TANT 17 - Local class field theory

Hello there! These are notes for the seventeenth class of the course "Topics in algebra and number theory" held in Block 4 of the academic year 2017/18 at the University of Copenhagen.

In the previous lecture we proved the global version of the theorem of Kronecker and Weber, and we have also seen how this theorem enables us to compute the abelian part of the absolute Galois group GQ. It is now natural to ask: can we do so for every number field K? More precisely, can we find an analogue of roots of unity which would enable us to characterize abelian extensions of a number field K, and thus to compute the group GabK?

It turns out that the second question has a positive answer, whereas the first question is rather hopeless (apart from one specific case, which we will see in the last week of the course). This last part of the course is thus devoted to compute the abelian group GabK for every number field K. This study is called "class field theory" and, as we will see later on, has its foundations in the laws of quadratic and cubic reciprocity envisioned by Gauß.

As we did for Q we will look first of all at the local world and then patch together all the information that we have there to get a global statement. In order to do this last step we will need a new object, the ring of adèles associated to a number field, which we will study for the next few lectures.


Profinite groups

Before we go on, we need a little introduction to profinite groups, since many of the groups that we will encounter in these lectures will be profinite. Recall that a topological group is a group G which is also a topological space such that the maps     G×GG(g,h)ghandGGgg1 are continuous. Profinite groups are a particular type of topological groups.

 Definition 1   A topological group G is called profinite if it is also a Stone space, i.e. Hausdorff, compact and totally disconnected.

It turns out that there is a canonical way to associate a profinite group to any topological group.

 Definition 2   Let G be a topological group. We define its profinite completion as the topological group ˆG:=limHG[G:H]<+GH where H runs over all open, normal subgroups of finite index in G. Thus, every finite factor of the inverse limit is endowed with the discrete topology.

The term "pro-finite" makes one think that such a topological group would always be a "limit" of finite groups. Which is precisely what happens.

 Theorem 3   Let G be a topological group. Then G is profinite if and only if GlimiIGi, where {Gi}iI is an inverse system of finite, discrete groups.

 Proof   Observe first of all that limiIGi is a closed subspace of the product iIGi. Indeed the inverse limit is defined as the set of sequences (ai)iIGi such that for every map ϕi,j:GiGj of the inverse system we have that ϕi,j(ai)=aj. Hence limiIGi is the intersection of the subspaces of iIGi obtained as (φi,j)1(Δj) where φi,j:kIGkGi×Gjϕi,j×IdGj×Gj and ΔjGj×Gj is the diagonal. Observe now that every finite, discrete group Gi is trivially a Stone space. This implies that limiIGi is a closed subset of a product of Stone spaces, and thus it is a Stone space itself. This follows from the fact that the product of Hausdorff spaces is Hausdorff (see this question on Math StackExchange), the product of compact spaces is compact (by Tychonoff's theorem) and the product of totally disconnected spaces is totally disconnected (see this question on Math StackExchange), and the fact that all these properties are preserved by closed sub-spaces. This shows that limiIGi is profinite, and thus that G is profinite if it is an inverse limit of finite, discrete groups.

Vice versa, suppose that G is profinite. Observe that we have a continuous group homomorphism GˆG given by sending each element gG to the "constant sequence" (πH(g))HˆG, where πH:GG/H is the quotient map. We would like to prove that this continuous map is a homeomorphism, and since both G and ˆG are compact and Hausdorff it is sufficient to show that this map is bijective (see Lemma B6.7 of these notes by Tom Leinster). First of all, we prove surjectivity. Let {gH}HG be a sequence of elements indexed on the finite-index, normal subgroups of  G such that (πH(gH))HˆG. Then by Exercise 4 the cosets gHHG are all non-empty closed subsets, and for every two finite-index, normal subgroups H,KG we have that gHHgKKgHK(HK). This implies that HgHH, by Cantor's intersection theorem. To conclude it is sufficient to observe that for every gHgHH we have that ¯g=πH(gH) for all normal subgroups of finite index HG. This proves indeed that the image of g under the map GˆG is equal to (πH(gH))H, and thus that the map is surjective.

To conclude, we want to prove injectivity. Observe that (πH(g))H=1 in ˆG if and only if gHH. Thus to conclude we have to prove that HH={1}. The proof of this result is rather technical, and we will omit it from these notes. If you are interested you can read Lemma 3.9 of these notes by Brian Osserman. Q.E.D.

The previous theorem tells us in particular that for every map of topological groups GP, where P is profinite, there exists a unique map of topological groups ˆGP such that the triangle on the right commutes (this is the universal property of the profinite completion).

 Exercise 4   Let G be a topological group, and HG be a subgroup. Prove that if H is open then H is also closed. Moreover, prove that if G is compact then H is open if and only if it is closed and has finite index.

The previous theorem tells us that a topological group is "pro-finite" if and only if it is the (inverse) limit of a system of finite groups considered with the discrete topology.

 Example 5   Let KL be a Galois extension of field. We have already seen that L is the direct limit of all the sub-extensions KML such that KM is finite. From this we get that Gal(L/K)limKML[M:K]<Gal(M/K) which implies that the group Gal(L/K) is a profinite topological group.

 Remark 6   Observe that taking only open normal subgroups of finite index in the definition of completion is necessary. Equivalently, we cannot expect that a profinite group is isomorphic to the profinite completion of itself regarded with the discrete topology. For example in the absolute Galois group GQ there are an infinite number of non-open normal subgroups of finite index (see Proposition 7.26 of these notes by James S. Milne). Thanks to a deep theorem of Nikolay Nikolov and Dan Segal (see their papers on this here and here) this implies that GQ is not topologically finitely generated, i.e. it is not the closure of a finitely generated subgroup (see also this question on MathOverflow, where the term "profinite completion" does not require the subgroups in the inverse limit to be open).

The local Artin map

Recall from the previous lecture that the local version of the theorem of Kronecker and Weber allows us to compute that GabQpZ×p׈Z. Observe in particular that Z×p׈Z^Q×p, because we have that Q×pZ×p×Z and Z×p is already profinite.

The main result of local class field theory is that this is true for every local field (see Definition 12 of the ninth lecture).

 Theorem 7   Let K be a local field. Then there exists a unique surjective homomorphism of topological groups θ:K×GabK (called local Artin map) such that:
  •  if K is non-Archimedean then for every finite, unramified extension KL (which is abelian by Corollary 2 of the thirteenth lecture) the composition of maps K×θGabKGal(L/K)Gal(κL/κK) sends every uniformizer πK× to the Frobenius automorphism κLκL defined by xxq, where q=#κK;
  • if KL is a finite extension we have that the diagram on the right is commutative, where NL/K:L×K× is the field norm and the map GabLGabK is given by restricting an automorphism σ:LabLab to Kab.
Moreover, θ induces an isomorphism of topological groups ^K×GabK.

This theorem can be proved in many ways: using the Brauer group of a field (as Helmut Hasse did), using Galois cohomology (as Emil Artin did), using formal groups (as Jonathan Lubin and John Tate did) or using explicit approaches (as Jürgen Neukirch and Michael Hazewinkel did). For now, we will only content ourselves with drawing some consequences from this theorem and proving the last part.

Observe first of all that if KL is a finite abelian extension we have a surjective map K×θGabKGal(L/K) whose kernel coincides with NL/K(L×)K×. Indeed observe that LKab, which implies, by Galois theory, that GabL=Gal(Kab/L). Thus we get an exact sequence {1}GabLGabKGal(L/K){1} where the first map is precisely the restriction map described in the second point of the theorem. Thus using the commutativity of the diagram above and the fact that θ is surjective we get that for every element xK× the automorphism θ(x)GabK becomes trivial in Gal(L/K) if and only if xNL/K(L×).

Norm groups

The Artin map that we mentioned in Theorem 6 is a surjective, continuous group homomorphism θ:K×GabK, but it cannot be an isomorphism because the group GabK is profinite, whereas K× is not. Indeed, we have that K×A×K×Z because for every uniformizer πK we can write K×=πZ×A×K.

Nevertheless, we can apply the universal property of the profinite completion to get a map ˆθ:^K×GabK, and we could hope that this map is an isomorphism. Since the Artin map is already continuous and surjective, we only need to prove that ˆθ is injective.
This is equivalent to show that every open, normal subgroup of K× of finite index corresponds to an open, normal subgroup of GabK of finite index. Galois theory tells us that these subgroups correspond to finite abelian extensions KL, and in turn we have seen in the previous paragraph that θ1(Gal(Kab/L))=NL/K(L×). Thus, proving that ˆθ is injective is equivalent to prove the following theorem.

 Theorem 8   Let K be a local field. For every open normal subgroup of finite index UK× there exists a unique finite abelian extension KL such that U=NL/K(L×).

 Proof   The uniqueness part is clear using Theorem 7 and Galois theory. If K is Archimedean, the statement is easy to prove (see Exercise 9 and Exercise 10), thus we will from now on assume that K is non-Archimedean. In this case, the proof of the existence part of the theorem uses the following facts:
  1. DK:=LNL/K(L×)={1}, where L runs over all the finite abelian extensions of K. To prove this, one proves first of all that DK is divisible, i.e. that nDK=DK for all nN1. This can be done using Kummer theory when K has characteristic zero (i.e. K is a finite extension of Qp), and needs some clever argument when K has positive characteristic (see Proposition 6 (page 174) and Propositon 16 of page 217 of the book "Local fields" by Jean-Pierre Serre). This implies that DK would be contained in every subgroup of finite index (if IK× has index n then nDK=DKnK×I) and thus in particular DKm,nU(m)KπnZ={1} where U(m)KA×K are the unit groups defined in Theorem 4 of the fourteenth lecture;
  2. if IK× is a subgroup of finite index which contains A×K, then there exists a finite abelian extension KL such that I=NL/K(L×). This is easier than the previous fact. Indeed, if I has index n and contains A×K we claim that N(Kn/K)(K×n)I, where KKn is the unique unramified extension of degree n (see Corollary 2 of the thirteenth lecture). This happens because the valuation map vK:K×Z is surjective and has kernel A×K, which implies that I=v1K(nZ), and finally we have that N(Kn/K)(K×n)v1K(vK(N(Kn/K)(K×n)))=v1K(nZ)=I where we use the facts that vK(N(Kn/K)(K×n))=nZ and A×KN(Kn/K)(K×n). The first fact is easy to prove, because any uniformizer πK is also a uniformizer for Kn, and thus every element of Kn can be written as πmx with mZ and xA×Kn, which implies that vK(N(Kn/K)(πmx))=vK(πnm)=nmnZ. The second fact can be proved using Hilbert's theorem 90 (see Proposition 1.2 of Chapter 3 of these notes by James S. Milne).
Let's see how that these two facts can be used to prove the theorem. Indeed, for every subgroup IK× of finite index we can use the first fact to see that LNL/K(L×)A×KLNL/K(L×)I and since I is open, A×K is compact and the subgroups NL/K(L×) are closed this implies that INL/K(L×)A×K for some finite abelian extension KL. Indeed, the quotient K×/I is finite, and the images of the subgroups NL/K(L×) in this quotient form a finite family of subgroups with trivial intersection. Since we have that NLE/K((LE)×)NL/K(L×)NE/K(E×) this implies that one of the subgroups NL/K(L×) becomes trivial in K×/I, i.e. that INL/K(L×)A×K for some finite abelian extension KL.
Actually, we have that  INL/K(L×)(A×K(NL/K(L×)I)) because if aNL/K(L×)(A×K(NL/K(L×)I)) then we can write a=bc with bA×K and cNL/K(L×)I, which implies that b=ac1NL/K(L×) and thus that aNL/K(L×)A×KI.
Since A×K(NL/K(L×)I) contains A×K we can use the second fact to find a finite abelian extension KE such that NE/K(E×)=A×K(NL/K(L×)I)I. Thus we can use Galois theory and the fact that Gal(E/K)K×/NE/K(E×) from Theorem 7 to find an intermediate extension KFE such that I=NF/K(F×), which is what we wanted to prove. Q.E.D.

 Exercise 9   Prove that C× has no non-trivial open subgroups. (Hint: use Exercise 4).

 Exercise 10   Prove that the only non-trivial open subgroup of R× is R>0. (Hint: use again Exercise 4).

Conclusions and references

In this lecture we managed to:
  • recall a few facts on profinite groups, including the definition of the profinite completion of a topological group;
  • state the main theorem of local class field theory (Theorem 7);
  • prove the existence theorem of local class field theory (Theorem 8) almost completely (we are missing the proof of the fact that DK is divisible and that A×KN(Kn/K)(K×n));
 References for this lecture include:

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