# 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 ${K}$.