# Local class field theory: a discussion

In this post we discuss local class field theory (specifically looking at $p$-adic fields) with a focus on the broader picture, and the multiple approaches.

# Goal of post

So, before I begin actually writing about local class field theory, I’d like to make clear what my goals for this post are. Namely, local class field theory is a subject which has been infinitely written about. Besides the various books on class field theory there are countless well-written, more conciseaccounts of class field theory (both local and global) available on the internet, covering the theory from essentially all angles.

For this reason, I feel obligated to explain why what I have written below adds anything, perhaps, to the canon of already available literature. My goal is, in some sense, to serve as a sort of ‘next step’ beyond the local portion of Bjorn Poonen’s valuable note A Brief Summary of the Statements of Class Field Theory. Namely, below I do a bit more than state the main statements of local class field theory. I try to motivate the need/desire for the subject, as well as try to explain where it sits in the larger picture of modern number theory. I then try to give vague outlines of the proofs of the subject. Focusing on trying to at least explain how the objects are defined, and deferring the proofs of the hard/non-instructive results to the actual books on the subject.

Being a ‘next step’, I assume a little more sophistication than Bjorn Poonen’s note. In particular, I assume below that the reader is well-acquainted with the standard material of a first course in algebraic number theory (including the theory of $p$-adic local fields). I will also assume, particularly in the motivational sections, as well as the Lubin-Tate portion some familiarity with the language of algebraic geometry.

Because this is not at all a comprehensive account of the theory, I have included at the end a list of the books I have found most helpful in each of the various topics we’ll be discussing below.

# Motivation

## Classical motivation to study local class field theory

Class field theory has its modern-historic origins in Hilbert’s 12th problem. Namely, at that point in history Kronecker and Weber had proven their famous theorem: all finite abelian extensions (Galois extensions with abelian Galois group) of $\mathbb{Q}$ were contained inside a cyclotomic extension (one of the form $\mathbb{Q}(\zeta_n)$ where $\zeta_n$ is a primitive $n^\text{th}$ root of unity). Hilbert then asked for a similar characterization of the abelian extensions of other number fields $K$ and, as a natural extension, what happens for the non-abelian extensions.

The desire to understand general number fields, and their extensions, came up naturally in the solving of Diophantine equations (a topic of great historical interest). Indeed, the first attempts to prove Fermat’s last theorem involved factorizing $X^n+Y^n$ in $\mathbb{Z}[\zeta_n][X,Y]$ and attempting to leverage facts about the ring $\mathbb{Z}[\zeta_n]$ to prove that any integral solution cannot be of the form $Z^n$ for $z\in\mathbb{Z}$. In fact, this technique can be made to work as long as $\mathbb{Z}[\zeta_n]$ is a PID. In fact, it works for $n=p$ a prime if $p\nmid h_K$, where $h_K:=|\mathrm{Cl}(K)|$ is the class number of $K$. Thus, suddenly the desire to solve Diophantine equations leads one to ask somewhat nuanced questions about various rings of integers and, in particular, an understanding of their class numbers.

It was a great insight of the movers and shakers of the time that there should be a connection between a number field $K$ having trivial class group (i.e. being a PID) and abelian extensions of $K$. Namely, one expects non-trivial elements of $\mathrm{Cl}(K)$ to be witnessed by finite abelian extensions of $K$. This can be made precise, for example, by the existence of the Hilbert class field of $K$ which is an abelian extension $H_K/K$, unramified everywhere (including the infinite places!) and such that $\mathrm{Gal}(H_K/K)=\mathrm{Cl}(K)$—it is characterized as being the maximal everywhere unramified abelian extension. But, this is only a posteriori motivation since one needs most of the class field theoretic machinery to justify the existence and basic properties of $H_K$.

So, how can we give an intuition about why one should expect a relationship between $\mathrm{Cl}(K)$ and abelian extensions without assuming the contents of the subject we are trying to motivate? Well, one motivation, I think, could be the following. Let us fix a number field $K$, and suppose (for simplicity), that $\mu_n(\overline{\mathbb{Q}})\subseteq K$. Consider then

$S=\{\mathfrak{p}\in\mathrm{Spec}(\mathcal{O}_K):\mathfrak{p}\mid n\}$

and consider

$U=\mathrm{Spec}(\mathcal{O}_K)-S=D(n)$

(where $D(n)$ is the usual scheme-theoretic non-vanishing set of $n$). Note then that, in fact, $U=\mathrm{Spec}(\mathcal{O}_K[\frac{1}{n}])$ and that we have a surjection

$\mathrm{Cl}(K)\to \mathrm{Cl}(U)=\mathrm{Pic}(U)$

Thus, a natural thing to do to see if $\mathrm{Cl}(K)$ is non-zero is to check whether $\mathrm{Cl}(U)$ is non-zero. But, note that $\mathrm{Pic}(U)=\mathrm{Cl}(U)$ has a very natural connection with abelian extensions of $K$. Namely, by general Kummer theory the fact that $U$ is a scheme over $\mathbb{Z}[\frac{1}{n}][\zeta]$ implies that we have a surjection

$\mathrm{Hom}_\mathrm{cont.}(\pi_1(U),\mathbb{Z}/n\mathbb{Z})\to \mathrm{Cl}(U)[n]$

But, $\mathrm{Hom}_{\mathrm{cont.}}(\pi_1(U),\mathbb{Z}/n\mathbb{Z})$ corresponds exactly to cyclic extensions of $K$ unramified outside of $S$ with degree dividing $n$.  Thus, a non-trivial element of $\mathrm{Pic}(U)[n]$ which comes from a non-trivial element of $\mathrm{Cl}(K)$ produces a non-trivial abelian extension with constrained ramification.

This gives an indication that the abelian extensions of $K$ (even those with constrained ramification) are capturing properties of the class number of $K$ and, intuitively, should bear witness to the non-PIDness of $\mathcal{O}_K$.

So, from the desire to study Diophantine equations we are pushed to study the abelian extensions of $K$. But, more particularly, we are prompted to study the abelian extensions with a view towards their ramification, or, lack thereof. But, the ramification properties of an abelian Galois extension are captured entirely by local data.

More specifically, let us assume that $L/K$ is a finite abelian extension and $\mathfrak{p}$ is a prime of $K$. Then, we can consider the decomposition group $D(\mathfrak{p})\subseteq\mathrm{Gal}(L/K)$ defined to be $D(\mathfrak{P}\mid\mathfrak{p})$ for any prime $\mathfrak{P}$ of $L$ lying over $\mathfrak{p}$ (in general two different choices of $\mathfrak{P}$ will result in conjugate subgroups of the Galois group, but this is unimportant since we’re abelian). Then, we know that $D(\mathfrak{p})$ and its inertia subgroup $I(\mathfrak{p})$ are abelian groups capturing all the ramification and inertial data at $\mathfrak{p}$. But, by basic number theory we have the following commutative rectangle of isomorphisms

$\begin{matrix} 0 & \to & I(\mathfrak{p}) & \to & D(\mathfrak{p}) & \to & \mathrm{Gal}(\ell_\mathfrak{P}/k_\mathfrak{p}) & \to & 0\\ & & \downarrow^{\approx} & & \downarrow^{\approx} & & \downarrow^{\approx} & &\\ 0 & \to & I_\mathfrak{p} & \to & \mathrm{Gal}(L_\mathfrak{P}/K_\mathfrak{p}) & \to & \mathrm{Gal}(\ell_\mathfrak{P}/k_\mathfrak{p}) & \to & 0\end{matrix}$

where $\ell_\mathfrak{P},k_\mathfrak{p}$ are the residue fields, and $\mathrm{Gal}(L_\mathfrak{P}/K_\mathfrak{p})$ is the Galois group of the completions.

Note then that $L_\mathfrak{P}/K_\mathfrak{p}$ is an abelian extension of local fields. Thus, our desire to understand Diophantine equations led us to consider the class group, which led us to consider abelian extensions of number fields with constrained ramification, which led us to study abelian extensions of local fields. This is the setting for local class field theory.

In particular, the goal of local class field theory is to try and understand the arithmetic of the abelian extensions $L/K$ (where now $K$ is local) by ‘internal to $K$ data’. This is not a term which I can define in a literal sense other than to say that one should be able to answer questions about the number theory of $L/K$ entirely in terms of objects involving objects sitting inside of/related to objects inside of $K$. For a concrete example of this internal type description, at least in the global case, see this post.

Explicitly, we shall try to understand the abelianized Galois group of $G_K^\mathrm{ab}:=G_K/\overline{[G_K,G_K]}$ in terms of the topological group $K^\times$. Of course, we will give a very explicit and rigorous definition of what we mean by ‘understand in terms of $K^\times$‘, but for now, let us start with an analogy.

Suppose that we wanted to understand the finite abelian extensions (or, for finite fields, equivalently, all finite extensions) of the field $\mathbb{F}_q$. Well, we know precisely what the lattice of finite subextensions of $\overline{\mathbb{F}_q}/\mathbb{F}_q$ looks like. For each $n$ we have a unique extension of degree $n$, and $\mathbb{F}_{q^n}\subseteq\mathbb{F}_{q^m}$ if and only if $n\mid m$. In other words, the lattice looks precisely like lattice associated to the divisibility ordering on $\mathbb{Z}$. But, this is precisely the same as the lattice of finite index open subgroups of $\mathbb{Z}$ (with the discrete topology) with $n$ corresponding to $n\mathbb{Z}$.

Thus, we have an ordering preserving bijection between the open finite index (here these adjectives are really unnecessary) subgroups of $\mathbb{Z}$ and the finite abelian extensions of $\mathbb{F}_q$. Moreover, we have the property that for any two comparable integers $n$ and $m$, say $n\mid m$, we know that

$\mathrm{Gal}(\mathbb{F}_{q^m}/\mathbb{F}_{q^n})\cong m\mathbb{Z}/n\mathbb{Z}$

and thus this bijection between the lattice of finite abelian extensions of $\mathbb{F}_q$ also preserves the relative ‘quotients’ of these objects.

That said, there is another group which has this precise property—the abelianized Galois group of $\mathbb{F}_q$! Indeed, $G_{\mathbb{F}_q}$ has a property that its open finite index subgroups correspond precisely with the finite abelian extensions of $\mathbb{F}_q$ and in such a way which preserves quotients. Thus, we see, at least insofar as finite abelian extensions go, the group $\mathbb{Z}$ ‘approximates’ $G_{\mathbb{F}_q}$. In fact, we can make this more explicit. We have an injection

$\mathbb{Z}\hookrightarrow G_{\mathbb{F}_q}:n\mapsto \mathrm{Frob}_q^n$

(here $\mathrm{Frob}_q$ is the arithmetic Frobenius $x\mapsto x^q$) sending open finite index subgroups to open finite index subgroups, and in fact, inducing a bijection of such objects. This is great since the group $\mathbb{Z}$ is patently simpler than the group $G_{\mathbb{F}_q}$, and when it comes to understanding finite abelian extensions the above says we can use interchangeably.

Similarly, for a general (assume $p$-adic for the sake of simplicity) local field $K$ we would like some sort of simplifying approximating subgroup $W_K^\mathrm{ab}\subseteq G_K^\mathrm{ab}$ which captures all of the data of finite abelian extensions. We would then like to relate this group $W_K^\mathrm{ab}$ to $K^\times$ giving us, as desired, an ‘internal to $K$‘ characterization of the finite abelian extension (the group $K^\times$ is certainly internal to $K$!).

This is precisely what class field theory gives us. Namely, we will define a continuous group map $W_K\subseteq G_K$ such that $W_K^\mathrm{ab}\hookrightarrow G_K^\mathrm{ab}$ induces a bijection on finite index open subgroups. We will then show that there is an isomorphism of topological groups, the Artin reciprocity map,

$\mathrm{Art}_K:K^\times\xrightarrow{\approx}W_K^\mathrm{ab}\subseteq G_K^\mathrm{ab}$

and then, finally, to round things off, we’ll give a characterization of the finite index open subgroups of $K^\times$.

## Motivation from the Langlands program

One often times hears the Langlands program described as ‘non-abelian class field theory’. In other words, a desire to understand the non-abelian extensions of a number field, and their ramification behavior, in terms of some sort of ‘reciprocity data’ similar to the Artin reciprocity map described above (or, more correctly, its global analogue). How this manifests itself concretely is that we’re interested in understanding so-called Artin representations of $G_K$. Namely, continuous Galois representations $G_K\to\mathrm{GL}_n(\mathbb{C})$.

In what way does the representation theory of $G_K$ help us understand the ramification behavior of extensions of number fields? Here is a concrete, example. Consider the polynomial $f(x)=x^5 +10x^3 -10x^2 +35x-18$. This defines an $A_5$-extension $L$ of $\mathbb{Q}$, and thus we have no hope of trying to understand the ramification behavior in terms of the (abelian) class field theory we discussed in the last section, and no hope of trying to understand it in terms of something as simple as congruence conditions (see, again, this post). That said, there exists a representation which essentially dictates the splitting behavior of this extension. Sounds complicated? It’s not. Choose a faithful embedding of $\rho_0:\mathrm{Gal}(L/\mathbb{Q})\hookrightarrow\mathrm{GL}_n(\mathbb{C})$ and consider the representation $\rho:G_{\mathbb{Q}}\to \mathrm{GL}_n(\mathbb{C})$ obtained by first quotienting to $\mathrm{Gal}(L/\mathbb{Q})$ and then composing with $\rho_0$. Then, note that for almost all primes $p$ we have that $p$ is unramified, and thus there is a well-defined element $\mathrm{Frob}_p:=\{\mathrm{Frob}_{\mathfrak{p}}\}_{\mathfrak{p}\mid p}$ and $p$ splits in $L$ if and only its image under $\rho$ is trivial.

But, this is unsatisfying in some sense. We have not reduced the problem at all since, almost definitionally, understanding when $\rho(\mathrm{Frob}_p)$ is zero is equally as hard as understanding when $\mathrm{Frob}_p=0$ (i.e. when $p$ is split). But, if we could understand the representation $\rho$ in terms of some other object, one in which we had a better grip on, then perhaps this representation theoretic point of view would be a little less vapid.

The idea is that, roughly, we want our ‘algebraic representation’ $\rho$ (as defined above) to be equal to an ‘analytic representation’. What one means by ‘algebraic’ and ‘analytic’ is very complicated, so let us suffice ourselves with an example. In the above case of $L$, what type of ‘analytic representation’ one might equate to our defined $\rho$? Well, it can be shown (see, for example Buhler’s Icosahedral Galois Representations) that there exists a weight $1$ cuspidal eigenform $g\in S_1(\Gamma_0(800))$ whose associated Galois representation $\rho_g$ is equal to our $\rho$. I won’t go into what the definition of $\rho_g$ is (it’s especially nuanced for weight $1$ forms), but let me explain the upshot of this conclusion. One can show that the character (nebentypus) $\chi$ of $g$ has conductor $100$, and thus one can conclude from the statement that $\rho=\rho_g$ that for (all but finitely many) primes $p$ of $\mathbb{Q}$, the prime $p$ splits in $L$ if and only if

$a_p(g)^2=4\chi(p)$

where $a_p(g)$ is the $p^\text{th}$-coefficient of $g$. Thus, we can understand/attack the splitting behavior of $L/\mathbb{Q}$ by appealing, essentially, to the analytic object $g$ which opens us up to a whole new universe of techniques.

Remark: The astute reader will notice a glaring issue in my above statements. Namely, I never fixed $\rho$ (I just said take $\rho_0$ to be some faithful representation). I am skipping over some important, but technical details. Namely, I am really considering a faithful representation of a $4$-fold cover of $A_5$, and specifying $\rho$ by its projectivization (along with some other properties). To see the full technical details, again, consult Buhler’s book.

Before we relate all of this back to (abelian) class field theory, let us quickly make a remark. Somewhere in the above progression of ideas we went from trying to understand the ramification behavior of $L/\mathbb{Q}$ (i.e. the ramification indicies/residual degrees of all primes) to just understanding which primes split (meaning split completely). Moreover, we were only able to do this for all but finitely many primes. Thus, we seem to have not fulfilled our goal. But, in fact, we have, in a very strong sense. Indeed, recall that for an extension of number fields $L/K$ one defines the set $\mathrm{Spl}(L/K)$ to be the set of primes of $K$ which split in $L$. It is then essentially a consequence of Chebotarev density that for two Galois extensions $L,L'/K$ one has that $L=L'$ (literal equality) if and only if $\mathrm{Spl}(L/K)$ and $\mathrm{Spl}(L'/K)$ agree at all but finitely many primes. Thus, we see that understanding the analytic set $\{a_p(g)\}$ literally determines $L$ as a (Galois) extension of $\mathbb{Q}$. So, we’ve lost nothing in restricting our attention to splitting behavior.

So, hopefully at this point one believes that understanding number fields, and their ramification data, can be well-understood by studying representations and when these representations are ‘analytic in nature’. But, it’s not at all obvious how this relates to global class field theory, let alone the local class field theory we are after. To make precise how class field theory is “the $n=1$ case of the Langlands conjecture” (where the Langlands conjecture is roughly the statement that all reasonable Galois representations are ‘analytic’) we would need to actually define the objects that come into play. So, instead, let us suffice ourselves with an understanding of how global/local class field theory is a statement about representation theory at all.

This is not nearly as difficult as it may sound. Namely, the “$n=1$” part of the statement that local class field theory is part of the Langlands program refers to the dimension of representation. Namely, we should be looking at characters of the Galois group. So, continuous homomorphisms $\chi:G_K\to\mathbb{C}^\times$. But, as is the case with all Galois representations, this character comes bundled with a bunch of ‘local data’. Namely, for all primes $\mathfrak{p}$ of $K$ we have an induced character

$G_{K_\mathfrak{p}}\to G_K\to \mathbb{C}^\times$

(the first embedding is only defined up to conjugacy, but this is unimportant since our target is an abelian group).

But, there are two things to notice about this local data $G_{K_\mathfrak{p}}\to\mathbb{C}^\times$. First, since $\mathbb{C}^\times$ is abelian (and the homorphism is continuous) this character factors through $G_{K_\mathfrak{p}}^\mathrm{ab}$. Moreover, since $G_K$ is profinite it’s not hard to see that the map $G_{K_\mathfrak{p}}\to\mathbb{C}^\times$ must also factor through a finite quotient. Thus, our character actually factors through a finite abelian quotient of $G_{K_{\mathfrak{p}}}^{\mathrm{ab}}$. But, by local class field theory, this should correspond to a character of $K^\times$. Thus, class field theory is morally (and, rigorously) equivalent to a bijection

$\left\{\text{continuous characters }K^\times\to\mathbb{C}^\times\right\}\longleftrightarrow\left\{\text{continuous characters }G_K\to\mathbb{C}^\times\right\}$

which makes the connection between local class field theory and representation theory. Moreover, while we haven’t defined ‘analytic representation’ it’s believable that representations of $K^\times$ might be ‘analytic’ in nature. Namely, $K^\times$ is a locally profinite group, and so has well-defined Haar measure, as well as a reasonable theory of analytic functions. This makes believable the jump from connecting local class field theory to representaiton theory to connecting it with the Langlands program.

## Lubin-Tate theory

So, we’ve discussed the historical origins/desire for something like local class field theory in studying Diophantine equations, and how this fits into the much broader goal of the (local) Langlands program. But, despite all of this discussion, we haven’t indicated whether or not we have actually completed Hilbert’s goal. Namely, remember the wording: we want to describe abelian extensions in a way similar to the Kronecker-Weber theorem. For Hilbert this meant describing the abelian extensions in terms of values of special functions (e.g. $\zeta_n$ are the rational values of $\exp(2\pi i t)$).

One places where this is possible is in the theory of complex multiplication, a key motivation for many aspects of the Langlands program and related areas. The theory of complex multiplication says that every abelian extension of an imaginary quadratic number field $K$ is contained inside one of the form $K(j(E))$ where $E/\mathbb{C}$ is an elliptic curve with complex multiplication by $K$. Thus, again, we obtain abelian extensions by adjoining values of special functions (here the special function being the $j$-function).

So, one might seek to extend this type of result to describe arbitrary abelian extensions of number fields and/or local fields. The great realization of Lubin and Tate was that one could actually create such a uniformization, at least in the local fields case, if one reinterprets what it should mean to analogize the Kronecker-Weber theorem, then one can, in fact, find a satisfactory solution. In particular, one can also think about the roots of unity as the torsion on the algebraic group $\mathbf{G}_m$. So, one might think that one can obtain the abelian extensions of a general local field by adjoining roots of other algebraic groups.

This is almost correct. What Lubin and Tate showed is that for every local field $K$ with uniformizer $\pi$ there is an associated formal Lie group $\mathbb{G}$ with a ring map $\mathcal{O}_K\to \mathrm{End}(\mathbb{G})$ such that one obtains all the ‘interesting’ abelian extensions by adjoining to $K$ all the $\pi^n$-torsion in $\mathbb{G}(\mathcal{O}_{\overline{K}})$.

There should be clarification on the phrase ‘interesting’. For any local field $K$ with residue field $\mathbb{F}_q$ one has an easy way to build abelian extensions of $K$. I speak, of course, of the unramified extensions. These are all obtained by adjoining roots of unity $\mu_n$ with $(n,q)=1$. Thus, we have sitting inside the maximal abelian extension $K^\mathrm{ab}$ the maximal unramified extension

$K^\mathrm{ur}=K(\{\mu_n:(n,q)=1\})$

and by ‘interesting’ extensions we essentially mean the ‘complement in $K^\mathrm{ab}$ of $K^\mathrm{ur}$‘. In other words, the ‘ramified part’ of the maximal abelian extension.

The last sentence in the last paragraph brings up an important, and somewhat subtle point. Namely, if what we’re looking for is a ‘complement’ of $K^\mathrm{ur}$ in $K^\mathrm{ab}$, why not just do the obvious? Namely, why not just define the ‘maximal abelian totally ramified extension’ of $K$, call it $K^\text{tot.ram.,ab}$ and take this as a complement? While this seems reasonable, it fails from a simple issue. Namely, the compositum of two totally ramified extensions need not be totally ramified. So, this ‘maximal totally ramified abelian extension’ doesn’t literally make sense. So, we have to be more sophisticated to come up with a complement, and to describe it explicitly.

With this technical subtleties highlighted, let us explain what we mean by a ‘complement’ to the unramified piece. To make this precise note that implicitly in our discussion above, we have claimed that $G_K^\mathrm{ab}\cong \widehat{K^\times}$ (the profinite completion of $K^\times$). This should not be shocking—$K^\times$ was supposed to be an analogy in $G_K^\mathrm{ab}$ of $\mathbb{Z}\subseteq G_{\mathbb{F}_q}=\widehat{\mathbb{Z}}$. But,

$K^\times\cong \mathcal{O}_K^\times\times\mathbb{Z}$

and since $\mathcal{O}_K^\times$ is already profinitely complete (we’ll remark more on this later), we may deduce that

$\mathrm{Gal}(K^\mathrm{ab}/K)=G_K^\mathrm{ab}\cong \mathcal{O}_K^\times\times\widehat{\mathbb{Z}}$

Now, it stands to reason that since $\mathrm{Gal}(K^\mathrm{ur}/K)\cong \widehat{\mathbb{Z}}$ that $K^\mathrm{ur}$ is the fixed field of $\mathcal{O}_K^\times$ in $K^\mathrm{ab}$ under this isomorphism. Thus, the fixed field of $\widehat{\mathbb{Z}}$ should be a linearly disjoint subextension $K_\pi\subseteq K^\mathrm{ab}$ with $\mathrm{Gal}(K_\pi/K)\cong \mathcal{O}_K^\times$ and $K^\mathrm{ab}=K^\mathrm{ur}K_\pi$. It is the field $K_\pi$ for which Lubin and Tate describe in terms of the torsion of a formal group.

Remark: Why have we denoted this extension by $K_\pi$—why $\pi$? Well, not shockingly, $\pi$ will stand for a uniformizer. The dependence on this uniformizer is contained in the fact that the decomposition

$K^\times\cong \mathcal{O}_K^\times\times\mathbb{Z}$

is not canonical. In fact, it depends precisely on the choice of a uniformizer $\pi$ of $\mathcal{O}_K$. This is where this mysterious $\pi$ comes from.

As an example of this procedure, let’s consider the case when $K=\mathbb{Q}_p$. Then, it will turn out that $\mathbb{G}=\widehat{\mathbf{G}_m}$ and thus the $\pi^n$-torsion is just $\mu_{p^n}$. Thus, $K_\pi$ is $\mathbb{Q}_p(\mu_{p^\infty})$ and thus, from this Lubin-Tate theory, we recover the classic statement of the local Kronecker-Weber theorem:

$\mathbb{Q}_p^\mathrm{ab}=(\mathbb{Q}_p)_\pi \mathbb{Q}_p^\mathrm{ur}=\mathbb{Q}_p(\mu_{p^\infty})\mathbb{Q}_p(\{\mu_n:(n,p)=1\})=\mathbb{Q}_p^\mathrm{cycl}$

where, as one expects, $\mathbb{Q}_p^\mathrm{cycl}$ is the maximal cyclotomic extension of $\mathbb{Q}_p$.

The last thing that should be commented on is how this Lubin-Tate theory fits into the general framework of the local Langlands program. It turns out that, despite the apparent ad hoc nature of (abelian) Lubin-Tate theory, it is precisely this approach to class field theory (opposed to the cohomological one described below) which generalizes to the non-abelian setting. Namely, the local class field theory is essentially the local Langlands conjecture for $\mathrm{GL}_1$ (in the ways indicated above). The local Langlands conjecture for $\mathrm{GL}_n$ is now a theorem (due, ultimately, to Michael Harris and Richard Taylor) in general. And, it turns out, that the main statement of their work is that the local Langlands conjecture for $\mathrm{GL}_n$ (and, incidentally, the Jacquet-Langlands correspondence) can be realized in the cohomology of the Lubin-Tate tower. Whatever this means, the appearance of the names Lubin and Tate is not a coincidence since this is a strict generalization of the Lubin-Tate theory for local class field theory.

# The statements of local class field theory

## Statement and general corollaries

Before we begin discussing some of the key elements to the proof of local class field theory, let us begin by stating rigorously its two main statements. For the remainder of the post we’ll be sticking, mostly for convenience, not necessity, to $p$-adic local fields. So, let’s fix a $p$-adic local field $K$, with integer ring $\mathcal{O}_K$, maximal ideal $\mathfrak{p}_K$, valuation $v_K$, and residue field $k$.

Theorem 1(Local class field theory):

1. $\text{(Local Artin reciprocity)}$There exists a unique continuous homomorphism $\mathrm{Art}_K:K^\times\to G_K^\mathrm{ab}$ such that the composition

$\mathrm{Art}_K:K^\times\to G_K^\mathrm{ab}\to \mathrm{Gal}(K^\mathrm{ur}/K)$

is the map $\alpha\mapsto \mathrm{Frob}_q^{v_K(\alpha)}$ and such that for every finite abelian extension $L/K$ the composition

$\mathrm{Art}_K:K^\times\to G_K^\mathrm{ab}\to \mathrm{Gal}(L/K)$

is surjective with kernel $N_{L/K}(L^\times)$.

2. $\text{(Local existence theorem)}$ Every finite open subgroup of $K^\times$ is of the form $\mathrm{N}_{L/K}(L^\times)$ for some finite abelian extension $L/K$.

We call $\mathrm{Art}_K$ the local Artin reciprocity map. Let us explain some consequences of this theorem.

First, let us explain some properties of the map $\mathrm{Art}_K$ which follow, not immediately, from its defining properties.

First, let us observe that $\mathrm{Art}_K$ has dense image. Indeed, it is an elementary excercise in topological group theory that for a profinite group $G$, a continuous homomorphism $H\to G$ (for $H$ a topological group) has dense image if and only if for all open finite index subgroups $K\subseteq G$ the composition $H\to G\to K$ is surjective. Now, all the finite index subgroups of $G_K^\mathrm{ab}$ are of the form $\mathrm{Gal}(K^\mathrm{ab}/L)$ for some finite abelian extension $L/K$. Thus, the second property of 1. implies precisely that $\mathrm{Art}_K$ surjects onto all finite topological quotients, and thus $\mathrm{Art}_K$ has dense image.

But, a much stronger statement is true:

Corollary 2: The local Artin reciprocity map induces an isomorphism $\widehat{K^\times}\to G_K^\mathrm{ab}$.

Before we prove this corollary, we need the following lemma:

Lemma 3: Let $K$ be a local field. Then, all finite index subgroups of $K^\times$ are open.

Proof: Note that if $M\subseteq K^\times$ is a finite index subgroup with $[K^\times:M]=n$, then $(K^\times)^n\subseteq M$. Thus, it suffices to prove that $(K^\times)^n$ is open for all $n$ since a subgroup of a topological group is open if and only if it contains an open subgroup. To prove this result we note, again, that $(K^\times)^n\subseteq (K^\times)^m$ for $n\geqslant m$, and so it suffices to prove this result for $m\gg 0$. Finally, note that since $(\mathcal{O}_K^\times)^n\subseteq (K^\times)^n$ it suffices to show that $(\mathcal{O}_K^\times)^n$ is open for $n\gg 0$.

But, for $n\gg 0$ the exponential function defines a topological group isomorphism between $1+\mathfrak{p}_K^n$ and $\mathfrak{p}_K^n$. But, note that for $n\gg0$ this latter group is divisible by $n$, and thus, since the exponential function is an isomorphism of topological groups, $1+\mathfrak{p}_K^n$ is contained in $(K^\times)^n$. But, $1+\mathfrak{p}_K^n$ is open, and thus $(K^\times)^n$ is open as desired. $\blacksquare$

Thus, we see that the term ‘open’ in the local existence theorem is superfluous.

Remark: This is one place where we are using that $K$ is $p$-adic. The result that all finite index subgroups are open fails in the case of function fields. Clearly the proof fails since there is no function field analogue of the exponential map, but the result is also false.

So, with this, we can prove the corollary:

Proof(Corollary 2): Let us recall that the profinite completion of a group $H$ is the unique group initial amongst those profinite groups with maps to $H$. To this end, let us suppose that $G$ is a profinite group and $f:K^\times\to G$ is a group map. We need to show that there is a continuous group map $G_K^\mathrm{ab}\to G$. Now, note that

$\displaystyle G\cong \varprojlim_U G/U$

as $U$ runs over the finite index open subgroups of $G$. Thus, our map $K^\times \to G$ corresponds to maps $K^\times\to G/U$. Now, since $G/U$ is finite this map must factor through a finite quotient. Using the local existence theorem (together with Lemma 3) we see that, in particular, we have a factorization

$K^\times\to K^\times/N_{L/K}(L^\times)\to G/U$

for some finite abelian $L/K$. Now, by the properties of the local Artin map we obtain a  map

$\mathrm{Gal}(L/K)\xrightarrow{\approx}K^\times/N_{L/K}(L^\times)\to G/U$

and thus a map $G_K^\mathrm{ab}\to G/U$. Moreover, note that this map is necessarily continuous since the map $\mathrm{Gal}(L/K)\to K^\times/N_{L/K}(L^\times)$, with the discrete topology, is continuous.

Now, note that by combining the properties of the local Artin reciprocity map, and the local existence theorem, we can see that $L\to N_{L/K}(L^\times)$ is an order reversing bijection between the finite abelian extensions of $K$ and the finite index subgroups of $K^\times$. From this, it’s easy to see that the maps $G_K^\mathrm{ab}\to G/U$ we’ve constructed form an inverse system, and thus we get a continuous map $G_K^\mathrm{ab}\to G$ factorizing the map $K^\times\to G$. Moreover, since $K^\times$ has dense image in $G_K^\mathrm{ab}$ it’s clear that this factorization is unique. The conclusion follows. $\blacksquare$

Let us note the following observation we made during the proof:

Corollary 4: The map $L\to N_{L/K}(L^\times)$ is an order reversing bijection

$\left\{\begin{matrix}\text{Finite abelian}\\ \text{extensions of }K\end{matrix}\right\}\longleftrightarrow\left\{\begin{matrix}\text{Finite index}\\ \text{subgroups of }K^\times\end{matrix}\right\}$

such that $\mathrm{Gal}(L_2/L_1)$ is isomorphic to $N_{L_1/K}(L_1^\times)/N_{L_2/K}(L_2^\times)$.

Now, before we continue, let us stop and wonder in amazement at Corollary 2. It tells us that if $K/\mathbb{Q}_p$ is an extension of degree $n$, then there is an isomorphism of topological groups:

$G_K^\mathrm{ab}\cong \widehat{\mathbb{Z}}\times\mathbb{Z}_p^n\times \mu_{p^\infty}\left(K\right)\times \mathbb{Z}/(q-1)\mathbb{Z}$

where $\mu_{p^\infty}(K)$ denotes the $p^\text{th}$-power roots of unity in $K$, and the last two groups on the right have the discrete topology.

As an example, it tells us that

$G_{\mathbb{Q}_p}^\mathrm{ab}\cong \widehat{\mathbb{Q}_p^\times}\cong \mathbb{Z}_p^\times\times \widehat{\mathbb{Z}}\cong \mathbb{Z}/(p-1)\mathbb{Z}\times\mathbb{Z}_p\times \widehat{\mathbb{Z}}$

where we have used that $\mathbb{Q}_p^\times\cong \mathbb{Z}_p^\times \times p^{\mathbb{Z}}$ and used the fact that $\mathbb{Z}_p^\times$ is profinitely complete, as follows from Lemma 3. This is an astounding fact. There is no reason why one would hope, let alone expect, that we could have such a simple description of the group which dictates the abelian extensions of $\mathbb{Q}_p$—wow!

One thing we have not used yet is the first property of the local Artin reciprocity map. But, that gets put to use, essentially, to prove the following:

Corollary 5: Under the isomorphism $\widehat{K}\cong \mathcal{O}_K^\times\times\widehat{\mathbb{Z}}$ the fixed field of $\mathcal{O}_K^\times$ is $K^\mathrm{ur}$. More generally, the filtration

$\mathcal{O}_K^\times\supseteq 1+\pi\mathcal{O}_K\supseteq 1+\pi^2\mathcal{O}_K\supseteq\cdots$

corresponds to the filtration of $G_K^\mathrm{ab}$ by abelianized higher ramification groups.

Note, again, we’ve implicitly used the fact that $\mathcal{O}_K^\times$ is profinitely complete, which is true, again, by Lemma 3.

Proof: Note that the fixed field of $\widehat{\mathbb{Z}}$ is the same as the fixed field of the image of $\mathcal{O}_K^\times$ under $\mathrm{Art}_K$. But, by assumption we have that the composition of $\mathrm{Art}_K$ with $G_K^\mathrm{ab}\to \mathrm{Gal}(K^\mathrm{ur}/K)$ is the valuation map. In particular, the kernel is $\mathcal{O}_K^\times$, which proves the proposition.

The second claim follows by precisely the same idea.$\blacksquare$

Thus, we are naturally led to ask what the fixed field of $\widehat{\mathbb{Z}}$ is—a field extension $K_\pi/K$ with $\mathrm{Gal}(K_\pi/K)\cong \mathcal{O}_K^\times$.

## The Weil group

We have certainly stated the fundamental results of local class field theory above, in particular having shown that $K^\times$ ‘approximates’ $G_K^\mathrm{ab}$ in the same way that $\mathbb{Z}$ approximates $\widehat{\mathbb{Z}}$ (in particular classifying the finite quotients). But, we have yet to describe the ‘canonical subgroup’ $W_K^\mathrm{ab}$ of $G_K^\mathrm{ab}$ which really is the ‘approximating subgroup’ and which the Artin map just provides an isomorphism with.

So, let us attempt to define this group. As the notation suggests, we will actually define a subgroup $W_K$ (important in the general formulation of Langlands) for which our group is just the (Hausdorff) abelianization. We will take a somewhat lowbrow approach in defining $W_K$, deciding to ignore the more general notion of such groups as defined in Tate’s Number Theoretic Background article in the Corvallis proceedings.

So, let us begin by thinking about what we might want $W_K\to G_K$ to do. Namely, if $W_K$ is supposed to be the ‘approximating subgroup’ of $G_K$, then we would hope/expect its image under any surjection to be an ‘approximating subgroup’ in the quotient. In particular, we’d hope that the image of $W_K$ in $G_k$ to be the approximating subgroup we’ve already mentioned for finite field Galois groups—the integers $\mathbb{Z}$. Thus, we expect there to be a continuous surjection

$v:W_K\to \mathbb{Z}$

and we expect the kernel to be the same as the kernel of the reduction map $G_K\to G_k$—the inertia group $I_K$.

So, we expect to have a short exact sequence

$0\to I_K\to W_K\to \mathbb{Z}\to 0$

which essentially forces $W_K$ to be the preimage of $\mathbb{Z}$ under $G_K\to G_k$. Ok, cool. But, we expect this above sequence to be a short exact sequence not just of groups, but of topological groups. So, in particular, since $\mathbb{Z}$ is discrete we should require that $I_K$, the kernel of this map, to be open. Of course, in the subspace topology $W_K\subseteq G_K$ this is not the case. If it were open in $W_K$ then the map $G_K\to G_k$ would have open kernel, making $G_k$ be discrete, which it isn’t.

Thus, we see that to make $W_K$ perform as desired, it will be a subgroup of $G_K$, but will have a different topology than the one inherited from $G_K$. In particular, we give $W_K$ the unique topology making $v$ continuous and for which $I_K$ has the topology inherited from $G_K$. We call this topological group $W_K$ the Weil group of $K$.

Note that while the map $W_K\to G_K$ is not an embedding of topological groups, it’s an injective continuous group map with dense image in $G_K$. Why does it have dense image? Well, recall that a subset $X\subseteq G_K$ is dense if and only if for all finite Galois extensions $L/K$ the image of $X$ in $\mathrm{Gal}(L/K)$ is everything. To see that this is true for $W_K$ merely note that while $v:W_K\to\mathbb{Z}\subseteq\widehat{\mathbb{Z}}$ is not surjective (i.e. Frobenius doesn’t generate everything in infinite extensions) it is true that this map composed with any finite quotient is surjective with kernel $I(L/K)$.

Said slightly more rigorously, let $L/K$ be any finite Galois extension. Then, it’s clear that we have the following commutative diagram of maps

$\begin{matrix} 0 & \to & I_K & \to & G_K & \to & G_k & \to & 0\\ & & \downarrow^{\text{surj.}} & & \downarrow^{\text{surj.}} & & \downarrow^{\text{surj.}} & &\\ 0 & \to & I(L/K) & \to & \mathrm{Gal}(L/K) & \to & \mathrm{Gal}(\ell/k) & \to & 0\\ & & \uparrow^{\text{surj.}} & & \uparrow & & \uparrow^{\text{surj.}} & & \\ 0 & \to & I_K & \to & W_K & \to & \mathbb{Z} & \to & 0 \end{matrix}$

where $I(L/K)$ is the inertia group in $\mathrm{Gal}(L/K)$ and $\ell$ is the residue field. This diagram allows us to easily conclude that $W_K$ surjects onto $\mathrm{Gal}(L/K)$ (since the outer upward pointing arrows are surjective) and thus prove that the image of $W_K$ in $G_K$ is dense.

Moreover, note that the local Artin reciprocity map $\mathrm{Art}_K$ takes its image in the image of $W_K^\mathrm{ab}$. Indeed, note that the first property of the Artin map implies that the composition $K^\times G_K^\mathrm{ab}\to G_k$ has image in the preimage of $\mathbb{Z}$ in $G_K^\mathrm{ab}$ which implies its image lies in $W_K^\mathrm{ab}$.

But, even more is true:

Theorem 6: The local Artin reciprocity map provides an isomorphism of topological groups $K^\times\xrightarrow{\approx}W_K^\mathrm{ab}$.

Note that one obtains, from the above observation, a group map $K^\times\to W_K^\mathrm{ab}$ from the local Artin reciprocity map. We need just explain why it’s continuous, bijective, and closed. Now, to see why it’s continuous, we need really only check that the preimage of $I_K^\mathrm{ab}$ is open since, after all, the map $K^\times\to G_K^{\mathrm{ab}}$ is continuous, and $W_K^\mathrm{ab}$ differs from the subspace topology only because $I_K^\mathrm{ab}$ is not open. But, note that the preimage of $I_K$ is the same thing as the kernel of the composition

$K^\times \to G_K^\mathrm{ab}\to G_k$

but, by the defining properties of the local Artin reciprocity map, this preimage is just $\mathcal{O}_K^\times$ which is, in fact, open. The bijectivity follows by just considering the map on finite levels (quotients by open subtroups). Finally, we can conclude that the map is a homeomorphism since our continuous bijection corresponds to, when we decompose our groups as products of $\mathbb{Z}$ by closed subgroups, to continuous bijections on these closed subgroups. But, then we may conclude that the maps are homeomorphisms since continuous bijections between compact Hausdorff spaces are always such. The conclusion then easily follows.

Now, even though the above was instructive, it’s unclear as to why we went through so much effort. How does the observation that the local Artin reciprocity map is really providing an isomorphism $K^\times\to W_K^\mathrm{ab}$ helpful? In fact, why even define $W_K^\mathrm{ab}$ in the first place—it doesn’t provide that much more content than the statement of local class field theory by itself.

The reason is one of foresight. Namely, even though the Weil group $W_K$ plays only a minor theoretical role in the discussion of (abelian class field theory), the discussion of the character theory of $G_K$, it plays a much more substantive role in higher dimensions. In fact, the true statement of the local Langlands conjecture for $\mathrm{GL}_n$ speaks not of Galois representations but, instead, of representations of $W_K$ (more rigorously, Weil-Deligne representations). So, having stated the above isomorphism is, once one understands what this really means, the proof that local class field theory is ‘local Langlands for $\mathrm{GL}_1$‘.

But, a less highfalutin reason is that it’s just plain nice to know what the image of $\mathrm{Art}_K$ is. To understand the approximating group $W_K^\mathrm{ab}$ without having to understand all of local class field theory. The content of Theorem 6, in this light, is just saying that the approximating group inside of $G_K^\mathrm{ab}$ does, in fact, have an ‘internal to $K$‘ description.

# Cohomological proof

We begin with the more classical cohomological proof of local class field theory. But, it should be remarked that our discussion is certainly not historical in the following sense. The first proof of local class field theory actually, somewhat counterintuitively, reduced to the global case first. There people did not think in modern terms (i.e. in terms of adeles) and instead phrased everything in terms of class fields (thus the name of the subject).

To make the following more palatable, we give a brief outline of the steps needed to complete a cohomological proof of local class field theory:

1. Show that there is a pairing $X(K)\times K^\times\to \mathrm{Br}(K)$, where $X(K)$ is the Pontryagin dual of $G_K^\mathrm{ab}$ and $\mathrm{Br}(K)$ is the Brauer group of $K$ (see below).
2. Compute $\mathrm{Br}(K)$ to be $\mathbb{Q}/\mathbb{Z}$ and, in the process, show that every element of $\mathrm{Br}(K)$ is in the image of $X(K)\times K^\times$.
3. From steps 1. and 2. we obtain a map $K^\times \to\mathrm{Hom}_{\text{cont.}}(X(K),\mathbb{Q}/\mathbb{Z})$, but the right hand object, by Pontryagin duality, is just $G_K^\mathrm{ab}$. This map $K^\times\to G_K^\mathrm{ab}$ is the local Artin reciprocity map.
4. Show that the local Artin reciprocity map has the desired properties, and show the existence theorem by hand.

The ‘cohomological’ part of the proof will come in step 2., in particular, in understanding the Brauer group of $K$.

As stated in the introduction, our goal is really just to have an understanding of how the objects of local class field theory are defined. So, we’ll focus entirely on steps 1-3, deferring the check that the defined map ‘works’ to the more serious treatments of the subject.

## Step 1: the cyclic algebra pairing

### The Brauer group

#### Basic definitions

The Brauer group is an object of deceivingly great importance in both number theory and algebraic geometry. It’s so deceiving, in fact, that upon first hearing its definition, one is liable to get glazed eyes, and completely forget the definitions. This is especially true if one only ever considers the Brauer group of a field, which belies the true geometric nature of the object. To get a sense of their more interesting geometric content see, for instance, this post.

Let us fix a field $K$ throughout this section. We do not assume that $K$ is local, or even of characteristic $0$.

So, let us define a central simple algebra $A/K$ to be a finite-dimensional unital (but not necessarily commutative) algebra $A/K$ such that its center $Z(A)$ is $K$, and which has no non-trivial two-sided ideals.

There is essentially only one example of a central simple algebra over $K$. Let us recall that a central division algebra over $K$ is a unital finite-dimensional $K$-algebra $D$ such that every non-zero element has a two-sided inverse. Then, for any central division algebra $D/K$ the algebra $\mathrm{Mat}_n(D)$ of $n\times n$-matrices over $D$ is a central simple algebra over $K$. This comes from the basic fact that all two-sided ideals of $\mathrm{Mat}_n(R)$ (for any ring $R$) are of the form $\mathrm{Mat}_n(I)$ for a two-sided ideal $I\subseteq R$.

Let us give some concrete examples:

• Over any field $K$, we have the trivial example of a central division algebra. Namely, $K$ itself. This gives rise to the ‘family’ of trivial central simple algebras over $K$ given by $\mathrm{Mat}_n(K)$.
• Over $\mathbb{R}$ one has the Hamiltonian quaternions $\mathbb{H}$ which gives rise to a non-trivial central division algebra over $\mathbb{R}$.
• Over $\mathbb{F}_q$, Wedderburn’s little theorem says that $\mathbb{F}_q$ is the only central division algebra over $\mathbb{F}_q$.
• Much simpler, if $K$ is algebraically closed, then all central division algebras over $K$ are trivial. Indeed, if $D/K$ is a central division algebra and $x\in D-K$, then $K[x]$ is a finite field extension of $K$ and so must be $K$ itself—contradiction.
• Generalizing the Hamiltonian quaternions, one can define for any two $a,b\in K^\times$ the quaternion algebra $(a,b)$.

When we said that central simple algebras of the form $\mathrm{Mat}_n(D)$ were representative of the general case of central simple algebras over $K$ we weren’t exaggerating:

Theorem 7(Artin-Wedderburn): Let $K$ be a field, and $A/K$ a central simple algebra over $K$. Then, $A\cong\mathrm{Mat}_n(D)$ for a unique central division algebra $D/K$.

So, as a corollary, we see that if $K$ is a finite field or an algebraically closed field, then all central simple algebras are of the form $\mathrm{Mat}_n(K)$.

Now, we can form a group $\mathrm{Br}(K)$, called the Brauer group of $K$, as follows. We call two central simple algebras $A/k$ and $B/k$ equivalent if $\mathrm{Mat}_n(A)\cong\mathrm{Mat}_m(B)$ for some $m,n\in\mathbb{N}$. Then, as a set, $\mathrm{Br}(K)$ is just the equivalence classes of central simple algebras over $K$.

To define the group structure, we need the following lemma which is, essentially, an exercise in non-commutative algebra:

Lemma 8: Let $A$ and $B$ be central simple algebras over $K$. Then, $A\otimes_K B$ is a central simple algebra over $K$.

This endows $\mathrm{Br}(K)$ with the obvious group operation of tensor product. This is well-defined since the following identity holds:

$A\otimes_K \mathrm{Mat}_n(B)\cong \mathrm{Mat}_n(A\otimes_K B)$

as one can easily check.

One can shift the complicatedness of the set $\mathrm{Br}(K)$ to complicatedness of the group operation. Namely, by the Artin-Wedderburn theorem, every central simple algebra over $K$ is equivalent to a unique central division algebra over $K$. Thus, as a set, $\mathrm{Br}(K)$ is nothing but the central division algebras over $K$. But, in this guise the group operation is not precisely obvious. Namely, given $D_1,D_2/K$ central simple algebras how should we define $D_1\cdot D_2$—a new central simple algebra.

Of course, the answer is not $D_1\otimes_K D_2$ since this is not, in general, a central division algebra over $K$. Instead, we use the fact that $D_1\otimes_K D_2$ is a central simple algebra over $K$, and thus of the form $\mathrm{Mat}_n(D_3)$ for some central division algebra $D_3$. We then define $D_1\cdot D_2=D_3$. One can check that the map

$D\mapsto [D]$

from the isomorphism classes of central division algebras over $K$ to $\mathrm{Br}(K)$ (where $[D]$ stands for the equivalence class of $D$) is an isomorphism of groups.

Note that our observation following the Artin-Wedderburn theorem now allows us to conclude that $\mathrm{Br}(K)$ is trivial in case $K$ is either finite or algebraically closed.

#### A different perspective

We now seek to give a different description, a cohomological description, of the Brauer group. To this end, let us record the following theorem:

Theorem 9: If $A/K$ is a central simple algebra then $A_L:=A\otimes_K L$ is a central simple algebra over $L$ for any field extension $L/K$.

Thus, we see that if $A/K$ is a central simple algebra, then $A\otimes_K \overline{K}$ is a central simple algebra over $\overline{K}$. Thus, from previous observation we may conclude that $A\otimes_K \overline{K}$ is of the form $\mathrm{Mat}_n(\overline{K})$ for a unique $n$, called the degree of $A$.

In fact, we actually have a converse to this observation:

Theorem 10: Let $A$ be an algebra over $K$. Then, $A$ is a central simple algebra over $K$ if and only if one of the following equivalent properties holds:

1. $A\otimes_K\overline{K}\cong\mathrm{Mat}_n(\overline{K})$ for some $n$.
2. There exists a finite extension $L/K$ such that $A\otimes_K L\cong\mathrm{Mat}_n(L)$.
3. There exists a finite Galois extension $L/K$ such that $A\otimes_K L\cong \mathrm{Mat}_n(L)$.

This allows us to get a different perspective on central simple algebras. Namely, the above indicates, that in fact, $A\otimes_K K^\mathrm{sep}$ is just $\mathrm{Mat}_n(K^\mathrm{sep})$. Thus, we can think about about central simple algebras in terms of Galois cohomology.

In particular, recall that for an object $X_0/k$ a twist of $X_0$ is an object $X/K$ such that $X_0\otimes_K K^\mathrm{sep}\cong X\otimes_K K^\mathrm{sep}$. It then follows from the general yoga of torsors that the twists of $X_0$ (up to $k$-isomorphism) are classified by $H^1(G_K,\mathrm{Aut}_{K^\mathrm{sep}}(X_0\otimes_K K^\mathrm{sep}))$.

While this is not totally rigorous (check the link for a more detailed account), this tells us that algebras $A/K$ which become isomorphic to $\mathrm{Mat}_n(K^\mathrm{sep})$, in other words degree $n$ central simple algebras over $k$, are classified by $H^1(G_K,\mathrm{PGL}_n(K^\mathrm{sep}))$ (the action being the entry-wise one). Here we have used that the algebra automorphisms of $\mathrm{Mat}_n(K^\mathrm{sep})$ are just $\mathrm{PGL}_n(K^\mathrm{sep})$—this is a special case of the Skolem-Noether theorem.

The rough intuition for this correspondence is as follows. Given a degree $n$ central simple algebra $A/K$, we know from above that $A\otimes_K K^\mathrm{sep}$ is $\mathrm{Mat}_n(K^\mathrm{sep})$. We want to be able to go back to $A$ by some sort of ‘quotient operation’, something similar to just taking $G_K$-invariants. The data needed to perform such an operation turns out to be precisely the data of a $1$-cocycle—an element of $H^1(G_K,\mathrm{PGL}_n(K^\mathrm{sep}))$.

Now, there is a natural map

$H^1(G_K,\mathrm{PGL}_n(K^\mathrm{sep}))\times H^1(G_K,\mathrm{PGL}_m(K^\mathrm{sep}))\to H^1(G_K,\mathrm{PGL}_{nm}(K^\mathrm{sep}))$

which takes a pair of $1$-cocycles valued in $n\times n$-matrices and $m\times m$-matrices (up to scalar multiplication in both cases) and outputs the $1$-cocycle valued in $nm\times nm$-matrices (up to scalar multplication) by taking the tensor product (i.e. the Kronecker product):

$\left(\sigma:g\mapsto M_g,\psi:g\mapsto N_g\right)\mapsto \eta:g\mapsto M_g\otimes N_g$

which gives

$H^1(G_K,\mathrm{PGL}_\infty(K^\mathrm{sep})):=\varinjlim H^1(G_K,\mathrm{PGL}_n(K^\mathrm{sep}))$

where the direct limit is taken with respect to the obvious inclusions

$\mathrm{PGL}_n(K^\mathrm{sep})\hookrightarrow \mathrm{PGL}_{nm}(K^\mathrm{sep}):M\mapsto M\otimes \mathrm{id}$

a group structure.

Now, one can soup up our observation about twists to the following theorem:

Theorem 11: The association $\mathrm{Br}(K)\to H^1(G_K,\mathrm{PGL}_\infty(K^\mathrm{sep}))$ taking a central simple algebra over $K$ to its associated $1$-cocycle (being a twist of $\mathrm{Mat}_n(K^\mathrm{sep})$) is an isomorphism of groups.

This theorem is nice because it opens up the study of the Brauer group to cohomological methods of attack. But, it has the serious downside that we’re taking cohomology of a non-commutative $G_K$-module. In particular, the individual objects $H^1(G_K,\mathrm{PGL}_n(K^\mathrm{sep}))$ are not groups, they are just pointed sets. We only obtain the group operation only in the limit.

Thus, it would be nice if we could trade in this cohomology group $H^1(G_K,\mathrm{PGL}_\infty(K^\mathrm{sep}))$ for a cohomology group valued in some commutative $G_K$-module. It turns out, somewhat surprisingly, that this is the case. The key being the following short exact sequence $G_K$-modules:

$1\to \left(K^\mathrm{sep}\right)^\times\to\mathrm{GL}_n(K^\mathrm{sep})\to \mathrm{PGL}_n(K^\mathrm{sep})\to 1$

which gives us the following exact sequence (coming from the long exact sequence in cohomology):

$H^1(G_K,(K^\mathrm{sep})^\times)\to H^1(G_K,\mathrm{GL}_n(K^\mathrm{sep}))\to H^1(G_K,\mathrm{PGL}_n(K^\mathrm{sep}))\to H^2(G_K,(K^\mathrm{sep})^\times)$

But, note that the first term of this is zero. This is essentially a strengthening of Hilbert’s theorem 90—this group should be classifying vector bundles on $K$ which become isomorphic to $(K^\mathrm{sep})^n$ over $K^\mathrm{sep}$, but there is only one such vector space: $K^n$!

Thus, we have obtained an injection

$H^1(G_K,\mathrm{PGL}_n(K^\mathrm{sep}))\to H^2(G_K,(K^\mathrm{sep})^\times)$

The non-trivial fact is then then following:

Theorem 12: The direct limit map

$\mathrm{Br}(K)\cong H^1(G_K,\mathrm{PGL}_\infty(K^\mathrm{sep}))\to H^2(G_K,(K^\mathrm{sep})^\times)$

from the map on finite pieces described above, is an isomorphism of groups.

Thus we have, as desired, reduced the study of the Brauer group to the study of the cohomology of an abelian $G_K$-module. This isomorphism is the only practical way to compute $\mathrm{Br}(K)$ for many fields $K$.

As an example, let us explain why $\mathrm{Br}(\mathbb{R})\cong\mathbb{Z}/2\mathbb{Z}$, with non-trivial element being represented by the Hamiltonian quaternions $\mathbb{H}$. By Theorem 12 we need only consider the following computation:

\begin{aligned}\mathrm{Br}(\mathbb{R}) &= H^2(G_\mathbb{R},\mathbb{C}^\times)\\ &= \mathbb{R}^\times/\mathbb{R}_{>0}\\ &=\mathbb{Z}/2\mathbb{Z}\end{aligned}

this being done by the usual resolution associated to a cyclic group.

This will also be the method by which we compute the Brauer group of a local field.

### The pairing: cyclic algebras

Now that we have discussed the general theory of the Brauer group, we can describe the cyclic algebra pairing:

$K^\times\times X(K)\to\mathrm{Br}(K)$

mentioned above. Again, here, $X(K)$ denotes the Pontryagin dual of $G_K^\mathrm{ab}$, meaning that it’s $\mathrm{Hom}_\mathrm{cont.}(G_K^\mathrm{ab},S^1)$. But, since $G_K$ is profinite we can describe $X(K)$ is an alternate and more useful way.

Namely, note that $S^1$ has an open neighborhood $U$ of $1$ which contains no non-trivial subgroups. Then, for any continuous homomorphism $f:G_K^\mathrm{ab}\to S^1$ we have that $f^{-1}(U)$ contains an open subgroup $H$ (since the group is profinite). Now, $f(H)$ is a subgroup of $U$ and so, by construction, trivial. Thus, $f$ factors through $G_K^\mathrm{ab}/H$, a finite group.

Remark: This is a trivial incarnation of the no small subgroups argument. For an interesting read on this topic, I suggest this post of Terry Tao.

From this, we see that the image of any element of $X(K)$ lies in the torsion subgroup of $S^1$, which is the roots of unity. This is, as an abstract group, just $\mathbb{Q}/\mathbb{Z}$. But, note that since $f:G_K^\mathrm{ab}\to\mathbb{Q}/\mathbb{Z}\subseteq S^1$ factors through a finite quotient, it’s continuous with $\mathbb{Q}/\mathbb{Z}$ having the subspace topology of $S^1$ if and only if its continuous with $\mathbb{Q}/\mathbb{Z}$ endowed with the discrete topology.

So, summarizing, we may conclude that $X(K)$ can be alternatively thought about as $\mathrm{Hom}_\mathrm{cont.}(G_K^\mathrm{ab},\mathbb{Q}/\mathbb{Z})$ where $\mathbb{Q}/\mathbb{Z}$ is given the discrete topology. This is the form which will be most useful to us.

So, now, we want to construct for any pair $(\alpha,\chi)\in K^\times\times X(K)$ a central simple algebra, denoted $A(\alpha,\chi)\in \mathrm{Br}(K)$. To define $A(\alpha,\chi)$, note that $\ker\chi=\mathrm{Gal}(K^\mathrm{ab}/L)$ for some finite abelian extension $L/K$. In fact, since the only finite subgroups of $\mathbb{Q}/\mathbb{Z}$ are cyclic, we actually have that $\chi$ produces an isomorphism

$\mathrm{Gal}(L/K)\xrightarrow{\approx}\frac{1}{n}\mathbb{Z}/\mathbb{Z}\subseteq\mathbb{Q}/\mathbb{Z}$

and we single out $\sigma\in\mathrm{Gal}(L/K)$ which maps, under $\chi$, to $\frac{1}{n}$.

We then define $A(\alpha,\chi)$ to be

$L\langle T\rangle/(T^n-\alpha)$

where $L\langle T\rangle$ is the non-commutative polynomial algebra with commutation relation $T\ell=\sigma(\ell)T$. This is clearly a central $K$-algebra since if $x\in Z(A(\alpha,\chi))$ then $x\in L$ (for obvious reasons!) and $x=\sigma(x)$ which forces $x\in K$. But, one can also show that $A(\alpha,\chi)$ is also simple, and thus defines an element $A(\alpha,\chi)\in\mathrm{Br}(K)$. We call an algebra of the form $A(\alpha,\chi)$ a cyclic algebra.

It is this pairing

$K^\times\times X(K)\ni (\alpha,\chi)\mapsto A(\alpha,\chi)\in\mathrm{Br}(K)$

which will play the pivotal role in our discussion of local class field theory. Sticking to customary notation, we shall write just $(\alpha,\chi)$ for the class of $A(\alpha,\chi)\in\mathrm{Br}(K)$ highlighting the fact that we’re dealing with a pairing $(-,-)$.

## Step 2: Brauer group computation

The next step in the cohomological proof of class field theory is to actually compute the group $\mathrm{Br}(K)$ where $K$ is a ($p$-adic) local field. Now, by what we discussed in the last section, this is equivalent to computing the Galois cohomology group $H^1(G_K,\mathrm{PGL}_\infty(\overline{K}))$ or, equivalently, $H^2(G_K,\overline{K}^\times)$. This is a highly non-trivial task.

The first key observation one must make involves how ‘deep’ into $\overline{K}$ one must go to split central simple algebras. More specifically, let us call a central simple algebra $A/K$ split if it’s isomorphic to a matrix algebra over $K$. Or, said differently, if the class $[A]\in\mathrm{Br}(K)$ is trivial. Let us then say that a field extension $L/K$ splits $A$ if $A\otimes_K L$ is split. Note that by Theorem 10 there is always a finite Galois extension $L/K$ splitting any central simple algebra $A$.

Let us denote the set of central simple algebras over $K$ split by $L$ by $\mathrm{Br}(L/K)$. This is clearly a subgroup of $\mathrm{Br}(K)$ and, in fact, is just the kernel of the map

$\mathrm{Br}(K)\to\mathrm{Br}(L):A\mapsto A\otimes_K L$

Moreover, whereas $\mathrm{Br}(K)=H^2(G_K,\overline{K}^\times)$, one can easily check that $\mathrm{Br}(L/K)$ is just $H^2(\mathrm{Gal}(L/K),L^\times)$.

So, the aforementioned observation which serves as a first step in the computation of $\mathrm{Br}(K)$ is:

Theorem 13: Every central simple algebra $A/K$ is split by $K^\mathrm{ur}$. Or, said differently, $\mathrm{Br}(K)=\mathrm{Br}(K^\mathrm{ur}/K)$.

While not entirely obvious, this part is essentially just some simple non-commutative algebra combined with a basic understanding of unramified extensions.

The true ‘meat’ of the computation comes from the following fact:

Theorem 14: For all $L/K$ unramified, $H^2(\mathrm{Gal}(L/K),\mathcal{O}_L^\times)=0$.

The proof of this is non-trivial and really relies on the general theory of Tate cohomology and in particular leveraging the cyclicness of $\mathrm{Gal}(L/K)$ for $L/K$ finite unramified in the usage of Tate’s theorem.

Once one has these two theorems, the proof that $\mathrm{Br}(K)=\mathbb{Q}/\mathbb{Z}$ is actually not overly complicated. Indeed, begin by noting that by Theorem 13 we are really just trying to compute

$H^2(\mathrm{Gal}(K^\mathrm{ur}/K),(K^\mathrm{ur})^\times)$

But, note that since $K^\mathrm{ur}$ is, well, unramified, we have a splitting of sequences $\mathrm{Gal}(K^\mathrm{ur}/K)$-modules:

$0\to \mathcal{O}_{K^\mathrm{ur}}^\times\to (K^\mathrm{ur})^\times\to\mathbb{Z}\to 0$

given by choosing a uniformizer. From this, we know that

\begin{aligned}H^2(\mathrm{Gal}(K^\mathrm{ur}/K),(K^\mathrm{ur})^\times) &=H^2(\mathrm{Gal}(K^\mathrm{ur}/K),\mathcal{O}_{K^\mathrm{ur}}^\times)\times H^2(\mathrm{Gal}(K^\mathrm{ur}/K),\mathbb{Z})\\ &=H^2(\mathrm{Gal}(K^\mathrm{ur}/K),\mathbb{Z})\end{aligned}

where the second equality follows from Theorem 14.

Thus, we are reduced to proving that $H^2(\mathrm{Gal}(K^\mathrm{ur}/K),\mathbb{Z})$ is $\mathbb{Q}/\mathbb{Z}$. But, note that, again by unramifiedness, $\mathbb{Z}$ is a trivial $\mathrm{Gal}(K^\mathrm{ur}/K)\cong\widehat{\mathbb{Z}}$-module. So, we’re just trying to compute the continuous group cohomology $H^2(\widehat{\mathbb{Z}},\mathbb{Z})$ where $\mathbb{Z}$ has a trivial structure.

The key to this computation is the following short exact sequence of groups:

$0\to \mathbb{Z}\to\mathbb{Q}\to\mathbb{Q}/\mathbb{Z}\to 0$

thought about as a short exact sequence of trivial $\widehat{\mathbb{Z}}$-modules. Passing to the long exact sequence in cohomology we get the piece

$H^1(\widehat{\mathbb{Z}},\mathbb{Q})\to H^1(\widehat{\mathbb{Z}},\mathbb{Q}/\mathbb{Z})\to H^2(\widehat{\mathbb{Z}},\mathbb{Z})\to H^2(\widehat{\mathbb{Z}},\mathbb{Q})$

But, since $\mathbb{Q}$ is uniquely divisible (i.e. $[n]:\mathbb{Q}\to\mathbb{Q}$ is an isomorphism) the modules $H^i(\widehat{\mathbb{Z}},\mathbb{Q})$ are uniquely divisible. But, for $i>0$ they’re totally torsion which, of course, implies they are zero. Thus, we obtain an isomorphism between $H^2(\widehat{\mathbb{Z}},\mathbb{Z})$ and $H^1(\widehat{\mathbb{Z}},\mathbb{Q}/\mathbb{Z})$.

But, this gives us precisely what we want. Indeed, since $\mathbb{Q}/\mathbb{Z}$ is a trivial $\widehat{\mathbb{Z}}$-module, we know that

$H^1(\widehat{\mathbb{Z}},\mathbb{Q}/\mathbb{Z})=\mathrm{Hom}_{\text{cont.}}(\widehat{\mathbb{Z}},\mathbb{Q}/\mathbb{Z})$

But, since $\widehat{\mathbb{Z}}$ is pro-cyclic, any continuous map out of it is determined by a topological generator—namely, Frobenius. Thus, the map

$\mathrm{Hom}_{\text{cont.}}(\widehat{\mathbb{Z}},\mathbb{Q}/\mathbb{Z})\to\mathbb{Q}/\mathbb{Z}:\varphi\mapsto \varphi(\mathrm{Frob})$

is an isomorphism, as desired.

Of course, this doesn’t give one a very explicit description of $\mathrm{Br}(K)$. Namely, what is the division algebra associated to an element $\displaystyle \frac{r}{s}\in\mathbb{Q}/\mathbb{Z}$? It turns out that there is a very explicit way of describing it, in fact. In particular, for such an $\displaystyle \frac{r}{s}$ with, say $r>0$ and the two coprime, we can create the cyclic algebra $(\pi^s,\sigma)$ where $\sigma:G_K^\mathrm{ab}\to \mathbb{Q}/\mathbb{Z}$ is the character which comes from composing the quotient map $G_K^\mathrm{ab}\to\mathrm{Gal}(L/K)$, where $L/K$ is the unique unramified extension of degree $r$, with the isomorphism

$\displaystyle \mathrm{Gal}(L/K)\xrightarrow{\approx}\frac{1}{r}\mathbb{Z}/\mathbb{Z}\subseteq\mathbb{Q}/\mathbb{Z}$

given by sending Frobenius to $\frac{1}{r}$. One can check that the group of such cyclic algebras creates a copy $\mathbb{Q}/\mathbb{Z}\subseteq\mathrm{Br}(K)$ which must be surjective since $\mathbb{Q}/\mathbb{Z}$ doesn’t contain any proper subgroups isomorphic to itself.

As an added bonus, we see that we have shown the the cyclic algebra pairing is actually surjective.

## Step 3: construction of the local Artin reciprocity map

With the setup we have developed this step is not particularly difficult and, in fact, was essentially given in the outline. But, for completeness, we give a slightly more detailed account here.

So, combining steps 1 and 2 we have created a pairing

$K^\times\times X(K)\to \mathrm{Br}(K)\cong\mathbb{Q}/\mathbb{Z}$

which, of course, gives us a map

$K^\times\to \mathrm{Hom}(X(K),\mathbb{Q}/\mathbb{Z})$

But, notice that by construction the kernel of any map $X(K)\to\mathbb{Q}/\mathbb{Z}$ obtained in this way is actually open (consider that the image is finite). So, the image of $K^\times$ actually lands in $\mathrm{Hom}_{\text{cont.}}(X(K),\mathbb{Q}/\mathbb{Z})$ where the latter has the discrete (equivalently subspace $\mathbb{Q}/\mathbb{Z}\subseteq S^1$) topology.

Then, by an argument similar to what happened with $X(K)$ we can see that

$\mathrm{Hom}_\text{cont.}(X(K),\mathbb{Q}/\mathbb{Z})=\mathrm{Hom}_\text{cont.}(X(K),S^1)$

One may then appeal to Pontryagin duality to conclude that the group on the right is just $G_K^\mathrm{ab}$. Thus, we have given a map $K^\times \to G_K^\mathrm{ab}$.

Now, as stated before, we won’t check that this map is actually the local Artin reciprocity map, but the task is easier than one may think. Indeed, given the semi-explicit nature of the map, checking the desired properties reduces to understanding cyclic algebras which is fairly routine.

# Proof using Lubin-Tate theory

We now move on to the proof of local class field theory using the work of Lubin and Tate in their seminal paper Formal Complex Multiplication in Local Fields.

Again, to make the following discussion clearer, we outline the steps of the proof as follows:

1. Show that to any local field $K$ with chosen uniformizer $\pi$ there is an associated canonical isomorphism class of formal $\mathcal{O}_K$-modules $\mathbb{G}$.
2. Study the properties of the $[\pi^n]$-torsion points of $\mathbb{G}(\mathcal{O}_{\overline{K}})$ and the fields $K_{\pi,n}$ they generate.
3. Show that for $K_\pi$, the compositum of the $K_{\pi,n}$, one has the equality $K^\mathrm{ab}=K^\mathrm{ur}K_\pi$.
4. Show using 2. and 3. that there is a natural map $K^\times\to G_K^\mathrm{ab}$ satisfying the properties of the local Artin reciprocity map.

Again, since our goal is just to explain the construction of the objects, not to show that they satisfy the desired properties, we only discuss steps 1 and 2.

## Step 1: Lubin-Tate formal modules

In this section we carry out part 1. of our outline. We begin with the recollection of the notion of a formal $\mathcal{O}_K$ module.

Let $A$ be an $\mathcal{O}_K$-algebra (thought of as a topological ring with the discrete topology). Then, a (one-dimensional) formal $\mathcal{O}_K$-module is a (one-dimensional) formal Lie group $\mathbb{G}$ over $A$ together with a ring map $\iota:\mathcal{O}_K\to\mathrm{End}(\mathbb{G})$ such that the derivative map

$\mathcal{O}_K\to\mathrm{Lie}(\mathbb{G})=A$

coincides with the ring map $\mathcal{O}_K\to A$ defining the $\mathcal{O}_K$-algebra structure on $A$.

Remark: Of course, there is no reason why one needs to assume one-dimensional in the above definition. We restrict ourselves to this case though, so the point is moot.

If one is willing to take the point of view of formal group laws, opposed to formal Lie groups, which is the view we’ll take here, one can phrase the above in more simple terms. Namely, the group $\mathbb{G}$ will correspond to a one-dimensional formal group law $F(X,Y)\in A[[X,Y]]$. Then, an endomorphism $f$ of $\mathbb{G}$ will be just an element $f(T)\in TA[[T]]$ such that

$F(f(X),f(Y))=f(F(X,Y))$

and the condition on the ring map $\mathcal{O}_K\to \mathrm{End}(\mathbb{G})$ reduces to the statement that if $a$ corresponds to the endomorphism $f(T)\in TA[[T]]$ then

$f(T)\equiv a T\mod T^2$

or, said differently, $f'(0)=a$.

We shall denote the element of $TA[[T]]$ associated to $a$ by $[a]$ or $[a](T)$.

Now, let us give some examples:

• Consider the formal group law associated to $\widehat{\mathbf{G}_a}$—namely $F(X,Y)=X+Y$. Then, note that we can make this into a formal $\mathcal{O}_K$-module by just defining $[a](T)=aT$. Clearly this is an endomorphism of $\widehat{\mathbf{G}_a}$, and one can check easily that it defines the structure of a formal $\mathcal{O}_K$-module
• A much less trivial example is the following. Consider the multiplicative group $\widehat{\mathbf{G}_m}/\mathbb{Z}_p$ corresponding to the group law $F(X,Y)=X+Y+XY$. Then, we can define a formal group law associating to $a\in\mathbb{Z}_p$ the series $(1+T)^a-1$. Of course, one should explain what $(1+T)^a$ means for $a\notin\mathbb{Z}$. For such $a$ we define it as $\displaystyle \sum_{n=0}^{\infty}{a\choose n}T^n$ where $\displaystyle a\choose n$ denotes the usual $\displaystyle \frac{a(a-1)\cdots (a-(n-1))}{n!}$ which actually lies in $\mathbb{Z}_p$ since this element of $\mathbb{Q}_p$ has non-negative $p$-adic valuation. Once again, it’s a simple check that this really does define a formal $\mathbb{Z}_p$-module structure on $\widehat{\mathbf{G}_m}/\mathbb{Z}_p$.
• It is shown in Silverman’s Arithmetic of Elliptic Curves that the formal group law $\widehat{E}$ associated to an elliptic curve $E/A$ has the structure of a formal $A$-module.

But, there is, in fact, a natural way to build formal $\mathcal{O}_K$-modules which was first observed by Lubin and Tate in there aforementioned paper. Let us fix a uniformizer $\pi$ of $\mathcal{O}_K$ and let $S_\pi$ be the set of power series $f(T)\in T\mathcal{O}_K[[T]]$ subject to the following two conditions:

\begin{aligned}&(1)\qquad f(T)\equiv \pi T\mod T^2\\ &(2)\qquad f(T)\equiv T^q\mod \pi\end{aligned}

where, recall, $q$ was the size of the residue field of $K$.

The first result towards a proof of local class field theory is then the following:

Theorem 15: Associated to any $f\in S_\pi$ is a unique formal $\mathcal{O}_K$-module $F_f$ such that $[\pi](T)=f(T)$. Moreover, for any two $f,g\in S_\pi$ the formal $\mathcal{O}_K$-modules $F_f$ and $F_g$ are isomorphic.

The key to this proposition is the following exercise in algebra:

Lemma 16: Let $f,g\in S_\pi$ and let $\ell(X_1,\ldots,X_n)\in \mathcal{O}_K[[X_1,\ldots,X_n]]$ be a homogenous linear polynomial. Then, there is a unique $\varphi\in \mathcal{O}_K[[X_1,\ldots,X_n]]$ such that

1. $\varphi\equiv \ell\mod (X_1,\ldots,X_n)^2$
2. $\varphi(f(X_1),\ldots,f(X_n))=g(\varphi(X_1,\ldots,X_n))$

Proof: We build $\phi$ iteratively by constructing it modulo $(X_1,\ldots,X_n)^m$ for increasing $m$. Well, by assumption we must have that $\varphi\equiv \ell \mod (X_1,\ldots,X_n)$. So, assume that we have a $\varphi_m$ so that $\varphi_m(f(X_1),\ldots,f(X_n))\equiv g(\varphi_m(X_1,\ldots,X_n))$ modulo $(X_1,\ldots,X_n)^m$. We must build a unique $\varphi_{m+1}$ satisfying the obvious property.

We begin by noting that $\varphi_m(f(X_1),\ldots,f(X_n))\equiv g(\varphi(X_1,\ldots,X_n))$ modulo $\pi$. Indeed, since $f,g\in S_\pi$ this is merely the statement that

$\varphi_m(X_1^q,\ldots,X_n^q)=\varphi(X_1,\ldots,X_n)^q\mod \pi$

which is true since we’re in characteristic $p$. Thus, we see that modulo $(X_1,\ldots,X_n)^{m+1}$ we have that $\varphi_m(f(X_1),\ldots,f(X_m))$ and $g(\varphi_m(X_1,\ldots,X_m))$ must differ by $p\psi_{m+1}$ for some homogenous polynomial $\psi_{m+1}$ of degree $m+1$. Divide $\psi_{m+1}$ by $1-\pi^m$ (a unit in $\mathcal{O}_K$) and call this polynomial $\psi'_{m+1}$. Finally, define $\varphi_{m+1}$ to be $\varphi_m+\psi'_{m+1}$.

We now must check that $\varphi_{m+1}$ satisfies the desired property. Begin by noting that $g(T)$ is just $\pi T$ modulo $(T)^2$ that

$g(\varphi_{m+1}(X_1,\ldots,X_n))=g(\varphi_m(X_1,\ldots,X_n))+\pi \psi'_{m+1}(X_1,\ldots,X_n)\mod (X_1,\ldots,X_n)^{m+2}$

Similarly, since $f(T)=\pi T \mod (T)^2$ one can see that

$\psi'_{m+1}(f(X_1),\ldots,f(X_n))=\psi'_{m+1}(\pi X_1,\ldots,\pi X_n)=\pi^{m+1}\psi'_{m+1}(X_1,\ldots,X_n)\mod (X_1,\ldots,X_n)^{m+2}$

where we have used the fact that $\psi'_{m+1}$ is homogenous of degree $m+1$. So, finally we check that

\begin{aligned}\varphi_{m+1}(f(X_1),\ldots,f(X_n)) &= \varphi_m(f(X_1),\ldots,f(X_n))+\pi^{m+1}\psi'_{m+1}\mod (X_1,\ldots,X_n)^{m+2}\\ &\equiv g(\varphi_m(X_1,\ldots,X_n))+(\pi-\pi^{m+1})\psi'_{m+1}+\pi^{m+1}\psi'_{m+1}\mod (X_1,\ldots,X_n)^{m+2}\\ &\equiv g(\varphi_{m+1}(X_1,\ldots,X_n))\mod (X_1,\ldots,X_n)^{m+2}\end{aligned}

Thus, we have defined power series $\varphi_m$ of degree $m$ for all $m$ satisfying

$\varphi_m(f(X_1),\ldots,f(X_m))\equiv g(\varphi_m(X_1,\ldots,X_n))\mod (X_1,\ldots,X_n)^{m+1})$

Letting $\varphi(X_1,\ldots,X_n)$ be the unique power series having $\varphi_m$ modulo $(X_1,\ldots,X_n)^m$ we see that $\varphi$ is the desired power series. Moreover, it’s clear in each step of our construction that $\varphi_m$, and thus $\varphi$ is unique. $\blacksquare$

So, now we can proceed with Theorem 15:

Proof(Theorem 15): Let $f\in S_\pi$. By Lemma 8 there exists a unique $F_f(X,Y)\in \mathcal{O}_K[[X,Y]]$ such that $F_f(f(X),f(Y))\equiv f(F_f(X,Y))$ and $F_f(X,Y)\equiv X+Y\mod (X,Y)^2$. We claim that this is a formal group law. But, since $F_f(X,Y)\equiv X+Y\mod (X,Y)^2$ we must only check associativity and commutativity. But, this follows from the unicity in Lemma 16.

Specifically, to see that $F_f(X,F(Y,Z))=F_f(X,F(Y,Z))$ note that modulo $(X,Y,Z)^2$ both sides are $X+Y+Z$. Moreover, it’s easy to see that they commute with $f$ in the sense of Lemma 16 which implies, by unicity, that they are equal. A similar technique works to show that $F_f$ is commutative.

Now, to define the $\mathcal{O}_K$-module structure on $F_f$ note that, again, by Lemma 16 there exists a unique $[a](T)\in \mathcal{O}_K[[T]]$ such that modulo $(T)^2$ it is $aT$, and commutes with $f$. In particular, one sees that $[\pi](T)$ must be $f(T)$ itself. Checking that this actually defines the structure of a formal $\mathcal{O}_K$-module is again, just an exercise in the unicity part of Lemma 8 and is left to anyone interested in writing it all out.

Finally, we need to explain why two $f,g\in S_\pi$ give isomorphic formal $\mathcal{O}_K$-modules $F_f$ and $F_g$. Again, by Lemma 16 there is a unique $u(T)\in \mathcal{O}_K[[T]]$ such that $u\equiv T\mod (T)^2$ and $F_f(u(X),u(Y))=u(F_g(X,Y))$. One can check, using unicity, that this is actually a homomorphism of formal $\mathcal{O}_K$-modules $F_f\to F_g$. Moreover, since it’s leading coefficient is $1$, a unit, it must be an isomorphism. $\blacksquare$

Let us make the following observation. There is always a simple element of $S_\pi$: the polynomial $f(T)=\pi T+X^q$. That said, elements of $S_\pi$ can take much more complicated forms. For example, the power series $f(T)=(1+T)^p-1$ is also an example of an element of $S_p$ (working with $K=\mathbb{Q}_p$ and $\pi=p$). Indeed,

$f'(T)=p(1+T)^{p-1}$

so $f'(0)=p$ showing that $f(T)=pT\mod (T)^2$. But, also modulo $p$ we evidently have that $f(T)=T^p$ (using the fact that Frobenius is a ring map modulo $p)$. This shows, as a consequence of Theorem 7, that the formal $\mathbb{Z}_p$-module associated to the choice $\pi=p$ is $\widehat{\mathbf{G}_m}$ since this is a formal group law with $[p](T)=(1+T)^p-1$.

Note that we cannot modify the case for $K=\mathbb{Q}_p$ and $\pi=p$ to the general setting. Namely, one is tempted to think that for any such pair $(K,\pi)$ that the polynomial $f(T)=(1+T)^\pi-1$ (which can be given meaning similar to the case over $\mathbb{Z}_p)$ is an element of $S_\pi$. Unfortunately, this fails precisely because we have no ‘$\pi$-Frobenius’, so that modulo $\pi$ there is no reason why this should be anything nice, let alone $X^q$.

In fact, wanting $f(T)=(1+T)^\pi-1$ to work for the pair $(K,\pi)$ one will quickly realize that the only hope of proving that it is in $S_\pi$ is for $\pi$ to be a power of $p$. But, this would force $\pi$ to be exactly $p$. In other words, we need $K$ to be an unramified extension of $\mathbb{Q}_p$. But, then $f(T)=(1+T)^p-1$ won’t work again, since modulo $p$ it’s $T^p$ where it needs to be $T^{p^{[K:\mathbb{Q}_p]}}$. This peculiarity has semi-deep meaning that we will mention later.

## Step 2: Studying the extensions

So, let us start putting these formal groups $F_f$, with $f\in S_\pi$, to some use.

So, let’s fix $K$ a $p$-adic local field, $\pi$ a uniformizer of $\mathcal{O}_K$, and let, $f\in S_\pi$. Denote the formal group law by $F_f$ and the associated formal Lie group by $\mathbb{G}_f$. We define then the extension $K_{\pi,n,f}$ (temporary) denote the field obtained by adjoining to $K$ the elements of the $\pi^n$-torsion in $\mathbb{G}_f(\mathcal{O}_{\overline{K}})$.

Let us make this slightly more precise. Recall that the formal group law $F_f$ defines a group structure on the topologically nilpotent elements of the adic ring $\mathcal{O}_{\overline{K}}$. But, these elements bear a simple description in this case. Namely, the topologically nilpotent elements of $\mathcal{O}_{\overline{K}}$ are just the elements of $\mathfrak{m}_{\overline{K}}$—the maximal ideal. This follows immediately from considering the fact that $a\in\mathcal{O}_{\overline{K}}$ goes to zero if and only if its absolute value does.

Thus, we have that $\mathfrak{m}_{\overline{K}}$ together with the group operation

$\mathfrak{m}_{\overline{K}}^2\ni (x,y)\mapsto F_f(x,y)\in\mathfrak{m}_{\overline{K}}$

is a group. Moreover, note that it actually carries the structure of an $\mathcal{O}_K$-module by defining

$ax:= [a](x)$

which commutes with the group operations on $\mathfrak{m}_{\overline{K}}$ definitionally. This is the $\mathcal{O}_K$-module which we are denoting $\mathbb{G}_f(\mathcal{O}_{\overline{K}})$.

Thus, with this said, it’s reasonable to consider the set

$\mathbb{G}_f[\pi^n](\mathcal{O}_{\overline{K}})=\left\{x\in\mathfrak{m}_{\overline{K}}:\pi^n x=0\right\}$

and then consider the field extension $K_{\pi,n,f}:=K(\mathbb{G}_f[\pi^n](\mathcal{O}_{\overline{K}}))$. But, from what we said, there is no reason to assume that this field extension $K_{\pi,n,f}$ has any reasonable properties. In particular, it’s not even clear that $K_{\pi,n,f}$ is a finite extension. So, our first point of order should be determining its basic properties.

The key observation is that the $\mathcal{O}_K$-modules $\mathbb{G}_f[\pi^n](\mathcal{O}_{\overline{K}})$ and and the field extensions $K_{\pi,n,f}$ do not actually depend on the chosen $f\in S_\pi$. One part of this is obvious. Namely, we showed that we have an isomorphism of formal $\mathcal{O}_K$-modules $\mathbb{G}_f\xrightarrow{\approx}\mathbb{G}_g$ for any two $f,g\in S_\pi$. Thus, it follows that we have an isomorphism of $\mathcal{O}_K$-modules $\mathbb{G}_f[\pi^n](\mathcal{O}_{\overline{K}})\to\mathbb{G}_g[\pi^n](\mathcal{O}_{\overline{K}})$.

To see why the fields $K_{\pi,n,f}$ and $K_{\pi,n,g}$ are equal is slightly more complicated. The important fact is that even though the set $\mathbb{G}_f[\pi^n](\mathcal{O}_{\overline{K}})$ was defined in terms of the zeros of a power series, namely the power series $[\pi^n](T)$, they are actually the zeros of a polynomial. This follows from the $p$-adic analogue of the Weierstrass preparation theorem (see Gouvea’s $p$-adic Numbers: an Introduction Theorem 6.3.6).

Thus, $K_{\pi,n,f}$ and $K_{\pi,n,g}$ are the result of adjoining the zeros of these polynomials. But, since the modules $\mathbb{G}_f[\pi^n](\mathcal{O}_{\overline{K}})$ and $\mathbb{G}_g[\pi^n](\mathcal{O}_{\overline{K}})$ are isomorphic they will have polynomials with the same roots—being minimal polynomials as $\mathcal{O}_K$-modules. So, we will drop the $f$ in the notation and just denote the module $\mathbb{G}_f[\pi^n](\mathcal{O}_{\overline{K}})$ As $\mathbb{G}[\pi^n](\mathcal{O}_{\overline{K}})$ and $K_{\pi,n,f}$ as $K_{\pi,n}$.

This is useful, not only psychologically (i.e. we don’t have to make an arbitrary choice of $f$), but practically since we can now, in the proof of our theorems, assume we’re working with the element $f(T)=\pi T+T^q$ in $S_\pi$.

So, let us begin discovering the basic properties of the extensions $K_{\pi,n}$:

Theorem 17: The module $\mathbb{G}[\pi^n](\mathcal{O}_{\overline{K}})$ is isomorphic, as an $\mathcal{O}_K$-module, to $\mathcal{O}_K/(\pi^n)$.

Proof:  As noted above, we can assume that we’re working with the polynomial $f(T)=\pi T+T^q$. In particular, it’s clear that $\mathbb{G}[\pi^n](\mathcal{O}_{\overline{K}})$ is finite. So, evidently a finitely generated torsion $\mathcal{O}_K$-module. So, by the structure theorem for modules over PIDs we know that $\mathbb{G}[\pi^n](\mathcal{O}_{\overline{K}})$ is of the form

$\mathcal{O}_K/(\pi^{d_1})\times\cdots\times\mathcal{O}_K/(\pi^{d_m})$

for some integers $d_i$ with $d_i\mid d_{i+1}$.

Now, we note that $f(T)$ has $q$-distinct roots ($0$ and the $q-1$ distinct roots of $T^{q-1}=-\pi$). Thus, we see that $\mathbb{G}[\pi](\mathcal{O}_{\overline{K}})$ has to have size $q$, and so is evidently $\mathcal{O}_K/\pi$. Now, for $n\geqslant 1$ we have a short exact sequence

$0\to\mathbb{G}[\pi](\mathcal{O}_{\overline{K}})\to \mathbb{G}[\pi^n](\mathcal{O}_{\overline{K}})\to \mathbb{G}[\pi^{n-1}](\mathcal{O}_{\overline{K}})\to 0$

where the second map is multiplication by $\pi$. One should justify why the multiplication by $\pi$ map is actually surjective. It suffices to check that it’s surjective as a map on $\mathbb{G}(\mathcal{O}_{\overline{K}})$. But, this comes down to the statement that if $\alpha\in\mathfrak{m}_{\overline{K}}$ we want a solution to $[\pi](T)=\alpha$. In other words, we want a solution to $T^p-\pi T-\alpha=0$. But, it’s clear that the roots of this polynomial in $\overline{K}$ have positive valuation and so lie in $\mathfrak{m}_{\overline{K}}$.

The fact that $\mathbb{G}[\pi^n](\mathcal{O}_{\overline{K}})$ has size $q^n$ and is cyclic now follows by a simple induction. $\blacksquare$

Note that we can conclude from the above that

$\mathrm{Aut}_{\mathcal{O}_K}(\mathbb{G}[\pi^n](\mathcal{O}_{\overline{K}}))=\left(\mathcal{O}_K/(\pi^n)\right)^\times$

which has size $q^{n-1}(q-1)$.

We now seek to understand the structure of the field extension $K_{\pi,n}/K$:

Theorem 18: The field extension $K_{\pi,n}/K$ is a totally ramified Galois extension with Galois group $\left(\mathcal{O}_K/\pi^n\right)^\times$.

Proof: Begin by noting that any root of $[\pi^n](T)$ is a root of $[\pi^m](T)$ for $m\leqslant n$. This follows, with our choice of $[\pi](T)=f(T)=\pi T+T^q$ since $[\pi^m](T)$ is the $m$-fold composition of $[\pi](T)$, which has zero constant term.

Thus, we can create a filtration of $K_{\pi,n}/K$ as follows. First take $\alpha_1$ to be a no-zero root of $[\pi](T)=f(T)$ and consider $K[\alpha_1]\subseteq K_{\pi,n}$. Next, take a non-trivial root $\alpha_2$ of $f(T)-\alpha_2$. This root is contained in $K_{\pi,n}$ since

$f(f(\alpha_2))=f(\alpha_1)=0$

and thus we have the containment $K[\alpha_1]\subseteq K[\alpha_2]\subseteq K_{\pi,n}$. Now, take a root $\alpha_3$ of $f(T)-\alpha_2$. Again, this is contained in $K_{\pi,n}$ since

$f(f(f(\alpha_3)))=f(f(\alpha_2))=f(\alpha_1)=0$

Repeating this process we get a filtration

$K\subseteq K[\alpha_1]\subseteq K[\alpha_2]\subseteq\cdots\subseteq K[\alpha_n]\subseteq K_{\pi,n}$

Note that evidently $K[\alpha_1]/K$ is degree $q-1$. And, for $i\geqslant 1$ we have that $K[\alpha_{i+1}]/K[\alpha_i]$ is degree $q$. This follows since they were constructed from degree $q$ Eisenstein polynomials. Thus, we may conclude that $[K_{\pi,n}:K]\geqslant q^{n-1}(q-1)$.

On the other hand, note that $\mathrm{Gal}(K_{\pi,n}/K)$ acts faithfully on $\mathbb{G}(\mathcal{O}_{\overline{K}})[\pi^n]$. Moreover, it acts $\mathcal{O}_K$-linearly since the coefficients of $[a](T)$, $a\in\mathcal{O}_K$ are fixed by $\mathrm{Gal}(K_{\pi,n}/K)$ and this group acts continuously (so that we can distribute over the limit). Thus, we get an injection

$\mathrm{Gal}(K_{\pi,n}/K)\hookrightarrow \mathrm{Aut}_{\mathcal{O}_K}(\mathbb{G}(\mathcal{O}_{\overline{K}})[\pi^n]=\left(\mathcal{O}_K/\pi^n\right)^\times$

So, since the right hand side has size $q^{n-1}(q-1)$ we conclude by Galoisness $[K_{\pi,n}:K]\leqslant q^{n-1}(q-1)$, and thus from the above, we may conclude that $[K_{\pi,n}:K]=q^{n-1}(q-1)$ and that $\mathrm{Gal}(K_{\pi,n}/K)\cong \left(\mathcal{O}_K/(\pi^n)\right)^\times$.

Finally, to see that $K_{\pi,n}/K$ is totally ramified, note that $K_{\pi,n}$ certainly contains the roots of

$\displaystyle \frac{[\pi^n](T)}{[\pi^{n-1}](T)}=([\pi^{n-1}](T))^{q-1}+\pi$

(which can be proven easily by induction). But, this is a $\pi$-Eisenstein polynomial of degree $q^{n-1}(q-1)$, and thus $K_{\pi,n}/K$ must be the adjunction to $K$ of a root of this polynomial. But, since it’s $\pi$-Eisenstein, this implies that $K_{\pi,n}/K$ is totally ramified as desired. $\blacksquare$

In particular, we see that if we define $K_\pi$ to be the compositum(=union) of the $K_{\pi,n}$‘s then the above shows that $K_\pi/K$ is totally ramified extension with Galois group $\mathcal{O}_K^\times$. This is precisely the type of field that we are after—the fixed field of $\widehat{\mathbb{Z}}$.

Note that also implicit in the proof was the fact that $\pi$ is a norm from $K_{\pi,n}$ to $K$. Working harder (doing an analysis of the units) one can, in fact, show the following:

Theorem 19: The image of the norm map from $K_{\pi,n}$ to $K$ is the subgroup of $K^\times$ generated by $\pi$ and $1+\pi^n\mathcal{O}_K$. The image of the norm map from $K_\pi$ to $K$ is $\pi^{\mathbb{Z}}$.

The first part of this proposition is a serious amount of elbow grease, and so we omit it here. The second part follows from the first since the norms from $K_\pi$ are the intersection of the norms from $K_{\pi,n}$ as $n$ varies. In particular, we see that the fields $K_\pi$ are distinct for different $\pi$.

Let us remark now why it’s not a coincidence that we had issue replicating the creation of a polynomial like $f(T)=(1+T)^p-1$ that worked for other field extensions. Note that, as we observed before, the formal $\mathbb{Z}_p$-module associated to this $f$ is $\widehat{\mathbf{G}_m}$. Thus, the $p^n$-torsion is just $\mu_{p^n}$ and so we see that $K_{p,n}$ is just $\mathbb{Q}_p(\mu_{p^n})$. So, we see that this choice of $f$ gives us $K_p$ is $\mathbb{Q}_p(\mu_{p^\infty})$. And using the fact that $K^\mathrm{ab}=K_\pi K^\mathrm{ur}$ we see that this gives us precisely the local Kronecker-Weber theorem. So, it’s not shocking that we had trouble replicating the creation of such an $f$ since it’s existence is integrally linked to the Kronecker-Weber theorem, which shouldn’t hold for other $K$!

# Good references

So, as stated above, I provide below what I believe to be good sources to learn the above material (more in-depth) from:

## Background on number theory:

1. J.P. Serre’s Local Fields
2. Jürgen Neukirch’s Algebraic Number Theory

## Background on non-commutative algebra:

1. Tamás Szamuely and Phillipe Gille’s Central Simple Algebras and Galois Cohomology
2. Falko Lorenz’s Algebra: Volume II: Fields with Structure, Algebras and Advanced Topics
3. T.Y. Lam’s A First Course on Non-Commutative Rings
4. J.P. Serre’s Galois Cohomology (for non-commutative group cohomology)

## Local class field theory from cohomological point of view

1. Kazuya Kato, Takeshi Saito, and Nobushige Kurokawa’s Number Theory 2: an Introduction to Class Field Theory
2. J.P. Serre’s Local Fields
3. Georges Gras’s Class Field Theory
4. James Milne’s Class Field Theory
5. Jürgen Neukirch’s Class Field Theory

## Local class field theory from Lubin-Tate point of view

1. Kenkichi Iwasawa’s Local Class Field Theory
2. Emily Riehl’s undergraduate thesis Lubin-Tate Formal Groups and Local Class Field Theory
3. James Milne’s Class Field Theory