$ is totally ramified in the $p$-cyclotomic extension and unramified in the others. Hence your first guess is correct. Most of your other questions can also be approached by this general description.

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