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 \(H^i_T\text{.}\)
SubsectionOverview
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 \(H^2(L/K)\text{.}\)
Proposition4.2.1.
For any finite Galois extension \(L/K\) of local fields, \(H^2(L/K)\) is cyclic of order \([L:K]\text{.}\) Moreover, this group can be canonically identified with \(\frac{1}{[L:K]}\ZZ/\ZZ\) in such a way that if \(M/L\) is another finite extension such that \(M/K\) is also Galois, the inflation homomorphism \(H^2(L/K) \to H^2(M/K)\) corresponds to the inclusion \(\frac{1}{[L:K]}\ZZ/\ZZ \subseteq \frac{1}{[M:K]}\ZZ/\ZZ\text{.}\)
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).
Remark4.2.3.
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.
SubsectionThe unramified case
Recall that unramified extensions are cyclic, since their Galois groups are also the Galois groups of extensions of finite fields.
Proposition4.2.4.
For any finite extension \(L/K\) of finite fields, the map \(\Norm_{L/K}\colon L^* \to K^*\) is surjective.
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, \(\Norm_{L/K}\colon L^* \to K^*\) is surjective.
Proposition4.2.5.
For any finite unramified extension \(L/K\) of local fields, the map \(\Norm_{L/K}\colon \gotho_L^* \to \gotho_K^*\) 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 \(u\text{.}\)
Corollary4.2.6.
For any finite unramified extensions \(L/K\) of local fields, \(H^i_T(\Gal(L/K), \gotho_L^*) = 1\) for all \(i \in \ZZ\text{.}\)
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.
Proposition4.2.7.
For any finite unramified extension \(L/K\) of local fields, \(H^2(L/K)\) is cyclic of order \([L:K]\text{.}\)
Proof.
As in the proof of Corollary 4.2.6, we use the fact that \(L^*\) splits as \(\gotho_L^* \pi_K^\ZZ\) to argue that
the first summand is zero by Corollary 4.2.6, while by periodicity (Theorem 3.4.1) the second summand is \(H^0_T(\Gal(L/K), \ZZ) \cong \ZZ/[L:K]\ZZ\text{.}\)
Remark4.2.8.
Continuing to assume that \(L/K\) is finite unramified, let us now make the description of \(H^2(L/K)\) more canonical. Consider the short exact sequence
What’s really going on here is that \(H^0_T(L/K)\) is a cyclic group generated by a uniformizer \(\pi\) (since every unit is a norm). Under the map \(H^0_T(L/K) \to \QQ/\ZZ\text{,}\) that uniformizer is being mapped to \(1/[L:K]\text{.}\)
SubsectionThe 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 \([L:K]\text{;}\) recall that this alone is the key input for the use of abstract class field theory (Remark 4.1.16).
Lemma4.2.9.
Let \(L/K\) be a finite Galois extension of local fields. Then there is an open, Galois-stable subgroup \(V\) of \(\gotho_L\) such that \(H^i(\Gal(L/K), V) = 0\) for all \(i>0\) (i.e., \(V\) is acyclic for cohomology).
Proof.
By the normal basis theorem, there exists \(\alpha \in L\) such that \(\{\alpha^g\colon g \in \Gal(L/K)\}\) 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 3.2.8, \(V = \Ind^G_{1} \gotho_K\) is induced and hence acyclic.
The following proof uses that we are in characteristic 0, but it can be modified to work also in the function field case.
Lemma4.2.10.
Let \(L/K\) be a finite Galois extension of local fields. Then there is an open, Galois-stable subgroup \(W\) of \(\gotho_L^*\) such that \(H^i(\Gal(L/K), W) = 0\) for all \(i>0\text{.}\)
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
(you need \(v_p(x) > 1/(p-1)\text{,}\) to be precise), and we may take \(W = \exp(V)\text{.}\)
Proposition4.2.11.
For \(L/K\) a cyclic extension of local fields, \(\# H^0_T(L/K) = [L:K]\text{.}\)
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 \(H^0_T(\Gal(L/K), L^*) = [L:K]\text{.}\)
Remark4.2.12.
Notwithstanding Proposition 4.2.11, at this stage we cannot yet check that \(H^0_T(L/K)\) is cyclic, because the groups \(H^1_T(\Gal(L/K), \gotho_L^*)\) are not guaranteed to vanish; see Exercise 3. That said, cyclicity will follow later from periodicity plus Lemma 4.2.20.
SubsectionThe general case
For those in the know, there is a spectral sequence underlying this next result; compare Exercise 3.
Let \(G\) be a finite group, let \(H\) be a normal subgroup, and let \(M\) be a \(G\)-module. If \(H^i(H, M) = 0\) for \(i=1, \dots, r-1\text{,}\) then the sequence
For \(r=1\text{,}\) the condition on \(H^i\) is empty. In this case, \(H^1(G,M)\) classifies crossed homomorphisms \(\phi\colon G \to M\text{.}\) 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
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 \(r \gt 1\text{,}\) we induct on \(r\) by dimension shifting. Recall (from Proposition 3.2.6) that there is an injective homomorphism \(M \to \Ind^G_1 \Res^G_1 M\) of \(G\)-modules. Let \(N\) be the \(G\)-module which makes the sequence
\begin{equation*}
0 \to M \to \Ind^G_{1} \Res^G_1 M \to N \to 0
\end{equation*}
exact. We will prove the claim by constructing a commutative diagram
Figure4.2.14. whose columns are isomorphisms.
In Figure 4.2.14, the second vertical arrow arises from the long exact sequence for \(G\)-cohomology; since \(\Ind^G_{1} \Res^G_1 M\) 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 \(\Ind^G_1 \Res^G_1 M\) is also an induced \(H\)-module; moreover, \(H^i(H, N) = 0\) for \(i=1, \dots, r-2\text{.}\)
Finally, using the hypothesis on the vanishing of \(H^i(H,M)\text{,}\) taking \(H\)-invariants yields another exact sequence
We may take the long exact sequence for \(G/H\)-cohomology to obtain the first vertical arrow; it is an isomorphism because \((\Ind^G_1 \Res^G_1 M)^H\) is an induced \(G/H\)-module. More generally, for any abelian group \(A\text{,}\) we have an isomorphism \((\Ind^G_1 A)^H \cong \Ind^{G/H}_1 A\) of \(G/H\)-modules. The point is that \(Z[G/H] \cong Z[G]^H\) via the map \([gH] \mapsto \sum_{h \in H} [gh]\text{.}\)
Corollary4.2.15.
If \(M/L/K\) is a tower of fields with \(M/K\) and \(L/K\) finite and Galois, the sequence
For any finite Galois extension \(L/K\) of local fields, the group \(H^2(L/K)\) has order at most \([L:K]\text{.}\)
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
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 \([L:K]\text{.}\) We do this by comparing with an unramified extension of the same degree.
Proposition4.2.17.
Let \(L/K\) be a finite Galois extension of local fields. Let \(M/K\) be an unramified extension of degree \([L:K]\text{.}\) Then the image of \(H^2(L/K)\) in \(H^2(ML/K)\) contains the image of \(H^2(M/K)\) in \(H^2(ML/K)\text{.}\) Consequently (by Proposition 4.2.7 and Corollary 4.2.15), the group \(H^2(L/K)\) contains a cyclic subgroup of order \([L:K]\text{.}\)
Proof.
Consider the diagram
Figure4.2.18. in which the bottom row is exact and the vertical arrow is injective, both by Corollary 4.2.15. 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 \(U/K\) be the maximal unramified subextension of \(L/K\text{;}\) then \(U/K\) is also a subextension of \(M/K\) (because the latter is cyclic and there is a unique unramified extension of \(K\) of any given degree). We can thus enlarge our previous diagram to
Figure4.2.19. where the maps out of \(H^2(U/K)\) are again injective by Corollary 4.2.15; moreover, the cokernel of \(H^2(U/K) \stackrel{\Inf}{\to} H^2(M/K)\) maps injectively via \(\Res\) into \(H^2(M/U)\text{.}\) This means that we can replace \(K\) by \(U\) before checking that the diagonal arrow is injective; in other words, we may safely assume that \(L/K\) is totally ramified.
At this point, \(L/K\) and \(M/K\) are totally disjoint, so we have a canonical isomorphism \(\Gal(ML/L) \cong \Gal(M/K)\) of cyclic groups. We may thus apply ordinary functoriality to the morphism \(M^* \to (ML)^*\) to obtain a map \(H^2(M/K) \to H^2(ML/L)\) which is exactly the one we want to be zero. By periodicity (Theorem 3.4.1) we may replace \(H^2 = H^2_T\) with \(H^0_T\) on both sides; we are then reduced to showing that the map
is zero. Now \(K^*/\Norm_{M/K} M^*\) is a cyclic group of order \([L:K]\) generated by \(\pi_K\text{,}\) a uniformizer of \(K\text{,}\) and \(L^*/\Norm_{ML/L} (ML)^*\) is a cyclic group of order \([L:K]\) generated by \(\pi_L\text{,}\) a uniformizer of \(L\text{.}\) But because \(L/K\) is totally ramified, \(\pi_K\) is a unit of \(\gotho_L\) times \(\pi_L^{[L:K]}\text{,}\) so the map in question is indeed zero.
SubsectionThe local invariant map
We have now proved the first assertion of Proposition 4.2.1 (by combining Proposition 4.2.16 and Proposition 4.2.17). We now turn to the second assertion. In the process, we will also see that \(H^2(\overline{K}/K) \cong \QQ/\ZZ\text{.}\)
Lemma4.2.20.
For any Galois extension \(L/K\) of local fields, the group \(H^2(L/K)\) can be canonically identified with \(\frac{1}{[L:K]}\ZZ/\ZZ\) in such a way that if \(M/K\) is another Galois extension containing \(L\text{,}\) the inflation homomorphism \(H^2(L/K) \to H^2(M/K)\) corresponds to the inclusion \(\frac{1}{[L:K]}\ZZ/\ZZ \subseteq \frac{1}{[M:K]}\ZZ/\ZZ\text{.}\)
Proof.
By Corollary 4.2.15 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 \(H^2(L/K)\) running over all finite extensions \(L\) of \(K\text{.}\) By Proposition 4.2.17, if \(L/K\) is a finite extension and \(M\) is the unramified extension of \(K\) of degree \([L:K]\text{,}\) then the images of \(H^2(L/K)\) and \(H^2(M/K)\) in \(H^2(ML/K)\) are the same. In particular, \(H^2(L/K)\) 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 \(L\text{,}\) we get that the map is indeed surjective, hence an isomorphism.
Remark4.2.21.
By combining Proposition 4.2.17 with Lemma 4.2.20, 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 4.2.22 below.
Proposition4.2.22.
For \(L/K\) a finite extension of local fields of degree \(n\text{,}\) the map \(\Res\colon H^2(K^{\unr}/K) \to H^2(L^{\unr}/L)\) corresponds, via the local reciprocity map, to the map from \(\QQ/\ZZ\) to itself given by multiplication by \(n\text{.}\)
Proof.
We compute the map following [37], Proposition III.1.8. Put \(e = e_{L/K}, f = f_{L/K}\text{.}\) We form a commutative diagram
Figure4.2.23. 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.
ExercisesExercises
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.
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 3.2.24).
3.
As in Exercise 2, take \(K = \QQ_p\) for some odd prime \(p\) and \(L = \QQ_p(\sqrt{p})\text{.}\) Compute the groups \(H^i_T(\Gal(L/K), \gotho_L^*)\) and see that they are not all zero; consequently, the conclusion of Corollary 4.2.6 does not apply to this ramified extension.
