Theorem 7.1.1.
For \(L/K\) a cyclic extension of number fields,
\begin{equation*}
h(C_L) = [L:K].
\end{equation*}
Section 7.1 Cohomology of the idèles I: the “First Inequality”
Reference.
[36] VII.2-VII.4; [37] VI.3, but see below. See also this blog post by Akhil Mathew.
1
amathew.wordpress.com/2010/05/30/the-first-inequality-cohomology-of-the-idele-classes/By analogy with local class field theory, we want to prove that for \(L/K\) a cyclic extension of number fields and \(C_K, C_L\) the respective idèle class groups of \(K\) and \(L\text{,}\)
\begin{equation*}
H^1(\Gal(L/K), C_L) = 1, \qquad H^2(\Gal(L/K), C_L) = \ZZ/[L:K]\ZZ.
\end{equation*}
Our first step is to calculate the Herbrand quotient.
Proof.
This will end up reducing to a study of lattices in a real vector space, much as in the proof of Dirichlet’s units theorem (Corollary 6.2.11).
From Theorem 7.1.1, we will deduce the so-called “First Inequality”.
Theorem 7.1.2. First Inequality.
For \(L/K\) a cyclic extension of number fields,
\begin{equation*}
\#H^0_T(\Gal(L/K), C_L) \geq [L:K]\text{.}
\end{equation*}
Proof.
The “Second Inequality” will be the reverse, which will be a bit more subtle (see Theorem 7.2.10).
Subsection Some basic observations
Definition 7.1.3.
Let \(L/K\) be a Galois extension of number fields with Galois group \(G\text{.}\) (We do not yet need \(G\) to be cyclic.) For any finite set \(S\) of places of \(K\) containing all infinite places, write \(I_{L,S}\) to mean the group \(I_{L,T}\) where \(T\) denotes the set of places of \(L\) lying over some place in \(S\text{.}\) Similarly, write \(\gotho_{L,S}\) to mean \(\gotho_{L,T}\text{.}\)
Note that each \(I_{L,S}\) is stable under the action of \(G\) and that \(I_L\) is the direct limit of the \(I_{L,S}\) over all \(S\text{.}\) Moreover, by Corollary 6.2.10, for \(S\) sufficiently large we have
\begin{equation*}
I_L = I_{L,S} L^*.
\end{equation*}
Proposition 7.1.4.
Let \(L/K\) be a Galois extension of number fields with Galois group \(G\text{.}\) For each \(i>0\text{,}\)
\begin{equation*}
H^i(G, I_L) = \bigoplus_v H^i(G_w, L_w^*),
\end{equation*}
where \(v\) runs over places of \(K\) and \(w\) denotes a single place of \(L\) above \(v\text{.}\) Similarly, for each \(i\text{,}\)
\begin{equation*}
H^i_T(G, I_L) = \bigoplus_v H^i_T(G_w, L_w^*).
\end{equation*}
Proof.
View \(I_L\) as the direct limit of the \(I_{L,S}\) over all finite sets \(S\) of places of \(K\) containing all infinite places and all ramified places; then \(H^i(G,I_L)\) is the direct limit of the \(H^i(G, I_{L,S})\text{.}\) The latter is the product of \(H^i(G, \prod_{w|v} L_w^*)\) over all \(v \in S\) and \(H^i(G, \prod_{w|v} \gotho_{L_w}^*)\) over all \(v \notin S\text{,}\) but the latter is trivial because \(v \notin S\) cannot ramify. By Shapiro’s lemma (Lemma 3.2.3), \(H^i(G, \prod_{w|v} L_w^*) = H^i(G_w, L_w^*)\text{,}\) so we have what we want. The argument for Tate groups is analogous.
Corollary 7.1.5.
Let \(L/K\) be a Galois extension of number fields with Galois group \(G\text{.}\) Then
\begin{equation*}
H^1(G, I_L) = 0, \qquad H^2(G, I_L) = \bigoplus_v \frac{1}{[L_w:K_v]} \ZZ/\ZZ.
\end{equation*}
Proof.
This follows by combining Proposition 7.1.4, the computation of cohomology of local fields (Lemma 1.2.3 and Proposition 4.2.1), and the equality
\begin{equation*}
H^2(\Gal(\CC/\RR), \CC^*) \cong H^0_T(\Gal(\CC/\RR), \CC^*) = \RR^*/\RR^+ \cong \ZZ/2\ZZ.
\end{equation*}
Remark 7.1.6. Sanity check.
The case \(i=0\) of Proposition 7.1.4 asserts something that is evidently true: an idèle in \(I_K\) is a norm from \(I_L\) if and only if each component is a norm.
Remark 7.1.7.
If \(S\) contains all infinite places and all ramified places, then
\begin{equation*}
\Norm_{L/K} I_{L,S} = \prod_{v \in S} U_v \times \prod_{v \notin S} \gotho_{K_v}^*
\end{equation*}
where \(U_v\) is open in \(K_v^*\text{.}\) The group on the right is open in \(I_K\text{,}\) so \(\Norm_{L/K} I_K\) is open.
By quotienting down to \(C_K\text{,}\) we see that \(\Norm_{L/K} C_K\) is open. In fact, the snake lemma on the diagram
implies that the quotient \(I_K/(K^* \times \Norm_{L/K} I_L)\) is isomorphic to \(C_K/\Norm_{L/K} C_L\text{.}\)
Subsection Cohomology of the units: first steps
Definition 7.1.9.
Let \(L/K\) be a cyclic extension of number fields with Galois group \(G\text{.}\) Apply Corollary 6.2.10 to choose a finite set \(S\) of places of \(K\) so that \(I_L = I_{L,S} L^*\text{.}\) From the exact sequence
\begin{equation*}
1 \to \gotho_{L,S}^* \to I_{L,S} \to I_{L,S}/\gotho_{L,S}^* = C_L \to 1
\end{equation*}
we have an equality of Herbrand quotients
\begin{equation*}
h(C_L) = h(I_{L,S})/h(\gotho_{L,S}^*).
\end{equation*}
By Corollary 7.1.5,
\begin{equation*}
h(I_{L,S}) = \prod_{v \in S} \#H^0_T(G_v, L_w^*)= \prod_{v \in S} [L_w:K_v].
\end{equation*}
(Since \(G\) is abelian, we write \(G_v\) instead of \(G_w\text{.}\)) To get \(h(C_L) = [L:K]\text{,}\) it will thus suffice to establish Lemma 7.1.10 below.
Lemma 7.1.10.
Let \(L/K\) be a cyclic extension of number fields. Let \(S\) be a finite set of places of \(K\) containing all infinite places. Then
\begin{equation*}
h(\gotho_{L,S}^*) = \frac{1}{[L:K]} \prod_{v \in S} [L_w:K_v].
\end{equation*}
Proof.
Subsection Cohomology of the units: a computation with \(S\)-units
At this point, we have reduced the computation of the Herbrand quotient \(h(I_L)\text{,}\) and by extension the First Inequality, to the computation of \(h(\gotho_{L,S}^*)\) for a suitable set \(S\) of places of \(K\text{.}\) We treat this point next, using similar ideas to the proof of Dirichlet’s units theorem (Corollary 6.2.11).
Definition 7.1.11.
Let \(L/K\) be a cyclic extension of number fields with Galois group \(G\text{.}\) Let \(S\) be a finite set of places of \(K\) containing all infinite places, and let \(T\) be the set of places of \(L\) lying above places of \(S\text{.}\) Let \(V\) be the real vector space consisting of one copy of \(\RR\) for each place in \(T\text{.}\) Define the map \(\gotho_{L,S}^* \to V\) by
\begin{equation*}
\alpha \to \prod_{w \in T} \log |\alpha|_w
\end{equation*}
with normalizations as in Definition 6.1.10. By the product formula (Proposition 6.1.11) and Dirichlet’s units theorem (Corollary 6.2.11), the kernel of this map consists of roots of unity, and the image \(M\) is a lattice in the trace-zero hyperplane \(H\) of \(V\text{.}\) Since \(G\) acts compatibly on \(\gotho_{L,S}^*\) and \(V\) (the latter by permuting the factors), it also acts on \(M\text{.}\)
Remark 7.1.12. Caveat.
At this point, we deviate from [37] due to an error therein. Namely, Lemma VI.3.4 is only proved assuming that \(G\) acts transitively on the coordinates of \(V\text{,}\) but in Definition 7.1.11 this is not the case: \(G\) permutes the places above any given place \(v\) of \(K\) but those are separate orbits. So we’ll follow [36] instead.
Definition 7.1.13.
Continuing from Definition 7.1.11, we can write down two natural lattices in \(V\text{.}\) One of them is the lattice generated by \(M\) together with the all-ones vector, on which \(G\) acts trivially. As a \(G\)-module, the Herbrand quotient of that lattice is \(h(M) h(\ZZ) = [L:K] h(M)\text{.}\) The other is the lattice \(M'\) in which, in the given coordinate system, each element has integral coordinates. To compute its Herbrand quotient, notice that the projection of this lattice onto the coordinates corresponding to the places \(w \in T\) above some \(v \in S\) form a copy of \(\Ind^G_{G_v} \ZZ\text{.}\) Thus
\begin{equation*}
h(G, M') = \prod_{v \in S} h(G, \Ind^G_{G_v} \ZZ) = \prod_{v \in S} h(G_v, \ZZ) = \prod_{v \in S} \#G_v = \prod_{v \in S} [L_w:K_v].
\end{equation*}
To sum up, the calculations from Definition 7.1.11 and Definition 7.1.13 reduce Lemma 7.1.10 to the following statement (Lemma 7.1.14).
Subsection Herbrand quotients of real lattices
We conclude the proof of the First Inequality with the following statement.
Lemma 7.1.14.
Let \(V\) be a real vector space on which a finite cyclic group \(G\) acts linearly, and let \(L_1\) and \(L_2\) be \(G\)-stable lattices in \(V\) for which at least one of \(h(L_1)\) and \(h(L_2)\) is defined. Then \(h(L_1) = h(L_2)\) (and both are defined).
Proof.
Note that \(L_1 \otimes_{\ZZ} \QQ\) and \(L_2 \otimes_{\ZZ} \QQ\) are \(\QQ[G]\)-modules which become isomorphic to \(V\text{,}\) and hence to each other, after tensoring over \(\QQ\) with \(\RR\text{.}\) By Lemma 7.1.15, this implies that \(L_1 \otimes_{\ZZ} \QQ\) and \(L_2 \otimes_{\ZZ} \QQ\) are isomorphic as \(\QQ[G]\)-modules.
From this isomorphism, we see that as a \(\ZZ[G]\)-module, \(L_1\) is isomorphic to some sublattice of \(L_2\text{.}\) Since a lattice has the same Herbrand quotient as any sublattice (the quotient is finite, so its Herbrand quotient is 1), that means \(h(L_1) = h(L_2)\text{.}\)
Lemma 7.1.15.
Let \(F/E\) be an extension of infinite fields. Let \(G\) be a finite group. Let \(V_1\) and \(V_2\) be two right \(E[G]\)-modules which are finite-dimensional as \(E\)-vector spaces. If \(V_1 \otimes_E F\) and \(V_2 \otimes_E F\) are isomorphic as \(F[G]\)-modules, then \(V_1\) and \(V_2\) are isomorphic.
Proof.
By hypothesis, the \(F\)-vector space
\begin{equation*}
W_F = \Hom_{F}(V_1 \otimes_{E} F, V_2 \otimes_{E} F),
\end{equation*}
on which \(G\) acts by the formula \(T^g(x) = T(x^{g^{-1}})^g\text{,}\) contains an invariant vector which, as a linear transformation, is invertible. Now \(W_F\) can also be written as
\begin{equation*}
W \otimes_E F, \qquad W = \Hom_{E}(V_1, V_2).
\end{equation*}
The fact that \(W_F\) has an invariant vector says that a certain set of linear equations has a nonzero solution over \(F\text{,}\) namely the equations that express the fact that the action of \(G\) leaves the vector invariant. But those equations have coefficients in \(E\text{,}\) so
\begin{equation*}
W^G \otimes_E F = W_F^G;
\end{equation*}
in particular, the space of invariant vectors in \(W\) is also nonzero.
It remains to check that some element of \(W^G\) corresponds to a map \(V_1 \to V_2\) which is actually an isomorphism; for this, we argue as in Exercise 3. Fix an isomorphism of vector spaces between \(V_2 \otimes_{E} F\) and \(V_1 \otimes_{E} F\) (which need not respect the \(G\)-action). By composing each element of \(W\) with this isomorphism and taking the determinant, we get a well-defined polynomial function on \(W\text{,}\) which we can restrict to \(W^G\text{.}\) By hypothesis, this function is not identically zero on \(W_F^G\text{,}\) so (because \(E\) is infinite) it cannot be identically zero on \(W^G\) either.
Subsection Splitting of primes
As a consequence of the First Inequality, we record the following fact which was previously stated as a consequence of the Chebotaryov density theorem (Theorem 2.4.11), but which will be needed logically earlier in the arguments. (See [37], Corollary VI.3.8 for more details.)
Corollary 7.1.16.
For any nontrivial extension \(L/K\) of number fields, there are infinitely many primes of \(K\) which do not split completely in \(L\text{.}\)
Proof.
Suppose first that \(L/K\) is cyclic. Suppose that all but finitely many primes split completely; we can then take a finite set \(S\) of places which contains all of them as well as all of the infinite places and all of the ramified places. For each \(v \in S\text{,}\) the group \(U_v = \Norm_{L_w/K_v} L_w^*\) is open of finite index in \(K_v^*\text{.}\) For any \(\alpha \in I_K\text{,}\) using the approximation theorem (Proposition 6.1.17) we can then find \(\beta \in K^*\) such that \((\alpha/\beta)_v \in U_v\) for all \(v \in S\text{.}\) For each place \(v \notin S\text{,}\) we have \(L_w = K_v\text{,}\) so \(\alpha/\beta \in \Norm_{L/K} I_L\text{.}\) We deduce that the class of \(\alpha\) in \(C_L\) is a norm; that is, \(C_K = \Norm_{L/K} C_L\text{,}\) whereas Theorem 7.1.2 asserts that \(H^0_T(\Gal(L/K), C_L) \geq [L:K]\text{,}\) contradiction.
In the general case, let \(M\) be the Galois closure of \(L/K\text{;}\) then a prime of \(K\) splits completely in \(L\) if and only if it splits completely in \(M\text{.}\) Since \(\Gal(M/K)\) is a nontrivial finite group, it contains a cyclic subgroup; let \(N\) be the fixed field of this subgroup. By the previous paragraph, there are infinitely many prime ideals of \(N\) which do not split completely in \(M\text{,}\) proving the original result.
Exercises Exercises
1.
Let \(K\) be a number field. Let \(L_1, \dots, L_r\) be cyclic extensions of \(K\) of the same prime degree \(p\) such that \(L_i \cap L_j = K\) for \(i \neq j\text{.}\) Using the First Inequality (Theorem 7.1.2), prove that there are infinitely many primes of \(K\) which split completely in \(L_2, \dots, L_r\) but not in \(L_1\text{.}\)
