4 Global Class Field Theory

In this chapter we let $l/k$ be a finite Galois extension of algebraic number fields. We shall consider the idele class group $\mathrm{Cl}_l = {\mathbb A}_l^\times / l^\times $ as a module for the Galois group $\mathrm{Gal}(l/k)$ and we shall describe the construction of fundamental classes in $H^2(l/k, \mathrm{Cl}_l)$. These classes give rise to a reciprocity isomorphism

\[ \mathrm{Gal}(l/k)^{\mathrm{ab}} \cong \mathrm{Cl}_k / N(\mathrm{Cl}_l). \]

We would therefore like to prove for all intermediate fields $l / m / k$:

$H^1(l/m, \mathrm{Cl}_l) = 0$,
$H^2(l/m, \mathrm{Cl}_l)$ is cyclic of order $[l:m]$.

We note the following consequence of Hilbert’s theorem 90:

Lemma 106

Let $l/k$ be a finite Galois extension of number fields (or even global fields). The map $\mathrm{Cl}_k \to \mathrm{Cl}_l ^{\mathrm{Gal}(l/k)}$ is an isomorphism.

Proof ▶

We have a short exact sequence

\[ 0 \to l^\times \to {\mathbb A}_l^\times \to \mathrm{Cl}_l \to 0. \]

Taking $\mathrm{Gal}(l/k)$-invariants gives a long exact sequence beginning with

\[ 0 \to k^\times \to {\mathbb A}_k^\times \to \mathrm{Cl}_l^{\mathrm{Gal}(l/k)} \to H^1(l/k, l^\times ). \]

The last term is zero by 78. This implies the result.

By the lemma, we may regard $H^0_{\mathrm{Tate}}(l/k,\mathrm{Cl}_l)$ as the quotient $\mathrm{Cl}_k / N_{l/k} \mathrm{Cl}_l$. Furthmore, the inflation map for a tower of Galois extensions $l/m/k$ takes the form

\[ H^\bullet (m/k, \mathrm{Cl}_m) \to H^\bullet (l/k, \mathrm{Cl}_l). \]

4.1 Choice of $S$

Let $S$ be a finite set of places of $k$, containing all of the infinite primes and all of the primes which ramify in $l$. We shall use the notation

\[ {\mathbb A}_{l,S} = \prod _{v \in S} \prod _{w | v} l_w \times \prod _{v \notin S} \prod _{w | v} {\mathcal O}_v. \]

We also write ${\mathcal O}_{l,S}$ for the ring $S$-integers in $l$:

\[ {\mathcal O}_{l,S} = \{ x \in l, \forall v \not\in S, \forall w | v, |x|_v \le 1\} = l \cap {\mathbb A}_{l,S}. \]

We regard ${\mathcal O}_{l,S}$ as a subring of ${\mathbb A}_{l,S}$ and ${\mathcal O}_{l,S}^\times $ as a subgroup of ${\mathbb A}_{l,S}^\times $.

The quotient ${\mathbb A}_l^\times / {\mathbb A}_{l,S}^\times $ is naturally isomorphic to the group of non-zero fractional ideals of ${\mathcal O}_{l,S}$. By adding more primes to $S$ if necessary, we may assume that ${\mathcal O}_{k,S}$ and ${\mathcal O}_{l,S}$ are both principal ideal domains (this can be achieved by adding to $S$ primes $P$ whose ideal classes generate the class groups of $k$ and $l$). With such a choice of $S$ we have:

\[ {\mathbb A}_{l}^\times = {\mathbb A}_{l,S}^\times l^\times , \qquad {\mathbb A}_k^\times = {\mathbb A}_{k,S}^\times k^\times , \]

and therefore

\[ \mathrm{Cl}_{l} = {\mathbb A}_{l,S}^\times / {\mathcal O}_{l,S}^\times , \qquad \mathrm{Cl}_{k} = {\mathbb A}_{k,S}^\times / {\mathcal O}_{k,S}^\times . \]

The big advantage of working with ${\mathbb A}_{l,S}^\times $ and ${\mathcal O}_{l,S}$ instead of ${\mathbb A}_l^\times $ and $l^\times $ is that ${\mathbb A}_{l,S}^\times $ and ${\mathcal O}_{l,S}$ have finite cohomology groups, whereas the cohomology groups of ${\mathbb A}_l^\times $ and $l^\times $ are infinite.

4.2 The Herbrand quotient of the $S$-ideles

Let $v$ be a place of $k$ and $\hat v$ a place of $l$ above $v$. We’ll write $D_{\hat v}$ the decomposition group at $\hat v$.

Lemma 107

Then there are isomorphisms

\[ \prod _{w | v} l_w^\times \cong \mathrm{coind}_{D_{\hat v}}^{\mathrm{Gal}(l/k)} l_{\hat v}^\times , \qquad \prod _{w | v} {\mathcal O}_w^\times \cong \mathrm{coind}_{D_{\hat v}}^{\mathrm{Gal}(l/k)} {\mathcal O}_{\hat v}^\times . \]

Proof ▶

We’ll write $x$ for an element of $\prod _{w | v} l_w^\times $ and $x_w$ for it’s component in $l_w$. Define a map $\Phi : \prod _{w | v} l_w^\times \to (\mathrm{Gal}(l/k) \to l_{\hat v}^\times )$ by

\[ \Phi (x) = (g \mapsto g \bullet x_{g^{-1} \hat v}). \]

Note that for $h \in D_{\hat v}$ we have

\[ \Phi (x) (hg) = h \bullet (g \bullet x_{g^{-1} \hat v}) = h \bullet \Phi (x) (g). \]

Therefore $\Phi (x)$ is in the subspace $\mathrm{coind}_{D_v} l_{\hat v}$, and it’s easy to check that $\Phi $ gives a group isomorphism $\prod _{w | v} l_w^\times \cong \mathrm{coind}_{D_v} l_{\hat v}^\times $. We’ll check that this map intertwines the actions of $G$. For $g \in G$, then element $g \bullet x$ has $w$-component $g \bullet x_{g^{-1}\bullet w}$. This implies

\[ \Phi ( g \bullet x)(h) = h\bullet (g \bullet x)_{h^{-1} \bullet \hat v} = h g \bullet x_{(hg)^{-1} \bullet \hat v} = \Phi (x) (hg). \]

The proof of the isomorphism for $\prod _{w | v} {\mathcal O}_w^\times $ is similar.

Lemma 108

There are isomorphisms for all $n {\gt} 0$

\[ H^n(l/k, {\mathbb A}_{S,l}^\times ) \cong \prod _{v \in S} H^n(l_{\hat v} / k_v, l_{\hat v}^\times ). \]

Proof ▶

We note that by 107 we have

\[ {\mathbb A}_{S,l}^\times \cong \prod _{v \in S} \mathrm{coind}_{D_{\hat v}} l_{\hat v}^\times \times \prod _{v \not\in S} \mathrm{coind}_{D_{\hat v}} {\mathcal O}_{\hat v}^\times \]

By Shapiro’s lemma (24) we have

\[ H^n(l/k,{\mathbb A}_{S,l}^\times ) \cong \prod _{v \in S} H^n(l_{\hat v}/k_v , l_{\hat v}^\times ) \times \prod _{v \not\in S} H^n(l_{\hat v}/k_v , {\mathcal O}_{\hat v}^\times ) \]

For $v \not\in S$, the extension $l_{\hat{v}}/l_v$ is unramified, so by 89 ${\mathcal O}_{\hat v}^\times $ has trivial cohomology.

Lemma 109

If $l/k$ is a cyclic extension then we have

\[ h(l/k, {\mathbb A}_{S,l} ^\times ) = \prod _{v \in S} |D_{\hat v}|. \]

Proof ▶

This follows from 108 and 84.

4.3 The Herbrand quotient of the $S$-units

Define the logarithmic space $V_S$ to be the following finite dimensional vector space over the real numbers:

\[ V_S = \prod _{v \in S} \prod _{w | v} {\mathbb R}. \]

We consider $V_S$ as a representation of $\mathrm{Gal}(l/k)$, where the Galois action permutes the places $w$ lying above each $v \in S$. As a Galois representation we have

\[ V_S \cong \prod _{v \in S} \mathrm{ind}_{D_{\hat v}}^{\mathrm{Gal}(l/k)} {\mathbb R}. \]

Contained in $V_S$ we have a lattice $L_S$ consisting of vectors whose components are all in ${\mathbb Z}$. Here we are using the word “lattice” to mean the ${\mathbb Z}$-space of a basis for $V_S$. We have an isomorphism

\[ L_S \cong \prod _{v \in S} \mathrm{ind}_{D_{\hat v}}^{\mathrm{Gal}(l/k)} {\mathbb Z}. \]

Lemma 110

If $l/k$ is a cyclic extension then $h(l/k,L_S) = \prod _{v \in S} |D_{\hat v}|$.

Proof ▶

This follows from Shapiro’s lemma (24) together with the calculation of the cohomology of a cyclic group with values in ${\mathbb Z}$ (53).

Lemma 111

Let $l/k$ be cyclic and let $M$ be any Galois-invariant lattice in $V_S$. Then $h(l/k,M) = \prod _{v \in S} |D_{\hat v}|$

Proof ▶

The representations $M \otimes {\mathbb Q}$ and $L_S \otimes {\mathbb Q}$ have the same character (this is just the character of the representation $V_S$). Therefore the representations $M \otimes {\mathbb Q}$ and $L_S \otimes {\mathbb Q}$ are isomorphic. Hence $M$ is isomorphic to subrepresentation of finite index in $L_S$, so by 59 and 58 they have the same Herbrand quotient. The result then follows from 110.

The vector $(1,1,\ldots ,1) \in V_S$ is fixed by all elements of $\mathrm{Gal}(l/k)$, so it spans a subrepresentation isomorphic to the trivial representation ${\mathbb Z}$. Recall the we have a logarithmic map

\[ \log _S : {\mathcal O}_S^\times \to V_S, \]

where the $w$-component of $\log _S(x)$ is $\log |x|_w$. The kernel of $\log _S$ is the finite group of roots of unity in $k$.

Theorem 112

✓

$\log _S({\mathcal O}_S^\times )$ has zero intersection with $\mathrm{Span}(1,1,\ldots ,1)$. The direct sum of these subrepresentations is a lattice in $V_S$.

Proof ▶

An equivalent statement is already in Mathlib as NumberField.Units.dirichletUnitTheorem.unitLattice_span_eq_top.

Corollary 113

Let $l/k$ be a cyclic extension. Then

\[ h(l/k,{\mathcal O}_{l,S}^\times ) = \frac{\prod _{v\in S} |D_{\hat v}|}{[l:k]}. \]

Proof ▶

Since $\log _S$ has finite kernel, the Herbrand quotient of ${\mathcal O}_{l,S}^\times $ is equal to that of $\log _S({\mathcal O}_S^\times )$. By Dirichlet’s unit theorem, $\log _S({\mathcal O}_S^\times ) \oplus {\mathbb Z}$ is a lattice in $V_S$. We know the Herbrand quotient of $\log _S({\mathcal O}_S^\times ) \oplus {\mathbb Z}$ from 111, and the Herbrand quotient of ${\mathbb Z}$ from 53.

Corollary 114

If $l/k$ is cyclic then $h(l/k,\mathrm{Cl}_l) = [l:k]$.

Proof ▶

Our choice of $S$ implies $\mathrm{Cl}_l \cong {\mathbb A}_{l,S}^\times / {\mathcal O}_{l,S}^\times $. We have calculated the Herbrand quotients of ${\mathbb A}_{l,S}^\times $ and ${\mathcal O}_{l,S}^\times $ in 109 and 113.

4.4 Dirichlet Density

Definition 115

Let $M$ be a set of primes of ${\mathcal O}_k$. We’ll say that $M$ has a Dirichlet density $c \in {\mathbb R}$ if

\[ \sum _{P \in M} N(P)^{-s} \stackrel{s \to 1+}\sim c \cdot \log \left(\frac{1}{s-1}\right). \]

where $s$ tends to $1$ through the real numbers $s{\gt}1$. The symbol $\sim $ denotes asymptotic equivalence, which means that the ratio of the left hand side to the right hand side converges to $1$ as $s$ tends to $1$ from above. Implied constants may depend of the set $M$ and the field $k$.

Lemma 116

Suppose $M_1$ and $M_2$ are disjoint sets of primes of ${\mathcal O}_l$. If two of the sets $M_1, M_2, M_1 \cup M_2$ have a Dirichlet density, then so does the third and we have

\[ \mathrm{density}(M_1 \cup M_2) = \mathrm{density}(M_1) + \mathrm{density}(M_2). \]

Proof ▶

This is trivial.

Lemma 117

The set of all primes of ${\mathcal O}_k$ has Dirichlet density $1$.

Proof ▶

Let $P$ be a prime. For $s {\gt} 1$ we have

\[ \left|N(P)^{-s} - \log \left( \frac{1}{1-N(P)^{-s}}\right)\right| \ll N(P)^{-2s} \]

Since $\sum N(P)^{-2s}$ is bounded in the region $s {\gt} 1$ (this follows from the convergence of the Dedekind zeta function), we have

\[ \sum _P N(P)^{-s} \sim \sum _P \log \left( \frac{1}{1-N(P)^{-s}} \right) = \log \zeta _l(s), \]

where $\zeta _l$ is the Dedekind zeta function (NumberField.dedekindZeta). By the analytic class number formula (NumberField.tendsto_sub_one_mul_dedekindZeta_nhdsGT) there is a positive real number $r$ such that

\[ \zeta _l(s) = \frac{r}{s-1} + O(1) \qquad (s {\gt} 1). \]

This implies

\[ \log (\zeta _l(s)) \sim \log \left( \frac{1}{s-1}\right). \]

Lemma 118

The set of primes of ${\mathcal O}_l$ of degree one has Dirichlet density $1$.

Proof ▶

Let $M$ be the set of primes of degree larger than one. It’s sufficient to prove that $M$ has Dirichlet density $0$. We have

\[ \sum _{P \in M} N(P)^{-s} = \sum _{n \in {\mathbb N}} A(n) n^{-s}, \]

Where $A$ is the number of primes of degree ${\gt}1$ with norm $n$. Note that $A(n) \le [l:{\mathbb Q}]$. Also $A(n)=0$ unless $n=m^r$ for positive integer $m$ and $1 \le r \le [l:Q]$. This implies

\[ \sum _{P \in M} N(P)^{-s} \le [l:{\mathbb Q}] \sum _{r=2}^{[l:{\mathbb Q}]} \sum _{m =1} ^\infty m^{-rs} \le [l:{\mathbb Q}] (\zeta _{\mathbb Q}(2) + \cdots + \zeta _{\mathbb Q}([l:{\mathbb Q}]) ). \]

Since the sum above is bounded on the region $s {\gt} 1$, the set $M$ has Dirichlet density $0$.

Lemma 119

Let $l/k$ be a finite Galois extension of number fields. Then the set of degree $1$ primes of $k$ which split completely in $l$ has density $\frac{1}{[l:k]}$.

Proof ▶

Let $M_k$ be the set of degree $1$ primes of $k$ and $M_l$ the set of degree $1$ primes of $l$. Every prime $Q$ in $M_l$ lies above some prime $P \in M_k$. If there is a prime $Q$ above $P$, then there are precisely $[l:k]$ of them; this happens when $P$ splits completely in $l$.

Let $M$ be the set of $P \in M_k$ which split completely in $l$. Then we have

\[ \sum _{P \in M} N(P)^{-s} = \frac{1}{[l:k]} \sum _{Q \in M_l} N(Q)^{-s} \sim \frac{1}{[l:k]} \log \left( \frac{1}{s-1}\right). \]

Here we have used 118.

4.5 Some $L$-functions

Lemma 120

$H^0_{\mathrm{Tate}}(l/k,\mathrm{Cl}_l)$ is finite.

Proof ▶

We have

\begin{align*} \mathrm{Cl}_{k} / N(\mathrm{Cl}_l) & \cong {\mathbb A}_{k,S}^\times / {\mathcal O}_{k,S}^\times N({\mathbb A}_{l,S}^\times ) \\ & \cong \left(\prod _{v \in S} k_v^\times / N(l_{\hat v}^\times )\right) / {\mathcal O}_{k,S}^\times , \end{align*}

where $\hat v$ is a place of $l$ lying above $v$. The result follows because each of the groups $k_v^\times / N(l_w)$ is finite (in fact by the local reciprocity isomorphism this is isomorphic to the abelianization of the decomposition group at $\hat v$).

Given any maximal ideal $P$ of ${\mathcal O}_{k,S}$, we let $\pi _P$ be an idele whose $P$-component is a uniformizer in $k_P$, and whose other components are all $1$. The coset of $\pi _P$ in $H^0(l/k,\mathrm{Cl}_l)$ does not depend on the choice of uniformizer since all the local units at $P$ are local norms from $l_{Q}$ for any $Q|P$ (because $P$ is unramified in $l$). Equivalently, if we choose a generator $P=(\pi )$ with $\pi \in {\mathcal O}_{k,S}$ then the coset of $\pi _P$ is the inverse of the image of $\pi $ in $\prod _{v \in S} k_v^\times / N(l_{\hat v}^\times )$. This clearly extends to a homomorphism

\[ \iota : \textrm{non-zero fractional ideals of ${\mathcal O}_{k,S}$} \to H^0_{\mathrm{Tate}}(l/k,\mathrm{Cl}_l), \]

whose kernel is the subgroup of ideals with a generator which is a local norm at $v$ for all $v \in S$.

Definition 121

Let $\chi : H^0_{\mathrm{Tate}}(l/k,\mathrm{Cl}_l) \to {\mathbb C}^\times $ be a character. For a non-zero ideal $I$ of ${\mathcal O}_{k,S}$, we shall write $\chi (I)$ in place of $\chi (\iota (I))$. We define the $L$-function of $\chi $ by

\[ L(s,\chi ) = \sum _{I} \chi (I) \cdot N(I)^{-s} = \prod _{P} \frac{1}{1-\chi (P) \cdot N(P)^{-s}}. \]

Here $s$ is a complex number with real part greater than $1$; both the product and the series converge absolutely in that region. In the sum, $I$ ranges over the non-zero ideals of ${\mathcal O}_{k,S}$, and in the product $P$ ranges over the maximal ideals of ${\mathcal O}_{k,S}$.

If $\chi $ is the trivial character, then $L(s,\chi )$ is (up to finitely many Euler factors for primes in $S$) equal to the Dedekind zeta function of $k$.

It’s known that $L(s,\chi )$ has a meromorphic continuation to ${\mathbb C}$ and is entire if $\chi $ is non-trivial (see for example Tate’s thesis, which is chapter XV of [ 2 ] ). We won’t need such a strong result here; we can make do with the following:

Lemma 122 Weak lemma

If $\chi $ is a non-trivial character then $L(s,\chi )$ is bounded on the interval $(1,2)$.

Lemma 123

Let $M$ be a set of primes of ${\mathcal O}_S$ whose image in $H^0_{\mathrm{Tate}}(l/k,\mathrm{Cl}_l)$ is zero. There exists a real number $c$ depending only on the fields $k$ and $l$, such that for all $s {\gt} 1$ we have:

\[ \sum _{p \in M} N(P)^{-s} \le \frac{1}{|H^0_{\mathrm{Tate}}(l/k,\mathrm{Cl}_l)|} \log \left(\frac{1}{s-1}\right) + c. \]

Proof ▶

Let $s {\gt} 1$. All the series in the following calculation converge absolutely in this region. The implied constants in the $O(1)$ terms do not depend on $s$.

\begin{align*} \sum _{P \in M} |N(P)|^{-s} & = \frac{1}{|H^0_{\mathrm{Tate}}(l/k,\mathrm{Cl}_l)|} \sum _P \sum _\chi \chi (P) N(P)^{-s}\\ & = \frac{1}{|H^0_{\mathrm{Tate}}(l/k,\mathrm{Cl}_l)|} \sum _P \sum _\chi -\log (1-\chi (P) N(P)^{-s}) + O(1)\\ & = \frac{1}{|H^0_{\mathrm{Tate}}(l/k,\mathrm{Cl}_l)|} \sum _P \sum _\chi -\log |1-\chi (P) N(P)^{-s}| + O(1)\\ & = \frac{1}{|H^0_{\mathrm{Tate}}(l/k,\mathrm{Cl}_l)|} \sum _\chi \sum _P -\log |1-\chi (P) N(P)^{-s}| + O(1)\\ & = \frac{1}{|H^0_{\mathrm{Tate}}(l/k,\mathrm{Cl}_l)|} \log \left(\prod _\chi |L(s,\chi )| \right) + O(1)\\ & \le \frac{1}{|H^0_{\mathrm{Tate}}(l/k,\mathrm{Cl}_l)|} \log \left(\frac{1}{s-1}\right) + O(1). \end{align*}

Interchanging the order of summation is justified because the series converge absolutely. We need to be slightly careful about which branch of the logarithm we are using here. In the expression $\log (1-\chi (P) N(P)^{-s})$ we shall mean the branch which is continuous on the ball of radius $1$, centred about $1$. The imaginary parts of $\log (1-\chi (P) N(P)^{-s})$ and $\log (1-\bar\chi (P) N(P)^{-s})$ cancel out; this justifies replacing $\log (1-\chi (P) N(P)^{-s})$ by $\log |1-\chi (P) N(P)^{-s}|$.

Remark

In fact the density of the set $M$ in this lemma is precisely $\frac{1}{|H^0_{\mathrm{Tate}}(l/k,\mathrm{Cl}_l)|}$. This can be proved by showing that each $L(s,\chi )$ has a continuation to a neighbourhood of $s=1$, and is non-zero at $s=1$. However, proving this is more difficult than the weak lemma above, and we only need the inequality of the lemma.

4.6 The first inequality

Theorem 124

For any finite Galois extension $l/k$ be have

\[ |H^0_{\mathrm{Tate}}(l/k, \mathrm{Cl}_{l}) | \le [l : k]. \]

Proof ▶

Let $M_1$ be the set of degree $1$ primes of ${\mathcal O}_{k,S}$ which split in $l$ and let $M_2$ be the set of primes of ${\mathcal O}_{k,S}$ whose image in $H^0_{\mathrm{Tate}}(l/k,\mathrm{Cl}_l)$ is zero. Given $P \in M_1$, the norm map $l_{\hat v}^\times \to k_v^\times $ is the identity map, and is in particular surjective. Hence the image of $P$ in $H^0_{\mathrm{Tate}}(l/k,\mathrm{Cl}_l)$ is zero. This shows that $M_1 \subseteq M_2$. This implies for $s{\gt} 1$:

\[ \sum _{P \in M_1} N(P)^{-s} \le \sum _{P \in M_2} N(P)^{-s} . \]

By 123 we have

\[ \sum _{P \in M_1} N(P)^{-s} \le \frac{1}{|H^0_{\mathrm{Tate}}(l/k,\mathrm{Cl}_l)|} \log \left(\frac{1}{1-s}\right) + O(1). \]

By 119 the left hand side is asymptotic to $\frac{1}{[l:k]}\log \left(\frac{1}{1-s}\right)$. Therefore $\frac{1}{[l:k]} \le \frac{1}{|H^0_{\mathrm{Tate}}(l/k,\mathrm{Cl}_l)|}$.

Corollary 125

If $l/k$ is cyclic then $|H^2(l/k, \mathrm{Cl}_{l})| = [l:k]$ and $H^1(l/k, \mathrm{Cl}_{l}) = 0$.

Proof ▶

This follows immediately from (a) the first inequality, (b) the periodicity of the cohomology for a cyclic group, and (c) the calculation of the Herbrand quotient of $\mathrm{Cl}_{l}$.

Theorem 126

If $l/k$ is any finite Galois extension then $H^1(l/k, \mathrm{Cl}_{l}) \cong 0$ and $|H^2(l/k, \mathrm{Cl}_{l})| \le [l:k]$.

Proof ▶

For each prime number $p$ dividing $[l:k]$ we let $k_p$ be the fixed field of a Sylow $p$-subgroup $S_p$ of $\mathrm{Gal}(l/k)$. By 49, it’s suffient to prove

\[ H^1(l/k_p, \mathrm{Cl}_{l}) = 0, \qquad |H^2(l/k_p, \mathrm{Cl}_{l})| \le [l:k_p]. \]

Since $S_p$ is solvable, this reduces us to the case that $\mathrm{Gal}(l/k)$ is solvable. We’ll prove the result by induction on $k$ starting with $k=l$ and working downwards in cyclic quotients.

Clearly the result holds for $k=l$. Assume the result for a subfield $m$ of $l$ and let $m/k$ be cyclic. We have an inflation restriction sequence:

\[ 0 \to H^1(m/k, \mathrm{Cl}_{m}) \to H^1(l/k, \mathrm{Cl}_{l}) \to H^1(l/m,\mathrm{Cl}_{l}). \]

The first term is zero by 125 and the last term is zero by the inductive hypothesis. Therefore $H^1(l/k,\mathrm{Cl}_l) \cong 0$.

Since $H^1(l/m,\mathrm{Cl}_l)$ is assumed to be zero, we also have an inflation-restriction sequence in dimension 2:

\[ 0 \to H^2(m/k, \mathrm{Cl}_m) \to H^2(l/k, \mathrm{Cl}_l) \to H^2(l/m,\mathrm{Cl}_l). \]

By the inductive hypothesis we have $|H^2(l/m,\mathrm{Cl}_l)| \le [l:m]$ and by 125 we have $|H^2(m/k, \mathrm{Cl}_m)| = [m : k]$. It follows that $|H^2(l/k,\mathrm{Cl}_l)| \le [l : k]$.

To complete the construction of fundamental classes and the reciprocity isomorphism, we need only show that there is an element in $H^2(l/k,\mathrm{Cl}_l)$ of order $[l:k]$. Such an element is constructed first for a cyclic cyclotomic extension $l'/k$ with the same degree as $l/k$. It’s then shown that the inflation of such a class to $ll'/k$ must split on $ll'/l$, and must therefore be the inflation of an element of order $[l:k]$ in $H^2(l/k,\mathrm{Cl}_l)$.

Class Field Theory

4 Global Class Field Theory

4.1 Choice of \(S\)

4.2 The Herbrand quotient of the \(S\)-ideles

4.3 The Herbrand quotient of the \(S\)-units

4.4 Dirichlet Density

4.5 Some \(L\)-functions

4.6 The first inequality