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Notes on class field theory

Kiran S. Kedlaya

Section 3.1 Cohomology of finite groups I: abstract nonsense

Reference.

[36], II.1. See [48] for a much more general presentation. (We will generalize ourselves from finite to profinite groups a bit later on; see Section 3.5).
For the broader context of homological algebra, the original reference is [14]. See [36], Appendix II.A for a summary.

Caveat.

This material may seem a bit dry. If so, don’t worry; only a small part of the theory will be relevant for class field theory. However, it doesn’t make sense to learn that small part without knowing what it is a part of!
The euphemism “abstract nonsense” in specific reference to category theory and/or homological algebra has been attributed to Norman Steenrod. It was used in a tongue-in-cheek manner without intending a negative connotation, although such a connotation has been imputed by later authors (a notable example being [32]).

Caveat.

The Galois cohomology groups used in [37] are not the ones we define here. Rather, they are the Tate groups to be introduced in Section 3.3.

Caveat.

Some authors (like Milne, and Neukirch for the most part) put group actions on the left and some (like Neukirch in chapter IV, and myself here) put them on the right. Of course, the theory is the same either way!

Subsection \(G\)-modules and their invariants

Definition 3.1.1.

Let \(G\) be a finite group. A (right) \(G\)-module is an abelian group \(A\) equipped with a right \(G\)-action. I’ll write this action using superscripts, i.e., the image of the action of \(g\) on \(m\) is \(m^g\text{.}\) Alternatively, \(A\) can be viewed as a right module for the group algebra \(\ZZ[G]\text{.}\)
A homomorphism of \(G\)-modules \(\phi: M \to N\) is a homomorphism of abelian groups that is compatible with the \(G\)-actions, i.e., \(\phi(m^g) = \phi(m)^g\text{.}\)

Remark 3.1.2.

For those keeping score, the category of \(G\)-modules is an example of an abelian category; that is, it is an additive category (meaning that the Hom sets have abelian group structures for which composition distributes over addition) with some extra properties related to kernels and cokernels of morphisms. The following discussion is specialized from the general theory of derived functors on abelian categories.

Definition 3.1.3.

Given a \(G\)-module \(M\text{,}\) let \(M^G\) be the abelian group of \(G\)-invariant elements of \(M\text{:}\)
\begin{equation*} M^G = \{m \in M: m^g = m \forall g \in G\}\text{.} \end{equation*}
The functor \(M \to M^G\) from \(G\)-modules to abelian groups is left exact but not right exact. That is, say we start with an exact sequence
\begin{equation*} 0 \to M' \to M \to M'' \to 0 \end{equation*}
of \(G\)-modules; that is, the kernel of each map is equal to the image of the previous map. Then
\begin{equation*} 0 \to (M')^G \to M^G \to (M'')^G \end{equation*}
is again an exact sequence, but this need not remain true if we add 0 at the end; that is, the map \(M^G \to (M'')^G\) may not be surjective.
More generally, if \(0 \to M' \to M \to M'' \to 0\) is an exact sequence, then \(0 \to (M')^G \to M^G \to (M'')^G\) is exact,

Example 3.1.4.

Take the sequence \(0 \to \ZZ/p\ZZ \to \ZZ/p^2\ZZ \to \ZZ/p\ZZ \to 0\) of \(G\)-modules for \(G = \ZZ/p\ZZ\text{,}\) which acts on the middle factor by \(a^g = a(1+pg)\text{.}\) Then \(M^G \to (M'')^G\) is the zero map but \((M'')^G\) is nonzero.)

Subsection Injective objects and resolutions

The topic of homological algebra provides a systematic way to quantify the difference between an exact functor and a left exact (or a right exact) functor. This rests on the following key concept.

Definition 3.1.5.

A \(G\)-module \(M\) is injective if for every inclusion \(A \subset B\) of \(G\)-modules and every \(G\)-module homomorphism \(\phi: A \to M\text{,}\) there is a homomorphism \(\psi: B \to M\) that extends \(\phi\text{.}\)

Proof.

Definition 3.1.7.

An injective resolution of a \(G\)-module \(M\) is a sequence
\begin{equation*} I^0 \stackrel{d^0}{\to} I^1 \stackrel{d^1}{\to} \dots \end{equation*}
in which the objects \(I^0, I^1, \dots\) are injective \(G\)-modules and the augmented sequence
\begin{equation*} 0 \to M \to I^0 \stackrel{d^0}{\to} I^1 \stackrel{d^1}{\to} \dots \end{equation*}
is exact.
From Lemma 3.1.6, it follows that injective resolutions always exist. To wit, first embed \(M\) into an injective \(G\)-module \(I^0\text{,}\) then embed \(I^0/M\) into an injective \(G\)-module \(I^1\text{,}\) and so on.

Definition 3.1.8.

Starting with an injective resolution of \(M\text{,}\) apply the functor of \(G\)-invariants; the result
\begin{equation*} 0 \to (I^0)^G \stackrel{d^0}{\to} (I^1)^G \stackrel{d^1}{\to} \dots \end{equation*}
is still a complex in the sense that any two consecutive maps compose to zero, but it is not necessarily exact. (That is, we still have inclusions \(\im(d^i) \subseteq \ker(d^{i+1})\text{,}\) but these need not be equalities.) We turn this failure into success by defining the \(i\)-th cohomology group as the quotient
\begin{equation*} H^i(G, M) = \ker(d^i)/\im(d^{i-1}), \end{equation*}
with the temporary proviso that this appears to depend not just on \(M\) but also on the injective resolution. By convention, we let \(d^{-1}\) be the map \(0 \to I_0^G\text{,}\) so \(H^0(G,M)=M^G\text{.}\)
Given a homomorphism \(f: M \to N\) and another injective resolution \(0 \to N \to J_0 \to J_1 \to \cdots\text{,}\) Lemma 3.1.6 again implies the existence of a commutative diagram as in Figure 3.1.9 with exact rows. When we apply the functor of \(G\)-invariants, we again get a commutative diagram, but the rows are only complexes rather than exact sequences. However, the vertical arrows in the resulting diagram induce maps \(H^i(f): H^i(G, M) \to H^i(G, N)\text{.}\)
Figure 3.1.9.

Subsection Right derived functors

Continuing the thread, we observe the following.

Proof.

This proof is a standard example of “abstract nonsense”. It suffices to check that if \(f=0\text{,}\) then the \(H^i(f)\) are all zero regardless of what the \(f_i\) are. In that case, it turns out one can construct maps \(g_i: I_{i+1} \to J_i\) (and by convention \(g_{-1} = 0\)) such that \(f_i = g_i \circ d_i + d_{i-1} \circ g_{i-1}\text{,}\) as illustrated in Figure 3.1.11. (Beware that this figure is not a commutative diagram!) Details left as an exercise (see Exercise 4).
Figure 3.1.11.

Remark 3.1.12.

The diagonal arrows depicted in Figure 3.1.11 with the property described in the proof of Lemma 3.1.10 (namely, that \(f_i = g_i \circ d_i + d_{i-1} \circ g_{i-1}\)), are collectively called a chain homotopy for the map \(f\text{.}\)

Definition 3.1.13.

We can now close the books on Definition 3.1.8 as follows. If \(M=N\) and \(f\) is the identity, we get a canonical map between \(H^i(G,M)\) and \(H^i(G,N)\) for each \(i\text{.}\) That is, the groups \(H^i(G,M)\) are well-defined independent of the choice of the injective resolution. Likewise, the map \(H^i(f)\) is also independent of the choice of resolutions, so each \(H^i\) defines a functor from \(G\)-modules to abelian groups. These are called the right derived functors of the functor of \(G\)-invariants.
I will not subject you to the proof of this, but rather mention the key step in its proof.
Figure 3.1.15.

Proof.

The key point is to define the map \(\delta\text{,}\) as the rest amounts to “diagram chasing”. To wit, given \(x \in \ker(f_2) \subseteq M_2\text{,}\) lift \(x\) to \(M_1\text{,}\) push it into \(N_1\) by \(f_1\text{,}\) then check that the image has a preimage in \(N_0\text{.}\) Verification that this is well-defined (and a homomorphism), and that everything is exact, is left to the reader.

Remark 3.1.17.

The snake lemma is depicted in the movie It’s My Turn
 1 
www.youtube.com/watch?v=etbcKWEKnvg
, but not in any more detail than we have given here.

Remark 3.1.18.

In modern algebra, it is common to define objects (e.g., tensor products of modules) in terms of “universal properties” that they satisfy. This can be done for derived functors via the theory of \(\delta\)-functors, as introduced in [14].

Subsection Additional comments

One important consequence of the long exact sequence is that if \(0 \to M' \to M \to M'' \to 0\) is a short exact sequence of \(G\)-modules and \(H^1(G, M') = 0\text{,}\) then \(0 \to (M')^G \to M^G \to (M'')^G \to 0\) is also exact.
More abstract nonsense:
  • If \(0 \to M' \to M \to M'' \to 0\) is a short exact sequence of \(G\)-modules and \(H^i(G, M) = 0\) for all \(i>0\text{,}\) then the connecting homomorphisms in the long exact sequence induce isomorphisms \(H^i(G, M'') \to H^{i+1}(G, M')\) for all \(i > 0\) (and a surjection for \(i=0\)). This sometimes allows one to prove general facts by proving them first for \(H^0\text{,}\) where they have a direct interpretation, then “dimension shifting”; however, getting from \(H^0\) to \(H^1\) typically requires some extra attention.
  • If \(M\) is an injective \(G\)-module, then \(H^i(G,M) = 0\) for all \(i>0\text{.}\) (Use \(0 \to M \to M \to 0 \to \cdots\) as an injective resolution.) This fact has a sort of converse: see next bullet.
  • We say \(M\) is acyclic if \(H^i(G,M) =0\) for all \(i>0\text{;}\) so in particular, injective \(G\)-modules are acyclic. It turns out that we can replace the injective resolution in the definition by an acyclic resolution for the purposes of doing a computation; see Exercise 5.
Of course, the abstract nature of the proofs so far gives us almost no insight into what the objects are that we’ve just constructed. We’ll remedy that next time by giving more concrete descriptions that one can actually compute with.

Exercises Exercises

1.

Let \(G\) be the one-element group. Show that a \(G\)-module (i.e., abelian group) is injective if and only if it is divisible, i.e., the map \(x \mapsto nx\) is surjective for any nonzero integer \(n\text{.}\)
Hint.
You’ll need Zorn’s lemma or equivalent in one direction.

2.

Let \(A\) be an abelian group, regarded as a \(G\)-module for \(G\) the trivial group. Prove that \(A\) can be embedded in an injective \(G\)-module.

3.

Prove Lemma 3.1.6.
Hint.
For \(M\) a \(G\)-module, the previous exercises show that the underlying abelian group of \(M\) embeds into a divisible group \(N\text{.}\) Now map \(M\) into \(\Hom_{\ZZ}(\ZZ[G], N)\) and check that the latter is an injective \(G\)-module.

4.

Prove Lemma 3.1.10 following the sketch given.
Hint.
Construct \(g_i\) given \(f_{i-1}\) and \(g_{i-1}\text{,}\) using that the \(J\)’s are injective \(G\)-modules.

5.

Prove that if \(0 \to M \to M^0 \to M^1 \to \cdots\) is an exact sequence of \(G\)-modules and each \(M_i\) is acyclic, then the cohomology groups of the complex \(0 \to M^{0G} \to M^{1G} \to \cdots\) coincide with \(H^i(G, M)\text{.}\)
Hint.
Construct the canonical long exact sequence from the exact sequence
\begin{equation*} 0 \to M \to M^0 \to M^0/M \to 0\text{,} \end{equation*}
then do dimension shifting using the fact that
\begin{equation*} 0 \to M^0/M \to M^1 \to M^2 \to \cdots \end{equation*}
is again exact. Don’t forget to be careful about \(H^1\text{!}\))
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