Section 4.5 Making the reciprocity map explicit
It is natural to ask whether the local reciprocity map can be described more explicitly. In fact, given an explicit cocycle \(\phi\) generating \(H^2(L/K)\text{,}\) we can trace through the arguments to get the local reciprocity map. However, the argument is somewhat messy, so I won’t torture you with all of the details; the point is simply to observe that everything we’ve done can be used for explicit computations. (This observation is apparently due to Dwork.)
If you find this indigestible, you may hold out 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 (see Section 5.2).
Subsection Initial setup
Put \(G = \Gal(L/K)\text{.}\) First recall that \(G^{\ab} = H^{-2}_T(G, \ZZ)\) is isomorphic to \(H^{-1}_T(G, I_G) = I_G/I_G^2\text{,}\) with \(g \mapsto [g]-1\text{.}\) Next, use the exact sequence
\begin{equation*}
0 \to M \to M[\phi] \to I_G \to 0
\end{equation*}
and apply the “snaking” construction: pull \([g]-1\) back to \(x_g \in M[\phi]\text{,}\) 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 \(g \in \Gal(L/K)\text{.}\)
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 \(K = \QQ_p\text{,}\) one can use roots of unity; for general \(K\text{,}\) one can use the Lubin-Tate construction. Alternatively, one can argue as in our proof that \(H^2(L/K)\) is cyclic of order \(n\text{;}\) see below.
Subsection An explicit cocycle via periodicity
Let \(M/K\) be unramified of degree \(n\text{;}\) then \(H^2(M/K) \to H^2(ML/K)\) is injective, and its image lies in the image of \(H^2(L/K) \to H^2(ML/K)\text{.}\)
Now \(H^2(M/K)\) is isomorphic to \(H^0_T(M/K) = K^*/\Norm_{M/K}M^*\text{,}\) which is generated by a uniformizer \(\pi \in K\text{.}\) 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 \(g\text{.}\) Recall the exact sequence we used to produce the isomorphism in Theorem 3.4.1:
\begin{equation*}
0 \to M \to M \otimes_{\ZZ} \ZZ[G] \to M \otimes_{\ZZ} \ZZ[G] \to M \to 0\text{.}
\end{equation*}
(Remember, \(G\) acts on both factors in \(M \otimes_{\ZZ} \ZZ[G]\text{.}\) The first map is \(m \mapsto \sum_{h \in G} m \otimes [h]\text{,}\) the second is \(m \otimes [h] \mapsto m \otimes ([gh] - [h])\text{,}\) and the third is \([h] \mapsto 1\text{.}\)) 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 \(x \in H^0_T(M/K) = M^G/\Norm_G(M)\text{,}\) lift it to \(x \otimes [1]\text{.}\) Now view this as a 0-cochain \(\phi_0: G \to M \otimes_{\ZZ} \ZZ[G]\) given by \(\phi_0(h) = x \otimes [h]\text{.}\) Apply \(d\) to get a 1-cocycle:
\begin{equation*}
\phi_1(h_0, h_1) = \phi_0(h_1) - \phi_0(h_0) = x \otimes ([h_1]- [h_0])
\end{equation*}
which actually takes values in \(A\text{.}\) Now snake again: pull this back to a 1-cochain \(\psi_1: G^2 \to M \otimes_{\ZZ} \ZZ[G]\) given by
\begin{equation*}
\psi_1(g^i, g^{i+j}) = x \otimes ([g^i] + [g^{i+1}] + \cdots + [g^{j-1}])
\end{equation*}
for \(i,j=0, \dots, \#G-1\text{.}\) 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}) \amp = \psi_1(g^i, g^{i+j}) - \psi_1(e, g^{i+j}) + \psi_1(e, g^i)\\
\amp = x \otimes ([e] + \cdots + [g^{i-1}] + [g^i] + \cdots + [g^{i+j-1}] - [e] - \cdots - [g^{i+j-1}])\\
\amp = \begin{cases} 0 \amp i+j \lt \#G\\
-x \otimes ([e] + \cdots + [g^{\#G-1}]) \amp i+j \geq \#G. \end{cases}
\end{align*}
This pulls back to a 2-cocycle \(\phi_2: G^3 \to M\) given by
\begin{equation*}
\phi_2(e, g^i, g^{i+j}) = \begin{cases}0 \amp i+j \lt \#G \\ -x \amp i+j \geq \#G. \end{cases}
\end{equation*}
If you prefer, you can shift by a coboundary to get \(x\) if \(i+j \lt \#G\) and 0 if \(i+j \geq \#G\text{.}\)
Subsection From a cocycle to reciprocity
Back to the desired computation. Applying this to \(\Gal(M/K)\) acting on \(M^*\text{,}\) 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 \(i+j \lt \#G\) and 1 otherwise. Now push this into \(H^2(ML/K)\text{;}\) the general theory says the image comes from \(H^2(L/K)\text{.}\) That is, for \(h \in \Gal(ML/K)\text{,}\) let \(f(h)\) be the integer \(i\) such that \(h\) restricted to \(\Gal(M/K)\) equals \(g^i\text{.}\) 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 \(\Gal(M/K)\text{.}\) Putting \(\sigma(h) = \rho(e, h)\text{,}\) we thus have
\begin{equation*}
\frac{\phi(e, h_1, h_2h_1) \sigma(h_2h_1)}{\sigma(h_2)^{h_1} \sigma(h_1)}
\end{equation*}
depends only on \(h_1, h_2\) modulo \(\Gal(ML/L)\text{.}\)
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 \(\Gal(ML/K)\text{,}\) then compute
\begin{equation*}
\prod_h \frac{\phi(e, h, gh) \sigma(gh)}{\sigma(g)^h \sigma(h)}
\end{equation*}
for \(h\) running over a set of lifts of the elements of \(\Gal(L/K)\) into \(\Gal(ML/K)\text{.}\)
