Cohomology of the Idèles: The Class Field Axiom

Abstract

This chapter starts global class field theory: cohomology of idèles, Kummer extensions, second inequality.

In this chapter, we begin investigating cohomological properties of the groups $I_k$and $C_k$ defined in the previous chapter for a global field k. The aim is to prove the global analogue of Proposition 8.2, where the idèle class group will replace the multiplicative group of a local field. It is this result that (like in local class field theory) is the first step in the computation of the Brauer group of a global field.

1 Cohomology of the Idèle Group

In this paragraph, we fix a global field k and a separable closure $\overline{k}$ of k. We will consider finite separable extensions K of k, that will always be implicitly assumed to be contained in $\overline{k}$. For any place v of k, we fix a separable closure $\overline{k_v}$ of $k_v$ and a k-embedding $i_v:\overline{k}\mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }\overline{k_v}$, which fixes a place $\overline{v}$ of $\overline{k}$ above v: for $\overline{v}$ we take the restriction of the place of $\overline{k_v}$ induced by v. This allows for any separable algebraic extension $K \subset \overline{k}$ of k to have a distinguished place $v^{\bullet }$ of K above v. For simplicity we denote by $K_v:=i_v(K)k_v$ the field $K k_v$ (for example $(\overline{k})_v=\overline{k}k_v$), which is the completion of K at $v^{\bullet }$ if [K : k] is finite. We denote by $U_{K, v}$ the group of units of $K_v^*$, and $G_v(K/k)$ (or even $G_v$ if there is no ambiguity) the decomposition group of $v^{\bullet }$ in $G=\mathrm {Gal}(K/k)$ if K is a Galois extension of k.

Let K be a finite Galois extension of k with group G. We can write the idèle group of K as the restricted product of the $I_K(v)$ for v a place of k with respect to the $U_K(v)$, where we have set

$$I_K(v):=\prod _{w \vert v} K_w^*; \quad U_K(v):= \prod _{w \vert v} U_{K, w}.$$

($U_{K, w}$ is the multiplicative group of the ring of integers of $K_w$ for w a finite place of K). Each of the $I_K(v)$ and the $U_K(v)$ is a $G$-module via the $k_v$-isomorphism $K_w \simeq K_{\sigma w}$ induced by each $\sigma \in G$. We can also view $I_K(v)$ as the group of invertible elements of the ring $K \otimes _k k_v$, on which the group G acts $k_v$-linearly (via its natural action on K).

Proposition 12.14 easily implies that $I_K(v)$ (resp.$U_K(v)$) identifies with the induced module $I_G^{G_v} (K_v^*)$ of $K_v^*$ (resp.to $I_G^{G_v}(U_{K, v})$).

Proposition 13.1

With the above notations:

• (a) we have $I_k=H^0(G, I_K)$.

• (b) for any $i \in \mathbf{Z}$, we have

$${\widehat{H}}^i(G,I_K)=\bigoplus _{v \in \varOmega _k}{\widehat{H}}^i(G_v, K_v^*).$$

Proof

• (a) We have a natural injection of $I_k$ into $I_K$ via the decomposition (12.1). Let $\alpha \in I_k$. For any $\sigma \in G$ and any place w of K above a place v of k, the component $(\sigma \alpha )_{\sigma w}$ of $\sigma \alpha$ at $\sigma w$ is $\sigma \alpha _w=\alpha _w=\alpha _{\sigma w}$, hence $\sigma \alpha =\alpha$. Conversely, if $\alpha =(\alpha _w)$ is in $H^0(G, I_K)$, then

  $$(\sigma \alpha )_{\sigma w}=\sigma \alpha _w=\alpha _{\sigma w}$$

  for any $\sigma \in G$. For $\sigma \in G_w$, this already gives $\alpha _w \in k_v^*$. The fact that G acts transitively on the places w above v (Proposition 12.14) then gives that the $\alpha _w$ for $w \vert v$ come from the same $\alpha _v \in k_v^*$ via embeddings $k_v^* \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }K_w^*$, hence $\alpha \in I_k$.

• (b) Shapiro’s lemma gives

  $${\widehat{H}}^i(G,I_K(v))={\widehat{H}}^i(G_v, K_v^*)$$

  and besides if v is unramified in the extension K/k, we have

  $${\widehat{H}}^i(G,U_K(v))={\widehat{H}}^i(G_v,U_{K, v})=0$$

  since $U_{K, v}$ is a cohomologically trivial $G_v$-module by Proposition 8.3. Consider the finite sets S of places of k, containing places ramified in K/k and the archimedean places. Let us set

  $$I_{K, S}=\prod _{v \in S} I_K(v) \times \prod _{v \not \in S} U_K(v).$$

  Then $I_K$ is the inductive limit of the $I_{K, S}$. We deduce (with the help of the analogue for modified groups of Proposition 1.25, cf.  also the comment after Corollary 2.7):

   $\square$

Corollary 13.2

We have $H^1(G, I_K)=H^3(G, I_K)=0$.

Proof

By the previous proposition, we need to see that for every place v of k, we have $H^1(G_v, K_v^*)=H^3(G_v, K_v^*)=0$. The first assertion follows from the Hilbert 90 theorem. The second follows from it for real v by Theorem 2.16. For v finite, Tate–Nakayama theorem (Theorem 3.14) gives an isomorphism between $H^1(G_v,\mathbf{Z})=0$ and $H^3(G_v, K_v^*)$.    $\square$

Note also the following lemma about the behaviour of the idèle groups with respect to the norm.

Lemma 13.3

Let K be a finite Galois extension of k. An idèle $\alpha =(\alpha _v) \in I_k$ is in $N_{K/k} I_K$ if and only if for every place v of k, its local component $\alpha _v \in k_v^*$ is a norm for the extension $K_v/k_v$ (recall that $K_v$ is the completion of K for a place w above v).

Proof

This follows immediately from Proposition 13.1(b) applied to $i=0$. We can also simply observe that the norm $N_{K/k}: I_K \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }I_k$ is obtained by taking the map $(\prod _{w \vert v} N_{K_w/k_v}): I_K(v) \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }k_v^*$ on each $I_K(v)$, where v ranges through the set of places of  k.    $\square$

We would like to “pass to the limit” in some of the above assertions. Denote by the idèle group of $\overline{k}$, the limit being taken over the finite separable extensions K of k (warning: note that it is not the restricted product of the $(\overline{k})_v^*$). We also call $C:=I/\overline{k}^*$ the idèle class group of $\overline{k}$, which is the inductive limit of the $C_K$. Proposition 13.1(a) and Hilbert 90 theorem yield:

Proposition 13.4

If K is a finite Galois extension of k with group G, we have $H^0(G, C_K)=C_k$. Similarly, we have $H^0(k, I)=I_k$ and $H^0(k, C)=C_k$.

By passing to the limit in Proposition 13.1(b), we obtain

$$H^i(k, I)=\bigoplus _{v \in \varOmega _k} H^i(G_v(\overline{k}/k),(\overline{k})_v^*)$$

for any $i \geqslant 1$. In this statement we would like to replace $(\overline{k})_v$ by the separable closure $\overline{k_v}$ of $k_v$. We effectively have that $(\overline{k})_v=\overline{k}k_v$ is also $\overline{k_v}$, but this statement is non-trivial; it uses the following lemma.

Lemma 13.5

(Krasner) Let F be a complete field for an ultrametric absolute value (not necessarily associated to a discrete valuation), with separable closure $\overline{F}$. Let $\alpha \in \overline{F}$, we denote by $\alpha _1=\alpha ,\dots ,\alpha _n$ its conjugates over F. Let $\beta \in \overline{F}$ satisfying

$$\vert \alpha -\beta \vert < \vert \alpha -\alpha _i \vert $$

for $i=2,\dots , n$, where $\vert . \vert$ is the unique extension of the absolute value of F to $\overline{F}$. Then we have the inclusion of fields $F(\alpha ) \subset F(\beta )$.

Proof

Let L be the Galois closure over $F(\beta )$ of the field extension $F(\alpha ,\beta )/F(\beta )$. Let $\sigma \in G:=\mathrm {Gal}(L/F(\beta ))$. Then $\sigma (\beta -\alpha )=\beta - \sigma (\alpha )$. We also have that $\sigma$ preserves $\vert . \vert$ by uniqueness of the extension of the absolute value in the complete case, which follows from the equivalence of the norms on finite dimensional vector spaces over a complete field (the existence for a finite extension of a complete field can for example be obtained using the formula (7.1) of Theorem 7.7). Thus we obtain

$$\vert \beta -\sigma (\alpha ) \vert =\vert \beta -\alpha \vert <\vert \alpha _i-\alpha \vert $$

for $i=2,\dots , n$. We deduce that

$$\vert \alpha -\sigma (\alpha ) \vert <\vert \alpha -\alpha _i \vert $$

for $i=2,\dots , n$ since $\vert \cdot \vert$ is ultrametric. Finally $\sigma (\alpha )=\alpha$, that is: $\alpha \in F(\beta )$, as desired.    $\square$

We deduce the announced result.

Proposition 13.6

Let k be a global field. Let v be a place of k. Then $(\overline{k})_v$ (with the conventions explained at the beginning of the paragraph) identifies with the separable closure $\overline{k_v}$ of the completion $k_v$ of k at v.

Proof

We have a natural inclusion $(\overline{k})_v \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }\overline{k_v}$. We can assume that v is a finite place (the archimedean case is trivial). Let $\alpha \in \overline{k_v}$ with minimal polynomial $f \in k_v[X]$. As k is dense in $k_v$, we can find a separable polynomial $g \in k[X]$ very close to f, which implies that $\vert g(\alpha ) \vert$ can be made as small as we wish. Write $g(X)=\prod (X-\beta _j)$ with $\beta _j \in \overline{k}\subset (\overline{k})_v$. Then g has a root $\beta$ which can be made as close as we wish to $\alpha$, hence in particular made to satisfy $\vert \beta -\alpha \vert < \vert \alpha _i -\alpha \vert$ for all the conjugates $\alpha _i \in \overline{k_v}$ of $\alpha$ other than $\alpha$. Krasner’s lemma then implies $\alpha \in k_v(\beta )=k(\beta )_v \subset (\overline{k})_v$.    $\square$

Corollary 13.7

For any $i \geqslant 1$, we have $H^i(k, I)=\bigoplus _{v \in \varOmega _k} H^i(k_v,\overline{k}_v^*)$, where $\overline{k}_v$ is the separable closure of $k_v$. In particular, $H^1(k, I)=0$.

Note that by what we have just seen, the notation $\overline{k}_v$ is not ambiguous.

2 The Second Inequality

The aim of this paragraph is to show:

Theorem 13.8

Let k be a global field. Let K be a Galois extension of k with Galois group G assumed to be cyclic of order n. Let $h(G, C_K)$ be the Herbrand quotient of the $G$-module $C_K$, where $C_K=I_K/K^*$ is the idèle class group of K. Then $h(G, C_K)=n$.

We deduce:

Corollary 13.9

(“the second inequality”) Under the assumptions of the preceding theorem, the cardinality of the group $C_k/N_{K/k} C_K=I_k/k^* N_{K/k} I_K$ is at least n.

Proof

Applying Hilbert 90 and Proposition 13.1(a), we have $H^0(G, C_K)=C_k$, hence ${\widehat{H}}^0(G, C_K)=C_k/N_{K/k} C_K=I_k/k^* N_{K/k} I_K$. The corollary then follows from the definition of the Herbrand quotient.    $\square$

The aim of this chapter is to show that we in fact have $\# {\widehat{H}}^0(G, C_K)=n$ in the above corollary. To do this, we will need the first inequality. Even though it can be proved using Corollary 13.9 (see next paragraphs), historically it was proved earlier using analytic methods (cf.  [9], Exp. VIII or Appendix B of this book, Corollary B.23). This explains the terminology we adopt here (some authors use the inverse convention, it is for example the case of [9], Exp. VII). Corollary 13.9 is mainly used to show that $K=k$ when one can verify that $I_k=k^* N_{K/k} I_K$.

The Proof of Theorem 13.8 uses the following lemma.

Lemma 13.10

Let G be a finite group.

• (a) Let M and $M'$ be two modules over the ring $\mathbf{Q}[G]$, of finite dimension over $\mathbf{Q}$, and such that the $\mathbf{R}[G]$-modules $M_{\mathbf{R}}:=M \otimes _{\mathbf{Q}} \mathbf{R}$ and $M'_{\mathbf{R}}$ are isomorphic. Then the $\mathbf{Q}[G]$-modules M and $M'$ are isomorphic.

• (b) Assume that G is cyclic. Let E be a finite dimensional $\mathbf{R}$-vector space endowed with an action of G. Let L and $L'$ be two lattices in E, stables under the action of G, and generating the $\mathbf{R}$-vector space E. Then if one of the Herbrand quotients h(L), $h(L')$ is defined, the other is too and we have $h(L)=h(L')$.

Proof

• (a) Each $\mathbf{Q}[G]$-homomorphism $\varphi : M \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }M'$ induces a $\mathbf{R}[G]$-homomorphism $\varphi \otimes 1: M_{\mathbf{R}} \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }M'_{\mathbf{R}}$ and the map $\varphi \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }\varphi \otimes 1$ induces an isomorphism of $\mathbf{R}$-vector spaces:

  $$({\mathrm{Hom}}_{\mathbf{Q}[G]}(M, M'))\otimes _{\mathbf{Q}} \mathbf{R}\mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }{\mathrm{Hom}}_{\mathbf{R}[G]} (M_{\mathbf{R}}, M'_{\mathbf{R}}).$$

  Fix bases of the $\mathbf{Q}$-vector spaces M and $M'$ (which have the same dimension). The determinant of an element of ${\mathrm{Hom}}_{\mathbf{Q}[G]}(M, M')$ or of ${\mathrm{Hom}}_{\mathbf{R}[G]} (M_{\mathbf{R}}, M'_{\mathbf{R}})$ in these bases is then well defined. Let then $(\xi _i)$ be a basis of the $\mathbf{Q}$-vector space ${\mathrm{Hom}}_{\mathbf{Q}[G]}(M, M')$. The above isomorphism shows that it is a basis of the $\mathbf{R}$-vector space ${\mathrm{Hom}}_{\mathbf{R}[G]} (M_{\mathbf{R}}, M'_{\mathbf{R}})$. The assumption that $M_{\mathbf{R}}$ and $M'_{\mathbf{R}}$ are isomorphic implies that the polynomial

  $$F(t_1,\dots , t_n):=\det (\sum _i t_i \xi _i) \in \mathbf{Q}[t_1,\dots , t_m]$$

  is not identically zero on the whole of $\mathbf{R}^m$, hence it is nonzero on $\mathbf{Q}^m$ since $\mathbf{Q}$ is infinite. Thus we have a $\mathbf{Q}[G]$-isomorphism between M and $M'$.

• (b) Let us set $M=L \otimes \mathbf{Q}$ and $M'=L' \otimes \mathbf{Q}$. Then $M_{\mathbf{R}}$ and $M'_{\mathbf{R}}$ are both $\mathbf{R}[G]$-isomorphic to E. By a), there exists a $\mathbf{Q}[G]$-isomorphism $\varphi : M \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }M'$. Then $\varphi$ induces an injective homomorphism $\varphi : L \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }(1/N)L'$ for a certain integer $N >0$. This implies that $f:=N \varphi$ is an injective homomorphism from L to $L'$. As L and $L'$ are lattices of the same rank, the cokernel of f is finite and we conclude by applying Theorem 2.20, (c).

   $\square$

Proof of Theorem 13.8

By Proposition 12.25, we can find a finite non-empty set S of places of k (containing the archimedean places and the places ramified in K/k) such that $I_K=K^* I_{K, S}$, where

$$I_{K,S}=\prod _{v \in S} I_K(v) \times \prod _{v \not \in S} U_K(v)= \prod _{v \in S} (\prod _{w \vert v} K_w^*) \times \prod _{v \not \in S} (\prod _{w \vert v} U_{K, w}^*).$$

Let T be the (finite) set of places of K which are above a place of S. We then have

$$C_K=I_K/K^*=I_{K,S}/E_{K, T},$$

where we recall that $E_{K,T}:=K^* \cap I_{K, S}$ is the group of T-units of K. By Theorem 2.20, we have, shortening $h(G,\cdot )$ to $h(\cdot )$,

$$h(C_K)=h(I_{K,S})/h(E_{K, T})$$

whenever the right hand side member is defined (the advantage to introduce S and T is precisely that it allows this condition to be satisfied, while $h(K^*)$ is not defined since ${\widehat{H}}^0(G, K^*)=K^*/NL^*$ is in general infinite).

We first compute $h(I_{K, S})$ purely “locally”. For $v \not \in S$, the place v is unramified in K/k, which implies as we have already seen (consequence of Shapiro’s lemma and of Proposition 8.3) that the $G$-module $U_K(v)$ is cohomologically trivial. Let now $v \in S$, denote by $n_v:=[K_v:k_v]$ the local degree of K/k at v. Shapiro’s lemma implies that for v in S, we have:

$$h(I_K(v))=h(G_v, K_v^*)=n_v$$

by local class field axiom (Proposition 8.2) hence finally

$$h(I_{K, S})=\prod _{v \in S} n_v.$$

We now compute $h(E_{K, T})$, which is the “global” part of the proof. We need to show that $n\cdot h(E_{K, T})=\prod _{v \in S} n_v$. To do this, we will use Lemma 13.10. Let V be the vector space of maps from  T to $\mathbf{R}$, which is isomorphic to $\mathbf{R}^t$ with $t=\# T$. The group G acts on V by the formula $(\sigma f)(w)=f(\sigma ^{-1}w)$ for all $f \in V$, $\sigma \in G$, $w \in T$. Let N be the lattice consisting of $f \in V$ whose image is contained in $\mathbf{Z}$. It is clear that N generates the $\mathbf{R}$-vector space V and is stable for the action of G. The $G$-module N is isomorphic to $\prod _{v \in S} (\prod _{w \vert v} \mathbf{Z}_w)$, with $\mathbf{Z}_w=\mathbf{Z}$. More precisely, if we set (for a fixed v in S) $N_v:=\prod _{w \vert v} \mathbf{Z}_w$, then $N_v$ is a sub-$G$-module of N and G acts on $N_v$ by permutation of the $\mathbf{Z}_w$. Shapiro’s lemma then yields, for any $q \in \mathbf{Z}$:

$${\widehat{H}}^q(G, N)=\prod _{v \in S} {\widehat{H}}^q(G_v,\mathbf{Z})$$

which implies immediately

$$h(N)=\prod _{v \in S} n_v.$$

For $a \in E_{K, T}$, denote by $f_a: T \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }\mathbf{R}$ the map defined by $f_a(w)=\log \vert a \vert _w$ and apply Lemma 12.26. It implies that the map

$$\lambda : E_{K, T} \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }V, \quad a \mathchoice{\longmapsto }{\mapsto }{\mapsto }{\mapsto }f_a$$

has a finite kernel, and that the image of f is a lattice $M^0$ which generates the $\mathbf{R}$-vector space $V^0 \subset V$ consisting of f such that $\sum _{w \in T} f(w)=0$. We have $h(E_{K, T})=h(M^0)$ by Theorem 2.20, (c). Let $g \in V$ be defined by $g(w)=1$ for any $w \in T$. Let us set $M=M^0+\mathbf{Zg}$. Then the lattice M generates the $\mathbf{R}$-vector space $V=V^0+\mathbf{Rg}$. As $M^0$ and $\mathbf{Z}\cdot g$ are both stable by G, we can write

$$h(M)=h(M^0)\cdot h(\mathbf{Z})=nh(M^0)=nh(E_{K, T}).$$

As M and N generate the same $\mathbf{R}$-vector space, Lemma 13.10 implies that $h(N)=h(M)$. Finally $n\cdot h(E_{K, T})=h(N)$, or

$$n\cdot h(E_{K, T})=\prod _{v \in S} n_v,$$

as desired.   $\square$

We deduce:

Proposition 13.11

Let K be a finite abelian extension of a global field k. We assume that there exists a subgroup D of $I_k$ satisfying: $D \subset N_{K/k} I_K$ and $k^* D$ is dense in $I_k$. Then $K=k$.

Proof

If F is a cyclic extension of k contained in K, we have $D \subset N_{K/k} I_K \subset N_{F/k} I_F$ by transitivity of the norms. Thus, if we can deal with the case of a cyclic extension, we obtain that every subextension F of K with F/k cyclic is trivial, which implies that K/k is trivial by Galois theory as K/k is assumed to be abelian. We can thus assume that K is a cyclic extension of k.

Now, Theorem 11.20, (a) tells us that each $N_{K_v/k_v} K_v^*$ is an open subgroup of $K_v^*$. Furthermore $N_{K_v/k_v} K_v^*$ contains $U_v:=\mathcal {O}_v^*$ for almost all places v since v unramified implies ${\widehat{H}}^0(G_v,U_{K, v})=0$ with $G_v=\mathrm {Gal}(K_v/k_v)$ (Proposition 8.3). We deduce, using Lemma 13.3, that $N_{K/k} I_K$ is an open subgroup of $I_k$, hence also $k^* N_{K/k} I_K$ (union of open sets). Thus $k^* N_{K/k} I_K$ is a closed subgroup (as it is open) and dense (as it contains $k^* D$) of $I_k$, hence $k^* N_{K/k} I_K=I_k$. Corollary 13.9 then implies [K : k]$=1$, i.e., $K=k$.    $\square$

Proposition 13.12

Let S be a finite set of places of k, containing all archimedean places if  k is a number field. Let K be a finite abelian extension of k, unramified outside S. Then the Galois group $G=\mathrm {Gal}(K/k)$ is generated by the Frobenii $F_{K/k}(v)$ for $v \not \in S$.

Proof

Note that since G is abelian, the Frobenius $F_v:=F_{K/k}(v)$ is well defined as an element of G (and not just up to conjugation). Let $G'$ be the subgroup of G generated by the $F_v$ for $v \not \in S$ and let E be its fixed field. Then the image of $F_v$ in $\mathrm {Gal}(E/k)=G/G'$ is trivial for $v \not \in S$, which gives $E_v=k_v$, and hence each element of $k_v$ is a norm of $E_v/k_v$ for $v \not \in S$. Denote then by D the subgroup of $I_k$ consisting of the idèles $(\alpha _v)$ such that $\alpha _v=1$ for all $v \in S$. Then by Lemma 13.3 and what we just saw, we have $D \subset N_{E/k} I_E$. On the other hand, strong approximation Theorem 12.18 (or even its “weak” form, which is the one without a condition outside S) yields that $k^* D$ is dense in $I_k$. Proposition 13.11 then says that $E=k$, or that $G'=G$.    $\square$

Corollary 13.13

Let K be a non-trivial abelian extension of k. Then there are infinitely many places v of k which are not totally split in K. If furthermore K/k is cyclic of degree $\ell ^m$ with $\ell$ prime, there are infinitely many places of k which are inert in K/k (that is, unramified and such that the associated decomposition group is the whole of $G:=\mathrm {Gal}(K/k)$).

Note that an inert place v corresponds to a Frobenius $F_{K/k}(v)$ that generates $\mathrm {Gal}(K/k)$, or to a place that is, unramified and there is only one place w of K above v.

Proof

The first assertion follows from Proposition 13.12 and the fact that places v totally split in K correspond to $F_{K/k}(v) = 1$. The second assertion is obtained by applying the first to the intermediate extension E/k of degree $\ell$, that corresponds to the unique subgroup of order $\ell ^{m-1}$ of $\mathrm {Gal}(K/k)$: we obtain infinitely many places v for which the Frobenius $F_{K/k}(v)$ has a non-trivial image in the quotient of order $\ell$ of G, hence generates G.    $\square$

For example, the above corollary implies that an element of $\mathbf{Z}$ that is, not a square can not become a square modulo all primes except finitely many. There is in fact a more general and precise statement than Corollary 13.13, the Čebotarev theorem (cf.  Chap. 18).

3 Kummer Extensions

In this paragraph, we establish general results on Kummer extensions, that will be useful in the next paragraph and also in the proof of the global existence theorem.

Fix a field k and an integer $n >0$ not divisible by the characteristic of k, such that k contains a primitive nth root of unity $\zeta$.

Definition 13.14

A Kummer extension of k is a field extension K of k of the form $K=k({}\root n \of {\varDelta })$, where $\varDelta$ is a subgroup of $k^*$, containing $k^{*n}$, and such that $\varDelta /k^{*n}$ is finite.¹

Note that as $\zeta \in k$, the notation $k(\root n \of {\varDelta })$ is not ambiguous, and the choice of a primitive nth root of unity in k identifies the Galois modules $\mu _n$ and $\mathbf{Z}/n$. As for any $a \in \varDelta$, the extension $k(\root n \of {a})$ is cyclic of degree dividing n (it corresponds to the element of $H^1(k,\mathbf{Z}/n) \simeq H^1(k,\mu _n)=k^*/k^{*n}$ given by the class of a), a Kummer extension is a finite abelian extension of k whose Galois group is of exponent dividing n (isomorphic to a subgroup of $(\mathbf{Z}/n)^r$, where r is the cardinality of $\varDelta /k^{*n}$). Conversely, a cyclic extension of k of degree dividing n corresponds to an element of $H^1(k,\mathbf{Z}/n) \simeq k^*/k^{*n}$, hence is of the form $k(\root n \of {a})$. More generally, we have:

Proposition 13.15

Let K be a finite abelian extension of k whose Galois G is of exponent n. Then $K=k(\root n \of {\varDelta })$ with $\varDelta =K^{*n} \cap k^*$. Furthermore, we have an isomorphism

$$\overline{u}: \varDelta /k^{*n} \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }{\mathrm{Hom}}(G,\mathbf{Z}/n).$$

Conversely, if $\varDelta '$ is a subgroup of $k^*$ containing $k^{*n}$ and if we set $K=k(\root n \of {\varDelta '})$, then we have $\varDelta '=K^{*n} \cap k^*$.

Thus, for an abelian extension K of k with Galois group of exponent n, we have one and only one subgroup $\varDelta$ of $k^*$ containing $k^{*n}$ such that $K=k(\root n \of {\varDelta })$.

Proof

We clearly have $k(\root n \of {\varDelta }) \subset K$. Conversely, the extension K/k is the composite of all its cyclic subextensions E/k as it is abelian (as a finite abelian group is a direct product of cyclic groups). Such an extension E is of degree dividing n, hence is of the form $E=k(\root n \of {a})$ with $a \in k^* \cap K^{*n}$, hence $E \subset k(\root n \of {\varDelta })$ and finally $K \subset k(\root n \of {\varDelta })$. We define a morphism $u: \varDelta \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }{\mathrm{Hom}}(G,\mathbf{Z}/n)$ by $u(a)=\chi _a$, where

$$\begin{aligned} \chi _a(\sigma )=\sigma (\root n \of {a})/\root n \of {a} \end{aligned}$$

(13.1)

is defined using the identification of $\mathbf{Z}/n$ with the subgroup of nth roots of unity of $k^*$. The kernel of u is $k^{*n}$ since $\chi _a$ is trivial if and only if $\root n \of {a} \in k^*$. It remains to show that u is surjective. Let $\chi \in {\mathrm{Hom}}(G,\mathbf{Z}/n)={\mathrm{Hom}}(G,\mu _n)$, then $\chi$ becomes a 1-coboundary (by Hilbert 90) when viewed as a map from  G to $K^*$. This means that we have $b \in K^*$ such that $\chi (\sigma )=\sigma (b)/b$ for any $\sigma \in G$. Then

$$\sigma (b^n)=(\sigma b)^n=\chi (\sigma )^n b^n=b^n$$

hence $a:=b^n \in k^* \cap K^{*n}=\varDelta$, and finally $\chi =\chi _a$.

Lastly, let $\varDelta '$ be a subgroup of $k^*$ containing $k^{*n}$ and $K:=k(\root n \of {\varDelta '})$. Let us set $\varDelta =K^{*^n } \cap k^*$ and let us show that $\varDelta =\varDelta '$. The inclusion $\varDelta ' \subset \varDelta$ is obvious. Denote by H the subgroup of $G=\mathrm {Gal}(K/k)$ consisting of  $\sigma$ such that $\chi _a(\sigma )=1$ for every $a \in \varDelta '$. That means that H is the orthogonal of the family $(\chi _a)_{a \in \varDelta '}$ for the duality between G and ${\mathrm{Hom}}(G,\mathbf{Z}/n)$. By biduality, the orthogonal of H is the subgroup of ${\mathrm{Hom}}(G,\mathbf{Z}/n)$ consisting of the $\chi _a$ for $a \in \varDelta '$, which means that the image of $\varDelta '/k^{*n}$ by the isomorphism $\overline{u}: \varDelta /k^{*n} \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }{\mathrm{Hom}}(G,\mathbf{Z}/n)$ is exactly ${\mathrm{Hom}}(G/H, \mathbf{Z}/n)$. Besides, the formula (13.1) shows that H fixes each element of $\root n \of {\varDelta '}$, hence fixes the whole of K (which is the extension of k generated by $\root n \of {\varDelta '}$). Galois theory then tells us that H is the trivial group. As $\overline{u}$ is an isomorphism, this implies that $\varDelta /k^{*n}=\varDelta '/k^{*n}$ and finally $\varDelta =\varDelta '$, as desired.    $\square$

Remark 13.16

The link between the last proposition and cohomological Kummer theory seen in Sect. 6.2 (Corollary 6.6) is the following: the group $H^1(G,\mathbf{Q}/\mathbf{Z})=H^1(\mathrm {Gal}(K/k),\mathbf{Z}/n)$ is (by restriction-inflation sequence) given by

$$H^1(G,\mathbf{Q}/\mathbf{Z})={{\,\mathrm{Ker}\,}}[H^1(k,\mathbf{Z}/n) \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }H^1(K,\mathbf{Z}/n)].$$

Identifying (via a choice of $\zeta$) the Galois modules $\mathbf{Z}/n$ and $\mu _n$, we obtain an isomorphism between $H^1(G,\mathbf{Q}/\mathbf{Z})$ and the kernel of the map $k^*/k^{*n} \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }K^*/K^{*n}$, that is, with $\varDelta /k^{*n}$.

We will need the following “local” result.

Lemma 13.17

Let K be a local field. Let $n >0$ be an integer not divisible by the characteristic of the residue field $\kappa$ of K, Assume that K contains primitive nth roots of unity (and hence n divides $q-1$, where $q=\# \kappa$). Then, for $x \in K^*$, the extension $L=K(\root n \of {x})/K$ is unramified if and only if $x \in U_K K^{*n}$.

Proof

Assume that $x=u\cdot y^n$ with $u \in U_K$ and $y \in K^*$. We want to show that $K(\root n \of {u})$ is unramified over K. By Hensel lemma, the polynomial $X^n-u$ factors as a product of linear factors over the unramified extension $K'$ of K whose residue field is the decomposition field of the reduction $X^n-\overline{u}$ over $\kappa$. Thus $\root n \of {u} \in K'$ and we have that $K(\root n \of {u})$ is unramified over K. Conversely, if $K(\root n \of {x})/K$ is unramified, let us write $x=u\cdot \pi ^r$ with $u \in U_K$ and $\pi$ a uniformiser of K. Then the valuation $v_L(\root n \of {u\cdot \pi ^r})$ is $\frac{1}{n} v_L(\pi ^r)= \frac{1}{n} v_K(\pi ^r)$, which shows that n divides r.    $\square$

Lastly, Kummer extensions are related to another “local” result, that we will use in the next paragraph.

Proposition 13.18

Let k be a global field. Let v be a place of k. Let $n >0$ be an integer not divisible by the characteristic of k (which is automatic if  k is a number field), and such that $\mu _n \subset k_v$. Then the cardinality of $k_v^*/k_v^{*n}$ is $n^2/\vert n \vert _v$ and if v is finite, the cardinality of $\mathcal {O}_v^*/\mathcal {O}_v^{*n}$ is $n/\vert n \vert _v$.

Proof

The case where v is archimedean is straightforward, observing that for real v we necessarily have $n=1$ or $n=2$, and for a complex v $\vert n \vert _v=n^2$ by convention. Let us now assume that v is finite. As $k_v^* \simeq \mathbf{Z}\times \mathcal {O}_v^*$, it is enough to compute the cardinality of $\mathcal {O}_v^*/\mathcal {O}_v^{*n}$. To do this, make the group $\mathbf{Z}/n$ acts trivially on $\mathcal {O}_v^*$, and consider the corresponding Herbrand quotient $h_n(\mathcal {O}_v^*)$. As $H^1(\mathbf{Z}/n,\mathcal {O}_v^*)={\mathrm{Hom}}(\mathbf{Z}/n,\mathcal {O}_v^*)$ is of cardinality n (because $k_v \supset \mu _n$), we have

$$h_n(\mathcal {O}_v^*)=\frac{\# \mathcal {O}_v^*/\mathcal {O}_v^{*n}}{n}.$$

We are reduced to showing that $h_n(\mathcal {O}_v^*)=\vert n \vert _v^{-1}$. If k is a function field of characteristic p, the assumption made on n implies that $\vert n \vert _v=1$. Besides, the subgroup of finite index $U_v^1$ of $U_v=\mathcal {O}_v^*$ is a pro-p-group (Theorem 7.18, a), hence satisfies $h_n(U_v^1)=1$ as p does not divide n. We conclude using Theorem 2.20. If k is a number field, the group $U_v^1$ has a subgroup of finite index isomorphic to the additive group $\mathcal {O}_v$ (Theorem 7.18, b). As $H^1(\mathbf{Z}/n,\mathcal {O}_v)=\mathcal {O}_v[n]=0$, Theorem 2.20 implies that

$$h_n(U_v)=h_n(U_v^1)=h_n(\mathcal {O}_v)=\# (\mathcal {O}_v/n \mathcal {O}_v) =\vert n \vert _v^{-1},$$

as desired.    $\square$

4 First Inequality and the Axiom of Class Field Theory

We start with a linear algebra lemma:

Lemma 13.19

Let $\ell$ be a prime number and let m be a non negative integer. We set $n=\ell ^m$. Let A be the ring $\mathbf{Z}/n \mathbf{Z}$. Let M be a free finite type A-module and $M_1$ a sub-A-module of M such that $M/M_1$ is free. Then $M_1$ is free of finite type.

Proof

As $M_2:=M/M_1$ is free, we have $M \simeq M_1 \oplus M_2$. As an abelian group, M is isomorphic to $(\mathbf{Z}/n \mathbf{Z})^r$ with $r \geqslant 0$. The theorem on the structure of finite abelian groups implies that the abelian n-torsion group $M_1$ is isomorphic to a direct sum of groups of the form $(\mathbf{Z}/n' \mathbf{Z})$, where $n'$ is a power of $\ell$. But the uniqueness part of this same theorem imposes that $n'=n$ (we could also have used a more advanced theorem which says that every projective finite type module over a local ring is free cf.  [36], Part. I, Th. 2.9).    $\square$

We are going back to global fields. In the remaining part of this paragraph, k is a global field; in all the statements before Theorem 13.23, we denote by n a power of a prime number $\ell$ different from the characteristic of k, such that k contains a primitive nth root of unity $\zeta$. The first aim will be to compute (by a method due to Chevalley) the norm group $N_{K/k} C_K$ for a Kummer extension K/k (in the sense of Definition 13.14) with Galois group $G=(\mathbf{Z}/n)^r$. We start by fixing a finite set S of places of k, containing all the archimedean places (if k is a number field), all places above $\ell$, all places ramified in K/k, and such that $I_k=I_{k, S} k^*$ (cf.  Proposition 12.25). We denote by s the cardinality of S.

Proposition 13.20

With the above notation, we have $s \geqslant r$. Furthermore, there exists a set T of places of k, disjoint from S and of cardinality $s-r$, such that $K=k(\root n \of {\varDelta })$, where $\varDelta$ is the kernel of the diagonal map

$$E_{k, S} \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }\prod _{v \in T} k_v^*/k_v^{*n},$$

with $\varDelta =K^{*n} \cap E_{k, S}$.

(Recall that $I_{k, S}$ is the group of S-idèles and $E_{k,S}=k^* \cap I_{k, S}$ is the group of S-units).

Proof

We start by showing that $K=k(\root n \of {\varDelta })$ with $\varDelta =K^{*n} \cap E_{k, S}$. Proposition 13.15 gives $K=k(\root n \of {D})$ with $D=K^{*n} \cap k^*$. For $x \in D$, the choice of S yields that $k(\root n \of {x})/k$ is unramified for $v \not \in S$, hence $x=u_v y_v^n$ with $u_v \in U_v:=\mathcal {O}_v^*$ and $y_v \in k_v^*$ (Lemma 13.17). Let us set $y_v=1$ for $v \in S$. We obtain an idèle $y=(y_v)$ (indeed, x is of valuation zero outside a finite number of places), that we can write (again, by the choice of S) $y=\alpha \cdot z$ with $\alpha \in I_{k, S}$ and $z \in k^*$. Then $x/z^n \in I_{k,S} \cap k^*=E_{k, S}$ hence $x/z^n \in \varDelta$. Thus $D \subset \varDelta \cdot k^{*n}$, and the reverse inclusion is obvious, hence $K=k(\root n \of {\varDelta })$.

Let us set $N=k(\root n \of {E_{k, S}})$, then $N \supset K$ as $E_{k, S} \supset \varDelta$. As $E_{k,S} k^{*n}/k^{*n}=E_{k,S}/E_{k, S}^n$ (observe that $k^{*n} \cap E_{k,S}=E_{k, S}^n$, where $E_{k, S}^n$ is the subgroup of nth powers in $E_{k, S}$). Proposition 13.15 yields

$$\mathrm {Gal}(N/k) \simeq {\mathrm{Hom}}(E_{k,S}/E_{k, S}^n,\mathbf{Z}/n).$$

On the other hand $E_{k, S}$ contains the nth roots of unity and hence is an abelian group isomorphic (by the Dirichlet unit theorem) to $\mathbf{Z}^{s-1} \times \mu _q$, where q is an integer such that n divides q. This implies that $E_{k,S}/E_{k, S}^n$ is a free $\mathbf{Z}/n$-module of rank s, hence it is also the case for $\mathrm {Gal}(N/k)$. As its quotient $G=\mathrm {Gal}(K/k)$ is by assumption a free $\mathbf{Z}/n$-module of rank r, we already obtain that $r \leqslant s$ and $\mathrm {Gal}(N/K)$ is a free $\mathbf{Z}/n$-module by Lemma 13.19.

Choose a $\mathbf{Z}/n$-basis $\sigma _1,\dots ,\sigma _{s-r}$ of $\mathrm {Gal}(N/K)$, and call $N_i$ the fixed field of $\sigma _i$ for $i=1,\dots , s-r$. Then $K=\bigcap _{1 \leqslant i \leqslant s-r} N_i$. By Corollary 13.13, we can find for each i a prime ideal $\mathfrak {p}_i$ of $N_i$, above a finite place $v_i$ of k, such that: the places $v_i$ are pairwise distinct, are not in S, and each $\mathfrak {p}_i$ is inert in the extension $N/N_i$ (Recall that the $N/N_i$ are cyclic of order n, with n a power of a prime number $\ell$). This means that the Frobenius associated to $\mathfrak {p}_i$ generates $\mathrm {Gal}(N/N_i)$, or that there is only one prime $\mathfrak {p}'_i=\mathfrak {p}_i \mathcal {O}_N$ of N above $N_i$. We will show that we can take $T=\{ v_1,\dots , v_{s-r} \}$.

Let us first show that $\mathrm {Gal}(N/N_i)$ is the decomposition group $D_i$ of $v_i$ in N/k. The place $v_i$ is unramified in the extension N/k by Lemma 13.17. In particular, $D_i$ is cyclic, generated by the Frobenius $F_{N/k}(v_i)$. On the other hand $D_i \supset \mathrm {Gal}(N/N_i)$ as every element of $\mathrm {Gal}(N/k)$ which induces the identity on $N_i$ fixes $\mathfrak {p}_i$, hence $\mathfrak {p}'_i$. As $\mathrm {Gal}(N/N_i)$ is of order n and $D_i$ is cyclic of exponent $\leqslant n$, the only possibility is $D_i=\mathrm {Gal}(N/N_i)$, as desired.

As $\mathrm {Gal}(N/K)$ is the direct product of the $\mathrm {Gal}(N/N_i)$, we obtain that K/k is the maximal subextension of N/k in which all the places $v_i$ are totally split: indeed, this property comes down to saying that K is the maximal subextension of N/k such that all the $F_{N/k}(v_i)$ are in $\mathrm {Gal}(N/K)$. Let then $x \in E_{k, S}$. We obtain:

$$x \in \varDelta \mathchoice{\Longleftrightarrow }{\Leftrightarrow }{\Leftrightarrow }{\Leftrightarrow }k(\root n \of {x}) \subset K \mathchoice{\Longleftrightarrow }{\Leftrightarrow }{\Leftrightarrow }{\Leftrightarrow }k_{v_i}(\root n \of {x})=k_{v_i},\ \forall i=1,\dots , s-r,$$

which means exactly that $\varDelta$ is the kernel of $E_{k, S} \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }\prod _{i=1}^{s-r} k_{v_i}^* /k_{v_i}^{*n}$.    $\square$

We are coming to the most difficult result of this paragraph, which includes in particular the “first inequality” $[C_k:N_{K/k} C_K] \leqslant [K:k]$  in the case of a Kummer extension.

Theorem 13.21

Under the assumptions and notation of Proposition 13.20, we set (for any non-archimedean place v of k) $U_v:=\mathcal {O}_v^*$ and

$$I_k(S, T):=\prod _{v \in S} k_v^{*n} \times \prod _{v \in T} k_v^* \times \prod _{v \not \in S \cup T} U_v.$$

Let $C_k(S,T)=I_k(S, T)\cdot k^*/k^*$. Then we have

$$N_{K/k} C_K \supset C_k(S, T)$$

and $[C_k:C_k(S, T)]= [K:k]$. If furthermore the extension K/k is cyclic, we have $C_k(S, T)=N_{K/k} C_K$ (and hence $[C_k:N_{K/k} C_K]= [K:k]$).

We start with a lemma.

Lemma 13.22

We have $I_k(S, T) \cap k^*=E_{k,{S \cup T}}^n$.

Proof

It is straightforward to see that we have the inclusion $E_{k,{S \cup T}}^n \subset I_k(S, T) \cap k^*$. Let conversely $y \in I_k(S, T) \cap k^*$. Let us set $M=k(\root n \of {y})$. To show that $M=k$, it is enough by Corollary 13.9 to see that $C_k=N_{M/k} C_M$. Let us first show that the diagonal map $f: E_{k, S} \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }\prod _{v \in T} U_v/U_v^n$ is surjective, where $U_v^n$ is the subgroup of nth powers in $U_v$. Its kernel is $\varDelta$ by Proposition 13.20, and we have

$$\# (E_{k,S}/\varDelta )=\frac{\# E_{k,S}/E_{k,S}^n}{\# \varDelta /E_{k, S}^n}.$$

As we have seen in the Proof of Proposition 13.20 (consequence of the Dirichlet unit theorem), the cardinality of $E_{k,S}/E_{k, S}^n$ is $n^s$. On the other hand, the cardinality of $\varDelta /E_{k, S}^n=\varDelta k^{*n}/k^{*n}$ is that of $G=\mathrm {Gal}(K/k)$ by Proposition 13.15, it is thus $n^r$. Finally the cardinality of $E_{k, S}/\varDelta$ is $n^{s-r}$, and to obtain the surjectivity it is enough to see that it is also the cardinality of $\prod _{v \in T} U_v/U_v^n$. This follows from Proposition 13.18, since v does not divide $\ell$ if $v \in T$ and n is a power of $\ell$.

Let now $\alpha =(\alpha _v) \in I_{k, S}$, with image $[\alpha ] \in C_k=I_{k, S} k^*/k^*$. We want to show that $[\alpha ] \in N_{M/k} C_M$. By surjectivity of f we find an $x \in E_{k, S}$ such that for every place v in T, we have $\alpha _v=x\cdot u_v^n$ with $u_v \in U_v$. Let us set $\alpha '=\alpha /x$, it is enough to see that $\alpha '$ is in $N_{M/k} I_M$, which can be checked component by component by Lemma 13.3. For $v \in S$, we have $y \in k_v^{*n}$ hence $M_v=k_v$, which obviously implies that $\alpha '_v$ is a norm of the extension $M_v/k_v$. For $v \in T$, it also works because $\alpha '_v=u_v^n$ is an nth power. Lastly, for $v \not \in S \cup T$, we have that $M_v/k_v$ is unramified and $\alpha '_v \in U_v$, which is enough to guarantee that $\alpha '_v$ is a local norm, by Proposition 8.3. Finally we have $M=k$, and therefore $y \in k^{*n} \cap I_k(S, T) \subset E_{k,{S \cup T}}^n$, as desired.    $\square$

Proof of Theorem 13.21 We use the exact sequence:

$$\begin{aligned} 1 \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }I_{k,S \cup T} \cap k^*/I_k(S,T) \cap k^* \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }&I_{k,S \cup T}/ I_k(S,T)\\&\quad \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }I_{k, S \cup T}\cdot k^*/I_k(S, T) k^* \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }1 \end{aligned}$$

and we calculate the cardinalities of various terms. As $I_k=I_{k, S\cup T}\cdot k^*$, the order of the group on the right is $[C_k:C_k(S, T)]$. By Lemma 13.22, the order of the group on the left is $[E_{k,{S \cup T}}: E_{k,{S \cup T}}^n]=n^{2s-r}$ as we have already seen (as the cardinality of $S \cup T$ is $2s-r$). Lastly, the cardinality of the middle group is that of $\prod _{v \in S} [k_v^*: k_v^{*n}]$, which is $\prod _{v \in S} n^2/\vert n \vert _v$ by Proposition 13.18. But as S contains all the archimedean places and those dividing $\ell$, by the product formula we have (recall that n is a power of $\ell$):

$$1=\prod _{v \in \varOmega _k} \vert n \vert _v=\prod _{v \in S} \vert n \vert _v$$

hence $\prod _{v \in S} n^2/\vert n \vert _v=n^{2s}$. We obtain

$$[C_k:C_k(S, T)]=n^r=[K:k].$$

as desired.

Let us now show that $C_k(S, T) \subset N_{K/k} C_K$. Let $\alpha \in I_k(S, T)$, it is enough to check (Lemma 13.3) that each component $\alpha _v$ of $\alpha$ is a local norm. For $v \in S$, we have $\alpha _v \in k_v^{*n}$, which is a local norm via local reciprocity isomorphism $k_v^*/NK_v^* \simeq \mathrm {Gal}(K_v/k_v)$ and the fact that $\mathrm {Gal}(K/k)$ is of exponent n. For $v \in T$, we have $K_v=k_v$ since $\varDelta \subset k_v^{*n}$, hence the required condition is automatically satisfied. Lastly, for $v \not \in S \cup T$, $\alpha _v$ is a unit and $K_v/k_v$ is unramified by Lemma 13.17 and hence it still works (Proposition 8.3).

If now K/k is cyclic, we have $r=1$ and the Corollary 13.9 implies

$$[K:k] \leqslant [C_k: N_{K/k} C_K] \leqslant [C_k: C_k(S, T)]=[K:k],$$

which proves the equality $N_{K/k} C_K=C_k(S, T)$.    $\square$

At last we deduce:

Theorem 13.23

(The axiom of the global class field theory) Let k be a global field. Let K be a finite Galois extension of k with cyclic Galois group G. Then ${\widehat{H}}^0(G, C_K)$ is of cardinality [K : k], and $H^1(G, C_K)=0$.

We will see later on that the triviality of $H^1(G, C_K)$ remains true if G is an arbitrary finite group. The assertion on ${\widehat{H}}^0(G, C_K)$ is also true (in a more precise form) when G is finite and abelian. This will be a consequence of results that we will see in the next two chapters. We will deduce that in Theorem 13.21, the equality $C_k(S, T)=N_{K/k} C_K$ remains true if we only assume that the extension K/k is abelian.

Proof

in the case where the characteristic of  k does not divide [K : k] We postpone to the last paragraph the case where k is a field of functions of characteristic $p>0$ dividing [K : k], which needs to be treated separately. We can do this using the method of [1], Chap. 6, but we thought it would be more insightful to use a geometric approach, based on the results of Sect. 12.4. For now, we assume that  k is a number field, or a function field whose characteristic is prime to [K : k]. As we know that the Herbrand quotient $h(G, C_K)$ is $n:=[K:k]$ by Theorem 13.8, it is enough to see that ${\widehat{H}}^{-1}(G, C_K)$ (which is also $H^1(G, C_K)$ by Theorem 2.16) is of cardinality 1. We proceed by induction on n. Consider a sub-extension M/k of K/k of prime degree $\ell$. If $\ell <n$, the induction assumption and the restriction-inflation exact sequence imply $H^1(G, C_K)=0$. Let us then assume that $n=\ell$ is prime. Denote by $\mu _{\ell }$ the group of $\ell$th roots of unity (in the algebraic closure of k) and let us set $k'=k(\mu _{\ell })$, $K'=K(\mu _{\ell })$. Then $K'/k'$ is a cyclic Kummer extension and Theorem 13.21 implies that the group ${\widehat{H}}^0(\mathrm {Gal}(K'/k'), C_{K'})$ is of cardinality $[K':k']$, hence $H^1(\mathrm {Gal}(K'/k'), C_{K'})=0$ (still by Theorem 13.8). As $[k':k] \leqslant \ell -1$, the induction assumption implies also that $H^1(\mathrm {Gal}(k'/k), C_{k'})=0$ hence $H^1(\mathrm {Gal}(K'/k), C_{K'})=0$ by the restriction-inflation sequence, and a fortiori $H^1(G, C_K)=0$ by the same sequence.    $\square$

Corollary 13.24

(Hasse norm principle)  Let k be a global field and let K be a finite cyclic extension of k. Then an element $x \in k^*$ is a norm of the extension K/k if and only if its image $x_v$ in $k_v^*$ is a norm in $K_v/k_v$ for every place v of k.

Note that this result is not true for an arbitrary abelian extension (see for example Exercise 5 of [9]).

Proof

Let $G\!=\!\mathrm {Gal}(K/k)$. As ${\widehat{H}}^{-1}(G, C_K)\!=\!0$, the exact modified cohomology sequence yields an injection ${\widehat{H}}^0(G, K^*)\mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }{\widehat{H}}^0(G, I_K)$, which allows us to conclude using Proposition 13.1, (b).    $\square$

Theorem 13.25

Let K/k be a finite Galois extension of global fields with group G. Then $H^1(G, C_K)=0$.

Proof

For G cyclic, the result is part of the class field axiom (Theorem 13.23). On the other hand, recall that (for any prime number p) the abelianisation of a non-trivial p-group is a non-trivial p-group. Hence every non-trivial p-group has a quotient of order p. We deduce that the theorem is true when G is a p-group by induction on the cardinality of G, taking a Galois degree p subextension E/k and using the restriction-inflation exact sequence

$$0 \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }H^1(\mathrm {Gal}(E/k), C_E) \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }H^1(G, C_K) \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }H^1(\mathrm {Gal}(K/E), C_K).$$

Finally, for an arbitrary G, we consider, for every p dividing $\# G$, a p-Sylow $G_p$ of G and its fixed fixed field $K_p$. Then, as $H^1(G, C_K)$ is the direct sum of the $H^1(G, C_K)\{p\}$, the map

$$H^1(G, C_K) \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }\prod _p H^1(\mathrm {Gal}(K/K_p), C_K)$$

obtained via the restrictions is injective by Lemma 3.2, which yields the result by the case where G is a p-group.    $\square$

Corollary 13.26

Let . Then $H^1(k, C)=0$.

5 Proof of the Class Field Axiom for a Function Field

In this paragraph, we consider a global field k of characteristic p, that we view as the field of functions $k=\kappa (X)$ of a smooth projective geometrically integral curve X over a finite field $\kappa$. We will follow the method of [39] (Part. I, App. A) to prove the axiom of the global class field theory (Theorem 13.23) in this situation. In a certain sense, class field theory for a function field is easier than for a number field, once one knows some geometric results from Sect. 12.4. Indeed, this approach will for example yield the axiom of class field theory (as well as Theorem 13.25 and Corollary 13.26) much easier than with the method of Kummer extensions that we have followed in the last paragraph. Furthermore it has the advantage to work in the case where p divides the degree of the extension as well. The geometric approach also admits vast generalisations to higher dimensions whose very complete exposition can be found in [48].

The notations of this paragraph are analogous to those of Sect. 12.4. For any finite extension $\kappa ' \subset \overline{\kappa }$ of $\kappa$, set $X_{\kappa '}=X \times _{\kappa } \kappa '$ and denote by $\kappa '(X)=\kappa (X) \otimes _{\kappa } \kappa '$ the field of functions of the curve $X_{\kappa '}$. Thus $\kappa '(X)$ is a finite extension of the global field k. Let $I_{\overline{\kappa }(X)}$ be the inductive limit over these $\kappa '$ of the idèle groups $I_{\kappa '(X)}$. We have a natural inclusion i of $\overline{\kappa }(X)^*$ in $I_{\overline{\kappa }(X)}$.

Proposition 13.27

The map $H^2(\kappa ,\overline{\kappa }(X)^*) \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }H^2(\kappa , I_{\overline{\kappa }(X)})$ induced by i is injective.

Proof

A closed point w of $X_{\kappa '}$ defines a discrete valuation ${{\,\mathrm{val}\,}}_w$ on the completion $\kappa '(X)_w$, hence by definition of $I_{\kappa '(X)}$ an induced map

$$\bigoplus _{w \in X_{\kappa '}^{(1)}}{{\,\mathrm{val}\,}}_w: I_{\kappa '(X)} \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }\bigoplus _{w \in X_{\kappa '}^{(1)}} \mathbf{Z}\cdot w=\text {Div}X_{\kappa '}.$$

We thus have, by definition of the map $\text {Div}$ (cf.  Sect. 12.4), a commutative diagram:

By passing to the limit on the $\kappa '$, we obtain a commutative diagram:

Proposition 12.31 and Corollary 12.34 tell us that the maps $H^2(\kappa , \overline{\kappa }(X)^*) \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }H^2(\kappa , {{\,\mathrm{Div}\,}}_0\overline{X})$ and $H^2(\kappa , {{\,\mathrm{Div}\,}}_0\overline{X}) \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }H^2(\kappa , \text {Div}\overline{X})$ induced by this diagram are injective. The result follows.    $\square$

Let us fix a separable closure $\overline{k}$ of $k=\kappa (X)$ containing $\overline{\kappa }$ and consider a finite Galois extension $k' \subset \overline{k}$ of k. Let $\kappa '$ be the algebraic closure of $\kappa$ in $k'$. Denote by $I_{k'}$ the group of idèles of the global field $k'$, and similarly by $I_{\kappa '(X)}$ that of $\kappa '(X)$. We have a commutative diagram:

hence, by passing to the limit over the $k'$, we obtain the commutative diagram:

$$\begin{aligned} \begin{array}{c} \end{array} \end{aligned}$$

(13.2)

Here $I:=I_{\overline{k}}$ is the idèle group of $\overline{k}$ as defined in Sect. 13.1. We also have the idèle class group $C=I/\overline{k}^*$ of  $\overline{k}$.

Theorem 13.28

The group ${{\,\mathrm{Br}\,}}k$ is isomorphic to $H^2(\kappa ,\overline{\kappa }(X)^*)$. The map ${{\,\mathrm{Br}\,}}k=H^2(k,\overline{k}^*) \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }H^2(k, I_{\overline{k}})$ induced by the diagram (13.2) is injective.

Proof

Consider the algebraic extensions (of infinite degree) of the field $k=\kappa (X) \subset \overline{\kappa }(X) \subset \overline{k}$. The Galois group of $\overline{\kappa }(X)/\kappa (X)$ identifies with $\Gamma _{\kappa }= \mathrm {Gal}(\overline{\kappa }/\kappa )$. As $H^1(\overline{\kappa }(X),\overline{k}^*)=0$ by Hilbert 90, we have by Corollary 1.45 (in its profinite version) an exact sequence

$$0 \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }H^2(\kappa ,\overline{\kappa }(X)^*) \mathchoice{\xrightarrow {\textstyle {{\,\mathrm{Inf}\,}}}}{{\mathop {\longrightarrow }\limits ^{{{\,\mathrm{Inf}\,}}}}}{}{} H^2(k,\overline{k}^*)={{\,\mathrm{Br}\,}}k \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }H^2(\overline{\kappa }(X),\overline{k}^*)={{\,\mathrm{Br}\,}}(\overline{\kappa }(X)).$$

But the field $\overline{\kappa }(X)$ is $C_1$ by Tsen theorem (Example 6.20, b), hence has a trivial Brauer group (Theorem 6.22), which yields an isomorphism

$$H^2(\kappa ,\overline{\kappa }(X)^*) \simeq {{\,\mathrm{Br}\,}}k.$$

The first statement follows.

Let us now show that the map

$${{\,\mathrm{Inf}\,}}: H^2(\kappa , I_{\overline{\kappa }(X)}) \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }H^2(k, I_{\overline{k}})$$

deduced from the diagram (13.2) is injective. Let $k' \subset \overline{k}$ be a finite Galois extension of k and $\kappa '$ be the algebraic closure of $\kappa$ in $k'$. It is enough to show that ${{\,\mathrm{Inf}\,}}: H^2(\mathrm {Gal}(\kappa '(X)/k), I_{\kappa '(X)}) \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }H^2(\mathrm {Gal}(k'/k), I_{k'})$ is injective (we then pass to the limit on $k'$). Let us set $K=\kappa '(X)$. We apply Proposition 13.1(b) to finite extensions of the global fields K/k and $k'/k$, which reduces to showing that if v is a place of k then the map

$${{\,\mathrm{Inf}\,}}: H^2(\mathrm {Gal}(K_v/k_v), K_v^*) \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }H^2(\mathrm {Gal}(k'_v/k_v), k^{\prime ^{\, *}}_v) $$

is injective. This is the case by Hilbert 90 and Corollary 1.45.

We deduce from diagram (13.2) a commutative diagram:

We have seen that the vertical left hand side map is an isomorphism and the vertical right hand side map is injective. Proposition 13.27 implies that the horizontal bottom map is also injective. The result follows.    $\square$

Corollary 13.29

(Class field axiom) We have $H^1(k, C)=0$. If K is a finite Galois extension of k with Galois group G, then denoting by $C_K$ the idèle class group of K, we have $H^1(G, C_K)=0$. If furthermore G is cyclic, the group ${\widehat{H}}^0(G, C_K)$ is of cardinality [K : k].

Proof

The long cohomology exact sequence associated to

$$0 \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }\overline{k}^* \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }I \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }C \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }0$$

and Theorem 13.28 along with the fact that $H^1(k, I)=0$ (Corollary 13.7) yield the equality $H^1(k, C)=0$. The restriction-inflation exact sequence then implies (using Proposition 13.4) that $H^1(G, C_K)=0$, and we then conclude that for G cyclic, the cardinality of ${\widehat{H}}^0(G, C_K)$ is [K : k] by Theorem 13.8.    $\square$

Note that we have proved the triviality of $H^1(k, C)$ and that of $H^1(G, C_K)$ directly without having to deal with the case the case where G is cyclic like we did in the case of a number field.

Remark 13.30

The exact sequence (12.3) shows that ${{\,\mathrm{Br}\,}}k$ is isomorphic to $H^2(\kappa , {{\,\mathrm{Div}\,}}_0 \overline{X})$ since we just saw that it is isomorphic to $H^2(\kappa , \overline{\kappa }(X)^*)$, while the groups ${{\,\mathrm{Br}\,}}\kappa$ and $H^3(\kappa ,\overline{\kappa }^*)$ are trivial since $\mathrm {cd}(\kappa ) \leqslant 1$. As $H^1(\kappa , {{\,\mathrm{Pic}\,}}\overline{X})=0$ (Corollary 12.34) and $\mathrm {scd}(\kappa ) \leqslant 2$, the exact sequence (12.4) then yields the exact sequence

$$0 \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }{{\,\mathrm{Br}\,}}k \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }H^2(\kappa , \text {Div}\overline{X}) \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }H^2(\kappa ,{{\,\mathrm{Pic}\,}}\overline{X}) \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }0.$$

By Shapiro’s lemma, we have

$$H^2(\kappa , \text {Div}\overline{X}) \simeq \bigoplus _{v \in X^{(1)}} H^2(\kappa (v),\mathbf{Z}) \simeq \bigoplus _{v \in X^{(1)}} H^1(\kappa (v),\mathbf{Q}/\mathbf{Z}), $$

and this last group is also isomorphic to $\bigoplus _{v \in X^{(1)}} {{\,\mathrm{Br}\,}}k_v$ by Theorem 8.9. On the other hand we have $H^i(\kappa ,{{\,\mathrm{Pic}\,}}^0 \overline{X})=0$ for $i \geqslant 2$ since as ${{\,\mathrm{Pic}\,}}^0 \overline{X}$ is divisible, we have for any $n>0$ a surjection from $H^i(\kappa , ({{\,\mathrm{Pic}\,}}^0 \overline{X})[n])$ onto $H^i(\kappa ,{{\,\mathrm{Pic}\,}}^0 \overline{X})[n]$. But $H^i(\kappa , ({{\,\mathrm{Pic}\,}}^0 \overline{X})[n])=0$ as $\mathrm {cd}(\kappa ) \leqslant 1$. Using the exact sequence (12.5), we deduce that $H^2(\kappa , {{\,\mathrm{Pic}\,}}\overline{X}) \simeq H^2(\kappa ,\mathbf{Z}) \simeq \mathbf{Q}/\mathbf{Z}$. Finally, we obtain an exact sequence

$$0 \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }{{\,\mathrm{Br}\,}}k \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }\bigoplus _{v \in X^{(1)}} {{\,\mathrm{Br}\,}}k_v \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }\mathbf{Q}/\mathbf{Z}\mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }0,$$

which is the Brauer–Hasse–Noether sequence that we will encounter a bit later (Theorem 14.11). Here again, we see that this result can be obtained directly in the case of a function field.

Nevertheless, the identification of the maps obtained by this method with those of Theorem 14.11 is not completely obvious.

6 Exercises

Exercise 13.1

Let K be a local field. Let n be an integer prime to the characteristic of K, and such that K contains the nth roots of unity. We set $L=K(\root n \of {K^*})$. Show that L is a finite abelian extension of K and that the group of norms $N_{L/K} L^*$ is exactly $K^{*n}$.

Exercise 13.2

Let k be a global field. Let K be a finite Galois extension of k. Show that we have an injection

$${{\,\mathrm{Br}\,}}(K/k) \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }\bigoplus _{v \in \varOmega _k} {{\,\mathrm{Br}\,}}(K_v/k_v).$$

Exercise 13.3

Let k be a global field. Let K be a finite non-trivial Galois extension of  k (not necessarily abelian). Show that there are infinitely many places v of k such that the extension K/k is not totally split at v. Using Exercise 12.2, show that this result (that will be refined in Exercise 18.7) still holds true for K/k which is not Galois.

Exercise 13.4

Let k be a global field. Let F be a finite non-trivial Galois extension of k with group G. For any place v of k, we denote by $F_v$ the completion of F at a place above v. Show that we have an exact sequence

$$k^*/ N_{F/k} F^* \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }\bigoplus _{v \in \varOmega _k} k_v^*/ N_{F_v / k_v } F_v^* \mathchoice{\longrightarrow }{\rightarrow }{\rightarrow }{\rightarrow }{\widehat{H}}^0(G, C_F).$$

Using Exercise 13.3, show that in this exact sequence, the middle term is infinite (for the sequel, see Exercise 15.1).