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\begin{document}

\begin{center}
\bf
Math 254B, UC Berkeley, Spring 2002 (Kedlaya) \\
The Adelic Reciprocity Law and Artin Reciprocity
\end{center}

We now describe the setup by which we create a reciprocity law in
global class field theory, imitating the ``abstract'' setup from local
class field theory but using the idele class group in place of the
multiplicative group of the field. We then work out why the reciprocity
law and existence theorem in the adelic setup imply Artin reciprocity
and the existence theorem (and a bit more) in the classical language.

\head{Convention note} We are going to fix an algebraic closure
$\overline{\QQ}$ of $\QQ$, and regard ``number fields'' as finite
subextensions of $\overline{\QQ}/\QQ$. That is, we are fixing the
embeddings of number fields into $\overline{\QQ}$. We'll use these embeddings
to decide how to embed one number field in another.

\head{The adelic reciprocity law and existence theorem}

Here are the adelic reciprocity law and existence theorem; notice that
they look just like the local case except that the multiplicative group
has been replaced by the idele class group.
\begin{theorem}[Adelic reciprocity law]
There is a canonical map $r_K: C_K \to \Gal(K^{\ab}/K)$ which
induces, for each finite extension $L/K$ of number fields, an
isomorphism $r_{L/K}: C_K/\Norm_{L/K} C_L \to \Gal(L/K)^{\ab}$.
\end{theorem}
\begin{theorem}[Adelic existence theorem]
For every number field $K$ and 
every open subgroup $H$ of $C_K$ of finite index, there exists
a finite (abelian) extension $L$ of $K$ such that
$H = \Norm_{L/K} C_L$.
\end{theorem}


In fact, using local class field theory, we can construct the map that
will end up being $r_K$. For starters, let $L/K$ be a finite abelian
extension and $v$ a place of $K$. Put $G = \Gal(L/K)$, and let $G_v$ be
the decomposition group of $v$, that is, the set of $g \in G$
such that $v^g = v$. Then for any place $w$ above $v$,
$G_v \cong \Gal(L_w/K_v)$, so we can view the local reciprocity map
$K_v^* \to \Gal(L_w/K_v)$ as a map $r_{K,v}: K_v^* \to G$. That is, if $v$
is a finite place. If $v = \CC$, then $\Gal(L_w/K_v)$ is trivial, so
we just take $K_v^* \to G$ to be the identity map. If $v = \RR$, then we
take $K_v^* = \RR^* \to \Gal(L_w/K_v) = \Gal(\CC/\RR)$ to be the map
sending everything positive to the identity, and everything negative
to complex conjugation.

Now note that
\[
(\alpha_v) \mapsto \prod_v r_{K,v}(\alpha_v) 
\]
is well-defined on ideles: for $(\alpha_v)$ an idele, $\alpha_v$ is
a unit for almost all $v$ and $L_w/K_v$ is unramified for almost all
$v$. For the (almost all) $v$ in both categories, $r_{K,v}$ maps
$\alpha_v$ to the identity.

The subtle part is the following. As noted below, before proving reciprocity,
we'll only be able to check this for the map obtained from $r_{K,v}$ by
projecting from $\Gal(K^{\ab}/K)$ to $\Gal(K(\zeta_\infty)/K)$, the Galois
group of the maximal cyclotomic extension; in that case, we can reduce to
$K=\QQ$ and do an explicit computation. The general case will actually only
follow after the fact from the construction of global reciprocity!
\begin{prop}
The map $r_{K,v}$ is trivial on $K^*$.    
\end{prop}
Thus it induces a map $r_K: C_K \to \Gal(L/K)$ for each $L/K$ abelian,
and in fact to $r_K: C_K \to \Gal(K^{\ab}/K)$ using the analogous
compatibility for local reciprocity.

Since each of the local reciprocity maps is continuous, so is the map $r_K$.
That means the kernel of $r_K: C_K \to \Gal(L/K)$, for $L/K$ abelian,
is an open subgroup of $C_K$. Now recall that the
quotient of $C_K$ by any open subgroup of finite index is a generalized
ideal class group. Thus $r_K$ is giving us a canonical isomorphism between
$\Gal(L/K)$ and a generalized ideal class group; could this be anything
but Artin reciprocity itself? 

Indeed, let $U$ be the kernel of $r_K$,  let $\gothm$ be a conductor
for the generalized ideal class group $C_K/U$, and let $\gothp$ be a
prime of $K$ not dividing $\gothm$ and unramified in $L$. Then
the idele $\alpha = (1,1, \dots, \pi, \dots)$ with a uniformizer $\pi$
of $\gotho_{K_\gothp}$ in the $\gothp$ component and ones elsewhere
maps onto the class of $\gothp$ in $C_K/U$. On the other hand,
$r_K(\alpha) = r_{K, \gothp}(\pi)$ is (because $L$ is unramified over $K$)
precisely the Frobenius of $\gothp$. So indeed, $\gothp$ is being mapped
to its Frobenius, so the map $C_K/U \to \Gal(L/K)$ is indeed Artin reciprocity.

In fact, we discover from this a little bit more than we knew already about
the Artin map. All we said before about the Artin map is that it factors
through a generalized ideal class group, and that the conductor $\gothm$
of that group is divisible precisely by the ramified primes (which
we see from local reciprocity). In fact, we can now say \emph{exactly}
what is in the kernel of the classical Artin map: it is generated by
\begin{itemize}
\item all principal ideals congruent to 1 modulo $\gothm$;
\item norms of ideals of $L$ not divisible by any ramified primes.
\end{itemize}


\head{What needs to be done}

Many of these steps will be analogous to the steps in local CFT.
We may have to suppress some of the proofs depending on time constraints.
\begin{itemize}
\item Verify that the map $r_K$ given above does indeed kill principal ideles.
That's too hard to do all at once, so we first consider the projection
$r_K: K^* \to \Gal(K(\zeta_\infty)/K)$ onto the Galois group of the 
cyclotomic extension. In that case, we can reduce to
$K = \QQ$, where all the local reciprocity maps
are pretty easy to compute. The fact that $r_K$ kills principal ideles
in general will have to be deduced after some sort of reciprocity law
is established.
\item Verify that for $L/K$ cyclic, the Herbrand quotient of
$C_L$ as a $\Gal(L/K)$-module is $[L:K]$. In particular,
that forces $\#H^0(\Gal(L/K), C_L) \geq [L:K]$ (the ``First Inequality'').
\item
For $L/K$ cyclic, determine that
\[
\#H^0(\Gal(L/K), C_L) = [L:K], \qquad \#H^1(\Gal(L/K), C_L) = 1
\]
(the ``Second Inequality''). This step is trivial in local CFT by Hilbert's
Theorem 90 but is actually pretty subtle in the global case. We'll
do it by reducing to the case where $K$ contains enough roots of unity,
so that $L/K$ becomes a Kummer extension and we can compute everything
explicitly. There is also an analytic proof given in Milne which I'll
very briefly allude to.
\item
For a Neukirch-style completion, write down $d$ (i.e., choose the
``unramified'' extensions) and the ``valuation'' $v$, and check that
they satisfy the conditions for abstract class field theory.
\item
For a Milne-style completion, check that $H^2(\Gal(L/K), C_L)$ is
cyclic of order $[L:K]$
in ``unramified'' cases. Then as in the local case, deduce the
same result in general by induction on degree. 
Get a reciprocity map from $H^{-2}_T(\Gal(L/K), \ZZ) = \Gal(L/K)^{\ab}$
to $H^0_T(\Gal(L/K), C_K/\Norm_{L/K} C_L)$
using the theorem of Tate that we used before.
\item
Prove the existence theorem, by showing that every open subgroup
of $C_K$ contains a norm group. Again, we can enlarge $K$ in order
to do this, so we can get into the realm of Kummer theory.
\end{itemize}
Note: in either completion, the fact that the reciprocity map we get
is the one we actually want follows from the fact that we are ``paralleling''
the local construction in the global construction. In fact, that will give
us the fact that the reciprocity map that we tried to define at the outset
actually does kill principal ideles!

\end{document}