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Notes on class field theory

Kiran S. Kedlaya

Section 4.2 Cohomology of local fields: some computations

Reference.

[36], III.1 and III.2; [37], V.1.

Notation convention.

If you catch me writing \(H^i(L/K)\) for \(L/K\) a Galois extension of fields, that’s shorthand for \(H^i(\Gal(L/K), L^*)\text{.}\) Likewise for \(H_i\) or \(H^i_T\text{.}\)

Subsection Overview

We now make some computations of \(H^i_T(L/K)\) for \(L/K\) a Galois extension of local fields. To begin with, recall that by “Theorem 90” (Lemma 1.2.3), \(H^1(L/K) = 0\text{.}\) Our next goal will be to supplement this fact with a computation of \(H^2(L/K)\text{.}\)

Proof.

Remark 4.2.2.

Before continuing, it is worth keeping in a safe place the exact sequence
\begin{equation*} 1 \to \gotho_L^* \to L^* \to L^*/\gotho_L^* = \pi_L^\ZZ \to 1\text{.} \end{equation*}
In this exact sequence of \(G = \Gal(L/K)\)-modules, the action on \(\pi_L^\ZZ\) is always trivial (since the valuation on \(L\) is Galois-invariant).

Remark 4.2.3.

Another basic fact to keep in mind is that any finite Galois extension of local fields is solvable. To wit, the maximal unramified extension is cyclic,; the maximal tamely ramified extension is cyclic over that; and the rest is a Galois extension whose degree is a power of \(p\text{,}\) and every finite \(p\)-group is solvable.
This will allow us to simplify some of the following arguments by writing a general Galois extension as a tower of successive cyclic extensions. Of course we will have no such shortcut in the global case, because the Galois group of a Galois extension of number fields can be any group whatsoever; in fact the inverse Galois problem asks whether this always occurs for an extension over \(\QQ\text{,}\) and no counterexample is known.

Subsection The unramified case

Recall that unramified extensions are cyclic, since their Galois groups are also the Galois groups of extensions of finite fields.

Proof.

One can certainly give an elementary proof of this using the fact that \(L^*\) is cyclic (see Exercise 1). But one can also see it using the machinery we have at hand. Because \(L^*\) is a finite module, its Herbrand quotient is 1. Also, we know \(H^1_T(L/K)\) is trivial by Lemma 1.2.3. Thus \(H^0_T(L/K)\) is trivial too, that is, \(\Norm_{L/K}: L^* \to K^*\) is surjective.

Proof.

Say \(u \in \gotho_K^*\) is a unit. By Proposition 4.2.4, we may pick \(v_0 \in \gotho_L^*\) such that in the residue fields, the norm of \(v_0\) coincides with \(u\text{.}\) Thus \(u/\Norm(v_0) \equiv 1 \pmod{\pi}\text{,}\) where \(\pi\) is a uniformizer of \(K\text{.}\) Now we construct units \(v_i \equiv 1 \pmod{\pi^i}\) such that \(u_i = u/\Norm(v_0\cdots v_i) \equiv 1 \pmod{\pi^{i+1}}\text{:}\) simply take \(v_i\) so that \(\Trace((1-v_i)/\pi^i) \equiv (1-u_{i-1})/\pi^i \pmod{\pi}\text{.}\) (That’s possible because the trace map on residue fields is surjective by the normal basis theorem.) Then the product \(v_0v_1\cdots\) converges to a unit \(v\) with norm \(u\text{.}\)

Proof.

Again, \(\Gal(L/K)\) is cyclic, so by Theorem 3.4.1 we need only check this for \(i=0,1\text{.}\) For \(i=0\text{,}\) it is Proposition 4.2.5. For \(i=1\text{,}\) note that because \(L/K\) is unramified, we can split the surjection \(L^* \to L^*/\gotho_L^*\) by choosing a uniformizer \(\pi_K\) of \(K\) and writing \(L^* = \gotho_L^* \pi_K^\ZZ\text{.}\) Hence \(H^1_T(\Gal(L/K), \gotho_L^*)\) is a direct summand of \(H^1_T(\Gal(L/K), L^*)\text{,}\) and the latter vanishes by Lemma 1.2.3.

Proof.

Using the Herbrand quotient, we get \(h(L^*) = h(\gotho_L^*) h(\ZZ)\text{.}\) Corollary 4.2.6 says that \(h(\gotho_L^*) = 1\text{,}\) and
\begin{align*} h(L^*/\gotho_L^*) \amp = h(\ZZ)\\ \amp = \#H^0_T(\Gal(L/K), \ZZ)/\#H^1_T(\Gal(L/K), \ZZ)\\ \amp = \#\Gal(L/K)^{\ab} / \#\Hom(\Gal(L/K), \ZZ)\\ \amp = [L:K]. \end{align*}
Since \(H^1_T(\Gal(L/K), L^*)\) is trivial, we conclude that \(H^0_T(\Gal(L/K), L^*)\) has order \([L:K]\text{.}\) Moreover, the long exact sequence of Tate groups gives an exact sequence
\begin{equation*} 1 \to H^0_T(\Gal(L/K), L^*) \to H^0_T(\Gal(L/K), \ZZ) = \Gal(L/K) \to 1, \end{equation*}
where the ends vanish by Corollary 4.2.6 again; so by periodicity (Theorem 3.4.1) we also get that \(H^0_T(\Gal(L/K), L^*) \cong H^2(L/K)\) is cyclic of order \([L:K]\text{.}\)
Let us now make the description of \(H^2(L/K)\) more canonical. Consider the short exact sequence
\begin{equation*} 0 \to \ZZ \to \QQ \to \QQ/\ZZ \to 0 \end{equation*}
of modules with trivial Galois action. The module \(\QQ\) is injective as a \(G\)-module for any group \(G\) (Exercise 2). Thus we get an isomorphism \(H^0_T(\Gal(L/K), \ZZ) \to H^{-1}_T(\Gal(L/K), \QQ/\ZZ)\text{.}\) But the latter is
\begin{equation*} H^1(\Gal(L/K), \QQ/\ZZ) = \Hom(\Gal(L/K), \QQ/\ZZ); \end{equation*}
since \(\Gal(L/K)\) has a canonical generator (Frobenius), we can evaluate there and get a canonical map
\begin{equation*} \Hom(\Gal(L/K), \QQ/\ZZ) \to \ZZ/[L:K]\ZZ \subset \QQ/\ZZ.\text{.} \end{equation*}
Putting it all together, we get a canonical map
\begin{equation*} H^2(\Gal(L/K), L^*) \cong H^0_T(\Gal(L/K), L^*) \cong H^1(\Gal(L/K), \QQ/\ZZ) \hookrightarrow \QQ/\ZZ\text{.} \end{equation*}
In this special case, this is none other than the local invariant map! In fact, by taking direct limits, we get a canonical isomorphism
\begin{equation*} H^2(K^{\unr}/K) \cong \QQ/\ZZ\text{.} \end{equation*}

Remark 4.2.8.

What’s really going on here is that \(H^0_T(\Gal(L/K), L^*)\) is a cyclic group generated by a uniformizer \(\pi\) (since every unit is a norm). Under the map \(H^0_T(\Gal(L/K), L^*) \to \QQ/\ZZ\text{,}\) that uniformizer is being mapped to \(1/[L:K]\text{.}\)

Subsection The cyclic case

Let \(L/K\) be a cyclic but possibly ramified extension of local fields. Again, \(H^1_T(L/K)\) is trivial by Lemma 1.2.3, so all there is to compute is \(H^0_T(L/K)\text{.}\) We are going to show again that it has order \([L:K]\text{.}\) (It’s actually cyclic again, but we won’t prove this just yet.)

Proof.

By the normal basis theorem, there exists \(\alpha \in L\) such that \(\{\alpha^g: g \in \Gal(L/K)\}\) is a basis for \(L\) over \(K\text{.}\) Without loss of generality, we may rescale to get \(\alpha \in \gotho_L\text{;}\) then put \(V = \sum \gotho_K \alpha^g\text{.}\) As in the proof of Theorem 3.2.9, \(V\) is induced: \(V = \Ind^G_{1} \gotho_K\text{,}\) so is acyclic.
The following proof uses that we are in characteristic 0, but it can be modified to work also in the function field case.

Proof.

Take \(V\) as in Lemma 4.2.9. If we choose \(\alpha\) sufficiently divisible, then \(V\) lies in the radius of convergence of the exponential series
\begin{equation*} \exp(x) = \sum_{i=0}^\infty \frac{x^i}{i!} \end{equation*}
(you need \(v_p(x) > 1/(p-1)\text{,}\) to be precise), and we may take \(W = \exp(V)\text{.}\)

Proof.

Take \(W\) as in Lemma 4.2.10. Since \(W\) has finite index in \(\gotho_L^*\text{,}\) we have \(h(\gotho_L^*/W) = 1\) and hence \(h(\gotho_L^*) = h(W) = 1\) by Lemma 4.2.10. So again we may conclude that \(h(L^*) = h(\gotho_L^*) h(\ZZ) = [L:K]\text{,}\) and so \(H^0_T(\Gal(L/K), L^*) = [L:K]\text{.}\)

Remark 4.2.12.

Notwithstanding Proposition 4.2.11, at this stage we cannot yet check that \(H^0_T(\Gal(L/K), L^*)\) is cyclic, because the groups \(H^1_T(\Gal(L/K), \gotho_L^*)\) are not guaranteed to vanish. See Exercise 3.

Remark 4.2.13. Note.

This is all that we need for “abstract” local class field theory. We’ll revisit this point later.

Subsection The general case

For those in the know, there is a spectral sequence underlying this next result; compare Exercise 3.

Proof.

For \(r=1\text{,}\) the condition on \(H^i\) is empty. In this case, \(H^1(G,M)\) classifies crossed homomorphisms \(\phi:G \to M\text{.}\) If one of these factors through \(G/H\text{,}\) it becomes a constant map when restricted to \(H\text{;}\) if that constant value itself belongs to \(M^H\text{,}\) then it must be zero and so the restriction to \(H\) is trivial. Conversely, if there exists some \(m \in M\) such that \(\phi(h) = m^h - m\) for all \(h \in H\text{,}\) then \(\phi'(g) = \phi(g) - m^g + m\) is another crossed homomorphism representing the same class in \(H^1(G,M)\text{,}\) but taking the value 0 on each \(h \in H\text{.}\) For \(g \in G, h \in H\text{,}\) we have
\begin{equation*} \phi'(hg) = \phi'(h)^g + \phi'(g) = \phi'(g)\text{,} \end{equation*}
so \(\phi'\) is constant on cosets of \(H\) and so may be viewed as a crossed homomorphism from \(G/H\) to \(M\text{.}\) On the other hand,
\begin{equation*} \phi'(g) = \phi'(gh) = \phi'(g)^h + \phi(h) = \phi'(g)^h \end{equation*}
so \(\phi'\) takes values in \(M^H\text{.}\) Thus the sequence is exact at \(H^1(G,M)\text{;}\) exactness at \(H^i(G/H,M^H)\) is similar but easier.
If \(r>1\text{,}\) we induct on \(r\) by dimension shifting. Recall (from Proposition 3.2.6) that there is an injective homomorphism \(M \to \Ind^G_1 M\) of \(G\)-modules. Let \(N\) be the \(G\)-module which makes the sequence
\begin{equation*} 0 \to M \to \Ind^G_{1} M \to N \to 0 \end{equation*}
exact. We construct a commutative diagram
Figure 4.2.15.
The second vertical arrow arises from the long exact sequence for \(G\)-cohomology; since \(\Ind^G_{1} M\) is an induced \(G\)-module, this arrow is an isomorphism. Similarly, the third vertical arrow arises from the long exact sequence for \(H\)-cohomology, and it is an isomorphism because \(\Ind^G_1 M\) is also an induced \(H\)-module; moreover, \(H^i(H, N) = 0\) for \(i=1, \dots, r-2\text{.}\) Finally, taking \(H\)-invariants yields another exact sequence
\begin{equation*} 0 \to M^H \to (\Ind^G_1 M)^H \to N^H \to H^1(H, M) = 0\text{,} \end{equation*}
so we may take the long exact sequence for \(G/H\)-cohomology to obtain the first vertical arrow; it is an isomorphism because \((\Ind^G_1 M)^H\) is an induced \(G/H\)-module. The induction hypothesis implies that the top row is exact, so the bottom row is also exact.

Proof.

We now prove the following.

Proof.

We’ve checked the case of \(L/K\) cyclic, so we may use it as the basis for an induction. If \(L/K\) is not cyclic, since it is solvable (Remark 4.2.3), we can find a Galois subextension \(M/K\text{.}\) Now the exact sequence
\begin{equation*} 0 \to H^2(M/K) \to H^2(L/K) \to H^2(L/M) \end{equation*}
from Corollary 4.2.16 implies that \(\#H^2(L/K) \leq \#H^2(M/K) \#H^2(L/M) = [M:K][L:M] = [L:K]\text{.}\)
To complete the proof that \(H^2(L/K)\) is cyclic of order \([L:K]\text{,}\) it now suffices to produce a cyclic subgroup of order \([L:K]\text{.}\)

Proof.

Consider the diagram
Figure 4.2.19.
in which the bottom row is exact and the vertical arrow is injective, both by Corollary 4.2.16. It suffices to show that the diagonal arrow \(H^2(M/K) \to H^2(ML/L)\) is the zero map, as this will imply an inclusion \(H^2(M/K) \subseteq H^2(L/K)\) within \(H^2(ML/K)\) and we already know that \(H^2(M/K)\) is cyclic of order \([M:K]=[L:K]\) by Proposition 4.2.7.
Let \(e = e(L/K)\) and \(f = f(L/K)\) be the ramification index and residue field degree, so that \([ML:L] = e\text{.}\) Let \(U\) be the maximal unramified subextension of \(L/K\text{;}\) then we have a canonical isomorphism \(\Gal(ML/L) \cong \Gal(M/U)\) of cyclic groups. By using the same generators in both groups, we can make a commutative diagram
Figure 4.2.20.
in which the vertical arrows are isomorphisms. (Remember from Definition 3.3.12 that restriction maps on Tate homology make sense in degree 0; that gives the first horizontal arrow in Figure 4.2.20.) The composition in the bottom row is the map \(H^2(M/K) \to H^2(ML/L)\) which we want to be zero; it thus suffices to check that the top row composes to zero.
Let us rewrite this composition concretely as
\begin{equation*} K^*/\Norm_{M/K} M^* \to U^*/\Norm_{M/U} M^* \to L^*/\Norm_{ML/L} (ML)^* \end{equation*}
where the maps are induced by the inclusions \(K^* \to U^* \to L^*\text{.}\) Now \(K^*/\Norm_{M/K} M^*\) is a cyclic group of order \(ef\) generated by \(\pi_K\text{,}\) a uniformizer of \(K\text{,}\) and \(L^*/\Norm_{ML/L} (ML)^*\) is a cyclic group of order \(e\) generated by \(\pi_L\text{,}\) a uniformizer of \(L\text{.}\) But \(\pi_K\) is a unit of \(\gotho_L\) times \(\pi_L^e\text{,}\) so the map in question is indeed zero.

Subsection The local invariant map

We have now proved the first assertion of Proposition 4.2.1 (by combining Proposition 4.2.17 and Proposition 4.2.18). We now turn to the second assertion. In the process, we will also see that \(H^2(\overline{K}/K) \cong \QQ/\ZZ\text{.}\)

Proof.

By Corollary 4.2.16 we have an injection \(H^2(K^{\unr}/K) \to H^2(\overline{K}/K)\text{,}\) and the former is canonically isomorphic to \(\QQ/\ZZ\text{;}\) so we have to prove that this injection is actually also surjective. Remember that \(H^2(\overline{K}/K)\) is the direct limit of \(H^2(M/K)\) running over all finite extensions \(M\) of \(K\text{.}\) What we just showed above is that if \([M:K] = n\) and \(L\) is the unramified extension of \(K\) of degree \(n\text{,}\) then the images of \(H^2(M/K)\) and \(H^2(L/K)\) in \(H^2(ML/K)\) are the same. In particular, that means that \(H^2(M/K)\) is in the image of the map \(H^2(K^{\unr}/K) \to H^2(\overline{K}/K)\text{.}\) Since that’s true for any \(M\text{,}\) we get that the map is indeed surjective, hence an isomorphism.

Remark 4.2.22.

By combining Proposition 4.2.18 with Lemma 4.2.21, we see that for any local field \(K\text{,}\) the map \(H^2(K^{\unr}/K) \to H^2(\overline{K}/K)\) is an isomorphism. We can use this to see the effect of changing \(K\) on this group; see Proposition 4.2.23 below.

Proof.

We compute the map following [36], Proposition III.1.8. Put \(e = e_{L/K}, f = f_{L/K}\text{.}\) We form a commutative diagram
Figure 4.2.24.
as follows. The left square comes from the valuation maps. The middle square comes from the connecting homomorphisms for the sequence \(0 \to \ZZ \to \QQ \to \QQ/\ZZ \to 0\) with the trivial actions; note that these connecting homomorphisms are isomorphisms by Exercise 2. The right square comes from evaluating crossed homomorphisms at Frobenius. Since \(ef = n\text{,}\) this yields the claim.

Exercises Exercises

1.

Give an elementary proof (without cohomology) that the norm map from one finite field to another is always surjective.
Hint.
Write everything in terms of a generator of the multiplicative group of the larger field.

2.

\times Let \(G\) be a finite group. Let \(M\) be a \(G\)-module whose underlying abelian group is a \(\QQ\)-vector space. Prove that \(M\) is an acyclic \(G\)-module.
Hint.
First show that the groups \(H^i(G,M)\) are divisible, say using the description in terms of cochains. Then combine with the fact that these group are killed by the order of \(G\) (Example 3.2.22).

3.

Give an example of a cyclic ramified extension \(L/K\) of local fields in which the groups \(H^i_T(\Gal(L/K), \gotho_L^*)\) are nontrivial.
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