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Notes on class field theory

Kiran S. Kedlaya

Section 5.2 The abstract reciprocity map

We next define the reciprocity map in abstract class field theory (Definition 5.2.6). As a bonus, this definition will give an explicit recipe for computing the reciprocity map in local class field theory, but we will not expand on this point.

Subsection Construction of the reciprocity map

In order to define a candidate \(r: \Gal(L/K) \to A_K/\Norm_{L/K} A_L\) for the reciprocity map, we must first give a partial definition using a different domain. See Remark 5.2.9 for motivation.

Definition 5.2.1.

Under Hypothesis 5.1.12, let \(L/K\) be a Galois extension of finite subextensions of \(\kbar/k\text{.}\) Let \(H\) be the semigroup of \(g \in \Gal(L^{\unr}/K)\) such that \(d_K(g)\) is a positive integer. Define the map
\begin{equation*} r': H \to A_K/\Norm_{L/K} A_{L} \end{equation*}
as follows. For \(g \in \Gal(L^{\unr}/K)\text{,}\) let \(M\) be the fixed field of \(g\text{,}\) so that
\begin{equation*} e(M/K) = e((M \cap L)/K), \qquad f(M/K) = d_K(g). \end{equation*}
We may set \(r'(g) = \Norm_{M/K} (\pi_M)\) for some uniformizer \(\pi_M\) for \(M\text{,}\) once we check that this doesn’t depend on the choice of \(\pi_M\text{.}\) To wit, if \(\pi'_M\) is another uniformizer for \(M\text{,}\) then \(\pi_M/\pi'_M \in U_L\) belongs to \(\Norm_{L^{\unr}/L} U_{L^{\unr}}\) by Corollary 5.1.15, so \(\Norm_{M/K} (\pi_M/\pi'_M)\) belongs to \(\Norm_{L^{\unr}/K} U_{L^{\unr}} \subseteq \Norm_{L/K} U_L\text{.}\) So at least \(r'\) is now a well-defined map, if not yet a semigroup homomorphism.

Remark 5.2.2.

Before getting into the weeds, let’s make some other observations about Definition 5.2.1.
First, \(r'\) is invariant under conjugation: if we replace \(g\) by \(h^{-1}gh\text{,}\) then its fixed field \(M\) is replaced by \(M^h\) and we can take the uniformizer \(\pi_M^h\text{.}\)
Next, if \(g \in H\) is actually in \(\Gal(L^{\unr}/L)\text{,}\) then \(r'(g) \in \Norm_{L/K} A_L\text{.}\) In that case, \(M\) contains \(L\text{,}\) so \(r'(g) = \Norm_{M/K}(\pi_M)\) can be rewritten as \(\Norm_{L/K} \Norm_{M/L} (\pi_M) \in \Norm_{L/K} A_L\text{.}\) That is, if \(r'\) were known to be multiplicative, it would induce a group homomorphism from \(\Gal(L/K)\) to \(A_K/\Norm_{L/K} A_L\text{.}\)
The remaining difficulty is to check that \(r'\) is multiplicative. The issue here is that the definition of \(r'(g)\) involves taking the norm from a field extension that depends on \(g\text{,}\) so it is hopeless to directly compare different values of \(g\text{.}\) Instead, we rewrite the definition in a more uniform manner.

Proof.

Put \(U = M \cap K^{\unr}\text{,}\) so that \(\Norm_{M/K} = \Norm_{U/K} \circ \Norm_{M/U}\text{.}\) The group \(\Gal(U/K)\) is of order \(n\) generated by \(\phi\text{,}\) so for \(y \in A_U\) we have \(\Norm_{U/K}(y) = y y^\phi \cdots y^{\phi^{n-1}}.\) Meanwhile, on \(A_M\) we can view \(\Norm_{M/U}\) as the restriction of \(\Norm_{L^{\unr}/K^{\unr}}\text{,}\) so \(\Norm_{M/U}(x) = \Norm_{L^{\unr}/K^{\unr}}(x)\text{.}\) By taking \(y = \Norm_{M/U}(x)\) and rewriting \(y^{\phi^i}\) as \(\Norm_{L^{\unr}/K^{\unr}}(x)^{\phi^i} = \Norm_{L^{\unr}/K^{\unr}}(x^{\phi^i})\text{,}\) we deduce the claim.
Using the formula from Proposition 5.2.3, we can now establish multiplicativity, following [37], Proposition IV.5.5.

Proof.

By hypothesis, \(x\) is the class of some \(u \in U_M\) for which there exist \(u_1,\dots,u_r \in U_M\) and \(\tau_1,\dots,\tau_r \in G\) such that
\begin{equation*} u^{\phi=1} = \prod_{i=1}^r u_i^{\tau_i - 1}. \end{equation*}
Put \(n = [M:K]\) and \(\sigma = \phi^n\text{,}\) and let \(\Sigma\) be the fixed field of \(\sigma\text{,}\) which contains \(M\text{.}\) Let \(\Sigma_n\) be the unramified extension of \(\Sigma\) of degree \(n\text{;}\) it is the fixed field of \(\sigma^n\text{.}\) Since \(H^0_T(\Gal(\Sigma_n/\Sigma), U_{\Sigma_n})\) vanishes by Proposition 5.1.14, we can find \(\tilde{u}, \tilde{u}_i \in U_{\Sigma_n}\) such that
\begin{gather*} u = \Norm_{\Sigma_n/\Sigma}(\tilde{u}) = \tilde{u} \tilde{u}^{\sigma} \cdots \tilde{u}^{\sigma^{n-1}}\\ u_i= \Norm_{\Sigma_n/\Sigma}(\tilde{u}_i) = \tilde{u}_i \tilde{u}_i^{\sigma} \cdots \tilde{u}_i^{\sigma^{n-1}}. \end{gather*}
Since \(H^{-1}_T(\Gal(\Sigma_n/\Sigma), U_{\Sigma_n})\) vanishes by Proposition 5.1.14 again, we can find \(\tilde{y} \in U_{\Sigma_n}\) with
\begin{equation*} \tilde{u}^{\phi-1} / \prod_{i=1}^r \tilde{u}_i^{\tau_i-1} = \tilde{y}^{\sigma-1}, \end{equation*}
and so (because \(\sigma = \phi^n\))
\begin{equation*} \tilde{u}^{\phi-1} = (\tilde{y} \tilde{y}^{\phi} \cdots \tilde{y}^{\phi^{n-1}})^{\phi-1} \prod_{i=1}^r \tilde{u}_i^{\tau_i-1}. \end{equation*}
Applying \(\Norm_G\) yields
\begin{equation*} \Norm_G(\tilde{u})^{\phi-1} = \Norm_G(\tilde{y} \tilde{y}^{\phi} \cdots \tilde{y}^{\phi^{n-1}})^{\phi-1}. \end{equation*}
That is, if we set
\begin{equation*} z = \Norm_G(\tilde{u}) / \Norm_G(\tilde{y} \tilde{y}^{\phi} \cdots \tilde{y}^{\phi^{n-1}}), \end{equation*}
we have \(z^{\phi-1} = 1\) and so \(z \in U_K\text{.}\) Put
\begin{equation*} y = \tilde{y} \tilde{y}^\sigma \cdots \tilde{y}^{\sigma^n-1} = \Norm_{\Sigma_n/\Sigma}(\tilde{y}) \in U_\Sigma; \end{equation*}
using Proposition 5.2.3 we obtain
\begin{align*} \Norm_G(u) &= \Norm_G(\tilde{u} \tilde{u}^\sigma \cdots \tilde{u}^{\sigma^{n-1}})\\ & = \Norm_G(y y^{\phi} \cdots y^{\phi^{n-1}}) z^n\\ & = \Norm_{\Sigma/K}(y) \Norm_{M/K}(u) \in \Norm_{M/K} U_M, \end{align*}
proving the claim. (Compare [37], Lemma IV.5.4.)

Proof.

Let \(g_1, g_2 \in H\) be arbitrary and put \(g_3 = g_1g_2\text{.}\) Let \(M_i\) be the fixed field of \(g_i\text{,}\) let \(\pi_i \in A_{M_i}\) be a uniformizer of \(M_i\text{,}\) and put \(\rho_i = r'(g_i) = \Norm_{M_i/K}(\pi_i) \in A_K\text{.}\) Put \(\rho = \rho_1 \rho_2/\rho_3\text{;}\) note that
\begin{equation*} v_K(\rho_i) = f(M_i/K) v_{M_i}(\pi_i) = f(M_i/K) = d_K(g_i), \end{equation*}
which implies that \(v_K(\rho) = 0\) and hence \(\rho \in U_K\text{.}\) Our goal is to check that \(\rho \in \Norm_{L/K} A_{L}\text{;}\) our plan is to rephrase this as a relation among units, to which Lemma 5.2.4 will apply.
We first make an adjustment at the level of group elements. Put \(G = \Gal(L^{\unr}/K^{\unr})\text{.}\) Choose \(\phi \in H\) such that \(d_K(\phi) = 1\text{.}\) Put \(d_i = d_K(g_i)\) and \(\tau_i = g_i^{-1} \phi^{d_i} \in G\text{;}\) then
\begin{equation*} \tau_3 = g_2^{-1} g_1^{-1} \phi^{d_1 + d_2} = g_2^{-1} \phi^{d_2} (\phi^{-d_2} g_1 \phi^{d_2})^{-1} \phi^{d_1}. \end{equation*}
It will be convenient to replace \(g_1\) and \(\tau_1\) with
\begin{equation*} g_1' = \phi^{-d_2} g_1 \phi^{d_2}, \qquad \tau_1' = (g_1')^{-1} \phi^{g_1}, \end{equation*}
so that \(\tau_1' \tau_2 = \tau_3\text{.}\) We correspondingly define \(M_1'\) to be the fixed field of \(g_1'\) and set \(\pi_1' = \pi_1^{\phi^{n_2}} \in A_{M_1'}\text{,}\) noting that \(\Norm_{M_1'/K}(\pi_1') = \Norm_{M_1/K}(\pi_1) = \rho_1\text{.}\)
Let \(N\) be a finite subextension of \(L^{\unr}/L\) containing \(M_1, M_2, M_3, M_1'\text{.}\) Set
\begin{equation*} \sigma_i = \pi_i \pi_i^\phi \cdots \pi_i^{\phi^{d_i-1}}, \qquad \sigma_1' = (\pi_1') (\pi_1')^\phi \cdots (\pi_1')^{\phi^{d_1-1}} \end{equation*}
and \(u = \sigma_1' \sigma_2 / \sigma_3 \in U_N\text{;}\) by Proposition 5.2.3 we have \(\rho = \Norm_G(u)\text{.}\) By defining
\begin{equation*} u_1 = (\pi_1')^{1-\tau_2}, u_2 = \pi_2/\pi_1', u_3 = \pi_3/\pi_1' \in U_N \end{equation*}
and using the equality \(\tau_1' \tau_2 = \tau_3\text{,}\) we compute that
\begin{equation*} u^{\phi-1} = (\pi_1')^{\tau_1' - 1} \pi_2^{\tau_2 - 1} / \pi_3^{\tau_3 - 1} = u_1^{\tau_1'-1} u_2^{\tau_2 - 1} / u_3^{\tau_3 - 1} \end{equation*}
vanishes in \(H_0(G, U_N)\text{;}\) by Lemma 5.2.4 we obtain \(\Norm_G(u) \in \Norm_{N/K} U_N\text{,}\) proving the claim.

Definition 5.2.6.

With notation as in Definition 5.2.1, by combining Definition 5.2.1, Remark 5.2.2, and Lemma 5.2.5, we obtain a semigroup homomorphism \(r': H \to A_K/\Norm_{L/K}A_L\) which kills \(\Gal(L^{\unr}/L)\) and thus induces a homomorphism \(r = r_{L/K}: \Gal(L/K) \to A_K/\Norm_{L/K}A_L\text{.}\) We call \(r_{L/K}\) the reciprocity map.
Here are some consequences of the construction whose proofs are left to the reader.

Proof.

Remark 5.2.9.

Note that Proposition 5.2.7 dictates the form of Definition 5.2.1: we want to be able to compute the map \(r: \Gal(L/K) \to A_K/\Norm_{L/K} A_L\) by first computing the map \(r: \Gal(ML/M) \to A_M/\Norm_{ML/M} A_{ML}\text{,}\) which should take Frobenius to a uniformizer (as \(ML/M\) is an unramified extension), and then applying \(\Norm_{M/L}\text{.}\) We will use this picture again in Lemma 5.3.2.

Proof.

Exercises Exercises

3.

Under Hypothesis 5.1.12, choose \(c \in \widehat{\ZZ}^*\) and let \(s_{L/K}\) be the reciprocity map defined using \(cd\) instead of \(d\text{.}\) Show that \(s_{L/K} = c^{-1} r_{L/K}\text{.}\)

4.

Under Hypothesis 5.1.12, choose \(c \in \widehat{\ZZ}^*\) which acts on \(\im(v)\text{,}\) and let \(s_{L/K}\) be the reciprocity map defined using \(cd, cv\) instead of \(d, v\text{.}\) Show that \(s_{L/K} = r_{L/K}\text{.}\)
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