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\begin{document}

\begin{center}
\bf
Math 254B, UC Berkeley, Spring 2002 (Kedlaya) \\
Local Class Field Theory, via Tate's Theorem
\end{center}

For $L/K$ a finite extension of local fields, we have now computed that
$H^1(L/K) = 0$ and $H^2(L/K)$ is cyclic of order $[L:K]$. From this we want
to deduce the existence of the local reciprocity map $\phi_K: K^*
\to \Gal(K^{\ab}/K)$ and the local existence theorem (every open subgroup
of $K^*$ of finite index is a group of norms). Recall that we're going to
get this from a canonical isomorphism $K^*/\Norm_{L/K}(L^*) = H^0_T(L/K)
\to \Gal(L/K)^{\ab} = H^{-2}_T(\Gal(L/K), \ZZ)$. This will follow from the
following theorem.
\begin{theorem}[Tate]
Let $G$ be a finite (solvable) group and let $M$ be a $G$-module. Suppose
that for all subgroups $H$ of $G$ (including $G$ itself), $H^1(H,M)=0$
and $H^2(H,M)$ is cyclic of order $\#H$. Then there are isomorphisms
$H^i_T(G, \ZZ) \to H^{i+2}_T(G, M)$ which are canonical up to a choice of
generator of $H^2(G, M)$.
\end{theorem}
\begin{proof}
Let $\gamma$ be a generator of $H^2(G, M)$. Since $\Cor \circ \Res
= [G:H]$, $\Res(\gamma)$ generates $H^2(H,M)$ for any $H$. We start
out by ``artificially'' making a $G$-module containing $M$ in which
$\gamma$ becomes a coboundary.

Choose a 2-cocycle $\phi: G^3 \to M$ representing $\gamma$; by the definition
of a cocycle, 
\[
\phi(g_0 g, g_1 g, g_2 g) = \phi(g_0, g_1, g_2)^g, \qquad
\phi(g_1, g_2, g_3) - \phi(g_0, g_2, g_3) + \phi(g_0, g_1, g_3)
- \phi(g_0, g_1, g_2) = 0.
\]
And $\phi$ is a coboundary if and only if it's of the form
$d(\rho)$, that is, $\phi(g_0, g_1, g_2) = \rho(g_1, g_2) -
\rho(g_0, g_2) + \rho(g_0, g_1)$. This $\rho$ must itself be $G$-invariant:
$\rho(g_0, g_1)^g = \rho(g_0g, g_1g)$. Thus $\phi$ is a coboundary if
$\phi(e, g, hg) = \rho(e,h)^g - \rho(e,hg) + \rho(e,g)$.

Let $M[\phi]$ be the direct sum of $M$ with the free abelian group
with one generator $x_g$ for each element $g$
of $G - \{e\}$, with the $G$-action
\[
x_h^g = x_{hg} - x_g + \phi(e, g, hg).
\]
(The symbol $x_e$ should be interpreted as $\phi(e,e,e)$.)
One may verify that this is indeed a $G$-action; by construction,
the cocycle $\phi$ becomes zero in $H^2(G, M[\phi])$ by setting
$\rho(e,g) = x_g$. (Milne calls $M[\phi]$ the \emph{splitting module} of
$\phi$.)

The map $\alpha: M[\phi] \to \ZZ[G]$ sending $M$ to zero and $x_g$
to $[g]-1$ is a homomorphism of $G$-modules. Actually it maps into the
augmentation ideal $I_G$, and the sequence
\[
0 \to M \to M[\phi] \to I_G \to 0
\]
is exact. For any subgroup $H$ of $G$, we can restrict to $H$-modules, then
take the long exact sequence:
\[
0 = H^1(H,M) \to H^1(H, M[\phi]) \to H^1(H, I_G)
\to H^2(H, M) \to H^2(H, M[\phi]).
\]
To make headway with this, view $0 \to I_G \to \ZZ[G] \to \ZZ \to 0$
as an exact sequence of $H$-modules. Since $\ZZ[G]$ is induced, its Tate
groups all vanish. So we get a dimension shift:
\[
H^1(H, I_G) \cong H^0_T(H, \ZZ) = \ZZ/\#H\ZZ.
\]
Also, the map $H^2(H, M) \to H^2(H, M[\phi])$ is zero because
we constructed this map so as to kill off the generator $\phi$. 
Thus $H^1(H, I_G) \to H^2(H,M)$ is surjective. But these groups are both finite
of the same order! So the map is also injective, and $H^1(H, M[\phi])$
is also zero.

Now apply the lemma below to conclude that $H^i_T(G, M) = 0$ for all $H$.
This allows us to use the four-term exact sequence
\[
0 \to M \to M[\phi] \to \ZZ[G] \to \ZZ \to 0
\]
(as in the proof of periodicity for the Tate groups when $G$ is cyclic)
to conclude $H^i_T(G, \ZZ) \cong H^{i+2}_T(G, M)$.
\end{proof}

Note: we only need the results of this section for $G$ solvable, because
in our desired application $G$ is the Galois group of a finite extension
of local fields. But one can remove this restriction: see the note after
this lemma.
\begin{lemma}
Let $G$ be a finite (solvable) group and $M$ a $G$-module. Suppose that
$H^i(H,M) =0$ for $i=1,2$ and $H$ any subgroup of $G$ (including $G$
itself). Then $H^i_T(G,M) = 0$ for all $i \in \ZZ$.
\end{lemma}
Caveat: there's a bug in Milne's proof of this (Theorem II.2.17); see below.
\begin{proof}
For $G$ cyclic, this follows by periodicity. We prove the general result
by induction on $\#G$. Since $G$ is solvable, it has a subgroup $H$ so that
$G/H$ is cyclic. By the induction hypothesis, $H^i_T(H,M) = 0$ for all
$i$. Thus by the inflation-restriction exact sequence (see previous handout),
\[
0 \to H^i(G/H, M^H) \to H^i(G, M) \to H^i(H, M)
\]
is exact for all $i>0$. The term on the end being zero, we have
$H^i(G/H, M^H) \cong H^i(G,M) = 0$ for $i=1, 2$. By periodicity,
$H^i_T(G/H, M^H) = 0$ for all $i$, so $H^i(G/H, M^H) = 0$ for all
$i>0$, and $H^i(G,M) = 0$ for $i>0$. As for $H^0_T(G,M)$, we have that
$H^0_T(G/H, M^H) = 0$, so for any $x \in M^G$, there exists $y \in M^H$
such that $\Norm_{G/H}(y) = x$. Since $H^0_T(H,M) = 0$, there exists
$z \in M$ such that $\Norm_{H}(z) = x$. Now
$\Norm_G(z) = \Norm_{G/H} \circ \Norm_H(z) = x$. Thus
$H^0_T(G,M) = 0$, as desired.

So far so good, but we want to kill off the Tate groups with negative indices
too, so we do dimension shifting. (This is where the bug in Milne is.
He uses the wrong exact sequence, so his dimension shift goes in the wrong
direction.)
Make the exact sequence
\[
0 \to N \to M \otimes_{\ZZ} \ZZ[G] \to M \to 0,
\]
in which $m \otimes g$ maps to $m^g$. The term in the middle is acyclic, so
$H^{i+1}_T(H', N) \cong H^{i}_T(H', M)$ for any subgroup $H'$
of $G$. In particular, $H^1(H', N) = 
H^2(H', N) = 0$, so the above argument gives $H^i_T(G, N) = 0$ for $i\geq 0$.
Now from $H^0_T(G, N) = 0$
we get $H^{-1}_T(G, M) = 0$; since the same argument applies to $N$,
we also get $H^{-2}_T(G, M) = 0$ and so on.
\end{proof}
To go from the solvable case to the general case, you show that
the $p$-primary component of $H^i(G,M)$ injects into $H^i(G_p, M)$,
where $G_p$ is the $p$-Sylow subgroup. (Apply $\Cor \circ \Res$ from
$G$ to $G_p$; the result is multiplication by $[G:H]$ which is prime to
$p$.)

\head{Making things explicit}

The case we want is the isomorphism $H^{-2}_T(\Gal(L/K), \ZZ) \to H^0_T(L/K)$.
If we have an explicit cocycle $\phi$ generating $H^2(L/K)$, we can trace 
through this argument and get the local reciprocity map! The result is kind
of messy, though, so I won't torture with all of the details. The point is
simply to observe that everything we've done can be used for explicit
computations. (I think this observation is due to Dwork.) If you find
this intractable, wait until we hit abstract class field theory: that
point of view will give a different (though of course related) mechanism
for computing the reciprocity map.

Put $G = \Gal(L/K)$.
First recall
that $G^{\ab} = H^{-2}_T(G, \ZZ)$ is isomorphic to $H^{-1}_T(G, I_G)
= I_G/I_G^2$, with $g \mapsto [g]-1$. Next, use the exact sequence
\[
0 \to M \to M[\phi] \to I_G \to 0
\]
and apply the ``snaking'' construction: pull $[g]-1$ back to
$x_g \in M[\phi]$, take the norm to get $\prod_h x_g^h = \prod_h
(x_{gh} x_h^{-1} \phi(e,h,gh))$ (switching to multiplicative notation).
The $x_{gh}$ and $x_h$ term cancel out when you take the product, so we
get $\prod_h \phi(e, h, gh) \in L^*$ as the inverse image of $g \in \Gal(L/K)$.

As noted above, one needs $\phi$ to make this truly explicit; one can get
$\phi$ using explicit generators of $L/K$ if you have them. For $K = \QQ_p$,
one can use roots of unity; for $K$ general, one can use the Lubin-Tate
construction. In general, one can at least do the following, imitating
our proof that $H^2(L/K)$ is cyclic of order $n$. Let $M/K$
be unramified of degree $n$; then $H^2(M/K) \to H^2(ML/K)$ is injective,
and its image lies in the image of $H^2(L/K) \to H^2(ML/K)$.

Now $H^2(M/K)$ is isomorphic to $H^0_T(M/K) = K^*/\Norm_{M/K}(M^*)$ which
is generated by a uniformizer $\pi \in K$. To ``explicate'' that isomorphism,
we recall generally how to construct the isomorphism
$H^0_T(G,M) \to H^2_T(G,M)$ for $G$ cyclic with a distinguished generator $g$.
Recall the exact sequence we used to produce the isomorphism:
\[
0 \to M \to M \otimes_{\ZZ} \ZZ[G] \to M \otimes_{\ZZ} \ZZ[G] \to M \to 0.
\]
(Remember, $G$ acts on both factors in $M \otimes_{\ZZ}
\ZZ[G]$. The first map is $m \mapsto \sum_{h \in G} m \otimes [h]$,
the second is $m \otimes [h] \mapsto m \otimes ([gh] - [h])$, and
the third is $[h] \mapsto 1$.)
Let $A = M \otimes_{\ZZ} I_G$ be the kernel of the third arrow, so
$0 \to M \to M \otimes_{\ZZ} \ZZ[G] \to A \to 0$
and $0 \to A \to M \otimes_{\ZZ} \ZZ[G] \to M \to 0$ are exact.

Given $x \in H^0_T(M/K) = M^G/\Norm_G(M)$, lift it to $x \otimes [1]$.
Now view this as a 0-cochain $\phi_0: G \to M \otimes_{\ZZ} \ZZ[G]$ given by
$\phi_0(h) = x \otimes [h]$. Apply $d$ to get a 1-cocycle:
\[
\phi_1(h_0, h_1) = \phi_0(h_1) - \phi_0(h_0) = x \otimes ([h_1]-\otimes [h_0])
\]
which actually takes values in $A$. Now snake again: pull this back to
a 1-cochain $\psi_1: G^2 \to M \otimes_{\ZZ} \ZZ[G]$ given by
\[
\psi_1(g^i, g^{i+j}) = x \otimes ([g^i] + [g^{i+1}] + \cdots + [g^{j-1}])
\]
for $i,j=0, \dots, \#G-1$.
Apply $d$ again: now we have a 2-cocycle $\psi_2: G^3 \to M \otimes_{\ZZ}
\ZZ[G]$ given by (again for $i,j=0, \dots, \#G-1$)
\begin{align*}
\psi_2(e, g^i, g^{i+j}) &= \psi_1(g^i, g^{i+j}) - \psi_1(e, g^{i+j})
+ \psi_1(e, g^i) \\
&= x \otimes ([e] + \cdots + [g^{i-1}] + [g^i] + \cdots + [g^{i+j-1}]
- [e] - \cdots - [g^{i+j-1}]) \\
&= \begin{cases} 0 & i+j < \#G \\
-x \otimes ([e] + \cdots + [g^{\#G-1}]) & i+j \geq \#G.
\end{cases}
\end{align*}
This pulls back to a 2-cocycle $\phi_2: G^3 \to M$ given by
\[
\phi_2(e, g^i, g^{i+j}) = \begin{cases} 0 & i+j < \#G \\
-x & i+j \geq \#G.
\end{cases}
\]
If you prefer, you can shift by a coboundary to get $x$ if $i+j < \#G$
and 0 if $i+j \geq \#G$.

Back to the desired computation. Applying this to $\Gal(M/K)$ acting on
$M^*$, with the canonical generator $g$ equal to the Frobenius,
we get that $H^2(M/K)$ is generated by a cocycle $\phi$ with
$\phi(e, g^i, g^{i+j}) = \pi$ if $i+j < \#G$ and 1 otherwise. Now push
this into $H^2(ML/K)$; the general theory says the image comes from 
$H^2(L/K)$. That is, for $h \in \Gal(ML/K)$, let $f(h)$ be the integer $i$
such that $h$ restricted to $\Gal(M/K)$ equals $g^i$. Then
there exists a 1-cochain $\rho: \Gal(ML/K)^2 \to (ML)^*$ such that
$\phi(e, h_1, h_2h_1) /(\rho(h_1, h_2h_1) \rho(e, h_2h_1)^{-1} \rho(e, h_1))$
belongs to $L^*$ and depends only on the images of $h_1, h_2$ in $\Gal(M/K)$.
Putting $\sigma(h) = \rho(e, h)$, we thus have
\[
\frac{\phi(e, h_1, h_2h_1) \sigma(h_2h_1)}{\sigma(h_2)^{h_1} \sigma(h_1)}
\]
depends only on $h_1, h_2$ modulo $\Gal(ML/L)$.

The upshot: once you compute such a $\sigma$ (which I won't describe how to
do, since it requires an explicit description of $L/K$), to
find the inverse image of $g \in \Gal(L/K)$ under the Artin map, 
choose a lift $g_1$ of $g$ into $\Gal(ML/K)$, then compute
\[
\prod_h \frac{\phi(e, h, gh) \sigma(gh)}{\sigma(g)^h \sigma(h)}
\]
for $h$ running over a set of lifts of the elements of $\Gal(L/K)$ into
$\Gal(ML/K)$.

\end{document}