Class Field Theory 1

Today we’ll to get to stating class field theory in terms of ray class groups. Let {L/K} be an extension of number fields. Recall that if {G=Gal(L/K)} is abelian and {\frak{p}\subset \mathcal{O}_K} is a prime unramified in {\mathcal{O}_L}, then the Frobenius element, {\Phi(\frak{q}/\frak{p})=Frob_{L/K}(\frak{q})}, is really an element and is independent of the {\frak{q}} over {\frak{p}}. The independence is just because we are assuming an abelian Galois group, and all the Frobenii are conjugate.

Let {S} be the set of places that ramify in {L}. Recall that {I_S(K)} are the fractional ideals relatively prime to {S}. Thus we can define a group homomorphism called the Artin map {\Psi_{L/K}: I_S(K)\rightarrow G} by sending a prime to its Frobenius element and then extending.

To have something concrete in your head take the example of {L=\mathbb{Q}(\sqrt{3})/\mathbb{Q}}. We have that {\mathcal{O}_L=\mathbb{Z}[\sqrt{3}]}, and the discriminant is {12}. Take a prime ideal downstairs, {p\mathbb{Z}}. There are three possibilities, it ramifies so {p\mathcal{O}_L=Q^2}, it splits so {p\mathcal{O}_L=Q_1Q_2}, or it is inert in which case {p\mathcal{O}_L=Q}.

From the discriminant we know the only primes that ramify are {p=2, 3}. To determine whether the other primes split or are inert depends on whether or not {x^2-3} has a root mod {p} by Hensel’s lemma. It turns out {p} splits if {3} is a square mod {p} and is inert if {3} is not a square mod {p}. Thus we can read off these two cases from the Legendre symbol.

In the case that it splits, {p\mathcal{O}_L=Q_1Q_2}. The decomposition group {D_{Q_1}=\{id\}} because the Galois group acts transitively on {Q_1} and {Q_2}, but there is only one non-trivial element which must switch the two primes. Thus only the identity fixes {Q_1} which is the definition {D_{Q_1}}. So we get {\Psi_{L/\mathbb{Q}}(p)=1} when {p} splits (i.e. 3 a square mod {p}). Likewise, in the inert case we get {D_{Q}=G}. Thus the Frobenius must be the non-trivial element of {G} since it generates {G}. Thus {\Psi_{L/\mathbb{Q}}(p)=\sigma} (the generator, i.e. raising to the {p} 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 {\frak{m}=(\frak{m}_f, w_1, \ldots, w_i)} depending on {L/K} such that:

a) {\text{Supp}(\frak{m})=S} where the support is the set of places that divide {\frak{m}_f} together with real embeddings {K\hookrightarrow \mathbb{R}}.

b) The Artin map {\Psi_{L/K}: I_S(K)\rightarrow G} factors through the ray class group, i.e. {\Psi_{L/K}: I_S(K)\rightarrow Cl_{\frak{m}}(K)\rightarrow G}.

c) {\Psi_{L/K}} is surjective.

There is a smallest such {\frak{m}} (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 {\frak{m}} and subgroup {H} of {Cl_\frak{m}(K)}, there is an abelian extension {E/K} such that {\frak{m}_{E/K}} (the conductor coming from the earlier statement) divides {\frak{m}} and the composition {Cl_\frak{m}(K)\rightarrow Cl_{\frak{m}_{E/K}}(K)\stackrel{\Psi}{\rightarrow} Gal(E/K)} is surjective and has kernel {H}.

In particular, you can take {H=\{e\}} and force the maps to be isomorphisms. In that case {E} is called the ray class field of {K} of conductor {\frak{m}_{E/K}}. One caution is that even in this nice situation you can’t force {\frak{m}_{E/K}=\frak{m}}. For example, take {\frak{m}} to be {\mathcal{O}_K} 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 {Cl(\mathcal{O}_K)\stackrel{\sim}{\rightarrow} Gal(E/K)} via the Artin map for the Hilbert class field. Now take {\frak{m}=\{\mathcal{O}_K, w_1\}}. Just throw in an embedding. The ideal class group {Cl(\mathcal{O}_K)} is the group of fractional ideals mod the principal ideals. This {Cl_\frak{m}(K)} is the group of fractional ideals mod principal ideals generated by something that is positive under {w_1}. These are isomorphic because you can always multiply the generator by {-1}. Thus starting with this {\frak{m}} and {H=\{e\}} we get the Hilbert class field again, but {\frak{m}_{E/K}=\{\mathcal{O}_K\}} without the embedding and hence in general we can’t force {\frak{m}=\frak{m}_{E/K}} 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 {K}, say {L_0}. The proof that it is unramified is just that {\text{Supp}(\mathcal{O}_K)=\emptyset}. Let {L/K} be any unramified abelian extension. Suppose {L} is not contained in {L_0}. Then {LL_0} is an unramified extension strictly containing {L_0}. Thus by the main theorem of class field theory {LL_0} has a conductor dividing {\frak{m}} and hence equal to {\mathcal{O}_K}. Thus {Cl(\mathcal{O}_K)\simeq Gal(L/K)\rightarrow Gal(LL_0/K)} 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 {\mathcal{O}_K} and get the standard definition that it is the maximal unramified abelian extension of {K}.

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