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I was reading up on class field theory, and I have a question. On wiki (, one formulation is that there's some modulus for which $I^c_K/i(K_{c,1})Nm_{L/K}(I^c_L)$ is isomorphic to $Gal(L/K)$.

Another formulation on the same page is: $C_K/N_{L/K}(C_L)$ is isomorphic to $Gal(L/K)$. (where $C_{number\,field}$ is the idele class group of that number field).

How does one relate these two formulations? Is it true that for some modulus, $c$, $I^c_K$ is $C_K$? I don't really see how this fits into one picture.

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up vote 15 down vote accepted

If you read the initial section or two of the chapter by Tate in Cassels and Frolich, he gives a nice explanation of how to pass from the classical formulation in terms of generalized ideal class groups w.r.t a modulus, and the more modern formulation in terms of idele class groups. As Tate explains, the two formulations are indeed equivalent, but it is not quite as simple as saying that $I^c_K = C_K$. Here is a sketch of the equivalence (of course it is the same is in Akhil Mathew's answer, just slightly more detailed):

Since $I_K^c$ has already been taken to denote the ideals prime to $c$ (at least, this is how I interpret your notation), let me use $J_K$ to denote the ideles for $K$. Then we can consider the subgroup $J_K^c$ of the ideles whose entries are all $1$ at any finite place dividing $c$, and at any infinite place.

Then there is a natural surjection $J_K^c \to I_K^c$ given by sending any element $(a_{\wp})$ of the former to the ideal $\prod_{\wp} \wp^{v_{\wp}(a_{\wp})}$ of the latter (where the product is over finite places, i.e. prime ideals, $\wp$).

Now one can show that $K^{\times} J_K^c$ is dense in $J_K$, so the image of $J_K^c$ is dense in $C_K$. Since $N_{L/K}(C_L)$ is open in $C_K$, we see that $J_K^c$ surjects onto $C_K/N_{L/K}(C_L)$. Now one checks that this map factors through the surjection $J_K^c \to I_K^c$ described in the preceding paragraph, and in fact induces an isomorphism $I_K^c/i(K_{c,1}) N_{L/K}(I_L^c) \buildrel \sim \over \longrightarrow C_K/N_{L/K}(C_L)$, as required.

In practice, suppose you want to compute the Artin map on an element of $J_K$: the algorithm is you first multiply by a principal idele so that the resulting element is in $J_K^c$ times $N_{L/K}(C_L)$. (You may not know exactly what this group is, but its not hard to at least identify an open subgroup of it: for example, at any complex infinite place $v$ the norm map is surjective, at any real place $v$ the image of the norm map at least contains the positive reals, and at any finite place $\wp$ the image of the norm map will contain elements which are congruent to $1$ modulo the power of $\wp$ dividing the relevant modulus $c$.) Now the Artin map on $J_K^c$ factors through the surjection $J_K^c \to I_K^c$, and is computed on the target using Frobenius elements.

Indeed, this was the argument via which local class field theory was originally proved; one took a local extension, embedded it into a global context (so that the original local situation was realized as $L_{\wp}/K_{\wp}$ for some abelian extension of number fields $L/K$), and then defined the Artin map via the above computation (which means concretely that one passes from the possibly ramified situation at $\wp$ to a consideration just at the unramified primes, where everything is easily understood just in terms of ideals and Frobenius elements). Of course, one then had to check that the resulting local Artin map was well-defined independent of the choice of "global context".

Nowadays, one can define the local Artin maps at all places (unramified or ramified) first. However, in generalizations to the non-abelian situation (i.e. local and global Langlands) one generally uses the old-fashioned technique of proving certain global results first, and then establishing the precise local results by passing to a well-chosen global context and reducing to a calculation at unramified primes. (This is a bit of an oversimplification, but I think it is correct in spirit.) So (if one has an eventual aim of understanding modern algebraic number theory and the Langlands program) it is well worth understanding the passage between the idelic and ideal-theoretic view-points on class field theory, and practicing how to use the algorithm described above.

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The generalized ideal and (normal) idele class groups (by the generalized ideal class group I really mean ideals prime to $\mathfrak{c}$ for $\mathfrak{c}$ an "admissible cycle," which more or less means divisible by enough primes such that anything congruent to 1 mod it is a local norm) are isomorphic. The short answer for why this is is that given an element in the idele class group, you can get an element of the (generalized) ideal class group since an idele always gives you a fractional ideal. By choosing a suitable representative, you can assume that the ideal you get is actually prime to $\mathfrak{c}$. It is then necessary (and nontrivial) to prove that the corresponding norm groups are mapped to each other, but this is done in Lang's "Algebraic number theory."

There are (at least) two approaches to class field theory:

  1. Define the Artin map globally, on the generalized ideal class group. This has the benefit that you only have to define it on ideals, so just on primes which are unramified (in which case you just use the Frobenius). In this case, it is a lot of work to show that it actually factors not just through the norm group, but also through a group of principal ideals.
  2. Define the Artin map on ideles by piecing together the local Artin maps. This then requires some construction of the local Artin maps (which can be done using more-or-less pure group cohomology theory and a little bit of analysis of local fields). One has to check again that it is trivial on principal ideles, which is very nontrivial -- since, again the Artin map was defined by piecing together local things.

So, in a sense, 1 is the one which gives you the ideal form of Artin reciprocity, and 2 gives you the idele one.

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Sorry for being slow, but do you think of the modulus statement as being somehow a statement about the ideal class group? – Nicole Dec 20 '10 at 2:43
@Nicole: If I understand correctly, the modulus statement is a statement about a generalized version of the ideal class group (where the ideals are required to be prime to the modulus), while the other statement is purely idelic (and independent of any modulus). @Arturo: No, I was referring to the ideal class group. – Akhil Mathew Dec 20 '10 at 2:48
I think you meant Ray Class Group. Surely you didn't mean: "the ideal class group is a generalized ideal class group". – Nicole Dec 20 '10 at 2:51
Also, thanks. I think I understand a little bit better now, but I have to keep thinking about it... – Nicole Dec 20 '10 at 2:51
@Nicole: My apologies. I was not being precise in my language, and I meant the parenthetical remark to be merely a clarification that the "ideal class group" I initially mentioned should actually have been called the "generalized ideal class group." – Akhil Mathew Dec 20 '10 at 3:01

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