Local Fields and their Abelian Extensions

Local Fields and their Abelian Extensions Final Degree Dissertation Degree in Mathematics Jorge Fari˜na Asategui Supervisor: Gustavo A. Fern´andez Alcober Leioa, 2019-2020 Contents Preface ii 1 Preliminaries 1 1.1 Topological groups and fields . . . . . . . . . . . . . . . . . . . 1 1.2 Profinitegroups........................... 3 1.3 Infinite Galois correspondence . . . . . . . . . . . . . . . . . . . 5 1.4 Character groups and Pontrjagin’s duality . . . . . . . . . . . . 6 2 Global and Local Fields 10 2.1 Discrete valuations . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.2 Completion and local fields . . . . . . . . . . . . . . . . . . . . 12 2.3 Prime ideals in finite extensions of global and local fields . . . . 14 3 The Brauer Group 21 3.1 Central simple algebras and the Brauer group . . . . . . . . . . 21 3.2 Group cohomology, crossed products and cyclic algebras . . . . 28 3.3 Brauer group of a local field . . . . . . . . . . . . . . . . . . . . 31 4 Class Field Theory 35 4.1 Finitefields............................. 35 4.2 Localfields ............................. 36 A Solved Problems 44 A.1 Preliminaries ............................ 44 A.2 Global and local fields . . . . . . . . . . . . . . . . . . . . . . . 44 A.3 TheBrauergroup ......................... 46 A.4 Classfieldtheory.......................... 51 Bibliography 53 i Preface “In most sciences one generation tears down what another has built and what one has established another undoes. In mathematics alone each generation adds a new story to the old structure.” —Hermann Hankel. April 8th, 1796. A young Carl Frieldrich Gauss (1777-1855) woke up and wrote the entry, “Numerorum primorum non omnes numeros infra ipsos residua quadratica esse posse demonstratione munitum.” 1to his diary [Kle03]. He had just devised the first correct proof of the quadratic reciprocity law. Previous efforts from Fermat, Euler and Legendre among others, had helped to establish this law and partial results on its veracity. Gauss was amazed by the beauty of this law, which he called Theorema Aureum (Golden Theorem), and he managed to provide seven more different proofs in his lifetime (three of them were published along the first one in his Disquisitiones Arithmeticae in 1801 [Gau01]). Today, more than two hundred different proofs of this law have been published. Quadratic reciprocity shows an impressive simmetry that allows to determine if a prime pis a square modulo a prime q, by looking whether qis a square modulo p. A natural generalization is to find higher reciprocity laws, i.e. cubic, cuartic, quintic, etc. This leads directly to extending the field of rationals to more elaborated number fields. Ernst Kummer’s (1810-1893) ideal numbers, a precursor for ideals later developed by Richard Dedekind (18311916), came to existence in search of these higher reciprocity laws. In 1900, David Hilbert (1862-1943) made up a list of twenty three problems concerning some of the most relevant unsolved questions of his time. Among those problems, Problem 9th deals with general reciprocity laws [Hil00]: 1Prime numbers below (modulo) all numbers may not be quadratic residues, possesses a tough proof. ii “Für einen beliebigen Zahlkörper soll das Reciprocitätsgesetz der lten Potenzreste bewiesen werden, wenn leine ungerade Primzahl bedeutet und ferner, wenn leine Potenz von 2 oder eine Potenz einer ungeraden Primzahl ist. Die Aufstellung des Gesetzes, sowie die wesentlichen Hülfsmittel zum Beweise desselben werden sich, wie ich glaube, ergeben, wenn man die von mir entwickelte Theorie des Körpers der lten Einheitswurzeln 1) und meine Theorie 2) des relativ-quadratischen Körpers in gehöriger Weise verallgemeinert.”2 Theory of ideals was further developed and Emil Artin (1898-1962) provided the first proof for his general reciprocity law in a series of papers around 1927, which implies all known reciprocity laws. This was deduced proving the main theorem of global class field theory, which describes abelian extensions of a global field in terms of its arithmetic intrinsic properties. A good account of this procedure is given in [Lan94] for example. Later, the approach turned about to the local-global principle. The local version of class field theory, i.e. for local fields, was first proved by determining the Brauer group of a local field (see Chapter 3), and the global version was obtained by considering all primes at once via ideles and adeles. This procedure turned out to be better understood in the language of group cohomology, and this is the way it is currently presented [AT09]. Nowadays, generalisations to non abelian extensions are being developed extending the theory for abelian extensions. In this work we shall seek to study local class field theory. Let us show the reader an intuitive motivation for class field theory. Given a field Kand a finite Galois extension L/K, the main theorem of Galois theory describes the intermediate field extensions in terms of the subgroups of the Galois group. However, this procedure needs the finite extension field L to be fixed first. We may wonder if we can give a description of all the finite extensions of a given field. This is not an easy feat to accomplish with all generality3, but it is easier if we restrict our attention to finite abelian extensions. In the case where the base field is either the field of complex numbers or the reals, this description is trivial since Cis algebraically closed and Rhas a unique abelian extension, namely C. 2For any number field the reciprocity law for lth residues should be proved, when lis an odd prime and when lis a power of 2 or of an odd prime. The list of laws, as well as complementaries for the proof itself should be obtained, in my opinion, from my welldeveloped lth cyclotomic field theory 1) and an appropiate generalization of my theory 2) of relative-quadratic fields. 3There are still open problems concerning the Galois group Gal(Qal/Q), such as whether each finite group occurs as a quotient of it [Mil20]. Let Eand Fbe two finite abelian extensions of K. Then, the composite EF is also Galois. What is more, Gal(EF/K)is isomorphic to a subgroup of the cartesian product Gal(E/K)×Gal(F/K), which is abelian; thus, EF/K is abelian too. But, what can be said about the composite of a countable number of abelian extensions? Let us see a motivating example. Let the base field be the field of rationals. Then, it is known from the course in Algebraic Equations that for each natural number n, the nth cyclotomic extension is abelian. Since the degree of the nth cyclotomic extension is precisely ϕ(n), these degrees are not bounded and the composite of them all is an infinite extension. We leave to the reader the details on why this composite is an abelian Galois extension of Q, as a preparation for the following explanations which shall generalize this example to a general field extension L/K. Let L/K be a general Galois infinite field extension and define the set L:= {E:Efinite Galois intermediate extension of L/K}. Note that for any pair E, F ∈ L their composite EF ∈ L. This makes Linto a directed poset with respect to the inclusion. Also, note that for any pair E, F ∈ L the natural inclusions ϕEF :E→Flift the elements in Eto F, whenever E⊆F. This makes Linto a direct system over itself. Note that the same natural liftings ϕE:E→Lexist for each E∈ L. What is more, these liftings are compatible with the ones in the direct system, i.e. , ϕE=ϕEF ϕF whenever E⊆F. Then, by the universal property of the direct limit (see Chapter 1), Lcan be regarded as the direct limit lim −→E=SE. Then, any element in Llies in a finite Galois extension of K,E, and there is a very natural way to describe the K-automorphisms of L: as coherent tuples (σE)E. An expected, but which the reader should prove, property of an infinite Galois extension L/K is that its Galois group is infinite (hint: show the degrees of intermediate fields are not bounded). We could expect the finite Galois correspondence to generalize immediately to the infinite case. Sadly, this is not the case. In particular, not all finite index subgroups of Gal(L/K)need correspond to finite intermediate extensions, i.e. they may not be of the form Gal(L/E)(see Problem A.1). However, this may be fixed by endowing the Galois group with a special topology where these subgroups are precisely open, turning Gal(L/K)into a topological group. Let Gdenote the Galois group of the extension L/K. Given an intermediate field EGalois over K, there is a natural projection from Gto Gal(E/K)by restriction of automorphisms. Also, for intermediate Galois field extensions K⊆E⊆F⊆L, there is a natural composition of restrictions G→Gal(F/K)→Gal(E/K)which is no more than the usual restriction G→Gal(E/K). What is more, for any pair of intermediate Galois field extensions K⊆E, F ⊆L, there exists another intermediate Galois extensions field, namely the composite EF , containing both Eand Fand whose corresponding subgroup Gal(L/EF)is precisely the intersection of both subgroups Gal(L/E)and Gal(L/F). Then, these subgroups form a filter of normal subgroups. This filter can be used to give a topology to G, which is called the Krull topology. Also, it gives us as the data of an inverse system, where the objects are the finite Galois groups Gal(E/K)endowed with the discrete topology and the connection homomorphisms are the above restrictions. Thus, we may identify Gwith the inverse limit lim ←−Gal(E/K)via the aforementioned projections and the universal property of the inverse limit (see Chapter 1). We shall see in Chapter 1 that both constructions coincide up to isomorphism and endow Gwith the same topology. Now, we restrict our attention to abelian extensions of a field K. Applying previous reasoning with the added condition the extensions are abelian, i.e. letting L:= {E:E/K is finite abelian}, the direct limit exists and it contains all the abelian extensions of K. This direct limit is called the maximal abelian extension of Kand denoted by Kab. Then, by the infinite Galois correspondence (see Chapter 1), studying the finite abelian extensions of Kis equivalent to studying open subgroups of Gal(Kab/K). The drawback is we have defined these open subgroups as the ones coming from finite intermediate extensions and a priori we have no way to distinguish among them. The special case where Kis a finite field is very well known and we shall study it first to come up with a motivation for other fields. And this is precisely the objective of class field theory: studying these open subgroups via an easier-to-study group. What is more, we shall show this easier-to-study group is related to the arithmetic of the base group K, making the data of abelian finite extensions of Kintrinsic to K, which, in my humble opinion, is a result of an astonishing beauty. We aim to provide a complete proof of local class field theory. To fulfill this goal, we take a more algebraic approach, rather than the modern cohomological perspective. We follow the outline in [KKS11] and fill in the details from other sources and ourselves, trying to make the proof as short and as easy as possible since most texts take a lot longer to provide a full proof of this theorem. Inevitably we will be missing many interesting concepts and theories that arise in our discussion, which could take a whole book by themselves. We have no space either to deduce the global version of class field theory, which is a beautiful application of the local-global principle. In the first chapter we introduce some preliminary concepts such as topological groups, profinite groups, character groups and Pontrjagin’s duality, which shall appear in the rest of the work. We follow mostly [RZ10]. In the second chapter, we present basic facts about local and global fields and describe ramification of prime ideals in their abelian extensions. Due to lack of space, concepts such as differents or discriminants are not even mentioned. We provide a short introduction to topological groups and fields, in order to state Pontrjagin’s duality, which is fundamental for the proof of local class field theory. We follow mostly [FV02] and [SG80] to state and prove results on completions. For the rest of the chapter we follow [KKS11]. The third chapter is devoted to determining the Brauer group of local fields. For that, we introduce the theory of division algebras, simple central algebras, crossed products and cyclic algebras, as well as a little introduction to classic group cohomology. We follow mostly [Jac85] for the theory of simple central algebras and [KKS11] for the determination of the Brauer group of a local field. The cohomological introduction is taken from [Mor96]. The last chapter is where a proof for local class field theory is given. We first show the finite field case as a precursor for local fields.Then, we prove local class field theory with all the tools from previous chapters. We follow and complete the proofs in [KKS11]. We have summarized some minor results that had no place on the main flow of the text in an appendix as solved exercises, just for completeness. Most problems are taken from the same sources as the main text, but some are taken from other sources. For example, Dedekind’s Independence Theorem has been taken from [Jac09], which is not used for the main text. A second appendix contains the essentials of the theory of inverse limits, direct limits and profinite groups, just in case the reader is unfamiliar with these notions, allowing the reader to follow the explanations in this work flawlessly. The reader is not assumed to have any prior knowledge apart from what is taught in this degree. Lastly, a little notation issue. Map composition will be written multiplicatively from left to right, i.e. for maps fand g, their composition (also called product) will be written fg where first we apply fand then g. Commutative rings will be assumed to have an identity, see [Poo14] for a short nice discussion on this. 1.2. Profinite groups may construct the inverse limit as the closed subset of the cartesian product Qi∈I Xiformed by all coherent tuples (at the level of elements), i.e. the tuples (xi)i∈I such that ϕij(xi) = xjwhenever ji. It is left to the reader to prove this construction satisfies the universal property of the inverse limit [RZ10]. This reduces many times properties of inverse limits of objects in Cto properties of the objects in C. In particular, if the objects are finite groups, then the resulting inverse limit will behave similarly to a finite group and many properties will be deduced by reducing it to the finite case. Again, the definition of the direct limit is the dual notion of the inverse limit and it is left to the reader the details of its definition (hint: reverse all arrows in the definition of the inverse limit). Intuitively, the direct limit is a union. In the category of abelian topological groups X= lim −→Xi=Siϕi(X) and X=SiXiif the projections are onto [RZ10]. In the introduction, we have seen how a field extension may be seen as a direct limit, and we shall see how taking its group of automorphisms dualizes it turning into an inverse limit and finally applying the hom functor gives us a direct limit again. This scheme applies in other situations too and it will be vital for us. We will be working with finite groups, which can be endowed with the discrete topology to turn them into topological groups. In this context, the inverse limit is compact, Hausdorff and totally disconnected [RZ10] and it is called a profinite group. Given a group Gwe may define the directed set of normal subgroups N:= {N≤fG:G/N finite}. Then, we define the profinite completion of G, denoted by b Gas the inverse limit lim ←−G/N where Nruns over N. Then, Gis naturally embedded into its profinite completion by the obvious map g7→ (gN)N. The topological closure of a subgroup of a profinite group can be obtained as follows. Lemma 1.3. Let Gbe a profinite group and Ha subgroup of G. Then, the topological closure of Hin Gcan be obtained as H=\ N HN ∼ =lim ←−HN/N, where Nruns over all open normal subgroups in G. We shall see in the next section that we are interested in the open subgroups of the Galois group Gal(Kab/K). Class field theory will be based upon an easier-to-study group whose profinite completion is precisely this Galois group (up to isomorphism). Then, the following proposition [RZ10] is vital for us to translate the Galois correspondence to this easier-to-study group. 4 Chapter 1. Preliminaries Proposition 1.4. Let Gbe a residually finite group, i.e. the intersection of all its normal subgroups of finite index is trivial. Then, there is a 1-to-1 correspondence between the open subgroups in Gand the open subgroups in its profinite completion b Ggiven by H7→ Hand K7→ K∩Grespectively. 1.3 Infinite Galois correspondence We state without proof the infinite version of the main theorem of Galois theory [Mor96] even if we just need the assertion on finite extensions. Theorem 1.5. Let L/K be an infinite Galois extension and Gits Galois group endowed with the Krull topology. Then, there is a one-to-one inclusion reversing correspondence betweeen the closed subgroups of Gand the intermediate extensions of L/K. What is more, the intermediate extension Eis normal if and only if its corresponding closed subgroup N:= Gal(L/E)is normal in G, in which case G/N ∼ =Gal(E/K)as topological groups. Also, if we restrict to finite extensions, this correspondence is one-to-one between finite intermediate extensions and open subgroups of G. We shall see the inverse limit lim ←−Gal(E/K)is isomorphic to Gwith the Krull topology as topological groups. Since the natural projections G→ Gal(E/K)are compatible with the connection homomorphisms for being the usual restrictions as above, by the universal property of the inverse limit, there exists a unique group homomorphism Φ : G→lim ←−Gal(E/K)compatible with the corresponding projections. By construction, this homomorphism is precisely the one given by σ7→ (σ|E)E. We shall see it is an isomorphism. The kernel is trivial since the only automorphism that restricts to the identity in all finite intermediate fields is the identity. For that, recall L= lim −→E=SEE where Eruns through all the finite intermediate extensions in L/K. Then, for each s∈L,s∈Efor some Eand since σE(s) = sfor all Eand all s∈L,σ is the identity as wanted. To show it is onto we shall see that each infinite coherent tuple (σE)Elifts to an automorphism in Gal(L/K)where coherence means that for intermediate fields E⊆Fand a tuple (σE)E, the component σFrestricts to σEin E. Then, since each s∈Lis contained in some finite Galois extension of K,E, we may define the automorphism σ∈Gas s7→ σ(s) := σE(s). It is well defined since if sis in another intermediate Galois extension field F,σE(s) = σF(s). To see that1, note that K(s)⊆E, F, and for being coherent, the restrictions of σEand σFto K(s)coincide; thus, their image on stoo. Note this automorphism is actually an automorphism and fixes K; thus, σ∈G. Clearly, all automorphisms in Gcan be obtained in this fashion, since their restriction to 1We could have used EF instead of K(s)too and apply coherence there. 5 1.4. Character groups and Pontrjagin’s duality each intermediate finite extension form coherent tuples (σE)E. Then, we obtain an isomorphism of groups between Gand the inverse limit lim ←−Gal(E/K). We need to make it into a homeomorphism. We shall copy the topology in the inverse limit to Gthrough the above isomorphism. We know that a fundamental system of open neighborhoods in lim ←−Gal(E/K)is given by the kernels of the projection homomorphisms (Lemma 2.1.1 in [RZ10]). Thus, since these are usual restrictions/projections of Ginto Gal(E/K)∼ = G/Gal(L/E)for intermediate finite Galois extension fields Ethe kernels are precisely the normal subgroups Gal(L/E). Then, {Gal(L/E)}Eforms a fundamental system of open neighborhoods in G, which is precisely the way we defined the Krull topology. 1.4 Character groups and Pontrjagin’s duality The character group or dual of a topological group, G∗, is the group of continuous group homomorphisms from Gto T, i.e. G∗= homcont(G, T), where Tis the multiplicative subgroup of complex numbers of unit norm. We will be dealing with character groups throughout the proof of local class field theory. Then, we shall fix some special notation and show a couple of results. For a field Kwe denote the character group of Gal(Kab/K) by X(K). Since we restrict to continuous homomorphisms, we see that the preimage of any atom in Tis open, since Tis given the discrete topology. But open in compact topological groups implies finite index. In particular, the kernel is of finite index, i.e. the image of the given homomorphism is of finite order. Thus, the image groups of these homomorphisms are mapped to Q/Zvia the usual isomorphism e2πix 7→ x+Z. Then, for a profinite group G∗= homcont(G, Q/Z). Also, if we define addition of homomorphisms via addition of their images, X(K)can be seen as an additive group and since for a homomorphism ϕ,o(ϕ) = lcm(o(ϕ(g)))g∈G, the additive order of any homomorphism is finite and X(K)is torsion. All characters of an abelian profinite group arise as charactes of a finite abelian group. Note that for any χ∈G∗,ker χis normal in Gand open too for being the preimage of 0 and Q/Zbeing discrete. Then, we can regard χas the natural composition G→G/ ker χ→Q/Z. A useful property of characters is that given a group homomorphism ϕ:G→Hwhere His abelian and finite, it will be onto if and only if the only character annihilating the image of ϕis the trivial character, i.e. if AnnH∗(ϕ(G)) = 0. Note the only if part is trivial. For the if part assume by 6 Chapter 1. Preliminaries contradiction that there exists an element h∈Hsuch that h /∈Im χ. We shall find a nontrivial character χ∈H∗such that χ(ϕ(G))=0. For that consider the non-trivial but finite quotient H/ϕ(G). Since the quotient is finite and abelian, Tχ∈(H/ϕ(G))∗ker χ= 0, there is at least one character χin (H/ϕ(G))∗which is non-trivial and we may lift it to a character in of Hby χ0:= πχ, getting a non trivial character of Hannhilating the image of Gby ϕ, arriving to a contradiction and proving the desired property. Putting everything together, let L/K be an infinite Galois extension. Let us recall the definition of the set Land how natural it is to construct the direct limit of this direct system, lim −→E=SE=L, which is no more than the infinite Galois extension L. Now, we shall consider the Galois groups of each finite Galois extension of Kin L. These groups Gal(E/K)together with the usual restrictions ϕEF : Gal(F/K)→Gal(E/K)give us the data of an inverse system over the same directed poset L. Now, it is natural again to consider the inverse limit lim ←−Gal(E/K)∼ =Gal(L/K). Note that Galois correspondence being inclusion reversing turns inclusions into restrictions, dualizing the construction, turning a direct limit into an inverse limit. This had been shown so far in our discussion. Now, consider the dual of each finite Galois group, i.e. the homomorphisms from Gal(E/K)to Q/Z. Note that a character χE: Gal(E/K)→Q/Z lifts to a character χF: Gal(F/K)→Q/Zwhenever E⊆Fand χF(σ) := χE(σ|E)for each σ∈Gal(F/K). Intuitively, we are plugging the bigger group Gal(F/K)in the left via a restriction. This allows us to lift the characters of the smaller group Gal(E/K)to the bigger group Gal(F/K). This gives us the data for a direct system over the same directed poset L. Naturally, we build the direct limit lim −→Gal(E/K)∗∼ =Gal(L/K)∗. To obtain that isomorphism, recall the image of each character is finite. Hence, any character can be obtained in this lifting fashion. To see that, note this kernel is a subgroup of the form Gal(L/E)where Eis a finite Galois extension of Kfor the kernel being open. Then, this character can be obtained from a character of the finite Galois group Gal(E/K)by plugging in the left Gal(L/K)through the usual restriction. Then the isomorphism follows from the universal property of the direct limit, since the liftings are compatible with the connection homomorphisms as they are also usual inclusions. Lastly, we want to build the bidual Gal(L/K)∗∗. For that, consider the bidual of each finite Galois group, Gal(E/K)∗∗. These form an inverse system over the same directed poset L. Just note a character χE: Gal(E/K)∗→Q/Z restricts to a character χF: Gal(F/K)∗→Q/Zsince the bidual of a finite group is known to be isomorphic to the original finite group through the evaluation homomorphism and we may define this restriction through these isomorphisms and the usual restriction in the Galois groups. Then, we consider 7 1.4. Character groups and Pontrjagin’s duality the inverse limit lim ←−Gal(E/K)∗∗ ∼ =Gal(L/K)∗∗. Note these evaluation isomorphisms are compatible with the restrictions by definition; thus, we obtain the evaluation isomorphism of inverse limits Gal(L/K)∗∗ ∼ =lim ←−Gal(E/K)∗∗ ∼ = lim ←−Gal(E/K)∼ =Gal(L/K). This is known with more generality for profinite groups as Pontrjagin’s duality and even in more generality for locally compact abelian groups. Theorem 1.6 (Pontrjagin’s duality). Given a locally compact abelian group Gand its character group G∗, the character group of G∗and the group Gare naturally isomorphic, where this isomorphism is given by the evaluation map. The interested reader is advised to check [Pon46] for a full proof of Pontrjagin’s duality for locally compact abelian groups and [RZ10] for profinite groups. Remark. The reader may be wondering why we are using restrictions instead of liftings when considering the Galois groups. In the case of characters we are able to do these liftings because we are considering just homomorphisms, not automorphisms. If we try to lift an automorphism in this fashion it will have a non-trivial kernel and thus it will not be injective, it will not be even a homomorphism since field homomorphisms are injective. This shows why in this case it is natural to consider restrictions and therefore the inverse limit whilst with characters it is more natural to lift them and consider the direct limit. Let us analyze the case K=Fq. Then, all finite intermediate fields are known to us to be cyclic and we have a more precise description of L= {Fqn:n∈N}. Then, Fab q=Fsep q= lim −→Fqn=SFqnand the Galois group Gal(Fab q/Fq)∼ =lim ←−Gal(Fqn/Fq)∼ =lim ←−Z/nZ=ˆ Z. Finally, since Q/Zis discrete and each finite group in Lis discrete, to know all continuous homomorphisms Gal(Fqn/Fq)→Q/Zis just to know all such group homomorphisms. For being the base group finite and cyclic, it is enough to provide the image of the Frobenius automorphism, and choose it to be an element in Q/Zof order divisor the order of the group, i.e. any element in h1/n +Ziwhere nis this order. Then, X(Fq)∼ =lim −→Gal(Fqn/Fq)∗∼ =lim −→h1/n+Zi=Sh1/n+Zi=Q/Z. Thus, X(Fq)∼ =Q/Zwhere this isomorphism is given by χ=χn7→ k/n +Z, where k/n is the image of the nth Frobenius automorphism under χn=πnχ. In other words, this isomorphism is obtained mapping each character χto its image in the Frobenius automorphism of Gal(Fab q/Fq). Note this argument is valid for any finite field, in particular for Fqf. Then, X(Fq)∼ =X(Fqf). But we know explicitly a very natural homomorphism between these two character groups (which is not an isomorphism, caution!). Since the Galois groups Gal(Fqn/Fq)are cyclic, whenever fdivides nthere is a unique subgroup of index f, namely Gal(Fqn/Fqf). What is more, an fth power of a Frobenius 8 Chapter 1. Preliminaries automorphism in Gal(Fqn/Fq)is a Frobenius automorphism in Gal(Fqn/Fqf). Thus, given a homomorphism Gal(Fqn/Fq)→Q/Z,σ7→ k/n +Zwe obtain a homomorphism Gal(Fqn/Fqf)by the multiplication-by-fmap in Q/Zand the usual power-by-fmap in the Galois groups, σ7→ fk/n +Z Gal(Fqn/Fq)Q/Z Gal(Fqn/Fqf)Q/Z restriction multiplication by f whenever ndivides f. Multiplication-by-fis an epimorphism in each term of the directed set, and if we consider the character groups and the natural isomorphisms mapping a character to its image in the corresponding Frobenius automorphisms, we get the following commutative diagram which we shall use in the proof of local class field theory. Lemma 1.7. The diagram X(Fq)Q/Z X(Fqf)Q/Z restriction multiplication by f is commutative. 9 Chapter 2 Global and Local Fields The first example of a field seen in an elementary algebra course is usually the field of rational numbers, Q. Finite extensions of Qare called number fields, and they are the main object of study in algebraic number theory. Along with number fields (and finite fields), function fields are the best known examples of fields. Of special interest in algebraic geometry are finite extensions of Fq(T), where Fq(T)is the field of rational functions in one variable with coefficients in Fq. We call the latter fields global function fields. These two types of fields may look rather different at first glance, but they share many properties. This analogy between both of them motivates the definition of a global field as either a number field or a global function field. Pursuing these analogies has been shown fruitful for both algebraic geometry and number theory. In this chapter we develop the basic theory of both global and local fields that is used in subsequent chapters. We will use the following equivalent [Mil17] definitions of a Dedekind domain throughout the chapter. Definition 2.1 (Dedekind domain). An integral domain Ais said to be a Dedekind domain if it is a field or any of the following equivalent conditions is satisfied. 1. Any proper ideal ain Afactors uniquely into a product of prime ideals. 2. Ais noetherian, integrally closed and every nonzero prime ideal is maximal. 2.1 Discrete valuations Definition 2.2 (Discrete valuation). Let Kbe a field. Let ν:K×→ Zbe a non-trivial surjective group homomorphism satisfying the additional condition 10 Chapter 2. Global and Local Fields (i) ν(a+b)≥min(ν(a), ν(b)),∀a, b ∈K×, and set ν(0) = ∞to extend νto the entire field K. Then, νis said to be a discrete valuation of K. A field with a discrete valuation is called a discrete valuation field. The first example of a discrete valuation of a field is the map ordp:Q×→Z defined as follows for a rational prime p. For any rational number a, let us write it as an irreducible fraction in the form a=pna0 a1 , n ∈Zand p-a0, a1. We define ordp(a) = nand set ordp(0) = ∞. It is easy to check that ordp is a discrete valuation of Q. This map is called the p-adic valuation. Similarly, for a Dedekind domain Aand its field of fractions K, we define the p-adic valuation,ordp:K×→Zfor a nonzero prime ideal pof A, by writing the fractional ideal (a)for any a∈K×as the unique product of prime ideals (a) = pnQqni iwith n, ni∈Zand qi6=pfor all i, and defining ordp(a) = n. Note that we set ordp(0) = ∞as before. Let Kbe a discrete valuation field for ν. Then, it is immediate from the definition of a discrete valuation that the set Oν={a∈K:ν(a)≥0}forms a subring of Kand it is called the valuation ring of K. Proposition 2.3. Let Kbe a field and νa valuation of K. (i) Let Oνbe the valuation ring of Kwith respect to ν. Then, Oνis a principal ideal domain and thus a Dedekind domain. The only nonzero prime (and maximal) ideal of Oνis p={a∈K:ν(a)≥1}, and νcoincides with ordp. Any element ain Ksuch that ν(a)=1 generates p; any ideal of Oνis of the form (an) = {b∈K:ν(b)≥n} for such an element aand n∈N, and any fractional ideal of Oνof the form (an) = {b∈K:ν(b)≥n}for same aand n∈Z. The group of units of Oνis exactly the set of the elements with null valuation, i.e. O× ν={a∈K:ν(a)=0}. (ii) Conversely, let Abe a Dedekind domain with a unique non-zero prime ideal p. Then, Acoincides with the valuation ring for the discrete valuation ordp. (iii) Given an integral domain A, the following conditions are all equivalent. 11 2.2. Completion and local fields (a) Ais the valuation ring of a discrete valuation of its field of fractions. (b) Ais a principal ideal domain with a unique nonzero prime ideal. (c) Ais a Dedekind domain with a unique nonzero prime ideal. An integral domain Asatisfying any of the last three equivalent conditions is called a discrete valuation ring. Proof. First note that an element c∈ Oνof null valuation is a unit in the valuation ring since 1 = cc−1in Kimplies taking valuations, 0 = ν(c) + ν(c−1); thus, ν(c−1)=0and c∈ O× ν. Similarly, if c∈ O× ν, then ν(c)=0. Let a be an element of Ksuch that ν(a) = 1. Now fix a nonzero ideal aof Oν. Let n= min{ν(b) : b∈a}. Then, by the definition of an ideal, a⊆b:= {b∈K:ν(b)≥n} ⊇ (an). To prove these inclusions are actually equalities, let first b∈b. Then, b=ancwhere c=a−nb∈K. Taking valuations, ν(c) = ν(a−n) + ν(b)≥0. Thus, c∈ Oνproving b∈(an). Let now b∈a such that ν(b) = n. Then, b=ancfor some c∈Kand taking valuations, ν(b) = ν(c) + ν(an). Thus, ν(c) = 0 and c∈ O× ν, so an=c−1b∈agetting both equalities. The other asssertions in (i) are straightforward to check now. For (ii) just note Ais trivially contained in the valuation ring. For the reverse inclusion note that for an element a∈ Oordp,(a) = pnwith n≥0; thus, a∈A. Now, (iii) follows from (i) and (ii). A generator πof the unique maximal ideal pof a discrete valuation ring is called a uniformizer or a prime element of Oνor K. The natural quotient field Oν/pis called the residue field of Oν. We shall see how prime ideals origin embeddings of global fields into what we call local fields. For that, we need to attach a special topology to these fields. 2.2 Completion and local fields From a discrete valuation ν, we can obtain a metric. Let cbe a real number such that 1<c<∞. Then, it is easily checked that the map dν:K×K→R defined as dν(x, y) := c−ν(x−y)for x6=yand dν(x, y) := 0 for x=y, defines a metric in K. This metric induces a Hausdorff topology where Vn,a ={b∈ K:ν(b−a)≥n}can be taken as a fundamental system of open (and closed) neighborhoods for each point ain K. 12 Chapter 2. Global and Local Fields As with respect to the usual metric in Q, a Cauchy sequence in Kwith respect to dνmay not converge in K. Thus, we may complete Kwith respect to this metric to make all Cauchy sequences converge. Lemma 2.4. Let Abe the set of all Cauchy sequences in K. Then, Ais a ring with respect to componentwise addition and multiplication. The set of all Cauchy sequences convergent to 0 form a maximal ideal of A,m. The field A/mis a discrete valuation field with discrete valuation ˆνdefined as ˆν((an) + m) = lim ν(an). Proof. Proving Ais a ring is straightforward; thus, it is left to the reader. To prove mis maximal, let m0be an ideal strictly containing m. We shall prove m0=A. Take a Cauchy sequence (an)in m0\m. Then, there exists a positive integer n0such that an6= 0 for n≥n0. Let (bn)be a sequence such that for n≥n0,bn=a−1 n. Then, (bn)is clearly Cauchy and (an)(bn) + m= (1) + m. Thus, since m0is an ideal, (1) is in m0and m0=Aas wanted. The fact ˆν is a discrete valuation for A/mfollows now from the properties of the usual limit. A discrete valuation field Kis called a complete discrete valuation field if every Cauchy sequence in Kconverges in K. A discrete valuation field ˆ K with valuation ˆνis called a completion of Kif it is complete, ˆν|K=νand K is a dense subfield of ˆ Kwith respect to dν. This completion is unique up to isomorphism. Proposition 2.5. Every discrete valuation field Khas a unique completion up to K-isomorphism. Proof. We shall prove that the field A/min previous lemma is the unique completion of K. For that, first note that Kis embedded in A/mby the natural map a→(a) + m. Now, for a Cauchy sequence (an)in Kand any real number M, there exists a positive integer n0such that for m, n ≥n0, ν(am−an)≥M. Thus, if we take (an0)which clearly converges in K, we get ˆν((an0)−(an)) ≥M, proving that Kis dense in A/m. To prove completeness, let ((a(m) n)n)mbe a Cauchy sequence in A/m(with respect to dˆν). Let n1, n2, . . . be an increasing sequence of positive integers such that for i, j ≥nm, ˆν(a(m) i−a(m) j)≥M. Then, (a(m) nm)mis a Cauchy sequence in Kand the limit of ((a(m) n)n)min A/m(with respect to dˆν). This proves A/mis a completion of K. For uniqueness, assume (ˆ K1,ˆν1)and (ˆ K2,ˆν2)to be two completions of K. Let 1Kbe the identity map in K. Then, we extend this isomorphism by continuity from K, as a dense subfield of ˆ K1, to ˆ K1. This means, for an element a∈ˆ K1, we take a Cauchy sequence (an)in Ksuch as lim1an=aand we map it to b= lim2an∈ˆ K2. This map is well defined. For that note that if we consider two distinct Cauchy sequences converging to a,(an)and (a0 n), by 13 2.3. Prime ideals in finite extensions of global and local fields Lastly, note that K×is generated by the prime elements in K. For that, note that a prime element is of valuation 1 and since a unit in the valuation ring has valuation 0, their product is again a prime; thus, all elements of valuation 0 can be obtained as division of primes and any nonzero element can be obtained from primes, as a power of a prime times an element of null valuation. This will be important in Chapter 4 to reduce proofs to the case of a prime element. 20 Chapter 3 The Brauer Group In the axiomatic definition of a field K, we assume Kto be a commutative ring. We may relax this definition not asking for commutativity. If we do so, we get a kind of non (necessarily) commutative fields. These will be called division algebras or skew fields and they are nothing but identity rings where division is possible, i.e. all non zero elements are invertible. Most of the division algebras seen in undergraduate courses are usually commutative and thus, usual fields. The first example of a non commutative division algebra was the so-called quaternion algebra, H, discovered by William Rowan Hamilton (1805-1865) in 1843 while walking along the Royal Canal in Dublin (he carved the defining formula for quaternions i2=j2=k2=ijk =−1into the stone of Broome Bridge in an impulse after years of thinking). Quaternions came to existence as an effort to understand rotations in a three dimensional space, just as complex numbers describe rotations in two dimensions. The theory presented in this chapter is further richer and more extensive than the one given in our presentation. We have developed just a minimal amount of theory due to space constraints. Thus, the interested reader is strongly encouraged to check [Jac85], for example, for more details. Even if not stated explicitly, all k-algebras in this work are assumed to be associative. 3.1 Central simple algebras and the Brauer group Let kbe a field and Aak-algebra. If the center of Ais exactly k,Ais said to be central over kand if the only (two-sided) ideals of Aare 0 and Aitself, Ais said to be simple. If Ais both central over kand simple, Ais called a central simple algebra over k. As an example of central simple algebras we have division algebras over their center. For instance, His a central simple 21 3.1. Central simple algebras and the Brauer group algebra over R. Our aim is to define a group, for a field k, whose elements will be some similarity classes defined upon simple central algebras over k. To define a group, we need an operation, and this operation will be based on the tensor product. Then, we need first to define the tensor product of two vector spaces. For that, let Aand Bbe two k-vector spaces. A balanced product of Aand Bis defined to be an abelian group Gtogether with a map f:A×B→G satifying for all a, a0∈A,b, b0∈Band λ∈k, 1. f(a+a0, b) = f(a, b) + f(a0, b), 2. f(a, b +b0) = f(a, b) + f(a, b0), 3. f(λa, b) = f(a, λb). It is denoted as (G, f). If (G0, f0)is another balanced product, a morphism from (G, f)to (G0, f0)is a group homomorphism η:G→G0such that f0=fη. Now, the tensor product of Aand Bis a balanced product (A⊗kB, ⊗)such that for any other balanced product (G, f), there exists a unique morphism from (A⊗kB, ⊗)to (G, f), i.e. (A⊗kB, ⊗k)is universal for this property. An explicit construction of the tensor product via the cartesian product A×Bwhere its elements are written as sums of the elementary tensors a⊗kbwith (a, b)∈A×Bis given in Problem A.6. Note dimk(A⊗kB) = dimk(A) dimk(B)and that A⊗kBcan be endowed with a natural product (a⊗kb)(a0⊗kb0)=(aa0⊗kbb0), making A⊗kBak-algebra. Now, we need to define the aforementioned similarity relation on central simple algebras over a field k. With that goal in mind, we state and prove a criterion to know under which conditions can an algebra over a field kbe factored as the tensor product of two k-subalgebras. We consider just the finite dimensional case. Proposition 3.1. Let Aand Bbe subalgebras of a finite dimensional algebra Dover a field k. Then, D∼ =A⊗kBif the following conditions are satisfied. 1. ab =ba for all a∈Aand b∈B. 2. D=AB and [D:k]=[A:k][B:k]. Proof. The first condition ensures the map ϕ:A⊗kB→Dmapping a⊗b→ ab is a ring homomorphism and k-linear. For the sake of illustration we shall show that the product is preserved thanks to first condition, other properties are easy to check from the definition of the tensor product and are left to the reader. ϕ((a⊗b)(a0⊗b0)) = ϕ(aa0⊗bb0) = aa0bb0=aba0b0=ϕ(a⊗b)ϕ(a0⊗b0). 22 Chapter 3. The Brauer Group The second condition implies ϕis surjective; thus, an isomorphism by dimension counting of vector spaces. Let us see a direct application of this criterion. Proposition 3.2. Let Abe a k-algebra. Then, Mn(A)∼ =Mn(k)⊗kA. In particular, Mmn(k)∼ =Mm(k)⊗kMn(k). Proof. It is straightforward to check that the subalgebras Mn(k)and A1n where 1nis the identity in Mn(A), satisfy the conditions in Proposition 3.1. Just note Ma1m=a1mMfor any M∈Mn(k)and any a∈Aand that any M∈Mn(A)can be expressed uniquely as an A-linear combination of elements of a given basis {eij}for Mn(k)since a k-basis for Mn(k)automatically gives an A-basis for Mn(A). Last assertion follows now from the first one by taking A:= Mm(k). Now, we are in position to define the similarity relation. Let Aand B be two finite dimensional central simple algebras over k. We will say Aand Bare similar and write A∼Bwhen Mn(A)∼ =Mm(B)for some positive integers n, m. This similarity relation is clearly reflexive and symmetric. To see transitivity, let Mn(A)∼ =Mm(B)and Ml(B)∼ =Mr(C), then Mnl(A)∼ =Mnl(k)⊗A∼ =Mn(k)⊗Ml(k)⊗A∼ =Ml(k)⊗Mn(A) ∼ =Ml(k)⊗Mm(B)∼ =Ml(k)⊗Mm(k)⊗B∼ =Mm(k)⊗Ml(B) ∼ =Mm(k)⊗Mr(C)∼ =Mm(k)⊗Mr(k)⊗C∼ =Mmr(k)⊗C ∼ =Mmr(C), where we have used associativity and commutativity of the tensor product and the formulas obtained in Proposition 3.2. Thus, ∼is an equivalence relation and we may consider equivalence classes [A] = {Bfinite dimensional simple central algebra : B∼A}. Now we shall define a binary operation via the tensor product for the set of these equivalence classes. Let A∼A0and B∼B0. We claim now A⊗B∼A0⊗B0. Since A∼A0and B∼B0we know by definition of ∼that Mn(A)∼ =Mm(A0)and Ml(B)∼ =Mr(B0), or equivalently by Proposition 3.2, A⊗Mn(k)∼ =A0⊗Mm(k)and B⊗Ml(k)∼ =B0⊗Mr(k)for positive integers n, m, l, r. Then Mnl(A⊗B)∼ =A⊗B⊗Mnl(k)∼ =A⊗Mn(k)⊗B⊗Ml(k) ∼ =A0⊗Mm(k)⊗B0⊗Mr(k)∼ =A0⊗B0⊗Mmr(k) ∼ =Mmr(A0⊗B0), proving our claim. This means that the binary operation [A] + [B] := [A⊗B] is well defined. 23 3.1. Central simple algebras and the Brauer group The opposite algebra of A, denoted Aop, is defined by dualizing the product in A, i.e. reversing the product in Aor, in other words, assigning the element ba to the product a·b. The enveloping algebra of A, denoted Ae, is defined as the tensor product Ae=A⊗kAop. To make this set of equivalence classes into an abelian group we need to prove first associativity, commutativity, existence of an identity and an inverse for all equivalence classes. Associativity and commutativity follows directly from associativity and commutativity of the tensor product. Note that Mn(A)∼ =A⊗Mn(k); thus, A∼A⊗Mn(k). Hence, [Mn(k)] = 0 acts as the identity for +. We shall see [A]+[Aop]=0and [Aop]acts as the inverse of [A]with respect to +, or equivalently by definition, that the enveloping algebra acts always as the identity. Primitive rings and the Density Theorem For an abelian group M, the set of endomorphisms of M,EndZMor End M, has a natural ring structure. With ring of endomorphisms, we mean a subring of End Mfor an abelian group M. We define a ring representation, as a ring homomorphism ρ:R→End Mfor an abelian group M. A representation ρof Racting on M, i.e. its image is in End M, yields a left R-module structure for Mby defining am =ρ(a)mfor any a∈Rand m∈M. Conversely, given a left R-module M,Mis an abelian group and we can define ρ=ρM:R→End M via the assignment a→aMfor any a∈R, where aM∈End Mis left multiplication by a. Thus, ρis a ring representation. For an R-module M,EndRM will denote the group of R-linear endomorphisms. An irreducible representation of a ring is a representation such the module Mis nonzero and whose only submodules are 0 and itself. We may say Mis R-irreducible if there is such a representation and it is completely reducible if it is the direct sum of irreducible R-modules. The kernel of a representation ρis called the annihilator of M,AnnR(M) := {r∈R:rm = 0,∀m∈M}. If AnnR(M) = ker ρ= 0, we say ρ(or sometimes the R-module M) is faithful. This kernel is obviously an ideal of R. A ring Ris called (left) primitive if it has an irreducible and faithful representation. The structure of primitive rings is totally determined by the important Density Theorem from Nathan Jacobson ([Jac85], p. 199). For our means we only need the partial result on finite dimensional case that given a primitive ring Racting on an abelian group M,Ris isomorphic to the finitely dimensional vector space1End∆Mwhere ∆ = EndRM. Thus, when refering to the Density Theorem we will be refering to this partial result. Note irreducibility 1We shall use the term vector space for modules over a division ring and not just over fields, just to agree with the terminology in [Jac85]. 24 Chapter 3. The Brauer Group of Mis just needed to ensure ∆ = EndRMis a division algebra via Schur’s Lemma. Then, if we can ensure this ring of endomorphisms is a division algebra, it is enough Mbeing completely reducible. For a more in detail discussion on this see [Jac85]. We have a natural module action of Aeon Adefined by (Pai⊗a0 i)x= Paixa0 i. Direct verification shows it is a well defined module action. Aesubmodules of Aare two-sided ideals of A; thus, if Ais simple Ais Aeirreducible. Regarding Aas a left (right) A-module in the natural way, the elements of EndAAare the right (left) multiplication maps x7→ xa (x7→ ax) since if an endomorphism fmaps 17→ a, then, f(x) = f(x·1) = xa. Note the reversing of left and right. Then, EndAeAis the set of maps that are both left and right multiplications. Just note that if fmaps 17→ athen, f(x) = f(1 ·1·x) = ax =xa =f(x·1·1) = f(x). These are precisely the ones x7→ cx where cis in the center of A. If Ais central over k, then x7→ αx, where α∈k. Theorem 3.3. Let Abe a finite dimensional central simple algebra over a field k. Then, Ae=A⊗kAop ∼ =Mn(k)where n= dimkA. Proof. Regarding Aas an Ae-module as above, Ais Ae-irreducible and EndAeA=kfor being simple and central over krespectively. Since Ais finite dimensional over k, by the Density Theorem Aemaps onto EndkA. Now, since both vector spaces are of dimension n2over k(note dimk(Ae) = dimk(A) dimk(Aop) = n2), it is an isomorphism Ae∼ =EndkA∼ =Mn(k). Ae∼ =Mn(k)is known to be simple (see [Gri07], Proposition 1.4, p. 360). Thus it is simple central over kand by Theorem 3.3, [Aop]acts as the inverse of [A]with respect to +. Thus, it is only left to check this set of equivalent classes is closed under the operation +. Theorem 3.4. Let Abe a finite dimensional central simple subalgebra of an algebra B. Then, B∼ =A⊗kCwhere Cis the centralizer of Ain B. The ideals of Bare in correspondence with the ideals of Cby the bijection a→Aa. Moreover, the center of Bcoincides with the center of C. Proof. We shall use Proposition 3.1. Since Aeis simple Bis a direct sum of irreducible Ae-modules and for Abeing Ae-irreducible, they are all isomorphic to A(note that two irreducible Ae-modules are always isomorphic since if Mis an irreducible R-module then M∼ =R/mfor a maximal ideal mand since Ris simple they are all isomorphic, see Problem A.7). Now, note that the generator of Aas an Ae-module, 1, satisfies the condition (a⊗k1)1 = a1=1a= 1(1⊗ka) and (a⊗k1)1 = 0 implies a= 0. Thus, since all irreducible Ae-modules are isomorphic we may choose an element cjin each irreducible Ae-module satisfying (a⊗k1)cj=acj=cja=cj(a⊗k1) and (a⊗k1)cj= 0 impliyng a= 0. 25 3.1. Central simple algebras and the Brauer group Applying this to Bas an Ae-module and noting the map from Ato each irreducible Ae-module mapping 17→ cjextends to an isomorphism by linearity, we may write B=LAcjwhere acj=cjafor all a∈Aand acj= 0 implies a= 0. Then, clearly cj∈Cand any element of Bcan be uniquely written as a finite sum Pajcjfor aj∈A. For any c∈C,c=Pajcj, but ac =ca implies aaj=ajafor any a∈A. Thus, aj∈k(for Abeing central over k) and c∈Pkcj. Hence, C=Pkcjand cjis a basis for Cand clearly B=AC and [B:k]=[A:k][C:k]. Thus, by Proposition 3.1, B∼ =A⊗kC, as wanted. Now, let abe an ideal in C. Then, Aais an ideal in B=AC. We claim that Aa∩C=a. Let βA={x1= 1, . . . , xn}be a k-basis for A. Since B∼ =A⊗kC, any element in Bcan be uniquely written as a C-linear combination of the k-basis βA. Thus, the elements of Aaare a-linear combinations of βA. In particular, elements both in Aaand in Care of the form c1x1=c1∈a. Hence, Aa∩C=aas claimed. This proves that the map a7→ Aais injective since Aa=Abimplies a=bby taking the intersection with C. To check surjectivity, let bbe an ideal of B. Then, bis an Ae-submodule of B. Hence, b=PAbj where bj∈a:= b∩C. This implies, b=Aa, proving surjectivity. Thus, we get a one-to-one correspondence between the ideals of Cand those of B. Lastly, we show the center of Bcoincides with the center of C. Clearly the center of Bis contained in Cand thus, in the center of C. For the converse, any element in the center of Ccommutes with every element of B=AC and thus, it is in the center of B. Thus, when Cis simple and central over k,Bis simple and central over k. Corollary 3.5. The tensor product of two finite dimensional central simple algebras over a field kis again a finite dimensional central simple algebra over k. In general, the tensor product of a finite number of finite dimensional central simple algebras over a field kis again a finite dimensional central simple algebra over k. Then, the set of equivalent classes is closed under +and it can be regarded as an abelian group. This group is called the Brauer group of kand it is denoted by Br(k). The Brauer group was first introduced by Richard Brauer (1901-1977) in 1929. Now, let Lbe a field extension of k. Then, for any k-algebra A,A⊗kLcan be regarded as an L-algebra. This L-algebra is denoted as ALand is called the algebra obtained from Aby extending the base field to L. If Ais finite dimensional central simple over k, it can be seen as a corollary to Theorem 3.4 that ALis finite dimensional central simple over L. Let Abe a finite dimensional central simple algebra over k. A field Lis called a splitting field for 26 Chapter 3. The Brauer Group Aif AL=A⊗kL∼ =Mn(L)for some positive integer n. Now, let Lbe a finite extension of k. Then, if Ais finite dimensional central simple over k,ALis finite dimensional central simple over L. We have (A⊗kB)L∼ =AL⊗LBLand Mn(k)L∼ =Mn(L). Thus, a group homomorphism may be defined from Br(k)to Br(L)mapping [A]7→ [AL]. The kernel of this homomorphism, i.e. the classes [A]in Br(k)such that AL∼0, are exactly the classes of k-algebras with splitting field L. This kernel forms a subgroup that is denoted by Br(L/k). Recall that for a field Fand an n-dimensional F-vector space V,EndF(V)∼ = Mn(F). If commutativity is not assumed, we should take the opposite ring when passing from endomorphisms to matrices (see Problem A.8). Lemma 3.6. Let Abe a central simple algebra over kand L/k a finite field extension such that Lis a subfield of Mn(A)∼ =EndAop Vfor Van Aop-vector space. Then, the centralizer of Lin Mn(A)coincides with EndAop⊗kLV. Proof. We shall see the centralizing Lcondition is equivalent to Aop ⊗kLlinearity regarding Vas an Aop ⊗kL-module via the action (d⊗kl)x=dlx = ldx. Let l∈L⊆EndAop V. Then, lc =cl for some endomorphism cmeans lc(v) = c(l(v)) for all v∈V. But cis Aop-linear; thus, c((l⊗ka)v) = c(lav) = lac(v)=(l⊗ka)c(v)and it is Aop ⊗kL-linear. The same reasoning proves the converse. Theorem 3.7. Let ∆be a finite dimensional central division algebra over k. Then, a finite extension L/k is a splitting field for ∆if and only if Lis a subfield of an algebra A=Mn(∆) such that CA(L) = L. Proof. We just need the only if part, for a proof of the converse see [Jac85]. Let Lbe a subfield of A=Mn(∆) which is self-centralised. Recall Amay be identified with End∆0Vfor an n-dimensional vector space Vover ∆0= ∆op. Then, we regard Vas an ∆0⊗kL-module as before. Since both ∆0and L are simple, by Corollary 3.5, ∆0⊗kLis simple too and this action is necessarily faithful. Now, since ∆0⊗kLis finite dimensional over k,Vis completely reducible as a ∆0⊗L-module. By previous lemma, the centralizer of Lin Ais End∆0⊗LVand this is Lby assumption, which is a field; thus, a division algebra. Then, we may apply the Density Theorem to obtain ∆0⊗L∼ =EndLV∼ =Mr(L)and Lis a splitting field for ∆0; hence, for ∆. 27 3.2. Group cohomology, crossed products and cyclic algebras 3.2 Group cohomology, crossed products and cyclic algebras Given a group Gand an abelian group M, we say Mis a (left) G-module if we can define a (left) G-action on M, i.e. a map G×M→M, such that for any g, h ∈Gand any n, m ∈Mwe have g(m+n) = gm +gn, g(hm)=(gh)m, 1m=m. Now, let Gbe a group and MaG-module. We define the group of ncochains, denoted as Cn(G, M)as the set of maps from Gnto Mfor any n∈N. For n= 0,C0(G, M) = Mby convention. This set can be endowed with a group operation by defining the sum of maps componentwise. This way, it becomes an abelian group. Let us define the maps δn:Cn(G, M)→Cn+1(G, M)via the assignments, δn(f)(σ1, ..., σn+1) =σ1f(σ2, ..., σn+1) + n X j=1 (−1)jf(σ1, ..., σjσj+1, ..., σn+1) + (−1)n+1f(σ1, ..., σn), for n∈Nand δ0(m)(σ) = σm −mfor n= 0. The maps δndefine group homomorphisms and δ2=δnδn+1 is the trivial map. Comprobation of these facts is straightforward but tedious; thus, it is left to the reader. The maps δn are called differentials. Since δ2= 0,Im δn−1is inside ker δn, and it makes sense to define the quotient group Hn(G, M) = Zn(G, M)/Bn(G, M), where Zn(G, M) := ker δn and its elements are called n-cocycles and Bn(G, M) := Im δn−1and its elements are called n-coboundaries. For n= 0 we define B0(G, M)=0. The group Hn(G, M)is called the nth cohomology group of Gwith coefficients in M. Two n-cocycles are called cohomologous if they are equal up to a coboundary, i.e. if they represent the same element of Hn(G, M). The G-invariant part of Mis defined as the subset of Mthat is invariant under the action of G, i.e. MG={m∈M:σm =m, σ ∈G}. Whenever Gis cyclic with generator σ, the norm group,N(G), is defined as the set N(G) := {Pσkm:m∈M}. In this cyclic case, the second cohomology group can be described by these constructions (Problem 16 in [Mor96], p. 105-106). Theorem 3.8. Let Gbe a cyclic group and MaG-module. Then, H2(G, M)∼ = MG/N(G). Proof. See Problem A.9. 28 Chapter 3. The Brauer Group We introduce now a way of constructing a k-algebra based on a finite Galois field extension. Let L/k be a finite Galois field extension and let {xσ} be a collection of symbols in 1-1 correspondence with the elements of G:= Gal(L/k). We regard the k-algebra Aas a vector space over Lwith basis {xσ}, i.e. A=LLxσ, and we define a product on Aby the relations xσxτ=κσ,τ xστ , xσl=σxσ,∀l∈L, where κσ,τ are elements of L×and Lis regarded as a G-module by the natural action of G, i.e. σl =σ(l)for any σ∈Gand any l∈L. Since {xσ}is an L-basis for A, any element of Acan be uniquely represented as a finite sum Plσxσwith coefficients in L. Thus, the product of two elements Plσxσand Pl0 σxσis defined as Pκσ,τ lσσl0 τxστ , by the above relations. To make Ainto an associative k-algebra, the κs,t must be chosen appropriately. Since (xσxτ)xρ=κσ,τ xστ xρ=κσ,τ κστ,ρxσστρ xσ(xτxρ) = xσκτ,ρxτρ = (σκτ,ρ)κσ,τρxστρ, and to ensure associativity we need, κσ,τ κστ,ρ = (σκτ,ρ)κσ,τρ, which is exactly the 2-cocycle condition we have seen before in multiplicative notation for the map κ:G×G→L×,(σ, τ)7→ κσ,τ . Distributive laws are straightforward to check from definition. Since by the 2-cocycle condition κ1,σ =κ1,1and κσ,1=σκ1,1, the element 1 = κ−1 1,1x1is the identity for the multiplication as a simple computation shows. Finally, since the elements of kare fixed by all k-automorphisms of L, the ring multiplication and the k-scalar multiplication as a vectorial space are compatible, i.e. λ(ab) = (λa)b=a(λb)for all λ∈k and a, b ∈A. Thus, Ais a k-algebra. We shall denote Ain the following as (L, G, κ). We shall see these k-algebras are simple central over kand identify the subgroup Br(L/k)with the second cohomology group H2(G, L×). Proposition 3.9. Let A= (L, G, κ). Then, Ais a simple central algebra over kand [A:k] = n2where n= [L:k]. Regarding Las a subfield of A(L1A)it coincides with its centralizer in A, i.e. CA(L) = L. Proof. The relation [A:k] = [A:L][L:k] = [L:k]2=n2holds from the definition of the crossed product. Now we shall see Ais simple. For that, we shall see first all xσare invertible. The product xσxσ−1=κσ,σ−1κ1,11is a nonzero element of L; thus, xσis invertible in A. Now, let abe a proper ideal of A. We shall see it is trivial. We write a=a+afor a∈A. Then, A=A/a6= 0 since it is proper. Hence, the usual projection restricted to L, l7→ lis a monomorphism into A, since Lis a field and this homomorphism 29 4.2. Local fields to the p-adic topology in K×are precisely those norm groups, and its profinite completion being isomorphic to the Galois group Gal(Kab/K). 4.2 Local fields Theorem 4.1 (Local Class Field Theory). Let Kbe a local field. Then, 1. There exists a unique continuous homomorphism ρK:K×→Gal(Kab/K), satisfying the following conditions. a) For a finite abelian extension Lof K,ρKinduces an isomorphism K×/NL/K(L×)∼ = →Gal(L/K). b) If Kis a complete discrete valuation field with finite residue field Fq, then the diagram K×Gal(Kab/K) ZGal(Fab q/Fq). ρK νK ρFq is commutative, where νKis the discrete valuation in Kand the map Gal(Kab/K)→Gal(Fab q/Fq)is the composition Gal(Kab/K)→Gal(Kur/K)∼ = →Gal(Fab q/Fq), where Gal(Kab/K)→Gal(Kur/K)is the restriction of automorphism of Kab to Kur. 2. There is a one-to-one correspondence through ρKbetween open subgroups of Gal(Kab/K)and open subgroups of finite index of K×, i.e. finite abelian extensions of Klie in one-to-one correspondence with open subgroups of finite index of K×. The remainder of the section is devoted to proving this important theorem. We will assume at some points Khas characteristic 0 for the sake of simplicity, but the results are still valid in positive characteristic even if the proofs tend to be more tedious. The cases K=Rand Care dealt separately (see Problem A.12). Now, let Kbe a complete discrete valuation field with finite residue field. 36 Chapter 4. Class Field Theory Proposition 4.2. Let Kbe a local field and La finite separable extension of K. Then, 1. The following diagram is commutative. Br(K)Q/Z Br(L)Q/Z. invK multiplication by [L:K] invL 2. The order of Br(L/K)is exactly [L:K]. Proof. First note second assertion follows from the first one. Since inv is an isomorphism, the kernel of the multiplication by [L:K]in Q/Zis isomorphic to the kernel of the restriction Br(K)→Br(L), which we denoted by Br(L/K). Since the first kernel is precisely {n/[L:K] + Z}and has order [L:K], Br(L/K)has order [L:K]too. Now we prove the first assertion. Let eand fbe the ramification index and residue degree of Lover Krespectively. By Proposition 2.9 we have [L:K] = ef. Let πbe a prime element in Land πe a corresponding prime element in Kand recall from Lemma 1.7 the following diagram is commutative X(Fq)Q/Z X(Fqf)Q/Z. restriction multiplication by f where the horizontal arrows are mapping χ7→ χ(σ)and χL7→ χ(σf), where χLis the restriction of χto X(Fqf)viewed as an element of X(L)and σ and σfare the Frobenius automorphisms in the corresponding absolute Galois groups. Now the first assertion follows from this since the isomorphism invKis mapping (χ, πe)7→ χ(σ)whilst invLis mapping (χ, π)7→ χ(σf). Then, since [L:K] = ef, we see (χ, πe)7→ χ(σ)7→ efχ(σ)and (χ, πe)7→ (χL, πe) = (χL, πe) = e(χL, π)7→ eχ(σf) = efχ(σ)coincide proving the commutativity of the diagram in the first assertion. Proposition 4.3. For each finite abelian extension Lof K, there is an isomorphism K×/NL/K(L×)∼ = →Gal(L/K)∗∗ ∼ = →Gal(L/K), given by the composition α7→ (χ→invK(χ, α)) 7→ σ. Proof. First note the map K×→Gal(L/K)∗∗ →Gal(L/K)given by the composition α7→ (χ7→ invK(χ, α)) 7→ σ, has kernel containing the norm 37 4.2. Local fields map. For that, since the last homomorphism is the evaluation isomorphism from Pontrjagin’s duality and since in finite abelian groups Tχ∈G∗ker χ= 1, it is enough to check that the cyclic algebras (χ, NL/K(L×)) are trivial in Br(K)for any character χ∈Gal(L/K)∗. Let Kχthe cyclic extension corresponding to the character χ. Then, by Theorem 3.12, (χ, NKχ/K(K× χ)) = 0 in Br(K). But, by the transitive property of the norm, we get NL/K(L×) = NKχ/K(NL/Kχ(L×)) ⊆NKχ/K(K× χ)and the induced map K×/NL/K(L×)→ Gal(L/K)is well defined. For injectivity it is enough to prove the inequality |K×/NL/K (L×)| ≤ |Gal(L/K)|once we prove it is onto. Note for two field extensions K⊆E⊆ L,NL/K =NL/ENE/K. Then, E×/NL/E(L×)NE/K →K×/NL/K(L×)is well defined and the sequence E×/NL/E(L×)NE/K →K×/NL/K(L×)→K×/NE/K(E×)→1 is exact. Then, |K×/NL/K(L×)|≤|E×/NL/E(L×)||K×/NE/K (E×)|, showing it is enough to consider finite extensions of prime degree by induction on the prime factors of [L:K]. But, in this case the extension is cyclic and combining Theorem 3.12 and previous proposition |K×/NL/K(L×)|=|Br(L/K)|= [L: K] = |Gal(L/K)|, which implies |K×/NL/K(L×)|≤|Gal(L/K)|for a separable extension L/K. Hence, we are only left to prove it is onto to obtain an isomorphism. For that, we have seen in Section 1.4 that this homomorphism will be onto if the only character annihilating its image is the trivial character. Such a character must satisfy (χ, K×)=0in Br(K)by definition. Then, by Theorem 3.12 we have Br(Kχ/K)=0and by the second assertion in Proposition 4.2, we have the formula [Kχ:K] = |Br(Kχ/K)|= 1; thus, Kχ=K and χ= 0 proving surjectivity. Now, note that for any finite Galois extensions M⊆Lof K, the diagram K×/NL/K(L×) Gal(L/K) K×/NM/K(M×) Gal(M/K) restriction commutes where the horizontal arrows are precisely the ones defined in the previous proposition and the left vertical arrow is a usual projection since NL/K(L×)⊆NM/K(M×). Then, the isomorphisms from the previous Proposition are compatible with the connection homomorphisms and by the universal property of the inverse limit they induce a homomorphism ρK:K×→ Gal(Kab/K),α7→ σ:= (σL)L. This is the celebrated homomorphism of local class field theory. We shall show it has the properties described in Theorem 4.1. 38 Chapter 4. Class Field Theory Proposition 4.4. Let Kbe a complete discrete valuation field with finite residue field. Then, ρKhas the property (b) in Theorem 4.1(1). Proof. Let us see commutativity of the diagram for a prime element πsince K×is generated by prime elements. The valuation of a prime element is 1 and 1 generates Zas a group and Gal(Fab q/Fq)is isomorphic to the procyclic profinite completion of Z,b Z, generated topologically by the Frobenius automorphism x7→ xq. Then, the image of the prime πthrough the composite νKρFqis the Frobenius automorphism in Gal(Fab q/Fq). On the other way, the image of πthrough ρKsatisfies χ(ρK(π)) = invK(χ, π)=1/n +Zfor each unramified character χ, where nis the index of the kernel of χin Gal(Kab/K). This element of the bidual is mapped via the evaluation isomorphism to an automorphism σin Gal(Kab/K)such that χ(σ) = 1/n +Zfor each unramified character, i.e. it induces generators in each cyclic unramified extension Kχ; thus, its restriction is a generator of Gal(Kur/K), which is mapped to the Frobenius automorphism in Gal(Fab q/Fq)by the canonical isomorphism Gal(Kur/K)∼ =Gal(Fab q/Fq), proving commutativity of the diagram. Proposition 4.5. For a local field K,ρKis continuous. Proof. Assume for simplicity char K= 0. Since Gal(Kab/K)is a profinite group, to check continuity of the map ρK:K×→Gal(Kab/K)it is enough to show continuity of each induced map K×→Gal(L/K), by the universal property of the inverse limit noting we are working over the category of topological groups. Since each finite Galois group is discrete, it is enough to check the kernel is open by Lemma 1.2. Since the kernel is of finite index for the image group being finite, it follows from Proposition 2.18 it is open. Remark. We assumed Kto be of null characteristic in order to apply Proposition 2.18. Now, we show the map ρKis unique in the sense of Theorem 4.1. Proposition 4.6. Let ˜ρ:K×→Gal(Kab/K)be an homomorphism satisfying, 1. Let Lbe a cyclic extension of K. Then, the composite map, Kטρ →Gal(Kab/K)→Gal(L/K), maps NL/K(L×)to {1}. 2. Let Lbe a finite unramified extension of K. Then, the image of a prime by the same composite map of (1), is a generator of the cyclic Galois group. Then, ˜ρ=ρK. 39 4.2. Local fields Proof. As before, it is enough to prove ρ0(π) = ρK(π)for prime elements π∈K×, as they generate the group of units. We shall see χ(ρ0(π)) = χ(ρK(π)) for all χ, which is easier to check and implies previous equality since the isomorphism in Pontrjagin’s duality is the evaluation isomorphism. Let nbe the order of χ(ρK(π)). Let Kn/K be the unique unramified extension of degree n. Then, ρK(π)restricts to a generator of Gal(Kn/K). Hence, there is some unramified character ψ∈X(K)such that ψ(ρK(π)) = χ(ρK(π)). Now, let L/K be the cyclic extension corresponding to the character ψ−χ. Then, the composite K×ρK →Gal(Kab/K)→Gal(L/K)maps πto 1. Thus, by Proposition 4.4, πis in the norm group NL/K(L×). By the first property of ρ0,ρ0(π) maps to 1 too, and we obtain (ψ−χ)(ρ0(π)) = 0. By the second property, ψ(ρ0(π)) = ψ(ρK(π)) and we have the equality χ(ρ0(π)) = ψ(ρ0(π)) = ψ(ρK(π)) = χ(ρK(π)), concluding the proof. Now, to conclude we need to prove there is a one-to-one correspondence between abelian extensions and open subgroups of finite index of the group of units of the base field. We have seen this is equivalent to having a one-toone correspondence between open subgroups of finite index in Gal(Kab/K) and open subgroups of finite index in K×. For a profinite abelian group G, we claim its open subgroups of finite index are in one-to-one correspondence with the finite subgroups of its character group G∗given via the assignments H≤G7→ ϕ(H) := {χ∈G∗:χH= 0}and H≤G∗7→ ψ(H) := Tχ∈Hker χ(see Problem A.13). Thus, if we set X(K×) := homcont(K×,Q/Z), it is sufficient to prove there is an isomorphism X(K)∼ =X(K×)given by the assignment χ7→ ρKχ. To prove injectivity, just note that the composite K×ρK →Gal(Kab/K)→Gal(L/K)is surjective for all abelian extensions L/K; thus, since taking the dual homcont(−,Q/Z)is a contravariant functor mapping an exact sequence K×→Gal(L/K)→0to an exact sequence 0→Gal(L/K)∗→X(K×), this inclusion is injective for the duals of each finite Galois groups; hence, for the dual of the Galois group Gal(Kab/K)too. Now, we prove surjectivity. First, we shall check Lab is an extension of Kab for any separable extension L/K. By definition, it is enough to see Kab/L is an abelian extension. Note that Gal(Kab/K)is abelian and Gal(Kab/L)is one of its subgroups; thus it is abelian too. This shows the restriction Gal(Lab/L)→Gal(Kab/K)given by the usual restriction of automorphisms is well defined. Then, we may define a natural map X(K)→X(L)by plugging the Galois group of the maximal abelian extension of Lthrough the usual restriction in the left hand side, i.e. χ7→ χL:= πL/Kχwhere πL/K denotes this restriction. 40 Chapter 4. Class Field Theory Proposition 4.7. Let Kbe a local field and La finite separable extension of K. Then, the diagram L×Gal(Lab/L) K×Gal(Kab/K). ρL NL/K ρK is commutative, where the right vertical map is the homomorphism obtained by restriction of automorphisms of Lab to Kab. Proof. Let Kbe a complete valuation field with finite residue field, since the real and complex cases are easy to check and thus left to the reader. Note that, as before, it is sufficient to prove invK(χ, NL/K(π)) = invL(χL, π) for all χ∈X(K), where χLdenotes the image of χby the inclusion map X(K)→X(L). Let fbe the residue degree of the extension L/K. Then, we may choose an unramified element ψ∈X(Fqf)⊆X(L)such that (ψ, π) = (χL, π). Note such an element exists since all the classes of the Brauer group contain such an element. Now, since the multiplication by fmap Q/Z∼ = X(Fq)→X(Fqf)∼ =Q/Zis surjective we may choose an unramified element ϕ∈X(Fq)⊆X(K)such that ϕL=ψ. Let L0/L be the cyclic extension corresponding to ϕL−χL. Since (ϕL−χL, π) = 0,πis a norm element by the main property of cyclic algebras, i.e. π=NL0/L(b)for some b∈(L0)×. Now, consider the cyclic extension K0/K corresponding to the character ϕ−χ. Since (ϕ−χ)L0= 0,K0⊆L0. Thus, by transitivity of the norm, NL/K(π) = NL/K(NL0/L(b)) = NL0/K(b) = NK0/K(NL0/K0(b)) ∈NK0/K((K0)×). Thus, (ϕ−χ, NL/K (π)) = 0 and we get invK(χ, NL/K(π)) = invK(ϕ, NL/K (π)) =νK(NL/K(π))(image of ϕby X(Fq)∼ = →Q/Z) =f·(image of ϕby X(Fq)∼ = →Q/Z) = (image of ϕLby X(Fqf)∼ = →Q/Z) = invL(ϕL, π) = invL(χL, π), where νKis the discrete valuation of Kand second and fifth equalities follow from Proposition 4.4, third from Proposition 2.17 and πbeing a prime element in Land fourth equality from the map X(Fq)→X(Fqf)being multiplication by f. This concludes the proof. Lemma 4.8. Let Kbe a local field of characteristic 0. Then, 1. For n∈N, let Xn(K) := {χ∈X(K) : nχ = 0}and Xn(K×) := {χ∈ X(K×) : nχ = 0}. If Kcontains a primitive nth root of unity, we have an isomorphism Xn(K)∼ = →Xn(K×), given by χ7→ ρKχ. 41 4.2. Local fields 2. Let Lbe a finite extension of Kand χ∈X(K×). If NL/Kχ∈X(L×)lies in the image of X(L)→X(L×),χlies in the image of X(K)→X(K×). Proof. Let us prove first the first assertion. We shall see the sequence of maps K×/(K×)n→Xn(K)→Xn(K×), is a sequence of injective maps and that K×/(K×)nand Xn(K×)are finite and have same order, implying the desired isomorphism Xn(K)∼ =Xn(K×). Note the second map is given by χ7→ ρKχ, i.e. it is obtained by plugging the group K×in the left hand side. Let Kcontain a primitive nth root of unity ζn. Then, K(n √a)is an abelian extension for any a∈K×. Thus, we may define a group homomorphism K×→Xn(K)via the assignment a7→ χawhere χa(σ) = r/n and ris chosen such that σ(n √a) = ζr nn √a. This is a well-defined group homomorphism and note that χa= 0 for all a∈(K×)n; thus, it induces a group homomorphism K×/(K×)n→Xn(K). We shall see it is injective, it is actually an isomorhism by Kummer Theory but we just need injectivity for our means. For that, we see the kernel is trivial. For a∈K×and χa= 0, we have n √ais fixed by the Galois group, i.e. it is in K×; hence, a∈(K×)nand the kernel is trivial proving injectivity. We show in Proposition 2.18 that [K×: (K×)n]is finite; thus, the quotient group is finite. Since Xn(K×)can be identified with X(K×/(K×)n) via χ(k)↔χ(k(K×)n)and the character group of a finite abelian group has same order as the group itself since they are isomorphic, although not naturally, we obtain the desired isomorphism. Now we prove the second assertion of the lemma. By transitivity of the norm, it is sufficient to consider intermediate fields of the finite abelian extension L/K, i.e. we may assume without loss of generality that L/K is cyclic. Let G:= Gal(L/K)and consider the action of Gover the groups X(L)and X(L×)defined by σχ : Gal(Lab/L)→Q/Z,τ7→ χ(˜σ−1τ˜σ), where ˜σis an element of Gal(Lab/K)whose image in Gal(L/K)is σfor χ∈X(L); and σχ =σ−1χfor χ∈X(L×)respectively for each σ∈G. Now, let χ1∈X(K×) and assume NL/Kχ1∈X(L×)is the image of χ2∈X(L). Recall from previous chapter that we write MGfor the elements of a G-module Minvariant under the G-action. Then, clearly NL/Kχ1∈X(L×)G, since Galois conjugates have the same norm. Note the map X(L)→X(L×)is a homomorphism of G-modules, this is easy and left to the reader, and injective; thus, neccesarily χ2∈X(L)Gnoting that σNL/Kχ1=NL/Kχ1and using injectivity to obtain σχ2=χ2. Now let us prove X(L)Gis contained in the image of the map X(K)→X(L),χ7→ χL. Let σbe a generator of Gand fix an element ˜σ. Then, if pis the order of Gany element of the group Gal(Lab/K)can be uniquely written as h˜σjfor some h∈Gal(Lab/L)and 42 Chapter 4. Class Field Theory some j= 0,1, . . . , p −1since Gal(Lab/K)∼ =Gal(Lab/L)×Gal(L/K). Now, for χ∈X(L)G, choose an element s∈Q/Zsuch that χ(σ) = ps and define the map χ0:= Gal(Lab/K)→Q/Zvia the assignment τ=h˜σj7→ χ(h) + js. This map is clearly a group homomorphism (note Gal(Lab/K)is abelian and thus it is easily verified χ0(τρ) = χ0(τ)χ0(ρ)) and it induces a map Gal(Kab/K)→ Q/Zand it can be regarded as an element of X(K)and its image χ0 Lcoincides with χ. For that, note that any automorphism τ∈Gal(Lab/L)has j= 0 in the above form; thus, χ0 L(τ) = χ(τ)and χ0 L=χ. Thus, the homomorphism X(K)→X(L)Gis surjective and there exists an element χ3∈X(K)such that (χ3)L=χ2. From previous proposition ρKχ3and χ1map to the same element in X(L×)via NL/K ; thus, χ1−ρKχ3annihilates NL/K(L×). Hence, the composition χ4: Gal(L/K)→Q/Zof χ1−ρKχ3:K×/NL/K(L×)→Q/Z and the induced isomorphism K×/NL/K(L×)∼ =Gal(L/K)can be seen as an element of X(K)and we have χ1=ρK(χ3+χ4), i.e. χ1lies in the image of X(K)→X(K×)concluding the proof. Now, assuming char K= 0 let χ∈X(K×). We shall see it lies in the image of X(K)→X(K×). Note both groups are torsion; thus, let nbe the order of χ. We shall assume Kcontains an nth primitive root of unity, since K(ζn)/K is a finite abelian extension, and by Lemma 4.8.2 all cases are reduced to this one. Then, by Lemma 4.8.1, the map Xn(K)→Xn(K×),χ07→ ρKχ0is an isomorphism for all natural nand surjectivity follows and we are done with the proof of the local class field theory. As usual, these isomorphisms are compatible with usual inclusions whenever ndivides mand we obtain an isomorphism of the direct limits, i.e. X(K)∼ =X(K×), concluding the proof of Theorem 4.1. To conclude, I would like to highlight again the astonishing beauty of local class field theory: how local fields encode the data of all their finite abelian extensions in their inner arithmetic in a rather unexpected but simple way. This is a nice representative of the charm of algebra and number theory: objects may seem to be so distant from each other but happen to be linked in an out of the blue but easy way. And, generation after generation more of these links are developed and we realize that even if we thought we fully understood a theory, we were just scratching the surface of a whole new world making you to keep learning constantly, which, for me, is the most captivating aspect of the queen of mathematics. 43 Appendix A Solved Problems A.1 Preliminaries Problem A.1. Show the Krull topology and the profinite topology need not coincide. Solution. We shall follow the procedure in [Mil20]. It is enough to show there is some Galois group with at least one subgroup of finite index non-open. Let Gal(Q/Q)and the intermediate field E:= Q(√−1,√2,√3,...,√p, . . . ). Then, it is an easy exercise to check G:= Gal(E/K) = lim ←−Gal(Q(√−1,√2, . . . , √p)/Q)and since each finite Galois group is a finite product of groups Z/2Z,Gis a closed subgroup of the direct product of a countable number of groups Z/2Z. Now, consider the subgroup Nof Gof tuples with only a finite number of non-trivial components, i.e. a direct sum of a countable number of groups Z/2Z. Also, it is clearly dense in Gand we may make the quotient Γ := G/N, which is a vector space over F2. Then, by Zorn’s Lemma Γcontains a maximal set of linearly independent vectors, which is necessarily a basis. Then, take nelements out of the basis and define the subspace spanned by the remaining set as Gn. Then, Γ/Gnis of dimension nover F2, i.e. of index 2nin Γ. If Gnwere open in Γ, it would be closed too, but that it is impossible since Nis dense in G. Then, Gnis of finite index and non-open, proving our claim. A.2 Global and local fields Problem A.2. Let Abe a complete valuation ring and mits unique maximal ideal. Then, A∼ =lim ←−nA/mnas topological rings. Proof. We shall see the canonical map ϕ:A→lim ←−nA/mnis an isomorphism. It is clearly a ring homomorphism. Thus, since the kernel is Tn≥1mn= 0, it is injective. To check surjectivity, note that an element s∈lim ←−nA/mnis 44 Appendix A. Solved Problems given by an infinite tuple s= (sn)where sn=a0+a1π+···+an−1πn−1, for aiare taken in a set of representatives of the cosets and πa uniformizer of A. Thus, (sn)is nothing but the image by ϕof the element Pn≥0anπn∈A. Hence, ϕis bijective and an isomorphism of rings. We are left to see it is continuous for it to be a homemorphism too. It is enough to check that the basis of neighborhoods mnof 0 in Aare mapped to a basis of neighborhoods of 0 in lim ←−nA/mn. Since the open sets Nn= Qk≥nA/mkform a basis of neighborhoods of 0 in Qn≥1A/mkand ϕ(mn) = Nn∩lim ←−nA/mn,ϕis continuous and thus an homemorphism. Problem A.3. Let Abe a complete valuation ring and mits unique maximal ideal. Then, mn/mn+1 ∼ =A/m. Proof. Note the elements a∈Amay be written as the sums a=X n≥0 anπn, where anare taken in a set of representatives of the cosets and πis a uniformizer of A. The elements of the ideal mnare the sums mn=X k≥n akπk. Thus, the elements in mn/mn+1 are of the form anπn+mn+1 and are in a clear one-to-one correspondence with the elements in the residue field (given by the canonical epimorphism) proving the result. Problem A.4. The residue field of a global field Kis finite. Proof. Let first Kbe a global function field. Then, K=Fq[t]and it is a principal ideal domain (PID) and a nonzero prime ideal is a maximal ideal given by a nonzero irreducible polynomial f. Then Fq[t] (f)={a0+a1t+···+an−1tn−1:ai∈Fq}∼ =Fqn, where n= deg f. Hence, the residue field is finite as it is a finite dimensional vector space over a finite field. Now, let Kbe a number field, i.e. a finite extension of Q. Then, OKis finite dimensional over Z. Thus, it is enough to show that for any positive prime integer p, and ν= ordp, the residue field Oν/pis finite where Oν=Z(p) (i.e. the localization of Zat the prime ideal (p)) and p=pZ(p). We shall see Oν/p=Z(p)/pZ(p)∼ =Fp. This follows directly from the following Lemma. 45 A.4. Class field theory index and finite subgroups of its character group via the assignments H≤G7→ ϕ(H) := {χ∈G∗:χ|H= 0}and H≤G∗7→ ψ(H) := Tχ∈Hker χ. Proof. First we see these maps are well defined. For that, let H≤Gbe open of finite index. Then, the characters in ϕ(H)are in one-to-one correspondence with the ones of G/H and since His of finite index |ϕ(H)|= |(G/H)∗|=|G:H|<∞. Now, let H≤G∗be finite. Then, ψ(H)is mapped into Lχ∈HG/ ker χand since each G/ ker χis finite and Htoo, G/H is finite proving it is of finite index. Also, it is the intersection of open sets; thus, open. Now, we shall see ψ(ϕ(H)) = Hand ϕ(ψ(H)) = H. First, let H≤Gopen of finite index. Clearly, ψ(ϕ(H)) ≤H. Should this inclusion not be an equality, the quotient ψ(ϕ(H))/H would be non-trivial and there would be a character χ∈G∗such that χ∈ϕ(H)but χ|ψ(ϕ(H)) 6= 0, which is a contradiction. Now, let H≤G∗finite. Clearly, H≤ϕ(ψ(H)). 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