Today we’ll to get to stating class field theory in terms of ray class groups. Let be an extension of number fields. Recall that if is abelian and is a prime unramified in , then the Frobenius element, , is really an element and is independent of the over . The independence is just because we are assuming an abelian Galois group, and all the Frobenii are conjugate.
Let be the set of places that ramify in . Recall that are the fractional ideals relatively prime to . Thus we can define a group homomorphism called the Artin map by sending a prime to its Frobenius element and then extending.
To have something concrete in your head take the example of . We have that , and the discriminant is . Take a prime ideal downstairs, . There are three possibilities, it ramifies so , it splits so , or it is inert in which case .
From the discriminant we know the only primes that ramify are . To determine whether the other primes split or are inert depends on whether or not has a root mod by Hensel’s lemma. It turns out splits if is a square mod and is inert if is not a square mod . Thus we can read off these two cases from the Legendre symbol.
In the case that it splits, . The decomposition group because the Galois group acts transitively on and , but there is only one non-trivial element which must switch the two primes. Thus only the identity fixes which is the definition . So we get when splits (i.e. 3 a square mod ). Likewise, in the inert case we get . Thus the Frobenius must be the non-trivial element of since it generates . Thus (the generator, i.e. raising to the power). This should make us think that quadratic reciprocity will turn out to be a special case of where we’re going.
Now let’s state the main theorem of class field theory (in our special case above). Artin actually proved this version of it. There is a conductor depending on such that:
a) where the support is the set of places that divide together with real embeddings .
b) The Artin map factors through the ray class group, i.e. .
c) is surjective.
There is a smallest such (under the partial order by dividing on the ideals and inclusion on the embeddings) called the conductor of the extension. There is an existence part of the theorem that gives a Galois correspondence. It says that for each conductor and subgroup of , there is an abelian extension such that (the conductor coming from the earlier statement) divides and the composition is surjective and has kernel .
In particular, you can take and force the maps to be isomorphisms. In that case is called the ray class field of of conductor . One caution is that even in this nice situation you can’t force . For example, take to be with no embeddings. The ray class field in this situation is called the Hilbert class field and the ray class group is just the ideal class group.
Thus we get via the Artin map for the Hilbert class field. Now take . Just throw in an embedding. The ideal class group is the group of fractional ideals mod the principal ideals. This is the group of fractional ideals mod principal ideals generated by something that is positive under . These are isomorphic because you can always multiply the generator by . Thus starting with this and we get the Hilbert class field again, but without the embedding and hence in general we can’t force in general.
To end today let’s just say a few more things about the Hilbert class field. It is the maximal abelian unramified extension of , say . The proof that it is unramified is just that . Let be any unramified abelian extension. Suppose is not contained in . Then is an unramified extension strictly containing . Thus by the main theorem of class field theory has a conductor dividing and hence equal to . Thus is surjective, a contradiction. This tells us we can take as our definition of the Hilbert class field to be the ray class field of conductor and get the standard definition that it is the maximal unramified abelian extension of .