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Notes on class field theory

Kiran S. Kedlaya

Section 3.4 Cohomology of cyclic groups

Reference.

[38], IV.7; [34], IX.1.
We next specialize attention to the case of a finite cyclic group, which will play a key role in many of our calculations. In this case, the cohomology, homology, and Tate groups satisfy a key periodicity property (Theorem 3.4.1) which allows us to define and manipulate a sort of “Euler characteristic”, the Herbrand quotient (Definition 3.4.4).

Subsection The periodicity theorem

In general, for any given \(G\) and \(M\text{,}\) it is at worst a tedious exercise (or a computer calculation) to compute \(H^i_T(G,M)\) for any single value of \(i\text{,}\) but try to compute all of these at once and you discover that they exhibit very little evident structure. Thankfully, there is an exception to that dreary rule when \(G\) is cyclic.

Proof.

Let \(m\) be the order of \(G\text{,}\) choose a generator \(g\) of \(G\text{,}\) and fix the identification of the group ring \(\ZZ[G]\) with the (commutative!) ring \(\ZZ[x]/(x^m-1)\) taking \(g\) to \(x\text{.}\) I can then view multiplication by \(x-1\) as a map of \(G\)-modules \(\ZZ[G] \to \ZZ[G]\text{,}\) then tensor over \(\ZZ\) with \(M\) to get a map
\begin{equation*} \Ind^G_1 \Res^G_1 M \stackrel{\times (x-1)}{\to} \Ind^G_1 \Res^G_1 M\text{.} \end{equation*}
The kernel of this map is the image of multiplication by \(x^{m-1} + \cdots + 1\text{,}\) while the image is the ideal generated by \(x-1\text{.}\) Consequently, if I extend this map to a sequence
\begin{equation*} 0 \to M \to \Ind^G_1 \Res^G_1 M \stackrel{\times (x-1)}{\to} \Ind^G_1 \Res^G_1 M \to M \to 0 \end{equation*}
by adding the adjunction maps on both sides, the result is exact.
By Shapiro’s lemma, the Tate groups of the two modules in the middle are all zero. The desired result now follows from the following general fact: if
\begin{equation*} 0 \to A \stackrel{f}{\to} B \stackrel{g}{\to} C \stackrel{h}{\to} D \to 0 \end{equation*}
is exact and \(B\) and \(C\) have all Tate groups zero, then there is a canonical isomorphism \(H^{i+2}_T(G, A) \to H^i_T(G, D)\text{.}\) To see this, form the long exact sequence associated to the short exact sequences
\begin{gather*} 0 \to A \to B \to \frac{B}{\im(f)} \to 0\\ 0 \to \frac{B}{\ker(g)} \to C \to D \to 0 \end{gather*}
and combine the resulting isomorphisms to get
\begin{equation*} H^{i+2}(G, A) \cong H^{i+1}\left(G, \frac{B}{\im(f)}\right) = H^{i+1}\left(G, \frac{B}{\ker(g)}\right) \cong H^i(G,D). \end{equation*}

Remark 3.4.2.

In particular, when \(G\) is cyclic, the long exact sequence of a short exact sequence \(0 \to M' \to M \to M'' \to 0\) of \(G\)-modules curls up into an exact hexagon as in Figure 3.4.3.
Figure 3.4.3.

Subsection Herbrand quotients

Definition 3.4.4.

Let \(G\) be a finite cyclic group and let \(M\) be a \(G\)-module. If the groups \(H^i_T(G, M)\) are finite, we define the Herbrand quotient as the ratio
\begin{equation*} h(M) = \#H^0_T(G,M) / \#H^{-1}_T(G, M). \end{equation*}
From the exactness of the hexagon in Figure 3.4.3, we see that if \(M', M, M''\) all have Herbrand quotients, then
\begin{equation*} h(M) = h(M') h(M''). \end{equation*}

Remark 3.4.5.

Expanding on the previous point, if two of \(M', M, M''\) have Herbrand quotients, then so does the third. For example, if \(M'\) and \(M''\) have Herbrand quotients (i.e., their Tate groups are finite), then we have an exact sequence
\begin{equation*} H^{-1}_T(G, M') \to H^{-1}_T(G, M) \to H^{-1}_T(G, M'') \end{equation*}
and the outer groups are all finite. In particular, the first map is out of a finite group and so has finite image, and modulo that image, \(H^{-1}_T(G,M)\) injects into another finite group. So it’s also finite, and so on.

Remark 3.4.6.

In general, we “almost” know that the groups \(H^i_T(G,M)\) are finite: by extended functoriality (Example 3.2.24), we know they are torsion with exponent dividing \(\#G\text{.}\) The only issue is that they could fail to be finitely generated.

Remark 3.4.7.

In practice, it will often be much easier to compute the Herbrand quotient of a \(G\)-module than to compute either of its Tate groups directly. The Herbrand quotient will then do half of the work for free: once one group is computed directly, at least the order of the other will be automatically known. The key example for our purposes will occur in Section 7.1.

Remark 3.4.8.

If \(M\) is finite, then the groups \(H^i_T(G,M)\) are finite and \(h(M) = 1\text{.}\) To wit, the sequences
\begin{gather*} 0 \to M^G \to M \to M \to M_G \to 0\\ 0 \to H^{-1}_T(G,M) \to M_G \stackrel{\Norm_G}{\to} M^G \to H^0_T(G,M) \to 0 \end{gather*}
are exact, where \(M \to M\) is the map \(m \mapsto m^g - m\text{;}\) thus \(M_G\) and \(M^G\) are finite groups of the same order, as then are \(H^{-1}_T\) and \(H^0_T\text{.}\)

Exercises Exercises

1.

The periodicity of the Tate groups for \(G\) cyclic means that there is a canonical (up to the choice of a generator of \(G\)) isomorphism between \(H^{-1}_T(G, M)\) and \(H^1_T(G,M)\text{,}\) i.e., between \(\ker(\Norm_G)/MI_G\) and the set of equivalence classes of 1-cocycles. What is this isomorphism explicitly? In other words, given an element of \(\ker(\Norm_G)/MI_G\text{,}\) what is the corresponding 1-cocycle?

2.

Put \(K = \QQ_p(\sqrt{p})\text{.}\) Compute the Herbrand quotient of \(K^*\) as a \(G\)-module for \(G = \Gal(\QQ_p(\sqrt{p})/\QQ_p)\text{.}\)
Hint.
Use the exact sequence \(1 \to \gotho_K^* \to K^* \to \ZZ \to 1\text{.}\)

3.

Let \(G = S_3\) (the symmetric group on three letters), let \(M = \ZZ^3\) with the natural \(G\)-action permuting the factors, and let \(N = M^G\text{.}\) Compute \(H^i(G, M/N)\) for \(i=1,2\) however you want: you can explicitly compute cochains, use the alternate interpretations given above, or use the exact sequence \(0 \to N \to M \to M/N \to 0\text{.}\) Better yet, use more than one method and make sure that you get the same answer.
Hint.
Part of the point of this exercise is that even in this relatively simple-looking situation, it is not all that easy to do the computation. One approach that minimizes the computational complexity is to use the Hochschild-Serre spectral sequence (see [37], Remark II.1.35) to reduce to working with the cyclic groups \(H = A_3 \cong \ZZ/3\ZZ\) and \(G/H = S_3/A_3 \cong \ZZ/2\ZZ\text{,}\) for which periodicity is applicable.
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