These are my live-TeXed notes for the course Math 223a: Algebraic Number Theory taught by Joe Rabinoff at Harvard, Fall 2012.

09/05/2012

TopIntroduction

This is a one-year course on class field theory — one huge piece of intellectual work in the 20th century.

Recall that a global field is either a finite extension of $\mathbb{Q}$ (characteristic 0) or a field of rational functions on a projective curve over a field of characteristic $p$ (i.e., finite extensions of $\mathbb{F}_q(t)$). A local field is either a finite extension of $\mathbb{Q}_p$ (characteristic 0) or a finite extension of $\mathbb{F}_q((t))$ (and sometimes we also include $\mathbb{C}$ and $\mathbb{R}$ as local fields) . The major goal of class field theory is to describe all abelian extensions of local and global fields $K$ (an abelian extension means a Galois extension with an abelian Galois group). Suppose $K^\mathrm{ab}\subseteq K^\mathrm{sep}$ is the maximal abelian extension of $K$, then $\Gal(K^\mathrm{ab}/K)\cong G_K^\mathrm{ab}$, the topological abelianization of the absolute Galois group $G_K=\Gal(K^\mathrm{sep}/K)$. Moreover, there is a bijection between abelian extensions of $K$ and closed subgroups of $G_K^\mathrm{ab}$. So we would like to understand the structure of $G_K^\mathrm{ab}$.

We also would like to know information about ramification of abelian extensions. For example, does $\mathbb{Q}$ have a degree 3 extension ramified only over 5? This can be nicely answered by class field theory. Class field theory also allows us to classify infinite abelian extensions via studying the topological group $G_K^\mathrm{ab}$. The course will start with lots of topological groups in the first week and one may be impressed by how seemingly unrelated to number theory at first glimpse.

Here some useful applications of class field theory.

Example 1 (Primes in arithmetic progressions) The famous Dirichlet theorem says that for an integer $m\ge2$, the primes $p\nmid m$ is equidistributed in $(\mathbb{Z}/m \mathbb{Z})^\times$. Notice that there is a canonical isomorphism between $\Gal(\mathbb{Q}(\zeta_m)/\mathbb{Q})$ and $(\mathbb{Z}/ m \mathbb{Z} )^\times$ and this isomorphism sends the Frobenius element associated to any $p\nmid m$ to $p\bmod{ m}$. So the classical theorem of Dirichlet can be viewed as a special case of the following Chebotarev's density theorem.
Theorem 1 Let $L/K$ be a finite Galois extension. Then the Frobenius elements (conjugacy classes) $\mathrm{Frob}_\mathfrak{p}$ for primes $\mathfrak{p} $ of $K$ are equidistributed on the conjugate classes of $\Gal(L/K)$.

Chebotarev's density theorem is proved via reducing to the case of cyclic extensions using a nice counting argument and then applying class field theory (cf. 34).

Example 2 (Artin $L$-functions) An Artin representation is a continuous representation $\rho: G_K\rightarrow GL(V)$ where $K$ is a number field and $V$ is a finite complex vector space. One can attach an Artin $L$-function to each Artin representation. When $V$ is one-dimensional, an Artin representation is simply a character of $G_K\rightarrow \mathbb{C}^\times$, which must factor through $G_K^\mathrm{ab}$. So a one-dimensional Artin representation is nothing but a character of $G_K^\mathrm{ab}$. By continuity, this character factors through $\Gal(L/K)$, where $L$ is a finite abelian extension of $K$.

Weber generalized the Dirichlet $L$-functions to Weber $L$-functions over any number fields. He proved that a Weber $L$-function has analytic continuation to the whole $\mathbb{C}$ and satisfies a functional equation. Class field theory then tells us that the Weber $L$-functions are exactly the one-dimensional Artin $L$-functions.

Example 3 Let $X$ be a smooth projective connected curve over a finite field and $K=K(X)$ be the function field of $X$. Then class field theory classifies all abelian covers of $X$. In particular, this gives the abelianization of the etale fundamental group $\pi_1(X)$ of $X$. The proof of Weil conjecture II uses it in an essential way.
Example 4 (Cohomology of $G_K$) In the second semester we will study the Tate global and local duality, Brauer groups and introduce all the cohomological machinery in order to prove class field theory.

Now we briefly turn to the main statements of class field theory. Class field theory gives Artin maps $\psi_K: \mathbb{A}_K^\times/K^\times\rightarrow G_K^\mathrm{ab}$ (in the global case) and $\psi_K:K^\times\rightarrow G_K^\mathrm{ab}$ and the kernel and image of the Artin maps can be described. The crucial thing is that the source of the Artin maps are intrinsic to the field $K$ (doesn't involve $K^\mathrm{sep}$). Moreover the Artin maps satisfy the local-global compatibility: the diagram $$\xymatrix{K_v^\times \ar[r] \ar[d]_{\psi_{K_v}}& \mathbb{A}_K^\times/K^\times \ar[d]^{\psi_K}\\ G_{K_v}^\mathrm{ab} \ar[r] & G_K^\mathrm{ab}}$$ commutes. In other words, class field theory is functorial in $K$. For a finite abelian extension $L/K$, the Artin map induces the relative Artin maps $\psi_{L/K}: \mathbb{A}_K^\times/K^\times\rightarrow \Gal(L/K)$ and $\psi_{L/K}: K^\times\rightarrow \Gal(L/K)$. They are surjective and the kernel is exactly the norm subgroups. We can furthermore read ramification data from the relative Artin maps. In the local case $\psi_{L/K}(\mathcal{O}_K^\times)$ is exactly the inertia group $L/K$ and $\psi_{L/K}(1+\mathfrak{m}_K^i)$ is exactly the $i$-th ramification group of $L/K$. In the global case, the ramification data can be extracted from the local-global compatibility. We will spend most of the first semester to state the class field theory and draw important consequences from it and devote the second semester to the proofs.

Here are a few words about the proofs of class field theory. The classical approach is to do the global case first, using cyclotomic extensions, Kummer extensions and Artin-Schreier extensions (in characteristic $p$) to fill up the absolute Galois group, and then derive the local case from it. The cohomological approach is to establish local class field theory using group cohomology and then "glue" the local Artin maps to obtain the global Artin maps. One of the advantage of the cohomological approach is that the local-global compatibility comes from the construction. We will take this approach in the second semester.

Finally we may also talk about explicit class field theory, i.e., finding explicit construction (e.g. as splitting field of polynomials) of abelian extensions. This is a highly open problem in general with several known cases:

  1. When $K=\mathbb{Q}$, $\mathbb{Q}^\mathrm{ab}=\mathbb{Q}^\mathrm{cyc}$, the union of all cyclotomic extensions of $\mathbb{Q}$. So the polynomials $X^n-1$ exhaust all abelian extensions of $\mathbb{Q}$. This the most satisfactory case.
  2. When $K=\mathbb{Q}(\sqrt{-d})$ is an imaginary quadratic field. The CM theory of elliptic curves assert that $K^\mathrm{ab}$ can be obtained essentially by adjoining all the torsion points on an elliptic curve with complex multiplication by $K$.
  3. When $K$ is a global function field, there is a theory of Drinfeld modules to obtain most abelian extensions of $K$ (apart from some ramification restriction).
  4. When $K$ is a nonarchimedean local field, Lubin-Tate theory tells that $K^\mathrm{ab}$ can be obtained by adjoining all torsion points of the Lubin-Tate formal groups.

Somehow adjoining torsion points of a group law is possibly the only known way to construct explicit class fields.

09/07/2012

TopGlobal fields

Today and next Monday we will review the basic notions we learned from Math 129, taking this opportunity to set up the notations.

Definition 1 A number field is a finite extension of $\mathbb{Q}$. It is an abstract field and have many embeddings into the complex numbers $\mathbb{C}$ (we do not specify one).
Definition 2 A global function field is a finite (separable) extension of $\mathbb{F}_q(t)$. It is a fact that the algebraic closure $k$ of $\mathbb{F}_p$ in a global function field $K$ is a finite field, called the field of constants. Equivalently, $K$ is the field of rational functions on a smooth projective geometrically connected curve $X$ over $k$, unique up to isomorphism. The geometrically connectedness ensures that $k$ is the field of constants.
Remark 1 If $K\rightarrow L$ is a finite homomorphism of function fields, then we have a corresponding map $X_L\rightarrow X_K$ of curves. In particular, a finite homomorphism $\mathbb{F}_q(t)\rightarrow K$ gives a map $X_K\rightarrow \mathbb{P}^1_{\mathbb{F}_q}$.
Definition 3 A global field is either a number field or a global function field.
Definition 4 Let $K$ be a global field. A place of $K$ is an equivalent class $v$ of nontrivial absolute values $|\cdot|$ on $K$. Two absolute values $|\cdot|$ and $|\cdot|'$ are equivalent if and only if they induce the same topology on $K$ under the metric $d(x,y)=|x-y|$, if and only if $|\cdot|'{}=|\cdot|^\alpha$ for $\alpha\in \mathbb{R}_{>0}$. The set of all places of $K$ is denoted by $V_K$. Suppose $v\in V_K$, then we have the completion $K_v$ of $K$ with respect to any absolute values $|\cdot|$ corresponding to $v$.
Definition 5 Let $|\cdot|$ be an absolute value.
  • $|\cdot|$ is called non-archimedean if $|x+y|\le \max(|x|,|y|)$
  • $|\cdot|$ is called archimedean if $|n|\rightarrow \infty$, when $n\rightarrow \infty$, $n\in \mathbb{Z}$.

It can be shown that these two properties are mutually exclusive.

Remark 2 If $K$ is a function field, then any absolute value $|\cdot|$ on $K$ is non-archimedean. If $K$ is a number field, let $r_1$ be the number of real embeddings $K\hookrightarrow \mathbb{R}$ and $2r_2$ be the number of complex embeddings $K\hookrightarrow \mathbb{C}$ (in conjugate pairs). Then there are exactly $r_1+r_2$ archimedean places.
Example 5 When $K=\mathbb{Q}$, there is only one archimedean place $\infty$ given by the usual absolute value and $\mathbb{Q}_\infty=\mathbb{R}$. For any prime number $p$, the $p$-adic absolute value $||\cdot||_p$ is defined on the generators of $\mathbb{Q}^\times\cong \{\pm1\}\times\bigoplus_p p^\mathbb{Z}$ by $$||p||_p=\frac{1}{p},\quad ||\ell||_p=1,\quad ||-1||_p=1.$$ In this way we have a bijection between the set of primes of $\mathbb{Q}$ and all non-archimedean places of $\mathbb{Q}$. The completion with respect to $||\cdot||_p$ is denoted by $\mathbb{Q}_p$.
Remark 3 Let $K\rightarrow L$ is a homomorphism of global fields, then the map $V_L\rightarrow V_K$ given by restriction is a surjection with finite fibers. For $w\in V_L$, we have $v=w|_K$ if and only $w\mid v$ ($w$ lies above $v$). In this case the homomorphism $K\rightarrow L$ extends uniquely to a continuous map $K_v\rightarrow L_w$.
Definition 6 Suppose $v$ is an archimedean place of $K$, then there exists an absolute value $|\cdot|_v$ such that $|n|_v=n$ for any $n\in \mathbb{Z}_{n\ge0}$. By the Gelfand-Mazur theorem, $K_v$ is either $\mathbb{R}$ or $\mathbb{C}$, so the absolute value $|\cdot|_v$ is the usual absolute value on $\mathbb{R}$ or $\mathbb{C}$. The normalized absolute value is often defined as $||\cdot||_v=|\cdot|_v$ (when $v$ is real) and $||\cdot||_v=|\cdot|_v^2$ (when $v$ is complex). This normalization simplifies many statements like the product formula.
Definition 7 Suppose $v$ is a non-archimedean place of $K$. Let $|\cdot|$ be a representative of $v$. Then $-\log|K^\times|\subseteq \mathbb{R}$ is discrete (Ostrowski's theorem), hence is equal to $\alpha \mathbb{Z}$ for some $\alpha\in \mathbb{R}_{>0}$. We normalized the valuation $\ord_v:K\rightarrow \mathbb{Z}\cup\{\infty\}$ such that $\ord_v(x)=-\frac{1}{\alpha}\log|x|$. This valuation is intrinsic to $v$ and is called the normalized valuation. It extends uniquely to $K_v$.
Definition 8 We denote $\mathcal{O}_v=\{x\in K_v: \ord_v(x)\ge0\}$, the valuation ring of $K_v$; $\mathfrak{m}_v=\{x\in \mathcal{O}_v: \ord_v(x)>0\}$, the maximal ideal of $K_v$ and $k_v=\mathcal{O}_v/\mathfrak{m}_v$, the residue field of $K_v$ (a finite field). We define the normalized $v$-adic absolute value to be $||x||_v=(\# k_v)^{-\ord_v(x)}$. It is equivalent to absolute value we started with.
Theorem 2 (Product formula) For $x\in K^\times$, $||x||_v=1$ for almost all places $v$ and $$\prod_{v\in V_K} ||x||_v=1.$$
Remark 4 In fact, the global fields can be characterized intrinsically using the product formula.
Theorem 3 (Artin-Whaples, 1945) Let $K$ be any field and $V$ be the set of places of $K$. Then $K$ is a global ield if and only if:
  1. There exists $||\cdot||_v$ for all places $v$ such that the product formula holds.
  2. The completion $K_v$ is locally compact for any $v$.

Notice that the second condition excludes the fields like $\mathbb{C}(t)$.

Definition 9 Let $S$ be a nonempty finite set of places of $K$ containing the set $S_\infty$ of archimedean places. The ring of $S$-integer is $\mathcal{O}_{K,S}=\{x\in K: ||x||_v\le1,\forall v\not\in S\}$. This is a Dedekind domain. For $v\not\in S$, we denote $\mathfrak{p}_v=\{x\in \mathcal{O}_{K,s}: ||x||_v<1\}=\mathcal{O}_{K,S}\cap \mathfrak{m}_v$, $\mathcal{O}_v=\varprojlim_n \mathcal{O}_{K,S}/\mathfrak{p}_v^n$ and $k_v=\mathcal{O}_v/\mathfrak{p}_v$.
Theorem 4 Every maximal ideal of $\mathcal{O}_{K,S}$ is of the form $\mathfrak{p}_v$ for $v\not\in S$.
Example 6 For $K=\mathbb{Q}$, $\mathcal{O}_{K,S}=\mathbb{Z}[\frac{1}{p}, p\in S]$. For any prime $p\not\in S$, $\mathcal{O}_p\cong \mathbb{Z}_p$.
Definition 10 Let $K$ be a number field. $\mathcal{O}_K=\mathcal{O}_{K,S_\infty}$ is called the ring of integers of $K$. It is the integral closure of $\mathbb{Z}$ in $K$.
Remark 5 Analogous to the case of $\mathbb{Q}$, we have a bijection between the maximal ideals of $\mathcal{O}_K$ and the non-archimedean places of $K$.
Remark 6 Let $K$ be a global field. Let $k_K$ be the constant field of $K$ and $X$ be the associated curve. We have a bijection between the places of $K$ and closed points of $X$. Any place is given by sending a rational function to the order of zero or pole at a fixed closed point. For a place $v$, the residue field $k_v$ is then the residue field $k(x)$ of the corresponding closed point $x$. When $S$ is empty, $\mathcal{O}_{K,\varnothing}$ is the constant field $k$. When $S$ is nonempty, $\mathcal{O}_{K,S}$ is the ring of regular functions on $X\setminus S$. In particular, $X\setminus S=\Spec\mathcal{O}_{K,S}$.
Example 7 Let $K=\mathbb{F}_q(t)$. Then taking the usual degree of rational function gives exactly the valuation associated to the closed point $\infty\in \mathbb{P}^1$. Moreover, $\mathcal{O}_{\mathbb{F}_q(t),\infty}=\mathbb{F}_q[t]$. A homomorphism $\mathbb{F}_q(t)\rightarrow K$ gives a map $\phi: X\rightarrow \mathbb{P}^1$. We know that $S_\infty=\phi^{-1}(\infty)$ and $\mathcal{O}_{K,S_\infty}$ is the integral closure of $\mathbb{F}_q[t]$ in $K$.

09/10/2012

Definition 11 Let $K$ be a global field and $S$ be a finite set of places of $K$ containing $S_\infty$. Then $\mathcal{O}_{K,S}$ is a Dedekind domain. We denote the group of fractional ideals of $\mathcal{O}_{K,S}$ by $I_{K,S}$. Then $I_{K,S}\cong\bigoplus_{\mathfrak{p}\not\in S} \mathfrak{p}^\mathbb{Z}$. Any $\alpha\in K^\times$ generates the principal fractional ideal $(\alpha)=\prod_{\mathfrak{p}_v}\mathfrak{p}_v^{\ord_v(\alpha)}$. Let $P_{K,S}\subseteq I_{K,S}$ be the subgroup of principal ideals. We define the $S$-class group of $K$ to be $\Cl_S(K)=I_{K,S}/P_{K,S}$.
Remark 7 When $K$ is a number field, $\Cl_S(K)$ a finite group and when $S=S_\infty$, we call $\Cl(K)=\Cl_{S_\infty}(K)$ the class group of $K$. In geometrical language, $\Cl_S(K)\cong \Pic(\Spec \mathcal{O}_{K,S})$. Since $\Cl(K)$ is finite (we will prove this fact using ideles), by killing the finitely many nontrivial ideal classes we know that $\Cl_S(K)=0$ for sufficient large $S$, i.e., $\mathcal{O}_{K,S}$ is a PID.
Remark 8 When $K$ is a global function field, $\Cl_S(K)=\Pic(X\setminus S)$. When $S=\varnothing$, $\Cl(K)=\Pic(X)$. The degree map of divisors induces a surjection $\Pic(X)\rightarrow \mathbb{Z}$, thus $\Pic(X)$ is not finite. Nevertheless, $\Pic^0(X)$ is always finite.

TopExtensions of global fields

Let $L/K$ be a finite extension of global fields and $S$ be a finite set of places of $K$ containing $S_\infty$. Let $T$ be the preimage of $S\subseteq V_K$. Then natural map $\mathcal{O}_{K,S}\rightarrow\mathcal{O}_{L,T}$ makes $\mathcal{O}_{L,T}$ a finite projective $\mathcal{O}_{K,S}$-module of rank $n=[L:K]$.

Fix a place $v\in V_K\setminus S$, we factorize $\mathfrak{p}_v\mathcal{O}_{L,T}=\prod_{i=1}^g \mathfrak{q}_i^{e_i}$. Then the $\mathfrak{q}_i$'s correspond to the places $w_i$ of $L$ restricting to $v$.

Definition 12 Denote $g=g(v)=g(L/v)$. The number $e_i$ is called the ramification index of $w_i/v$, denoted by $e(w_i/v)=e(w_i/K)=e(\mathfrak{q}_i)$. The number $f_i=[k_{w_i}: k_v]=f(w_i/v)$ is called the residue degree of $w_i/v$. We have $n=[L:K]=\sum_{i=1}^g e_if_i$.
Definition 13 $L/K$ is called unramified at $v$ if $e_i=1$ for all $i$'s, totally split at $v$ if $g=n$ and inert at $v$ if $g=1$ and $e_1=1$.
Remark 9 When $L/K$ is Galois, the $e_i$'s and $f_i$'s are the same, we write $e=e_i$ and $f=f_i$. So $n=efg$.

Now fix an archimedean place $v$. Then $[L_{w_i}: K_v]$ is either 1 or 2.

Definition 14 $w_i/v$ is called unramified if $[L_{w_i}: K_v]=1$ and ramified otherwise. Similarly we have $n=\sum_{i=1}^g[L_{w_i}:K_v]$.
Remark 10 For any place $v$, we have an isomorphism $L\otimes_K K_v\cong \prod_{w|v} L_w$ given by sending $x\otimes 1$ and $1\otimes x$ to the diagonal element $(x)_w$ under the embeddings $L\hookrightarrow L_w$ and $K_v\hookrightarrow L_w$.

Suppose $L/K$ is Galois with $G=\Gal(L/K)$, then $G$ acts on $V_L$ given by $g|\cdot|=|g^{-1}(\cdot)|$. This action is transitive on the fibers of $V_L\rightarrow V_K$.

Definition 15 The decomposition group $D(w)=D(w/v)=D(w/K)$ of $w$ is defined to be the stabilizer of $w$ in $G$. We have $gD(w)g^{-1}=D(gw)$. In particular, when $G$ is abelian the $D(w)$'s for all $w|v$ coincide, we simply denoted it by $D(v)$.
Remark 11 For $g\in D(w)$, $g$ acts on $L$ by isometries with respect to $||\cdot||_w$. So we obtain a map $D(w)\rightarrow \Aut_\mathrm{cont}(L_w/K_v)$. Its composition with the restriction map $\Aut(L_w/K_v)\rightarrow G$ is the identity map, thus $D(w)\cong\Aut_\mathrm{cont}(L_w/K_v)$. A simple counting shows that $[L_w:K_v]=ef=\#D(w)$, hence $L_w/K_v$ is also Galois.
Definition 16 When $v$ is archimedean and $w/v$ is ramified, we denote $\Gal(L_w/K_v)=\{1, \sigma_{w/v}\}$, where $\sigma_{w/v}$ is the complex conjugation. These complex conjugations are related by $g\sigma_{w/v}g^{-1}=\sigma_{gw/ v}$.
Definition 17 When $v$ is non-archimedean, $D(w)$ acts on the residue extension $k_w/k_v$ and gives a map $D(w)\rightarrow\Gal(k_w/k_v)$. This map is surjective and the kernel is called the inertia subgroup, denoted by $I(w)=I(w/v)=I(w/K)$. Similarly we have $gI(w)g^{-1}=I(gw)$. Counting shows that $\#I(w)=e$. So $w/v$ is unramified if and only if $I(w)=\{1\}$, if and only if $D(w)\cong\Gal(k_w/k_v)$.
Definition 18 Suppose $w/v$ is unramified. The generator $x\mapsto x^{\#k_v}$ of the cyclic group $D(w)\cong\Gal(k_w/k_v)$ is called the Frobenius, denoted by $\Frob_{w/v}$ (you may think it as the analogue of the complex conjugation $\sigma_{w/v}$. Similarly we have $g\Frob_{w/v}g^{-1}=\Frob_{gw}$. In particular, when $G$ is abelian we have a unique Frobenius attached to $v$, denoted by $\Frob_v$.
Remark 12 If $K\subseteq L\subseteq L'$ are finite Galois extension over $K$. Suppose $w'\in V_{L'}$, $w=w'|L$, $v=w'|K$. Then natural maps $\Gal(L'/K)\rightarrow \Gal(L/K)$, $D(w')\rightarrow D(w)$ and $I(w')\rightarrow I(w)$ are surjections and maps $\Frob_{w'/v}$ to $\Frob_{w/v}$ (when it makes sense). So the construction of the Frobenius element is functorial in the field extensions of $K$.

TopValued fields

Definition 19 A valued field is a pair $(K,|\cdot|)$, where $K$ is a field and $|\cdot|$ is a nontrivial absolute value on $K$. We endow $K$ with the metric $d(x,y)=|x-y|$. We say that $(K,|\cdot|)$ is complete if $K$ is complete under this metric.
Remark 13 Suppose $(K,|\cdot|)$ is a complete valued field and $L/K$ is a finite extension. Then there exists a unique extension of $|\cdot|$ to $L$, given by $|x|=|N_{L/K}(x)|^{1/[L:K]}$. Moreover, $(L,|\cdot|)$ is a complete valued field. In general, when $K$ is complete and $L/K$ is any algebraic extension, there is a unique extension of $|\cdot|$ to $L$. But $(L,|\cdot|)$ may not be complete.
Definition 20 A valued field extension is a map $(K,|\cdot|)\rightarrow (L,|\cdot|')$, i.e., a homomorphism $K\rightarrow L$ such that $|\cdot|'|_K=|\cdot|$.
Theorem 5 (Gelfand-Mazur) If $(K,|\cdot|)$ is a complete archimedean field. Then $K$ is isomorphic to either $\mathbb{R}$ or $\mathbb{C}$.
Definition 21 Let $(K,|\cdot|)$ be a non-archimedean valued field. The ring $\mathcal{O}_K=\{x\in K: |x|\le1\}$ is called the ring of integers. It is a valuation ring with valuation group contained in $\mathbb{R}$. Denote its fraction field by $K$. We similarly define the maximal ideal $\mathfrak{m}_K$ and the residue field $k=\mathcal{O}_K/\mathfrak{m}_K$.
Definition 22 We say $(K,|\cdot|)$ is discretely valued if $|K^\times|\subseteq \mathbb{R}_{>0}$ is discrete (equivalently, equal to $c^\mathbb{Z}$ for some $c\in(0,1)$). Any element $\pi\in K^\times$ such that $|\pi|=c$ is called a uniformizer. Let $\ord_K: K\rightarrow \mathbb{Z}\cup\{\infty\}$ be the unique valuation such that $\ord_K(\pi)=1$.

The famous Hensel's lemma holds for any valued fields (but the proof in this generality is different from the discrete valued case).

Theorem 6 (Hensel's lemma) Suppose $f\in \mathcal{O}_K[x]$ is monic and $\alpha_1\in \mathcal{O}_K$ such that $|f(\alpha_1)|<1$ and $|f'(\alpha_1)|=1$. Then there exists $\alpha\in \mathcal{O}_K$ such that $|\alpha-\alpha_1|<1$ and $f(\alpha)=0$.

09/12/2012

Definition 23 Let $(K,|\cdot|)$ be a complete discretely valued field and $(L,|\cdot|)$ be a complete discretely valued field extension of $(K,|\cdot|)$. Let $\pi_K$ be a uniformizer of $K$. We define $e=e(L/K)=\ord_L(\pi_K)$ to be the ramification index of $L/K$ and $f=[l: k]$ the residue degree. We say that $L/K$ is unramified if $e=1$ and $l/k$ is separable (so $L/K$ is separable), or equivalently, $\mathcal{O}_L/\mathcal{O}_K$ is etale. We say that $L/K$ is totally ramified if $f=1$. Notice that $e$ and $f$ are multiplicative in towers $K\subseteq L\subseteq L'$. So $L'/K$ is unramified (resp., totally ramified) if and only if $L'/L$ and $L/K$ are unramified (resp. totally ramified).
Remark 14 The functor $L\mapsto l$ restricts to an equivalence of categories between the category of finite unramified extensions $L/K$ and the category of finite separable extensions $l/k$. In particular we have $\Aut(L/K)=\Aut(l/k)$, thus $L/K$ is Galois if and only if $l/k$ is Galois. In particular, fixing $L/K$ a finite unramified extensions, the the separable closure of $k$ in $l$ corresponds to $L^\mathrm{ur}$, the maximal unramified sub-extension of $L/K$. $L^\mathrm{ur}$ contains all other unramified sub-extensions of $L/K$.
Remark 15 If $L/K$ is a finite totally ramified extension, then $L=K(\pi_L)$ and the minimal polynomial of $\pi_L$ is Eisenstein, i.e., it is of the form $x^e+a_{e-1}x^{e-1}+\cdots+a_0$, where $\ord_K(a_1),\cdots \ord_K(a_{e-1})\ge1$ and $\ord_K(a_0)=1$. When $\Char(k)\nmid e$, we can choose $\pi_L$ such that the minimal polynomial is $x^e-\pi_K$, i.e, $\pi_L^e\in K$ is the uniformizer of $K$.
Definition 24 We say $L/K$ is tamely ramified if $\Char(k)\nmid e(L/K)$ and $l/k$ is separable, and is wildly ramified otherwise. In particular, every unramified extension is tamely ramified.
Remark 16 If $L'/K$ is tamely ramified, then $L'/L$ and $L/K$ are both tamely ramified. Fixing an $L$, there exists a maximal tamely ramified sub-extension $L^\mathrm{t}\subseteq L$. In particular, $L^\mathrm{ur}\subseteq L^\mathrm{t}$ and the residue field of $L^\mathrm{t}$ is the separable closure of $k$ in $l$. The ramification index $e(L^\mathrm{t}/K)$ is then the prime-to-$p$ part of $e(L/K)$.

TopLocal fields

Definition 25 A local field is a locally compact complete valued field $(K,|\cdot|)$. So $(K,+)$ and $(K,\times)$ are locally compact Hausdorff topological group.
Remark 17 In fact the completeness can be deduced from the locally compactness.
Remark 18 When $K$ is archimedean, then $K$ is either $\mathbb{R}$ or $\mathbb{C}$ by Gelfand-Mazur's theorem.
Remark 19 When $K$ is non-archimedean complete field. Then the requirement that $K$ is locally compact holds if and only $K$ is discretely valued with finite residue field. When $\Char(K)=0$, $K$ must be a finite extension of $\mathbb{Q}_p$. When $\Char(K)>0$, the residue field $k$ embeds into $K$ and choosing a uniformizer of $K$ gives an isomorphism $K\cong k((t))$.
Remark 20 Let $L/K$ be a finite extension of non-archimedean local fields with $n=[L:K]$. Then for any $x\in K$, $$||x||_L=(\#l)^{-\ord_L(x)}=(\#k)^{-fe\ord_K(x)}=||x||_K^n.$$ So $||\cdot||_L=|\cdot|^n$, where $|\cdot|$ is the absolute value of $L$ extending $K$. In other words, the normalized absolute value is not functorial with respect to inclusion of fields. On the other hand, for $x\in L$, $$||x||_L=|x|^n=(||N_{L/K}(x)||_K^{1/n})^n=||N_{L/K}(x)||_K.$$ Thus the normalized absolute value is functorial with respect to the norm. The similar statement is true for archimedean local fields.
Remark 21 Let $K$ be a non-archimedean local field. Then we have an exact sequence $$0\rightarrow \mathcal{O}_K^\times\rightarrow K^\times\rightarrow \mathbb{Z}\rightarrow0.$$ Any choice of a uniformizer of $K$ splits this exact sequence and gives a decomposition $K^\times\cong \mathcal{O}_K^\times \times \mathbb{Z}$. For archimedean local fields, we have the decompositions $\mathbb{R}^\times\cong\{\pm1\}\times \mathbb{R}_{>0}$ and $\mathbb{C}^\times\cong S^1\times \mathbb{R}_{>0}$.
Definition 26 Let $K$ be a non-archimedean local field and $L/K$ be a finite Galois extension with $G=\Gal(L/K)$. $G$ acts on $L$ by isometries, so $G$ preserves $\mathfrak{m}_L^i$ for any $i$ and the action induces a map $G\rightarrow\Aut(\mathcal{O}_L/\mathfrak{m}_L^{i})$. We define $G_i$ to be the kernel of the map $G\rightarrow\Aut(\mathcal{O}_L/\mathfrak{m}_L^{i+1})$, called higher ramification groups (in the lower numbering). The higher ramification groups give a filtration $$G=G_{-1}\supseteq G_0\supseteq G_1\supseteq\cdots\supseteq \{1\}.$$
Definition 27 $G_0=\ker G\rightarrow \Aut(l/k)$ is called the inertia group. We have $G/G_0=\Gal(L^\mathrm{ur}/K)$, $G_0=\Gal(L/L^\mathrm{ur})$. In particular, $\#G_0=e(L/K)$ and $L/L^\mathrm{ur}$ is totally ramified.
Definition 28 The fixed field of $G_1$ is the maximal tamely ramified sub-extension $L^\mathrm{t}$ and $e(L/L^\mathrm{t})$ is the prime-to-$p$ part of $e(L/K)$. $G_0/G_1\cong \Gal(L^\mathrm{t}/L^\mathrm{ur})$ is called the tame inertia group and $G_1\cong\Gal(L/L^\mathrm{t})$ is called the wild inertia group. $G_0/G_1$ is cyclic of order prime to $p=\Char(k)$ and $G_1$ is the unique $p$-Sylow subgroup of $G_0$.
Remark 22 For $i\ge1$, $G_i/G_{i+1}$ is an elementary $p$-group. It follows that $G$ is solvable.
Definition 29 For $u\ge-1$ we define a function $\phi(u)=\int_0^u\frac{dt}{[G_0:G_t]}$, where $G_u=G_{\lceil u\rceil}$, $[G_0:G_{-1}]:=[G_{-1}:G_0]^{-1}$ and $[G_0:G_t]:=1$ for $t\in (-1,0]$. Then $\phi(u)$ is a piecewise linear continuous increasing convex-up function. We define $\psi(v)$ to be the inverse function of $\phi(u)$ and define the higher ramification groups (in the upper numbering) $G^v=G_{\phi(v)}$. Then $G^{-1}=G_{-1}=G$ and $G^0=G_0$, and $G^v=1$ for $v\gg 0$. For more details, see Chapter 4 of Serre's local fields.
Remark 23 Supoose $L'$ is a sub-extension of $L$, then the restriction map $\Gal(L/K)\rightarrow \Gal(L'/K)$ maps $G_u(L/K)$ to $G_v(L'/K)$, where $v=\psi(u)$. So the upper numbering behaves better than the lower numbering under extensions: Herbrand's theorem says that for any $H\lhd G$, we identify $\Gal(L^H/K)=G/H$, where $L^H$ is the fixed field of $H $, then $(G/H)^v=G^vH/H$.

TopTopological groups

Definition 30 A continuous map $f:X\rightarrow Y$ of topological space is called proper if $f^{-1}(C)$ is compact for any $C $ compact, open if $f(U)$ is open for any $U$ open, closed if $f(C)$ is closed for $C $ any closed.
Exercise 1
  1. A continuous map between locally compact Hausdorff spaces is proper if and only if it is closed with compact fiber.
  2. If $f:X\rightarrow Y$ and $f':X'\rightarrow Y'$ are proper and $Y\times Y'$ is Hausdorff, then $f\times f':X\times X'\rightarrow Y\times Y'$ is proper.
  3. If $f:X\rightarrow Y$ and $f':X'\rightarrow Y'$ are open, then $f\times f':X\times X'\rightarrow Y\times Y'$ is open.
Remark 24 The last statement is not true when "open" is replaced by "closed": this is one place where the properness can help us.

09/14/2012

Definition 31 A topological group is a group with a topology under which the multiplication and the addition are continuous.

The following basic properties of topological groups are easy exercises.

Exercise 2 Let $G$ be a topological group.
  1. If $H<G$ is a subgroup, the the closure $\bar H$ is also a subgroup. If $H $ is normal, so is $\bar H$.
  2. Any open subgroup of $G$ is closed.
  3. Any finite index closed subgroup of $G$ is open.
  4. An open subgroup of a compact group has finite index.
  5. $G$ is Hausdorff if and only if $\{e\}$ is closed.
Definition 32 For any subgroup $H<G$, we endow the coset space $G/H$ the quotient topology, i.e., $\bar U\subseteq G/H$ is open if and only if $\pi^{-1}(\bar U)$ is open. By definition, if $f: G\rightarrow G'$ is continuous homomorphism, then $G/\ker(f)\rightarrow G'$ is continuous.
Remark 25 If $H $ is furthermore normal, then $G/H$ is a topological group. If $H $ is normal and closed, then $G/H$ is Hausdorff. If $G$ locally compact Hausdorff and $H $ is normal and closed, then $G/H$ is also locally compact Hausdorff. In this case, $G\rightarrow G/H$ is proper if and only if it is closed and $H $ is compact.
Exercise 3
  1. Let $f: G\rightarrow G'$ be a surjective continuous homomorphism. Then $f$ is a quotient map if and only if $f$ is open.
  2. Let $f: G\rightarrow G'$ be a closed continuous homomorphism of Hausdorff topological groups, then $f$ is open if and only if $f(G)$ is open.
  3. Let $f: G\rightarrow G'$ be a surjective continuous homomorphism and $H=\ker (f)$. Suppose $s: G'\rightarrow G$ is a continuous homomorphism such that $f\circ s=\Id_{G'}$. Then $(h,g')\mapsto h\cdot s(g'): H\times G'\rightarrow G$ is a homeomorphism and $f$ is a quotient map. In particular, when $G$ is abelian, we have an isomorphism of topological groups $H\times G'\cong G$.
Definition 33 Let $G$ be a locally compact Hausdorff topological group. A Haar measure on $G$ is a nonzero Radon measure $\mu$ on $G$ such that $\mu(aU)=\mu(U)$ for any $a\in G$ and any measurable subset $U$ of $G$.
Theorem 7 (Haar) The Haar measure exists and is unique up to scalar multiplication.
Remark 26 Recall that the Borel $\sigma$-algebra on $G$ is the smallest set of subsets of $G$ that containing all the closed sets and open sets and is closed under countable union and intersection. A Haar measure takes a Borel set $U$ in the Borel $\sigma$-algebra to $\mu(U)\in [0,\infty]$, called the measure of $U$, satisfying
  1. $\mu(K)$ is finite if $K$ is compact.
  2. $\mu(U)>0$ if $U$ is open.
  3. $\mu$ is additive with respect to countable disjoint union.
Remark 27 If $G\rightarrow \mathbb{C}$ is a nice (e.g. continuous with compact support) function, then the integral $\int_Gf(x)d\mu(x)$ makes sense. In particular, $$\int_G\chi_U(x)d\mu(x)=\mu(U),$$ for the characteristic function $\chi_U$ of $U$. The left-invariance of the Haar measure implies that $$\int_Gf(ax)d\mu(x)=\int_Gf(x)d\mu(a^{-1}x)=\int_Gf(x)d\mu(x).$$
Example 8 The Lebesgue measure on $\mathbb{R}$ satisfies $\mu([a,b])=b-a$ and is a Haar measure. The standard measure on $\mathbb{C}\cong\mathbb{R}^2$ is a Haar measure.
Example 9 Let $K$ be any local field and $\mu$ be a Haar measure on $(K,+)$. For $a\in K^\times$, set $\mu_a(U)=\mu(aU)$, then $\mu_a$ is also a Haar measure. It follows from Haar's theorem that $\mu_a=c\cdot \mu$, where $c$ is a constant.
Lemma 1 $\mu(aU)=||a||_K\cdot \mu(U)$.
Proof The archimedean case is obvious. Suppose $K$ is non-archimedean. Since $\mathcal{O}_K$ is compact open, we know that $\mu(\mathcal{O}_K)$ is a positive real number. Thus it is enough to show that $\mu(a\cdot \mathcal{O}_K)=||a||_K\cdot\mu(\mathcal{O}_K)$, Replacing $a$ by $a^{-1}$, we may assume that $a\in \mathcal{O}_K$. It easy to see (via the filtration $\{\mathfrak{m}_K^n\}$) that $\mathcal{O}_K/(a)=(\#k)^{\ord(a)}=||a||_K^{-1}$. Hence $\mathcal{O}_K$ is a disjoint union of $||a||_K^{-1}$ cosets of $\mathcal{O}_K/(a)$. Therefore $\mu(\mathcal{O}_K)=||a||_K^{-1}\mu(a \mathcal{O}_K)$ using the left-invariance of $\mu$.
Corollary 1 Let $K$ be local field and $\mu$ be a Haar measure. Then $d\mu/||\cdot||_K$ is a Haar measure on $K^\times$, i.e., $f\mapsto \int_{K^\times}f(x)\frac{d\mu(x)}{||x||_K}$ is left-invariant.
Remark 28 Let $0\rightarrow K\rightarrow G\rightarrow H\rightarrow0$ be an exact sequence of locally compact Hausdorff topological groups, where $H $ is endowed with the quotient topology and $K$ is endowed with the subspace topology. Let $\mu_H$ and $\mu_K$ be Haar measures on $H $ and $K$ respectively. Then there exists a unique Haar measure on $G$ such that $$\int_Gf(x)d\mu(x)=\int_H\int_K f(zy)d\mu_K(y)d\mu_H(\bar z),$$ in other words, $\bar z \mapsto \int_Kf(zy)d\mu_K(y)$ is a measurable function on $H $, independent on the choice of a lift $z$ of $\bar z\in H$.

TopProfinite groups

Definition 34 A profinite group is a (filtered) inverse limit of finite groups $G=\varprojlim G_i$. We endow $\prod G_i$ with the product topology, which makes it a compact and Hausdorff topological group. Then $G\subseteq \prod G_i$ is a closed subspace, hence is also compact and Hausdorff and becomes a topological group under the subspace topology (equivalently, the weakest topology such that the projections $\pi_i: G\rightarrow G_i$ are continuous). Moreover, if $g\ne g'$ then there exists a projection $\pi_i(g)\ne \pi_i(g')$, hence $\pi_i^{-1}(\pi_i(g))$ and $\pi_i^{-1}(\pi_i(g))$ are disjoint open and closed subsets, we conclude that $G$ is totally disconnected.

In fact, we can characterize the profinite groups topologically as follows.

Theorem 8 A topological group $G$ is profinite if and only if it is compact and totally disconnected. In this case, $G\cong \varprojlim G/U$, where $U$ runs over all open normal subgroups of $G$
Remark 29 So any open subgroup of a profinite group $G$ has finite index. Any closed subgroup $H\subseteq G$ is compact and totally disconnected, hence is also profinite. If $H $ is furthermore normal, then $G/H$ is also profinite.
Remark 30 If $f: G\rightarrow G'$ is a continuous homomorphism where $G$ is profinite and $G'$ is Hausdorff. Then $f$ is closed (since $G$ is compact). Hence $f$ is open if and only if $f(G)$ is open. If $f$ is surjective, then $f$ is a quotient map and $G'$ is also profinite.
Remark 31 If $X$ is compact and Hausdorff, then weaker notion of totally connectedness that subsets all have one point, is equivalent to the stronger notion that every 2 points are separated by disjoint clopen subsets.
Definition 35 Let $G$ be any group. Then the profinite completion of $G$ is defined to be $\hat G:=\varprojlim G/H$, where $H $ runs over all normal subgroup of finite index of $G$. $\hat G$ is profinite and the natural map $G\rightarrow \hat G$ has dense image. Any homomorphism of $G\rightarrow G'$, for $G'$ profinite, factors through $G\rightarrow \hat G$.
Example 10 Let $K$ be a field, then $G_K=\Gal(\bar K/K)$ is a profinite group. Hence any homomorphism $\mathbb{Z}\rightarrow G_K$ factors as $\mathbb{Z}\rightarrow \hat{\mathbb{Z}}\rightarrow G_K$.
Example 11 $\hat{\mathbb{Z}}\cong\prod_p \mathbb{Z}_p$.
Example 12 For any non-archimedean local field $K$, $\mathcal{O}_K$ and $\mathcal{O}_K ^\times$ are profinite.
Example 13 If $\{G_i\}_{i\in I}$ are profinite, then the product $\prod_{i\in I}G_i$ is also profinite.

09/17/2012

TopInfinite Galois theory

Definition 36 Let $K$ be a field. We say $K$ is separably closed if there is no finite separable extension of $K$. We define the separable closure of $K$ to be an algebraic field extension of $K$ which is separably closed. Any two separable closure are isomorphic.
Definition 37 A separable extension $L/K$ is Galois if any of the following equivalent conditions hold
  1. Any two embeddings from $L$ to $K^\mathrm{sep}$ has the same image.
  2. The image of any embedding $L\rightarrow K^\mathrm{sep}$ is stable under $\Aut(K^\mathrm{sep}/K)$.
  3. The minimal polynomial of any $x\in L$ splits completely in $L$.
  4. $L$ is the union of $L'$'s, where every $L'/K$ is finite Galois.

In this case, the Galois group is $\Gal(L/K)=\Aut(L/K)$.

Definition 38 The absolute Galois group of $K$ is defined to be $G_K=\Gal(K^\mathrm{sep}/K)$.
Remark 32 If $L/K$ is Galois and $L'$ is any Galois finite subextension, then we have a surjection $\Gal(L/K)\rightarrow\Gal(L'/K)$. Since $L=\cup L'$, we know that $\Gal(L/K)=\varprojlim \Gal(L'/K)$. In particular, any Galois group is profinite. We call the profinite topology on a Galois group the Krull topology.
Theorem 9 (infinite Galois theory) Suppose $L/K$ is Galois. Then there is a bijection between subextension $L'$ and closed subgroups of $\Gal(L/K)$ under the Krull topology given by $L'\mapsto \Gal(L/L')$ and $L^H\mapsfrom H$. Moreover, $L'/K$ is Galois if and only if $\Gal(L/L')\lhd \Gal(L/K)$ is normal. In this case, we have $\Gal(L/K)/\Gal(L/L')\cong\Gal(L'/K)$ as topological groups.
Definition 39 We say $L/K$ is abelian if it is Galois with abelian Galois group $\Gal(L/K)$. In this case, it is easy to see that any subextension of $L/K$ is abelian.
Definition 40 Let $G$ be a topological group. Then $\overline{[G,G]}$ is a normal topological subgroup. The topological abelianization of $G$ is defined to be the Hausdorff topological group $G^\mathrm{ab}:=G/\overline{[G,G]}$, i.e., the maximal Hausdorff abelian quotient of $G$.
Remark 33 Suppose $L/K$ is Galois with $G=\Gal(L/K)$. Let $L'$ be the fixed field of $\overline{[G,G]}$. Then $L'/K$ is Galois with Galois group $G^\mathrm{ab}$. Suppose $L''$ is abelian, then the kernel of $G\rightarrow \Gal(L''/K)$ is closed and contains $[G,G]$. Thus $L''\subseteq L'$, i.e., $L'$ is the maximal abelian extension of $K$. When $L=K^\mathrm{sep}$, we write this maximal abelian extension as $K^\mathrm{ab}$.
Remark 34 Suppose $K\subseteq K'$ and $L/K$ and $L'/K'$ are Galois extensions. Let $L\rightarrow L'$ be a $K$-algebra homomorphism. Then the restriction gives a continuous map $\Gal(L'/K')\rightarrow \Gal(L/K)$. In particular, a continuous map $G_K\rightarrow G_{K'}$ depending on the embedding $K^\mathrm{sep}\rightarrow (K')^\mathrm{sep}$. This continuous map is determined up to conjugation by an element of $G_K$. By passing to the abelianization, the map $G_K^\mathrm{ab}\rightarrow G_K^\mathrm{ab}$ is unique. So the association $K\mapsto G_K^\mathrm{ab}$ is functorial with respect to $K$ and is intrinsic to $K$. In view of the Langlands program, this explains the study of the nonabelian structure of $G_K$ via representation theory, which is insensitive to conjugation. The geometrical situation is analogous: the (etale) fundamental group depends on the choice of the base point, hence is determined only up to conjugation. On the other hand, its abelianization (the first homology group) is intrinsic.

Now let $K$ be a global field and $L/K$ be a (possibly infinite) Galois extension.

Lemma 2 For any $v\in V_K$, there exists a place $w$ of $L$ such that $w|v$ and $\Gal(L/K)$ acts transitively on such $w$.
Proof The finite case is known. Assume that $L/K$ is infinite. We have a $K$-algebra homomorphism $L\rightarrow (K_v)^\mathrm{sep}$ induced by $K^\mathrm{sep}\rightarrow (K_v)^\mathrm{sep}$. There is a unique absolute value $|\cdot|$ on $(K_v)^\mathrm{sep}$ extending $||\cdot||_v$ and restricting $|\cdot|$ to $L$ gives a $w$. Now suppose $w$ and $w'$ are two such places of $L$. Then $\{g\in \Gal(L'/K)\colon g(w|_{L'})=w'|_{L'}\}$ is nonempty. So the inverse limit with respect to finite Galois extensions $L'$ is again nonempty.
Definition 41 The decomposition group at $w$ of $L/K$ is defined to be the stabilizer $D(w/v)=\{g\in \Gal(L/K): gw=w\}=\varprojlim D(w|_{L'}/v)$. This is a closed subgroup of $\Gal(L/K)$.
Definition 42 Suppose $v$ is non-archimedean. Let $k_w$ be the the residue field of $(L,|\cdot|)$ representing $w$. This is an algebraic extension of $k_v$. Then $D(w/v)$ acts on $k_w/k_v$ and give a continuous homomorphism $D(w/v)\rightarrow \Gal(k_w/k_v)$. This map at each finite level is surjective, hence the image is dense. But $D(w/v)$ is compact, so this map is actually surjective. The kernel $I(w/v)$ of this map is called the inertia group. It is a closed subgroup of $D(w/v)$ and is equal to $\varprojlim I(w|_{L'}/v)$. We say $v$ is unramified in $L$ if $I(w/v)=\{1\}$ for any $w|v$. In this case, by surjectivity, we have a Frobenius element $\Frob_{w/v}\in D(w/v)$ whose image is $x\mapsto x^{\# k_v}\in \Gal(k_w/k_v)$. When $L/K$ is abelian, this doesn't depend on $w$ and we have a Frobenius element $\Frob_v\in D(v)$.
Remark 35 Any $g\in D(w/v)$ acts by isometries on $L$, hence we have a homomorphism $D(w/v)\rightarrow \Aut_\mathrm{cont}(L_w/K_v)$. But in general, $L_w/K_v$ is not algebraic (and Galois groups don't make sense). For example, when $L=\overline{\mathbb{Q}}$, $L_v=\mathbb{C}_p$ is not algebraic over $\mathbb{Q}_p$. In other words, $\overline{\mathbb{Q}_p}$ is not complete.
Lemma 3 Define $L_w^\mathrm{alg}=\{x\in L_w: x \text{ is separable and algebraic over } K_v\}$. Then $L_w^\mathrm{alg}=\cup K_{v'}'$, where $K'$ runs over all finite separable extensions of $K$. Moreover, $L_w^\mathrm{alg}/K_v$ is Galois and the composition $$D(w/v)\hookrightarrow \Aut_\mathrm{cont}(L_w/K_v)\twoheadrightarrow \Gal(L_w^\mathrm{alg}/K_v)$$ is an isomorphism of topological groups, with all maps being bijective.
Proof Let $x\in L_w^\mathrm{alg}$, there exists $x_1,x_2,\ldots \in L$ such that $x_i\rightarrow x$. By Krasner's lemma, for $i\gg0$, $K_v(x)\subseteq K_v(x_i)$. Set $K'{}=K(x_i)$. Then $K_v'\supseteq K_v(x_i)\supseteq K_v(x)$. Now since $L=\cup K'$, where $K'$ runs over finite Galois subextensions, we know that any $K'_{v'}/K_v$ is Galois, hence $L_w^\mathrm{alg}/K_v$ is Galois. The composition map $D(w/v)\rightarrow \Gal(L_w^\mathrm{alg}/K_v)$ is an inverse limit of isomorphisms, hence is a topological isomorphism. The injectivity of $\Aut_\mathrm{cont}(L_w/K_v)$ follows from the fact that $L_w^\mathrm{alg}\supseteq L$ is dense in $L_w$.

09/19/2012

Theorem 10 Suppose $L=K^\mathrm{sep}$. Then $L_w^\mathrm{alg}=K_v^\mathrm{sep}$.
Proof By definition, $L_w^\mathrm{alg}\subseteq K_v^\mathrm{sep}$. Let $f\in K_v[X]$ be a monic irreducible separable polynomial and $\alpha$ be a root of $f$. Let $g\in K[X]$ with coefficients close to $f$ and the same degree as $f$. Then $|g(\alpha)|=|g(\alpha)-f(\alpha)|$ is small. On the other hand, write $g(X)=\prod (X-\beta_j)$, where $\beta_j\in \overline{K_v}$. Then $|g(\alpha)|=\prod|\alpha-\beta_j|$ can be made as small as possible. Now Krasner's lemma implies that $K_v(\alpha)\subseteq K_v(\beta_j)$. Comparison of degrees shows that $K_v(\alpha)=K_v(\beta_j)$. So $g$ is separable and irreducible over $K_v$, hence over $K$. We conclude that $K(\beta_j)\subseteq K^\mathrm{sep}$, thus $\alpha\in K_v(\beta_j)\subseteq L_w^\mathrm{alg}$.
Corollary 2 $D(w/v)\cong G_{K_v}$.
Remark 36 The inverse isomorphism can be described as follows, by extending continuously, we obtain $G_{K_v}\rightarrow \Aut_\mathrm{cont}(L_w/K_v)$, then restricting gives $\Aut_\mathrm{cont}(L_w/K_v)\rightarrow \Gal(K^\mathrm{sep}/K)$. The composition maps $G_{K_v}$ isomorphically to $D(w/v)\subseteq G_K$.
Remark 37 Suppose $L\subseteq K^\mathrm{sep}$ is Galois. Then the surjection $G_K\rightarrow \Gal(L/K)$ carries $D(w'/v)$ (resp. $I(w'/v)$) surjectively to $D(w/v)$ (resp. $I(w/v)$), where $w=w'|_L$ and $w'$ is a place of $K^\mathrm{sep}$.
Remark 38 Since $G_{K_v}\cong D(w'/v)$ maps surjectively to $D(v)\subseteq G_K^\mathrm{ab}$, we know that $G_{K_v}^\mathrm{ab}$ maps surjectively to $D(v)\subseteq G_K^\mathrm{ab}$. Is this injective? The injectivity simply means that every abelian extension of a local field is obtained as the completion of an abelian extension of some global field (the same statement without the adjective "abelian" follows from the previous corollary). The answer is yes, but not obvious (we will need the power of class field theory).
Exercise 4 Let $L/K$ be a finite separable extension of local fields. Then
  1. $K^\times\hookrightarrow L^\times$ is a homeomorphism onto a closed subgroup of $L^\times$.
  2. For $n\ge1$, $x\mapsto x^n$ is continuous, proper, and it is open if $\Char(K)=0$ or $\Char(K)\nmid n$.
  3. The norm map $N_{L/K}: L^\times\rightarrow K^\times$ is continuous, proper, and it is open if $\Char(K)=0$ or $\Char(K)\nmid [L:K]$.
Remark 39 In fact the following general theorem holds, which is much harder to prove.
Theorem 11 $N_{L/K}$ is always open.
Proof $N_{L/K}$ is proper, hence is closed. It is enough it show that $N_{L/K}(L^\times)$ is open is $K^\times$ by Exercise 3. Since we already know that the image is closed, it suffices to show that $N_{L/K}(L^\times)$ is of finite index. This follows from the norm-index formula. In fact, homological methods can show that $[K^\times: N_{L/K}(L^\times)]\mid [L:K]$ (and equality holds if and only $L/K$ is abelian).

The unramified case is relatively simpler.

Lemma 4 If $L/K$ is unramified, then $N_{L/K}(\mathcal{O}_L^\times)=\mathcal{O}_K^\times$. Hence $[K^\times: N_{L/K}(L^\times)]=[L:K]$.
Proof $N(\mathcal{O}_L^\times)\subseteq \mathcal{O}_K^\times$ is clear. Since the image is closed in $\mathcal{O}_K^\times$, it suffices to show that the image is dense. Any $g\in \Gal(L/K)$ acts by isometry on $L$, $g(1+\mathfrak{m}_L^r)\subseteq (1+\mathfrak{m}_L^r)$. So $N(1+x)=\prod_g g(1+x)\in (1+\mathfrak{m}_L^r)\cap K^\times=1+\mathfrak{m}_K^r$, where the last equality holds because $L/K$ is unramified. The norm map induces a map $N: \mathcal{O}_L^\times/(1+\mathfrak{m}_L)\cong l^\times\rightarrow \mathcal{O}_K^\times/(1+\mathfrak{m}_K)\cong k^\times$, which is coincides with the norm map on the residue fields. By a counting argument we know that the norm map on finite fields is surjective. The norm map also induces a map $N: (1+\mathfrak{m}_L^r)/(1+\mathfrak{m}_L^{r+1})\cong l\rightarrow (1+\mathfrak{m}_K^r)/(1+\mathfrak{m}_K^{r+1})\cong k$, which coincides with the trace map, so it is nonzero and $k$-linear (trace is always surjective for finite separable extension). This concludes that the norm map has dense image.

TopAdeles

Definition 43 Let $(X_v){v\in V}$ be a family of topological spaces, $U_v\subseteq X_v$ be open subset defined for almost all $v$. We define $$X=\{(x_v)_{v\in V}: x_v\in U_v \text{ for a.a. }v \}\subseteq \prod X_v,$$ and endow $X$ with the topology given by the base of open subsets $$\Big\{\prod_v Y_v: Y_v\subseteq X_v \text{ open}; Y_v= U_v \text{ for a.a. } v\Big\}.$$ The topological space $X$ is called the restricted topological product of $(X_v)$ with respect to $(U_v)$. Notice that this topology is different from the subspace topology induced from the product topology.

The following lemma is easy to check.

Lemma 5 Let $S\subseteq V$ be a finite set of indices containing all $v$'s such that $U_v$ is not defined. Then $$X_S=\prod_{v\in S}X_v\times \prod_{v\not\in S}U_v$$ is open in $X$ and the subspace topology on $X_S\subseteq X$ is the product topology on $X_S$.

The following proposition partially explains the reason of introducing the restricted product.

Proposition 1 If $X_v$'s are locally compact Hausdorff and the $U_v$'s are compact, then $X$ is locally compact Hausdorff.
Proof Notice that $X_S$ is locally compact: it is a product of a locally compact Hausdorff space and a compact space. Then result then follows from the fact that $X=\bigcup_S X_S$.
Definition 44 Let $K$ be a global field. The ring of ideles $\mathbb{A}_K$ is defined to be the restricted product of $\{K_v\}_{v\in V_K}$ with respect to $\{\mathcal{O}_v\}$ for $v$ non-archimedean. It is a subring of $\prod_v K_v$ and is a locally compact Hausdorff topological ring.
Remark 40 We have a natural diagonal embedding $K\rightarrow \mathbb{A}_K,\ x\mapsto (x)_v$. It is well-defined since $||x||_v\le1$ for almost all $v$. It is a topological ring homomorphism so we can regard $K$ as a subring of $\mathbb{A}_K$.
Example 14 Since $\hat{\mathbb{Z}}\cong\prod_p \mathbb{Z}_p$, we have $\mathbb{A}_\mathbb{Q}=\mathbb{R}\times (\hat{\mathbb{Z}}\otimes _\mathbb{Z}\mathbb{Q})$. The similar identification works for general number fields when replacing $\hat{\mathbb{Z}}$ by $\widehat{\mathcal{O}_K}$. We also have $\mathbb{A}_K=\mathbb{A}_\mathbb{Q}\otimes K$.
Exercise 5 Suppose $L/K$ is a finite extension. Then we have an diagonal embedding $K_v\hookrightarrow \prod_{w|v} L_v$ for $v\in V_K$, and hence an embedding $\mathbb{A}_K\hookrightarrow\mathbb{A}_L$.
  1. $\mathbb{A}_K\hookrightarrow \mathbb{A}_L$ is a homeomorphism onto a closed subring of $\mathbb{A}_L$.
  2. The natural map $L\otimes_K\mathbb{A}_K\rightarrow \mathbb{A}_L$ is an isomorphism of topological rings, where $L\otimes_K \mathbb{A}_K$ is given by the product topology viewing $L$ as $[L:K]$ copies of $K$.
  3. $\mathbb{A}_L/L\cong(\mathbb{A}_K/K)^{[L:K]}$ as topological groups.
Theorem 12 $K\subseteq \mathbb{A}_K$ is a discrete closed subgroup and the quotient $\mathbb{A}_K/K$ is compact.

09/21/2012

Remark 41 Noticing the discreteness of the embedding and the compactness of the quotient, the exact sequence $$0\rightarrow K\rightarrow \mathbb{A}_K\rightarrow\mathbb{A}_K/K\rightarrow0$$ can be thought as an analogy to the exact sequence $$0\rightarrow \mathbb{Z}\rightarrow \mathbb{R}\rightarrow \mathbb{R}/\mathbb{Z}\rightarrow 0.$$ Moreover, one can show that in fact $\mathbb{A}_K/K$ is the Pontryagin dual of $K$ like the familiar fact that $\mathbb{R}/\mathbb{Z}$ is the Pontryagin dual of $\mathbb{Z}$.
Proof Using Exercise 5, we reduce to the case $K=\mathbb{Q}$ or $K=\mathbb{F}_p(t)$ since any global field is a finite extension of one of them. We shall do the case $K=\mathbb{Q}$, the function field case is similar.

For discreteness of $\mathbb{Q}\subseteq \mathbb{A}_\mathbb{Q}$, we need to find a neighborhood of 0 in $\mathbb{A}_\mathbb{Q}$ such that $U\cap \mathbb{Q}=\{0\}$. The open neighborhood $$U=\{(x_v): ||x_\infty||_\infty<1, ||x_v||_v\le 1\}$$ suffices since the only units of $\mathbb{Z}$ are $\pm1$. Since every discrete subgroup of a Hausdorff topological group is closed, we know that $\mathbb{Q}$ is also closed in $\mathbb{A}_\mathbb{Q}$. In particular, the quotient $\mathbb{A}_\mathbb{Q} /\mathbb{Q}$ is Hausdorff. To prove the compactness of this quotient, it suffices to show that the natural continuous map $[-\frac{1}{2},\frac{1}{2}]\times\hat{\mathbb{Z}}\rightarrow \mathbb{A}_\mathbb{Q}/\mathbb{Q}$ is surjective. Let $x\in \mathbb{A}_\mathbb{Q}$. Since $\mathbb{Z}$ is dense in $\mathbb{Z}_p$, we can choose $r_p\in \mathbb{Z}[1/p]$ such that $||x_p-r_p||_p\le1$ and $r_p=0$ for almost all $p$. Let $r=\sum r_p$, then $||x_p-r||_p\le1$ for any $p$. Now choose $s\in \mathbb{Z}$ such that $x_\infty-r-s\in[-\frac{1}{2},\frac{1}{2}]$. Then $x-r-s\in [-\frac{1}{2},\frac{1}{2}]\times \hat{\mathbb{Z}}$ satisfying that $x-r-s\equiv x\pmod{\mathbb{Q}}$, hence maps to $x$ in $\mathbb{A}_\mathbb{Q}/\mathbb{Q}$.

We will omit the tedious measure-theoretic check of the following the lemma.

Lemma 6 There exists a Haar measure $\mu_K$ on $\mathbb{A}_K$ such that $\mu_K(\prod U_v)=\prod \mu_v(U_v)$ where $\mu_v$ is the normalized Haar measure on $K_v$ such that $\mu_v(\mathcal{O}_v)=1$.
Remark 42 From the exact sequence of locally compact Hausdorff groups $$0\rightarrow K\rightarrow \mathbb{A}_K\rightarrow \mathbb{A}_K/K\rightarrow0$$ we can construct another canonical measure on $\mathbb{A}_K$. Let $\bar \mu$ be the Haar measure on $\mathbb{A}_K/K$ with total measure 1 and $\nu$ be the counting measure on $K$. By the general construction, there exists $\mu$ on $\mathbb{A}_K$ such that $$\int_{\mathbb{A}_K} f(x) d\mu(x)=\int_{\mathbb{A}_K/K}\int_K f(yz)d \nu(y)d\bar\mu(\bar z).$$ The above two Haar measures differs by a scalar and it is an interesting question to figure out this scalar.

TopIdeles

Definition 45 The group of ideles is defined to be $\mathbb{A}_K^\times=\{ x\in \mathbb{A}_K: x_v\ne0; ||x_v||_v=1\text{ for a.a. } v\}$. The embeddings $K_v^\times\hookrightarrow \mathbb{A}_K^\times$ defined by $x\mapsto (1,\ldots1, x,1,\ldots,1)$ gives an embedding $K^\times\hookrightarrow \mathbb{A}_K^\times$. The quotient $\mathbb{A}_K^\times/ K^\times$ is the called the idele class group of $K$.
Exercise 6 Let $R $ be a topological ring. We have an embedding $R^\times\hookrightarrow R\times R$ given by $x\mapsto (x,x^{-1})$. Then
  1. $R^\times$ is a topological group with respect to the subspace topology coming from $R\times R$.
  2. When $R=\mathbb{A}_K$, $\mathbb{A}_K^\times$ is the restricted product of $(K_v^\times)_v$ with respect to $(\mathcal{O}_v^\times)_v$. Hence $\mathbb{A}_K^\times$ is a locally compact Hausdorff topological group.
  3. $\mathbb{A}_K^\times$ is not a topological group under the subspace topology coming from $\mathbb{A}_K$. But $\mathbb{A}_K^\times\hookrightarrow \mathbb{A}_K$ is continuous under the above topology.
  4. When $L/K$ is finite and separable, $\mathbb{A}_K^\times\hookrightarrow \mathbb{A}_L^\times$ is a homeomorphism onto a closed subgroup.
Definition 46 The idelic norm is defined to be the homomorphism $||\cdot||_K: \mathbb{A}_K^\times\rightarrow \mathbb{R}_{>0}$ given by $||x||_K=\prod_{v\in V_K} ||x_v||_v$. The unit ideles is defined to be the subgroup $(\mathbb{A}_K^\times)^1=\{x\in \mathbb{A}_K^\times: ||x||_K=1\}\subseteq \mathbb{A}_K^\times$.
Exercise 7
  1. $||\cdot||_K$ is continuous. Hence $(\mathbb{A}_K^\times)^1$ is a closed subgroup.
  2. When $K$ is a number field, $||\cdot||_K$ is surjective and open, hence is a quotient map.

When $K$ is a global function field, the image of $||\cdot||_K$ can be described as follows.

Theorem 13 Let $K$ be a global function field with constant field $k$ and $q=\#k$. Then $||\mathbb{A}_K^\times||=q^\mathbb{Z}$.
Proof We will use the following lemma, to be proved when we discuss the Chebotarev's density theorem (Corollary 10).
Lemma 7 If $L/K$ is finite separable of global fields and almost all places of $K$ split completely in $L$, then $L=K$.

Obviously $||\mathbb{A}_K^\times||\subseteq q^\mathbb{Z}$, hence the image is of the form $q^{m \mathbb{Z}}$. We need to show that $m=1$. Notice that $||K_v^\times||_v=(\#k_v)^\mathbb{Z}\subseteq q^{m \mathbb{Z}}$. There fore $\#k_v=(\#k)^{m\cdot*}$. Hence $m\mid[k_v:k]$. If $l$ is the degree $m$ extension of $k$, then $l\subseteq k_v$ for any $v$. Hence $k_v\otimes_k l=\prod_{i=1}^m k_v$. Write $L=K\otimes_kl$. Then $L/K$ is finite separable. Now $$\prod_{w|v} L_w=K_v\otimes_KL=K_v\otimes_K(K\otimes_k l)=K_v\otimes_{k_v}(k_v\otimes_kl)=K_v\otimes_{k_v}\prod_{i=1}^m k_v= K_v^m$$ shows that there are exactly $m$ places over $v$, i.e., $v$ splits completely. By the lemma, we know that $L=K$ and $m=1$.

Remark 43 Geometrically the theorem means that there exists a degree 1 divisor on the curve $X$ with $K=K(X)$. In fact, suppose $x_v\in |X|$ is a closed point corresponding to $v\in V_K$. Then $\deg ((x_v))=[k_v:k]$, hence $||\pi_v||_v=q^{-\deg x_v}$. So for any divisor $D=\sum n_vx_v$, $q^{-\deg D}=||\pi_v^{n_v}||_K$. (Question: is it true over an arbitrary constant field)?
Lemma 8 For $x\in \mathbb{A}_K^\times$, then multiplication by $x$ scales $\mu_K$ by $||x||_K$.
Proof It follows from the case of local fields (Lemma 1) and the Lemma 6.
Corollary 3 $\frac{d\mu_K}{||\cdot||_K}$ is a Haar measure on $\mathbb{A}_K^\times$.

From this we can obtain a slick proof of the product formula.

09/24/2012

Proof (Theorem 2) Consider the measure $\mu$ in Remark 42. Suppose $w\in K^\times$, then $$||w||_K\int_{\mathbb{A}_K}f(x)d\mu(x)=\int_{\mathbb{A}_K}f(wx)d\mu(x)=\int_{\mathbb{A}_K/K}\int_K f(wyz)d\nu(y)d\bar\mu(\bar z).$$ Since $d\nu $ is invariant under multiplication by $w$, this integral is equal to $$\int_{\mathbb{A}_K/K}\int_K f(yz) d\nu(y)d\bar\mu(\bar z)=\int_{\mathbb{A}_K }f(x)d\mu(x).$$ Hence $||w||_K=1$.
Remark 44 When $K$ is a number field, we have an exact sequence $$0\rightarrow (\mathbb{A}_K^\times)^1\rightarrow \mathbb{A}_K^\times\rightarrow \mathbb{R}_{>0}\rightarrow 0.$$ Sending the positive real numbers back to an archimedean place of $\mathbb{A}_K^\times$ gives a (non-canonical) isomorphism of topological groups $\mathbb{A}_K^\times\cong (\mathbb{A}_K^\times)^1\times \mathbb{R}_{>0}$. Similarly, $\mathbb{A}_K^\times/K^\times\cong (\mathbb{A}_K^\times)^1/K^\times\times \mathbb{R}_{>0}$ as topological groups. We will show that $(\mathbb{A}_K^\times)^1/K^\times$ is compact, so this can be seen as the analogy to $\mathbb{C}^\times\cong S^1\times \mathbb{R}_{>0}$.
Remark 45 When $K$ is a global function field, we have a similar (non-canonical splitting) $\mathbb{A}_K^\times\cong (\mathbb{A}_K^\times)^1\times q^\mathbb{Z}$ and $\mathbb{A}_K^\times/K^\times\cong(\mathbb{A}_K^\times)^1/K^\times\times q^\mathbb{Z}$, where $(\mathbb{A}_K^\times)^1$ is compact.
Example 15 Let $x\in (\mathbb{A}_\mathbb{Q}^\times)^1$. Then the fractional ideal $\prod_{p}(p)^{\ord_p(x_p)}=(y)$ where $y\in \mathbb{Q}^\times$ determined up to sign. Hence $x/y\in \mathbb{R}^\times\times\hat{\mathbb{Z}}^\times$. But $||x/y||_K=1$ implies that $||x_\infty/y||_{\infty}=1$. Replacing $y$ by $-y$ if neccesary, there exists a unique $y\in \mathbb{Q}^\times$ such that $x_\infty/y=1$ as well. Hence $\{1\}\times\hat{\mathbb{Z}}^\times$ is a fundamental domain for $(\mathbb{A}_\mathbb{Q}^\times)^1/\mathbb{Q}^\times$. Therefore $(\mathbb{A}_\mathbb{Q}^\times)^1/\mathbb{Q}^\times\cong \hat{\mathbb{Z}}^\times$ as topological groups. Hence $\mathbb{A}_\mathbb{Q}^\times/\mathbb{Q}^\times\cong \mathbb{R}_{>0}\times \hat{\mathbb{Z}}^\times$ as topological groups. Now $\mathbb{R}_{>0}\times \hat{\mathbb{Z}}^\times$ also embeds into $\mathbb{A}_\mathbb{Q}^\times$, hence $\mathbb{A}_\mathbb{Q}^\times\cong \mathbb{Q}^\times\times \mathbb{R}_{>0}\times \hat{\mathbb{Z}}^\times$ as topological groups (this can also be seen directly by extracting all $p$-powers of $x_p$).

TopAdelic Minkowski's theorem

The classical Minkowski's theorem says that for a compact convex and symmetric around 0 region $S\subseteq \mathbb{R}^n$, $\Vol(S)>2^n$ implies that there exists a nonzero $x\in \mathbb{Z}^n$ such that $x\in S$. The following is a reformulation.

Theorem 14 (Minkowski) Let $V$ be a $n$-dimensional $\mathbb{R}$-vector space and $\Lambda\subseteq V$ be a lattice. Suppose $\mu$ is a Haar measure on $V$ constructed from the counting measure on $\Lambda$ and the volume one measure on $V/\Lambda$. If $S\subseteq V$ is compact convex and symmetric around 0, then $\mu(S)>2^n$ implies that $S\cap\Lambda\ne \{0\}$.

The idea of the proof of the following adelic version is essentially the same as the classical version.

Theorem 15 (Adelic Minkowski) Let $K$ be a global field and $x\in \mathbb{A}_K^\times$. Then $$S_x:=\{y\in \mathbb{A}_K: ||y_v||_v\le ||x_v||_v\}$$ is compact. There exists $c>0$ depending only on $K$ such that if $||x||_K>c$, then $S_x\cap K\ne \{0\}$.
Proof Let $\mu$ be the Haar measure on $\mathbb{A}_K$ as before. Let $$Z=\{x\in \mathbb{A}_K: ||x_v||_v\le1, v\nmid\infty;||x_v||_v\le1/2, v\mid\infty\}.$$ Then for any $z,z'\in Z$, $||z_v-z'_v||\le1$ for any $v$. Then $Z$ is compact and contains a neighborhood of 0. So $\mu(Z)\in (0,\infty)$, which is an intrinsic quantity of $K$. Let $c=1/\mu(Z)$. We claim that this $c $ works. Suppose $x\in \mathbb{A}_K^\times$ with $||x||_K> c$. Then $\mu(xZ)=||x||_K\cdot\mu(Z)>1$.

We claim that the natural map $\pi: xZ\rightarrow \mathbb{A}_K/K$ has some fiber with at least two elements. Write $\chi$ be the characteristic function of $xZ$. Then $$\mu(xZ)=\int_{\mathbb{A}_K}\chi(x)d\mu(x)=
\int_{\mathbb{A}_K /K}\int_K \chi(yz) d\nu(y)d\bar\mu(\bar z)=\int_{\mathbb{A}_K/K}\#\pi^{-1}(\bar z)d\bar\mu(\bar z).$$ So if every fiber has size at most one, then the above integral is less than $\bar\mu(\mathbb{A}_K/K)=1$, a contradiction to $\mu(xZ)>1$.

Hence there exists $z, z'\in xZ$ such that $a=z-z'\in K^\times$. Write $z=xy$ and $z'{}=xy'$, where $y,y'\in Z$. Then $||a||_v=||x_v||_v||y_v-y_v'||_v\le||x_v||$ by the definition of $Z$. Therefore $a\in S_x$.

Theorem 16 (Strong approximation) Let $v_0\in V_K$ and $\mathbb{A}_K^{v_0}=\mathbb{A}_K /K_{v_0}$. Then the diagonal embedding $K\hookrightarrow \mathbb{A}_K^{v_0}$ has dense image.
Remark 46 In contrast, if we keep all the places, the image is discrete as in Theorem 12.
Remark 47 Concretely the strong approximation means the following: let $x\in \mathbb{A}_K$ and $S$ be a finite set of places not containing $v_0$. Then there exists $y\in K$ such that $||x_v-y||<\varepsilon$ for all $v\in S$ and $||x_v-y||\le1$ for $v\not\in S$ and $v\ne v_0$. For comparison, the weak approximation says that for $S$ a finite set of places. Then for $x\in \mathbb{A}_K$, there exists $y\in K$ such that $||x_v-y||_v<\varepsilon$ for any $v\in S$ (without restriction on other places).
Proof We claim that there exists $w\in \mathbb{A}_K^\times$ such that $S_w\twoheadrightarrow \mathbb{A}_K/K$. This follows from the fact that $\mathbb{A}_K\rightarrow \mathbb{A}_K/K$ is open (Exercise 3), $\mathbb{A}_K=\bigcup_w S_w$ and $\mathbb{A}_K /K$ is compact.

For such a $w$, let $x\in \mathbb{A}_K$, $S\subseteq V_K\setminus\{v_0\}$ and $\varepsilon>0$. Then for

  • $v\in S$, we choose $y_v\in K_v^\times$ such that $||y_v||_v<\frac{\varepsilon}{||w_v||_v}$.
  • $v\not\in S$ and $v\ne v_0$, we let $y_v=1/w_v$.
  • $v=v_0$, we choose $y_{v_0}$ such that $y=(y_v)$ has $||y||_K>c$, where $c$ is the constant in the Adelic Minkowski's Theorem 15.

By Adelic Minkowski's Theorem 15, there exists $0\ne z\in S_y\cap K$. Write $x/z=\alpha+\beta$ such that $\alpha\in K$ and $\beta\in S_w$. So $y=z\alpha$, then $2\beta=x-y$. Then $x-z\alpha=z\beta$, where $z\alpha\in K$ and

  • $||z\beta_v||_v<\varepsilon$ for any $v\in S$ by construction.
  • $||z\beta_v||_v\le1$ for $v\not\in S$ and $v\ne v_0$.

09/26/2012

Lemma 9 $(\mathbb{A}_K^\times)^1$ is closed in $\mathbb{A}_K$ and the subspace topology on $(\mathbb{A}_K^\times)^1$ from $\mathbb{A}_K$ coincides with the subspace topology from $\mathbb{A}_K^\times$.
Proof First we show that $(\mathbb{A}_K^\times)^1$ is closed in $\mathbb{A}_K$. For $x\in \mathbb{A}_K$, the products $\nu_N=\prod_{i=1}^N||x_{v_i}||_{v_i}$ will eventually be decreasing. So $||x||_K=\lim_{N\rightarrow\infty} \nu_N$ is well-defined. Suppose $x\in \mathbb{A}_K \setminus (\mathbb{A}_K^\times)^1$, then $||x||_K\ne1$. There are two cases:
  1. $||x||_K<1$. Let $S\supseteq S_\infty$ be a finite set of places such that $\prod_{v\in S}||x_v||_v<1$ and $x_v\in \mathcal{O}_v$ for all $v\not\in S$. For $\varepsilon>0$ and $v\in S$, we let $N_{v,\varepsilon}=\{y_v\in K_v: ||y_v||_v< ||x_v||_v+\varepsilon\}\subseteq K_v$ and $$N_\varepsilon=\prod_{v\in S}N_{v,\varepsilon}\times \prod_{v\not\in S}\mathcal{O}_v.$$ Then $N_\varepsilon$ is an open neighborhood of $x$ in $\mathbb{A}_K$. For $\varepsilon$ small enough, $N_\varepsilon\cap (\mathbb{A}_K^\times)^1=\varnothing$.
  2. $||x||_K=P>1$. Let $S\supseteq S_\infty$ be a finite set of places such that
    1. $x_v\in \mathcal{O}_v$ for $v\not\in S$.
    2. $q_v=\# k_v>2P$ for $v\not\in S$. This implies that if $\xi_v\in K_v$, and $||\xi_v||_v<1$, then $||\xi_v||\le 1/q_v<1/2P$.
    3. $\prod_{v\in S}||x_v||_v\in (1,2P)$.

For $v\in S$, let $N_v$ be a small open neighborhood of $x_v$ small enough such that $\prod_{v\in S}||\xi_v||_v\in (1,2P)$ for $(\xi_v)\in \prod_{v\in S}N_v$. Then $$N=\prod_{v\in S}N_{v}\times \prod_{v\not\in S}\mathcal{O}_v$$ is a neighborhood of $x$ in $\mathbb{A}_K$. Let $\xi\in N$. If $||\xi_v||_v<1$ for some $v\not\in S$, then $||\xi||_K< 2P\cdot\frac{1}{2P}<1$. If $||\xi_v||_v=1$ for all $v\not\in S$, then $||\xi||_K>1$. Hence $N\cap (\mathbb{A}_K^\times)^1=\varnothing$. This concludes that $(\mathbb{A}_K^\times)^1$ is closed in $\mathbb{A}_K$.

Now $(\mathbb{A}_K^\times)^1\hookrightarrow \mathbb{A}_K$ is continuous and has closed image. So we need to show that any neighborhood of $x\in (\mathbb{A}_K^\times)^1$ in $(\mathbb{A}_K^\times)^1$ contains $W\cap (\mathbb{A}_K^\times)^1$ for some $W$ a neighborhood of $x$ in $\mathbb{A}_K$. By homogeneity, we may assume $x=1$. The basic open neighborhood of $x=1$ in $(\mathbb{A}_K^\times)^1$ is of the form $N\cap (\mathbb{A}_K^\times)^1$, where $$N=\prod_{v\in S} N_v\times \prod_{v\not\in S}\mathcal{O}_v^\times$$ and $N_v$ is an open neighborhood of 1 in $K_v^\times$. Shrinking the $N_v$'s we may assume $||\xi||_K=\prod_{v\in S}||\xi_v||_v<2$ for any $\xi\in N$. We claim that $$W=\prod_{v\in S} N_v\times \prod_{v\not\in S} \mathcal{O}_v$$ works, i.e., $W\cap (\mathbb{A}_K^\times)^1=N$. If for some $v\not\in S$, we have $||\xi_v||_v<1$, then $||\xi_v||_v\le1/2$, thus $$1=||\xi||_K<2\cdot 1/2=1,$$ a contradiction.

Theorem 17 Let $K$ be a global field. Then $K^\times\hookrightarrow \mathbb{A}_K^\times$ is a discrete closed subgroup and $(\mathbb{A}_K^\times)^1/K^\times$ is compact.
Proof The assertion that $K^\times$ is discrete and closed follows form the case of $K\hookrightarrow \mathbb{A}_K$ (Theorem 12) since the topology on $\mathbb{A}_K^\times$ is finer than $\mathbb{A}_K$. It remains to prove the compactness of $(\mathbb{A}_K^\times)^1/K^\times$. By the previous lemma, if $W\subseteq \mathbb{A}_K$ is compact, then $W\cap (\mathbb{A}_K^\times)^1$ is compact in $(\mathbb{A}_K^\times)^1$. So it suffices to show that there is a surjection $W\cap (\mathbb{A}_K^\times)^1\twoheadrightarrow (\mathbb{A}_K^\times)^1/K^\times$ for some $W\subseteq \mathbb{A}_K$ compact. Let $c>0$ be as in the Adelic Minkowski's Theorem 15 and choose $x\in \mathbb{A}_K^\times$ such that $||x||_K>c$. We claim that the compact set $W=S_x$ works. Let $y\in (\mathbb{A}_K^\times)^1$, then $||x/y||_K=||x||_K>c$. It follows from the Adelic Minkowski's Theorem 15 that there exists $z\in K^\times\cap S_{x/y}$, i.e., $||z||_v\le{||x/y}||_v$ for any $v$. Now $||yz||_v\le||x_v||_v$, hence $yz\in (\mathbb{A}_K^\times)^1\cap W$. This concludes the surjectivity of $W\cap (\mathbb{A}_K^\times)^1\twoheadrightarrow (\mathbb{A}_K^\times)^1/K^\times$.

TopClassical finiteness theorems

Definition 47 Let $S\supseteq S_\infty$ be a finite set of places. We denote $$\mathbb{A}_{K,S}=\prod_{v\in S} K_v\times\prod_{v\not\in S}\mathcal{O}_v,$$ an open subring of $\mathbb{A}_K$. Similarly, we denote $$\mathbb{A}_{K,S}^\times==\prod_{v\in S} K_v^\times\times\prod_{v\not\in S}\mathcal{O}_v^\times,$$ an open subgroup of $\mathbb{A}_K^\times$. We have $\mathcal{O}_{K,S}=K\cap \mathbb{A}_{K,S}$ and $\mathcal{O}_{K,S}^\times=K^\times\cap \mathbb{A}_{K,S}^\times$.

Recall that the $S$-class group (Definition 11) is $\Cl_S(K)=I_{K,S}/P_{K,S}$, the fractional ideals of $\mathcal{O}_{K,S}$ quotient by the principal ideals. We have a natural surjection $\Phi: \mathbb{A}_K^\times\rightarrow I_{K,S}$ with kernel $\mathbb{A}_{K,S}^\times$ and $\Phi(K^\times)=P_{K,S}$. Hence we have an isomorphism $$\Phi: \mathbb{A}_K^\times/K^\times \mathbb{A}_{K,S}^\times\cong \Cl_S(K).$$

Theorem 18 (Finiteness of the class groups and finite generation of units) Let $S\supseteq S_\infty$ be a nonempty finite set of places.
  1. $\#\Cl_S(K)<\infty$.
  2. $\mathcal{O}_{K,S}^\times$ is a finitely generated abelian group of rank $\#S-1$.
Proof Write $(\mathbb{A}_{K,S}^\times)^1=\mathbb{A}_{K,S}^\times\cap (\mathbb{A}_K^\times)^1$. Then we have an exact sequence $$1\rightarrow (\mathbb{A}_{K,S}^\times)^1/\mathcal{O}_{K,S}^\times\cong K^\times (\mathbb{A}_{K,S}^\times)^1/K^\times\rightarrow (\mathbb{A}_K^\times)^1/K^\times\rightarrow (\mathbb{A}_K^\times)^1/K^\times (\mathbb{A}_{K,S}^\times)^1\rightarrow 1.$$ Notice that the first inclusion has open image, hence the quotient $(\mathbb{A}_K^\times)^1/K^\times(\mathbb{A}_{K,S}^\times)^1$ is discrete. But $(\mathbb{A}_K^\times)^1/K^\times$ is compact, hence $(\mathbb{A}_{K,S}^\times)^1/\mathcal{O}_{K,S}^\times$ is compact and $(\mathbb{A}_K^\times)^1/K^\times(\mathbb{A}_{K,S}^\times)^1$ is finite.

Now the natural homomorphism $(\mathbb{A}_K^\times)^1/K^\times(\mathbb{A}_{K,S}^\times)^1\rightarrow \mathbb{A}_K^\times/K^\times\mathbb{A}_{K,S}^\times\cong \Cl_S(K)$ has cokernel $||\mathbb{A}_K^\times||_K/||\mathbb{A}_{K,S}^\times||_K$, which is always finite: in the number field case, both are $\mathbb{R}_{>0}$ ; in the global function field case, both are nontrivial subgroup of $q^\mathbb{Z} $. It follows that $\Cl_S(K)$ is finite.

It remains to prove that $\mathcal{O}_{K,S}^\times$ is finitely generated. Consider the log map $\mathcal{L}: \prod_{v\in S}K_v^\times\rightarrow \mathbb{R}^N$ given by $(x_v)\mapsto (\log||x_v||_v)$, where $N=\# S$. Then $(\prod_{v\in S}K_v^\times)^1$ maps into the hyperplane $H=\{(x_v): \sum x_v=0\}\subseteq \mathbb{R}^N$. One can easily (and classically) show that $\mathcal{L}(\mathcal{O}_{K,S}^\times)$ is discrete in $H $ (only finitely many polynomial with bounded integral coefficients) and the kernel is the roots of unity in $\mathcal{O}_{K,S}^\times$, which is finite. This shows that $\mathcal{O}_{K,S}^\times$ is finitely generated with rank at most $\# S-1$. To show $\mathcal{L}(\mathcal{O}_{K,S}^\times)$ is of full rank in $H $, it suffices to show $(\prod_{v\in S}K_v^\times)^1/\mathcal{O}_{K,S}^\times$ is compact (hence of full rank) as the image $\mathcal{L}((\prod_{v\in S}K_v^\times)^1)$ is isomorphic to $\mathbb{R}^{N_1}\times \mathbb{Z}^{N_2}$. This follows from the surjection $(\mathbb{A}_{K,S}^\times)^1/\mathcal{O}_{K,S}^\times\rightarrow (\prod_{v\in S}K_v^\times)^1/\mathcal{O}_{K,S}^\times$ and the compactness of $(\mathbb{A}_{K,S}^\times)^1/\mathcal{O}_{K,S}^\times$.

09/28/2012

TopIdele class groups

Let $L/K$ be a finite extension of global fields. We know that $L\otimes \mathbb{A}_K\cong \mathbb{A}_L$ as topological rings. In particular, $\mathbb{A}_K\hookrightarrow \mathbb{A}_L$ is a closed embedding (i.e., a homeomorphism onto a closed subgroup). Hence $\mathbb{A}_K^\times\hookrightarrow \mathbb{A}_L^\times$ is also a closed embedding.

Proposition 2 The natural map $\mathbb{A}_K^\times/K^\times\hookrightarrow \mathbb{A}_L^\times/L^\times$ is a closed embedding.
Proof For any $x\in \mathbb{A}_L^\times$, $||x||_L=||N_{L/K}(x)||_K$. Hence $(\mathbb{A}_L^\times)^1 /L^\times\cap \mathbb{A}_K^\times/K^\times=(\mathbb{A}_K^\times)^1 /K^\times$. When $K$ is a global function field, we have two exact sequences $$\xymatrix{0 \ar[r] & (\mathbb{A}_K^\times)^1/K^\times \ar[r]\ar@{^{(}->}[d] & \mathbb{A}_K^\times/K^\times\ar[r] \ar@{^{(}->}[d] &q^\mathbb{Z} \ar[r] \ar[d]^{[L:K]}& 0\\ 0 \ar[r] & (\mathbb{A}_L^\times)^1/L^\times \ar[r] & \mathbb{A}_L^\times/L^\times \ar[r]  & q^\mathbb{Z}\ar[r]& 0.} $$ Since $(\mathbb{A}_K^\times)^1/K^\times$ is compact, the first vertical map is a closed embedding. The last vertical map is also a closed embedding. Now the middle term is infinite union of these closed embeddings, hence is also a closed embedding. The number field case is similar: $$\xymatrix{0 \ar[r] & (\mathbb{A}_K^\times)^1/K^\times \ar[r]\ar@{^{(}->}[d] & \mathbb{A}_K^\times/K^\times\ar[r] \ar@{^{(}->}[d] & \mathbb{R}_{>0} \ar[r] \ar[d]^{[L:K]}& 0\\ 0 \ar[r] & (\mathbb{A}_L^\times)^1/L^\times \ar[r] & \mathbb{A}_L^\times/L^\times \ar[r]  &  \mathbb{R}_{>0}\ar[r]& 0.} $$ The same argument shows that the middle map is a closed embedding.
Exercise 8
  1. For any $v\in V_K$, $K_v^\times\hookrightarrow \mathbb{A}_K^\times/K^\times$ is a closed embedding.
  2. If $S\subseteq V_K$ is a finite set of places and $\# S>1$. Then $\prod_{v\in S} K_v^\times\hookrightarrow \mathbb{A}_K^\times/K^\times$ is not a closed embedding.

Notice that $\mathbb{A}_L=\mathbb{A}_K^{[L:K]}$ is a finite $\mathbb{A}_K$-module, we obtain a norm map $N_{L/K}: \mathbb{A}_L^\times \rightarrow \mathbb{A}_K^\times$, compatible with the norm $N_{L/K}: L^\times\rightarrow K^\times$. Using $L\otimes_K K_v\cong \prod_{w|v}L_w$, we know that for any $w|v$, the norm map is also compatible with the local norm $N_{L_w/K_v}: L_w^\times\rightarrow K_v^\times$. Moreover, $||N_{L/K}(x)||_K=||x||_L$ for any $x\in \mathbb{A}_L^\times$.

Theorem 19 $N_{L/K}: \mathbb{A}_L^\times /L^\times\rightarrow \mathbb{A}_K^\times /K^\times$ is continuous, open and proper.
Proof It is continuous since each local norm is continuous and $N_{L_w/K_v}^{-1}(\mathcal{O}_{K_v}^\times)=\mathcal{O}_{L_w}^\times$ (hence the inverse image of a basic open subset is open). For the properness, we use the splitting $\mathbb{A}_L^\times/L^\times$ and $\mathbb{A}_K^\times/K^\times$. Then $N_{L/K}: \mathbb{A}_L^\times /L^\times\rightarrow \mathbb{A}_K^\times /K^\times$ is the norm on the compact factors $(\mathbb{A}_L^\times)^1/L^\times$ and identity on $q^\mathbb{Z} $ or $\mathbb{R}_{>0}$, so it is a product of two proper maps, hence is proper. To prove the openness, we use the fact that local norms are open (Theorem 11) and the local norm is surjective for unramified local extensions (Lemma 4) to conclude that the image of a basic open subset is open.
Exercise 9 The map $\mathbb{A}_K^\times/K^\times\rightarrow \mathbb{A}_K^\times/K^\times: x\mapsto x^n$ is continuous and proper.

Our next goal is to describe the connected component of 1 in ideles class group $\mathbb{A}_K^\times/K^\times$ (which turns out to be exactly the kernel of the global Artin map by class field theory).

First suppose $K$ is a number field. Then the connected component of 1 in $\prod_{v|\infty} K_v^\times$ is $$\left(\prod\nolimits_{v|\infty} K_v^\times\right)^0\cong \mathbb{R}_{>0}^{r_1}\times (\mathbb{C}^\times)^{r_2},$$ where $r_1$ and $r_2$ are the numbers of real and complex places of $K$. It is divisible, i.e., $x\mapsto x^n$ is surjective for any $n$. Let $D_k$ be its closure in $\mathbb{A}_K^\times/K^\times$. Then $D_K$ is a closed connected divisible subgroup (the divisibility follows from the fact that $x\mapsto x^n$ is proper, thus closed).

Lemma 10 Suppose $K$ is a number field. Then $D_K$ is the connected component of 1 in $\mathbb{A}_K^\times/K^\times$ and $(\mathbb{A}_K^\times/K^\times)/D_K$ is profinite. Moreover, $D_K$ is the set of all divisible elements in $\mathbb{A}_K^\times/K^\times$.
Proof Notice that $\mathbb{A}_K^\times/K^\times\mathbb{A}_{K,S_\infty}^\times\cong \Cl(K)$ is finite and $[(\prod_{v|\infty} K_v^\times: (\prod_{v|\infty} K_v^\times)^0]<\infty$, we know that the natural map $\prod_{v\nmid\infty} \mathcal{O}_v^\times\rightarrow (\mathbb{A}_K^\times/K^\times)/D_K$ has finite cokernel. Let $H $ be the image of this last map. Since $\prod_{v\nmid \infty}\mathcal{O}_v^\times$ is profinite, we know that the image $H $ is also a profinite group (Remark 30). Since $H $ is of finite index, it is also open in $(\mathbb{A}_K^\times/K^\times)/D_K$. Combining the fact that $H $ is compact and totally disconnected, we find that $(\mathbb{A}_K^\times/K^\times)/D_K$ is also compact and totally disconnected, thus is profinite.

Let $E $ be the connected component of 1, then $E $ is killed under the map to $(\mathbb{A}_K^\times/K^\times)/D_K$ by totally disconnectedness, therefore $E\subseteq D_K$. But $D_K$ is already connected, this shows that $D_K=E$. Every divisible element maps to 1 in $(\mathbb{A}_K^\times/K^\times)/D_K$ since profinite group has no nontrivial divisible elements, so must lie in $D_K$. But $D_K$ is already divisible, hence it consists of all divisible elements of $\mathbb{A}_K^\times/K^\times$.

Remark 48 If $\#S_\infty=1$ (i.e., $K=\mathbb{Q}$ or an imaginary quadratic field), then $K_\infty^\times\hookrightarrow \mathbb{A}_K^\times/K^\times$ is a closed embedding (Exercise 8), hence $D_K=\mathbb{R}_{>0}$ or $\mathbb{C}^\times$. In general we have (for a proof, see Artin-Tate Ch. IX) $$D_K\cong ((S^1)^\times)^{r_2}\times (\mathbb{A}_\mathbb{Q}/\mathbb{Q})^{r_1+r_2-1}\times \mathbb{R}_{>0}.$$ Notice that $$\mathbb{A}_\mathbb{Q} /\mathbb{Q}\cong (\mathbb{R}\times \hat{\mathbb{Z}})/\mathbb{Z}=\varprojlim_m \mathbb{R}/m \mathbb{Z}=\varprojlim_{x\mapsto x^m} S^1,$$ is the inverse limit of $m$-fold covers of $S^1$ (called a solenoid).

Now consider the global function field case.

Lemma 11 Suppose $K$ is a global function field. Then $\mathbb{A}_K^\times/K^\times$ is totally disconnected and has no nontrivial divisible elements.
Proof Then $\prod\mathcal{O}_v^\times$ is an open neighborhood of 1 in $\mathbb{A}_K^\times$, hence $\prod\mathcal{O}_v^\times$ is an open neighborhood of 1 in $\mathbb{A}_K^\times/K^\times$. Hence $\mathbb{A}_K^\times/K^\times$ is totally disconnected, thus $\mathbb{A}_K^\times/K^\times$ is profinite. Hence $\mathbb{A}_K^\times/K^\times$ has no divisible elements.

TopCyclotomic extensions

Let $K$ be any field and $\Char(K)\nmid m$. Let $\zeta_m\in K^\mathrm{sep}$ be a primitive $m$-th root of unity. Then $K(\zeta_m)$, the splitting field of $X^m-1$, is separable, hence is Galois. Let $\bbmu_m=K(\zeta_m)^\times[m]\cong\langle \zeta_m\rangle$. Then any $g\in \Gal(K(\zeta_m)/K)$ is determined by its action on $\zeta_m$, thus we obtain a injection $$\alpha_m: \Gal(K(\zeta_m)/K)\hookrightarrow\Aut(\bbmu_m)\cong (\mathbb{Z}/m \mathbb{Z})^\times,\quad g(\zeta)=\zeta^{\alpha_m(g)}.$$ This in particular shows that $K(\zeta_m)/K$ is abelian. The map $\alpha_m$ is functorial, i.e., for any field extension $L/K$, we have a commutative diagram $$\xymatrix{\Gal(L(\zeta_m)/L) \ar[r]^-{\alpha_m} \ar[d]& (\mathbb{Z}/m \mathbb{Z})^\times\\ \Gal(K(\zeta_m)/K) \ar[ru]^-{\alpha_m} &}.$$

Definition 48 For $d\mid m$, $\zeta_d=\zeta_m^{m/d}$ and $K(\zeta_d)\subseteq K(\zeta_m)$. So $\{K(\zeta_d)\}_{\Char(K)\nmid m}\}$ forms a filtered directed system. We define the maximal cyclotomic extension to be $K^\mathrm{cyc}=\cup_{\Char(K)\nmid m} K(\zeta_m)$.
Remark 49 Combining all $\alpha_m$'s gives an injection $$\alpha_K: \Gal(K^\mathrm{cyc}/K)\cong\varprojlim_{\Char(K)\nmid m} \Gal(K(\zeta_m)/K)\hookrightarrow \varprojlim_{p\ne \Char(K)} \mathbb{Z}_p^\times.$$ The right hand side is simply $\hat{\mathbb{Z}}^\times$ when $\Char(K)=0$.

10/01/2012

Example 16 When $K=\mathbb{Q}$ or $\mathbb{Q}_p$, the Kronecker-Weber theorem says that $K^\mathrm{ab}=K^\mathrm{cyc}$. However, this is not true for general number fields. For example, for $K=\mathbb{Q}(\sqrt{2})$, the extension $K(\sqrt[4 ]{2})$ is abelian over $K$ but is not even Galois over $\mathbb{Q}$, hence $K$ cannot be contained in a cyclotomic extension of $K$ since $K(\zeta_m)=\mathbb{Q}(\zeta_m)\cdot K$ is abelian over $\mathbb{Q}$ for any $m$.
Example 17 Suppose $K=k=\mathbb{F}_q$, where $q=p^r$. Suppose $p\nmid m$ and $f$ is the order of $q\in (\mathbb{Z}/m \mathbb{Z})^\times$. Since $\Frob(\zeta_m)=\zeta_m^q$, we know that $\alpha_m(\Frob)=q\mod{m}$. Therefore $\Gal(k(\zeta_m)/k)\cong \mathbb{Z}/f \mathbb{Z}$ and $k(\zeta_m)$ is a degree $f$ extension of $k$. We have $k^\mathrm{cyc}=\bar k$, $G_k\cong\hat{ \mathbb{Z} }$ and $\alpha_k: \hat{\mathbb{Z}}\rightarrow \prod_{\ell\ne q} \mathbb{Z}_{p}^\times$ is given by $1\mapsto q$.
Example 18 Suppose $K=\mathbb{R}$ and $m>2$, then $K(\zeta_m)=\mathbb{C}$ and $\alpha_m$ sends the complex conjugation to $-1\mod{m}$ as $\bar\zeta_m=\zeta_m^{-1}$.
Example 19 Suppose $K$ is a non-archimedean local field. If $p\nmid m$, then $K(\zeta_m)/K$ is unramified. Indeed, if $L/K$ is any finite extension, Hensel's lemma implies that $\zeta_m\in L$ if and only if $\zeta_m\in l$. Hence $K(\zeta_m)$ is the unramified extension with residue field $k(\zeta_m)$. We have $\alpha_m(\Frob)=q\mod{m}$ as the case of finite fields.
Exercise 10 Suppose $K$ is a global function field, then $K^\mathrm{cyc}=K\otimes _k\bar k$. However, it is not clear whether this is the maximal abelian extension (indeed, not in general).
Example 20 Let $K$ be a number field and $m>1$. Then $K(\zeta_m)$ is ramified at most over finite places $v\mid m$ and ramified at the real places if $m>2$. For $v\nmid m$, $\alpha_m(\Frob_v)=q_v\mod{m}$, where $q_v=\# k_v$, due to the compatibility of $\alpha_m$ with respect to the inclusion $K\rightarrow K_v$, and the decomposition group $D(v)=\langle \Frob_v\rangle$ maps to $\langle q_v\rangle \subseteq (\mathbb{Z}/m \mathbb{Z})^\times$. For $v|\infty$, $D(v)=\{1,\sigma\}$ maps to $\{\pm1\}\subseteq (\mathbb{Z}/m \mathbb{Z})^\times$ for $m>2$. The question remaining is what $\alpha_m$ takes $\Frob_v$ for $v|m$ (which can be answered by Artin reciprocity law).

Now consider the case $K=\mathbb{Q}$, we have $\alpha_m:\Gal(\mathbb{Q}(\zeta_m)/\mathbb{Q})\cong (\mathbb{Z}/m \mathbb{Z})^\times$. $\mathbb{Q}(\zeta_m)$ is ramified at $p$ if and only if $p|m$ (and $m/2$ is even for $p=2$). In fact, $p$ is totally ramified in $\mathbb{Q}(\zeta_{p^r})$ since $||\zeta_{p^r}-1||^{\phi(p^r)}_p=||p||_p$. Write $m=p^rm'$, then $\mathbb{Q}(\zeta_m)$ is the compositum field of $\mathbb{Q}(\zeta_{m'})$ (where $p$ is unramified) and $\mathbb{Q}(\zeta_{p^r})$ (where $p$ is totally ramified). By Chinese remainder theorem, we know that $$\Gal(\mathbb{Q}(\zeta_m)/\mathbb{Q})\cong\Gal(\mathbb{Q}(\zeta_{m'})/\mathbb{Q})\times\Gal(\mathbb{Q}(\zeta_{p^r})/\mathbb{Q}),$$ hence these two fields are linear disjoint. Comparing ramification index shows that the prime above $p$ of $\mathbb{Q}(\zeta_{m'})$ is totally ramified in $\mathbb{Q}(\zeta_m)$ and the prime above $p$ in $\mathbb{Q}(\zeta_{p^r})$ is unramified in $\mathbb{Q}(\zeta_m)$. So $\mathbb{Q}(\zeta_{m'})$ is the maximal unramified subextension at $p$ in $\mathbb{Q}(\zeta_{m'})$. Hence $I(p)\subseteq D(p)\subseteq \Gal(\mathbb{Q}(\zeta_m)/\mathbb{Q})$ maps isomorphic to $(\mathbb{Z}/p^r \mathbb{Z})^\times$ under $\alpha_m$ and $$D(p)/I(p)\cong\langle\Frob_p\rangle\hookrightarrow \Gal(\mathbb{Q}(\zeta_{m'})/\mathbb{Q})\cong (\mathbb{Z}/m' \mathbb{Z})^\times$$ given by $\Frob_p\mapsto p\mod{m'}$. Therefore $D(p)$ maps isomorphically to $(\mathbb{Z}/p^r \mathbb{Z})^\times\times \langle p\mod{m'}\rangle$ under $\alpha_m$.

Proposition 3 Let $K\subseteq \mathbb{Q}^\mathrm{cyc}$ be a finite subextension. Then there exists a unique smallest $f=\mathfrak{f}_{K/\mathbb{Q}}$ (the notation comes from German word Führer) such that $K\subseteq \mathbb{Q}(\zeta_f)$. Moreover, $p|f$ if and only if $p$ is ramified in $K$.
Proof Since $K/\mathbb{Q}$ is finite, there exists an $m$ such that $K\subseteq \mathbb{Q}(\zeta_m)$. Since $\mathbb{Q}(\zeta_m)\cap \mathbb{Q}(\zeta_{m'})=\mathbb{Q}(\zeta_{(m,m')})$, the gcd of all such $m$'s is the smallest $f$. If $p\nmid f$, then $p$ is unramified in $\mathbb{Q}(\zeta_f)\supseteq K$. Conversely, if $p$ is unramified in $K$, we write $f=p^rf'$, then the restriction map on inertia groups is $I(\mathbb{Q}(\zeta_{f})/p)\cong (\mathbb{Z}/p^r \mathbb{Z})^\times\rightarrow I(K/p)=\{1\}$. Hence $K$ is contained in the fixed field of $\mathbb{Q}(\zeta_f)$ of $(\mathbb{Z}/p^r \mathbb{Z})^\times$, i.e., $\mathbb{Q}(\zeta_{f'})$.
Remark 50
  1. $\mathfrak{f}_{K/\mathbb{Q}}$ is roughly the Artin conductor of $K$ (after adding a possible factor at $\infty$), which "controls" the abelian extension of $\mathbb{Q}$.
  2. $\mathbb{Q}(\zeta_m)$ and the maximal totally real subextension $\mathbb{Q}(\zeta_m)^+$ are examples of ray class fields, which form a cofinial system of finite abelian extensions of $K$.

TopArtin maps

Recall the following commutative diagram $$\xymatrix{\alpha: \Gal(\mathbb{Q}^\mathrm{cyc}/\mathbb{Q})\ar[r]^-{\cong}\ar[d] & \hat{\mathbb{Z}}^\times \ar[d]\\\alpha_m: \Gal(\mathbb{Q}(\zeta_m)/\mathbb{Q}) \ar[r]^-{\cong} & (\mathbb{Z}/m \mathbb{Z})^\times. }$$ We know that for $p\mid m$, $\mathbb{Z}_p^\times$ maps surjectively to the inertia group $(\mathbb{Z}/p^r \mathbb{Z})^\times\subseteq (\mathbb{Z}/m \mathbb{Z})^\times\cong\Gal(\mathbb{Q}(\zeta_m)/\mathbb{Q})$ and $1+p \mathbb{Z}_p\subseteq \mathbb{Z}_p^\times$ maps surjectively to the wild inertia subgroup, i.e., the $p$-Sylow subgroup of $(\mathbb{Z}/p^r \mathbb{Z})^\times$. For $p\nmid m$, the element $u_p'{}=(p,\ldots,p, 1,p, p,\ldots)\in \hat{\mathbb{Z}}^\times$ (with 1 at the place $p$) maps to $p\in (\mathbb{Z}/m \mathbb{Z})^\times$, which is equal to $\alpha_m(\Frob_p)$.

Definition 49 Let $K\subseteq \mathbb{Q}(\zeta_m)$ be a finite abelian extension $\mathbb{Q}$. Define the Artin map $$\Psi_{K/\mathbb{Q}}: \mathbb{A}_\mathbb{Q}^\times/\mathbb{Q}^\times\twoheadrightarrow \Gal(K/\mathbb{Q})$$ by sending $\mathbb{R}_{>0}$ to $\{1\}$ and $\hat{\mathbb{Z}}^\times\rightarrow \Gal(K/\mathbb{Q})$ as the restriction $\Gal(\mathbb{Q}^\mathrm{cyc}/\mathbb{Q})\rightarrow \Gal(\mathbb{Q}(\zeta_m)/\mathbb{Q})\rightarrow \Gal(K/\mathbb{Q})$. It is a continuous surjection. It follows that $\mathbb{Z}_p^\times$ maps surjectively to the inertia group $I(K/p)$ and $p\in \mathbb{Q}_p^\times$ maps to $\Frob_p^{-1}$.

10/03/2012

Lemma 12 $\Psi_{K/\mathbb{Q}}(\mathbb{Q}_p^\times)=D(K/p)$, $\Psi_{K/\mathbb{Q}}(\mathbb{Z}_p^\times)=I(K/p)$. When $p$ is unramified is $K$, $\Psi_{K/\mathbb{Q}}(u_p)=\Frob_p^{-1}$ (opposite to the usual Artin map), where $u_p$ is the image of $p\in \mathbb{Q}_p^\times\hookrightarrow \mathbb{A}_\mathbb{Q}^\times/\mathbb{Q}^\times$.
Proof We already know that $\Psi_{K/\mathbb{Q}}(\mathbb{Z}_p^\times)=I(K/p)$. Without loss of generality we may assume $K=\mathbb{Q}(\zeta_m)$. Suppose $p\nmid m$ is unramified, then $$u_p=(1,\ldots,1,p,1,\ldots,,1)=(1/p,\ldots,1/p,1,1/p,\ldots,1/p)\in \mathbb{A}_\mathbb{Q}^\times/\mathbb{Q}^\times$$ maps to $p^{-1}\mod{m}\in (\mathbb{Z}/m \mathbb{Z})^\times$, i.e., $\Frob_p^{-1}$. For arbitrary $m$, we write $m=p^rm'$. Then $\Psi_{\mathbb{Q}(\zeta_{m'})/\mathbb{Q}}(u_p)=p^{-1}\in (\mathbb{Z}/m' \mathbb{Z})^\times$. Because any prime $\ell\ne p$ is unramified in $\mathbb{Q}(\zeta_{p^r})$, we also know that $\Psi_{\mathbb{Q}(\zeta_{p^r})/\mathbb{Q}}(u_p)=1\in (\mathbb{Z}/p^r \mathbb{Z})^\times$. Hence $$\Psi_{K/\mathbb{Q}}(u_p)=(p^{-1},1)\in (\mathbb{Z}/m' \mathbb{Z})^\times\times (\mathbb{Z}/p^r \mathbb{Z})^\times.$$ We know that $$\Psi(\mathbb{Q}_p^\times)=\Psi(\mathbb{Z}_p^\times)\langle\Psi(u_p)\rangle=(\mathbb{Z}/p^r \mathbb{Z})^\times\times \langle p\mod{m'}\rangle\cong D(K/\mathbb{Q}).$$ This completes the proof.
Remark 51 In general, to compute the image of an idele under the Artin map, one needs scale it by a global element (via strong approximation) to obtain an idele with units at all ramified places (these are mapped into the inertia groups) and compute the product of the images of components at the unramified places (i.e., powers of the Frobenius elements).
Definition 50 Now for any $K\subseteq \mathbb{Q}^\mathrm{cyc}$ with infinite degree, $\Psi_{K/\mathbb{Q}}$ makes sense by taking the inverse limit over $K'\subseteq K$ finite over $\mathbb{Q}$. $\Psi_{K/\mathbb{Q}}$ is obviously continuous by construction. Since $\hat{\mathbb{Z}}^\times$ is compact and the image is dense (surjective on every finite $\Gal(K/\mathbb{Q})$, we find the Artin map $\Psi_\mathbb{Q}:=\Psi_{\mathbb{Q}^\mathrm{cyc}/\mathbb{Q}}$ is surjective.
Remark 52 Write $m=\prod p_i^{e_i}$ and $K=\mathbb{Q}(\zeta_m)$. Then $$\ker (\Psi_{K_m/\mathbb{Q}})=\mathbb{R}_{>0}\times \prod_{\ell\nmid m} \mathbb{Z}_{\ell}^\times\times\prod_{i=1}^r(1+p_i^r \mathbb{Z}_p)=: U_m.$$ Under the isomorphism $\mathbb{R}_{>0}\times  \hat{\mathbb{Z} }^\times\cong \mathbb{A}_\mathbb{Q}^\times/\mathbb{Q}^\times$, $U_m$ corresponds to the subgroup $\mathbb{Q}^\times U_m/\mathbb{Q}^\times$ of the idele class group $\mathbb{A}_\mathbb{Q}^\times/\mathbb{Q}^\times$. These form a cofinal system of subgroups of $\mathbb{A}_\mathbb{Q}^\times/\mathbb{Q}^\times$. Moreover, for any $K\subseteq K_m$, $\ker\Psi_{K/\mathbb{Q}}\supseteq \ker \Psi_{K_m/\mathbb{Q}}=\mathbb{Q}^\times U_m/\mathbb{Q}^\times$ is an open subgroup.

The following proposition summarizes easy properties of the Artin map $\Psi_\mathbb{Q}$. We will see how they generalize for any global field.

Proposition 4
  1. $K\mapsto \ker(\Psi_{K/\mathbb{Q}})$ is a bijection between the finite subextension of $\mathbb{Q}^\mathrm{cyc}$ and open subgroups (of finite index) of $\mathbb{A}_\mathbb{Q}^\times/\mathbb{Q}^\times$. Indeed, any open subgroup $U$ contains some $\mathbb{Q}^\times U_m/\mathbb{Q}^\times$, hence $\mathbb{A}_\mathbb{Q}^\times/\mathbb{Q}^\times U$ is a subgroup of $\mathbb{A}_\mathbb{Q}^\times/\mathbb{Q}^\times U_m\cong\Gal(\mathbb{Q}(\zeta_m)/\mathbb{Q})$, hence corresponds to a finite extension $K\subseteq K_m$.
  2. $\Psi_\mathbb{Q}$ is continuous (and surjective onto $\Gal(\mathbb{Q}^\mathrm{cyc}/\mathbb{Q})$).
  3. $\ker(\Psi_\mathbb{Q})=\mathbb{R}_{>0}$, the connected component of $\mathbb{A}_\mathbb{Q}^\times/\mathbb{Q}^\times$.

Now let us turn to the local case $K=\mathbb{Q}_p$.

Suppose $p\nmid m$, then $\mathbb{Q}_p(\zeta_m)$ is unramified over $\mathbb{Q}_p$. Recall that $\alpha_m:\Gal(\mathbb{Q}_p(\zeta_m)/\mathbb{Q}_p)\hookrightarrow (\mathbb{Z}/m \mathbb{Z})^\times$ sending $\Frob_p$ to $p\mod{m}$, so $\Gal(\mathbb{Q}_p(\zeta_m)/\mathbb{Q}_p)\cong \langle p\rangle=\mathbb{Z}/f_m \mathbb{Z}$, where $f_m$ is the order of $p$ in $(\mathbb{Z}/m \mathbb{Z})^\times$. On the other hand, $\mathbb{Q}_p(\zeta_{p^r})$ is totally ramified of degree $\phi(p^r)$. Hence $\alpha_{p^r}: \Gal(\mathbb{Q}_p(\zeta_{p^r})/\mathbb{Q}_p)\cong (\mathbb{Z}/p^r \mathbb{Z})^\times$ is an isomorphism. Hence for general $m=m'p^r$, we have $\Gal(\mathbb{Q}_p(\zeta_m)/\mathbb{Q}_p)\cong \mathbb{Z}/f_{m'} \mathbb{Z}\times (\mathbb{Z}/p^r \mathbb{Z})^\times$ and the inertia subgroup $I_m\cong (\mathbb{Z}/p^r \mathbb{Z})^\times$.

Taking the inverse limit of the exact sequence $$0\rightarrow I_m\rightarrow \Gal(\mathbb{Q}_p(\zeta_m)/\mathbb{Q}_p)\rightarrow \Gal(\mathbb{F}_{p^{f_{m'}}}/\mathbb{F}_p)\rightarrow 0$$ we obtain two isomorphic exact sequences $$\xymatrix{0\ar[r] &\I\ar[r] \ar[d]^\cong& \Gal(\mathbb{Q}_p^\mathrm{cyc}/\mathbb{Q}_p)\ar[r]\ar[d]^\cong &\Gal(\overline{\mathbb{F}_p}/\mathbb{F}_p)\ar[r]\ar[d]^\cong& 0 \\ 0\ar[r] & \mathbb{Z}_p^\times\ar[r] & \hat{\mathbb{Z}} \times\mathbb{Z}_p^\times\ar[r] & \hat{\mathbb{Z}}\ar[r] & 0.}$$

Definition 51 We define the Artin map $\Psi_{\mathbb{Q}_p}: \mathbb{Q}_p^\times\cong q^\mathbb{Z}\times \mathbb{Z}_p^\times\rightarrow \Gal(\mathbb{Q}_p^\mathrm{cyc}/\mathbb{Q}_p)$ sending $$(p,x)\mapsto (-1, x)\in \mathbb{Z}\times \mathbb{Z}_p^\times\subseteq \hat{\mathbb{Z}}\times \mathbb{Z}_p^\times\cong\Gal(\mathbb{Q}_p^\mathrm{cyc}/\mathbb{Q}_p).$$ This is a continuous map with dense image and maps $\mathbb{Z}_p^\times$ surjective onto the inertia group.
Remark 53 The image $W$ of $\Psi_{\mathbb{Q}_p}$ consists of $g\in \Gal(\mathbb{Q}^\mathrm{cyc}/\mathbb{Q}))$ which induces an integral power of $\Frob_p$ on $\overline{\mathbb{F}_p}$. $\mathbb{Q}_p^\times\rightarrow W$ is continuous but not homeomorphism for the subspace topology on $W$. Indeed $\Psi_{\mathbb{Q}_p}(\mathbb{Z}_p^\times)=\{0\}\times \mathbb{Z}_p^\times$ is not open in $\hat{\mathbb{Z}}\times \mathbb{Z}_p^\times$ (in other words, the subspace topology $\mathbb{Z}\subseteq \hat{\mathbb{Z}}$ is not discrete).

10/05/2012

Remark 54 Suppose $K\subseteq \mathbb{Q}_p^\mathrm{cyc}$, we define the Artin map $\Psi_{K/\mathbb{Q}_p}: \mathbb{Q}_p^\times\rightarrow\Gal(K/\mathbb{Q}_p)$ by composing with the natural quotient map $\Gal(\mathbb{Q}_p^\mathrm{cyc}/\mathbb{Q}_p)\rightarrow \Gal(K/\mathbb{Q}_p)$. In particular, when $K/\mathbb{Q}$ is finite, $\Psi_{K/\mathbb{Q}_p}$ is surjective since the cosets of $\ker(\Gal(\mathbb{Q}_p^\mathrm{cyc}/\mathbb{Q}_p)\rightarrow\Gal(K/\mathbb{Q}_p))$ are open. Moreover, $\Psi_{K/\mathbb{Q}_p}(\mathbb{Z}_p^\times)=I(K/p)$ is the inertia group and $\Psi_{K/\mathbb{Q}_p}(1+p \mathbb{Z}_p)$ is the wild inertia group.

The following important proposition is left as an exercise.

Proposition 5 (Local-global compatibility) We have the following commutative diagram $$\xymatrix{ \mathbb{Q}_p^\times \ar[r]^-{\Psi_{\mathbb{Q}_p}}\ar[d] & \Gal(\mathbb{Q}_p^\mathrm{cyc}/\mathbb{Q}_p) \ar[d]\\ \mathbb{A}_\mathbb{Q}^\times/\mathbb{Q}^\times\ar[r]^-{\Psi_{\mathbb{Q}}} & \Gal(\mathbb{Q}^\mathrm{cyc}/\mathbb{Q}) .}$$
Remark 55 We have similar (trivial) local-global compatibility for $\mathbb{R}$ and $\mathbb{C}$.

TopWeil groups

Let $K$ be a nonarchimedean local field (resp. a global function field) and $k$ be the residue field (resp. the constant field). The residue field (resp. constant field) of $K^\mathrm{sep}$ is $\bar k$. Sow we have a continuous surjection $\pi: \Gamma=\Gal(K^\mathrm{sep}/K)\rightarrow \hat{\mathbb{Z}}$. Denote its kernel by $I$ (which is the inertia group in the local case). As a group, the Weil group is simply $W=\pi^{-1}(\mathbb{Z})\subseteq \Gamma$, i.e., the elements in $\Gamma$ which induces integral powers of Frobenius on $\bar k$. However, as we have seen in Remark 53, $I$ may not be open in $W$ under the subspace topology from $\Gamma$. But there exists a finer topology on $W$ such that $I$ is open in $W$. under this finer topology, $\Psi_K: K^\times\rightarrow W^\mathrm{ab}$ (in the local case) or $\Psi_{K}: \mathbb{A}_K^\times/K^\times\rightarrow W^\mathrm{ab}$ (in the global function field case) will be isomorphisms of topological groups.

More precisely, suppose we have a short exact sequence of profinite groups $$0\rightarrow I\rightarrow \Gamma\xrightarrow{\pi} \hat{\mathbb{Z}}\rightarrow0$$ and $W=\pi^{-1}(\mathbb{Z})$.

Lemma 13 There exists a unique topology on $W$ such that $I$ is open in $W$, $I$ has the subspace topology under $I\hookrightarrow \Gamma$ and any splitting of $W\rightarrow \mathbb{Z}$ induces an isomorphism $W\cong I\times \mathbb{Z}$ of topological groups.
Proof See the handouts for details.

Under this topology on the Weil group $W$, $W\hookrightarrow \Gamma$ is continuous and has dense image (but is not a homeomorphism onto its image). Moreover, the topology is compatible under abelianization as in following proposition.

Proposition 6 Let $\pi^\mathrm{ab}: \Gamma^\mathrm{ab}\rightarrow \hat{\mathbb{Z}}$ be the abelianization of $\pi$ and $W'=(\pi^\mathrm{ab})^{-1}(\mathbb{Z})$: $$\xymatrix{0\ar[r] & I\ar[r]\ar[d] & W\ar[r]\ar[d] & \mathbb{Z}\ar[r]\ar[d]^{\cong} &0\\ 0\ar[r] & I' \ar[r] & W'  \ar[r]& \mathbb{Z}\ar[r] &0.}$$ Then $W'=W^\mathrm{ab}$ with$W\rightarrow W'$ the abelianization as topological groups (but not for $I\rightarrow I'$).
Remark 56 Notice that the abelianization of topological groups usually does not preserve injectivity!

The following theorem allows us to use Weil groups as a replacement of to classify finite abelian extensions of local or global function fields.

Theorem 20 The maps $H\mapsto H\cap W$ and $\bar H' \mapsfrom H'$ gives a bijection between open subgroups of $\Gamma$ and open finite index subgroup of $W$. For any $H \subseteq\Gamma$ an open subgroup, $W/(W\cap H)$ is isomorphic to $\Gamma/H$ as discrete coset spaces.
Exercise 11 This bijection extends to a bijection between closed subgroups $H $ of $\Gamma$ such that $\pi(H)=\{0\}$ or $\pi(H)$ has finite index in $\hat{\mathbb{Z}}$, and closed subgroups of $W$. This will allows us to partially classify infinite abelian extensions of local fields and global function fields.

TopStatement of global class field theory

Theorem 21 (Global class field theory) Let $K$ be a global field. There exists a continuous homomorphism, called the Artin map, $\Psi_K: \mathbb{A}_K^\times/K^\times\rightarrow G_K^\mathrm{ab}$ with dense image, satisfying:
  1. For any $L/K$ finite abelian and $v\in V_K$. The composition $$K_v^\times\hookrightarrow \mathbb{A}_K^\times/K^\times\xrightarrow {\Psi_K}G_K^\mathrm{ab}\rightarrow\Gal(L/K)$$ satisfies:
    1. When $v$ is nonarchimedean, it kills $\mathcal{O}_v^\times$ if and only if $v$ is unramified in $L$, in which case it sends any uniformizer to $\Frob_v$.
    2. When $v$ is archimedean, it kills the connected component of $1\in K_v^\times$. When $v$ is unramified, it kills all of $K_v^\times$; when $v$ is ramified, it sends $-1$ to the complex conjugation.
  2. (Existence theorem) Every finite index open subgroup $H\subseteq \mathbb{A}_K^\times/K^\times$ arises from a finite extension $L/K$ as the kernel of $\Psi_{L/K}: \mathbb{A}_K^\times/K^\times \xrightarrow{\Psi_K}G_K^\mathrm{ab}\rightarrow \Gal(L/K)$. $L$ is called the class field to the subgroup $H $.
Remark 57 Since $G_K^\mathrm{ab}$ is the inverse limit of $\Gal(L/K)$ for finite abelian extensions $L/K$, we know that $\Psi_K$ is determined by the maps $\Psi_{L/K}$. Moreover, $\Psi_{L/K}$ is surjective since $\Psi_K$ has dense image.
Lemma 14 Property (a) in Theorem 21 uniquely characterizes $\Psi_K$.
Proof Observe that if $S$ is a finite set of places of $K$. Then $X_S=K^\times\{x\in \mathbb{A}_K^\times : x_v=1, \forall v\in S\}$ is dense in $\mathbb{A}_K^\times$ by the weak approximation (Remark 47). If $L/K$ is a finite abelian extension, we let $S$ contain $S_\infty$ and all the ramified places. Then $\Psi_{L/K}|_{X_S}$ is determined by (a), hence by continuity $\Psi_{L/K}$ is determined by (a).
Remark 58 In particular, $\Psi_\mathbb{Q}$ is the opposite to the one defined in Definition 50.

Assuming Theorem 21, let us prove the following "real version" of the existence theorem (and please hope for the "real real version").

Theorem 22 (Existence Theorem) $L\mapsto \ker(\Psi_{L/K})$ is an inclusion reverse bijection between finite abelian extensions $L/K$ and open subgroups of finite index in $\mathbb{A}_K^\times/K^\times$.
Proof It suffices to show the injectivity. Suppose $L$ and $L'$ are finite abelian extensions such that $\ker(\Psi_{L/K})=\ker(\Psi_{L'/K})$. Let $H_L$ and $H_{L'}$ be the corresponding open subgroup in $G_K^\mathrm{ab}$, then $\Psi_K^{-1}(H_L)=\Psi_K^{-1}(H_{L'})$. Hence $H_L\cap \Im (\Psi_K)=H_{L'}\cap \Im(\Psi_K)$. Suppose $g\in H_L\setminus H_{L'}$, then there is an neighborhood $U(g)\subseteq H_L\setminus H_{L'}$ since $H_{L}$ and $H_{L'} $are both open and closed. But $\Im(\Psi_K)$ is dense, this would contradict $H_L\cap \Im (\Psi_K)=H_{L'}\cap \Im(\Psi_K)$. We conclude that $H_L=H_{L'}$, thus $L=L'$.

10/10/2012

Suppose $K$ is a number field and $D_K$ be the connected component of $\mathbb{A}_K^\times/K^\times$. We have shown that $(\mathbb{A}_K^\times/K^\times)/D_K$ is profinite (Lemma 10). $G_K^\mathrm{ab}$ is profinite, thus contains no divisible elements, hence $D_K\subseteq \ker(\Psi_K)$. But the image of $\Psi_K$ is dense, we know that $\Psi_K$ is surjective in this case. On the other hand, $(\mathbb{A}_K^\times/K^\times)/D_K$ is profinite implies that the intersection of all open subgroups $H\subseteq (\mathbb{A}_K^\times/K^\times)/D_K$ is trivial. We find that the intersection of all open subgroups (of finite index) of $H\subseteq \mathbb{A}_K^\times/K^\times$ is $D_K$, as any subgroup of finite index contains divisible elements $D_K$. Therefore $\ker(\Psi_K)=D_K$ by the existence theorem. Namely, we have shown

Theorem 23 Suppose $K$ is a number field, then $\Psi_K$ induces an isomorphism of topological groups $(\mathbb{A}_K^\times/K^\times)/D_K\cong G_K^\mathrm{ab}$.

We thus deduce a stronger version of existence theorem for number fields.

Corollary 4 The map $L\mapsto H=\ker(\Psi_{L/K})$ is a bijection between all abelian extensions $L/K$ and closed subgroups of $H\subseteq \mathbb{A}_K^\times/K^\times$ such that $H\supseteq (\prod_{v|\infty}K_v^\times)^0$. Under this bijection, $L$ is called the class field of $H $.
Remark 59 Suppose $H\subseteq \mathbb{A}_K^\times/K^\times$ is open. Write $H^1=H\cap (\mathbb{A}_K^\times)^1$, then we have an isomorphism $((\mathbb{A}_K^\times)^1/K^\times)/H^1)\cong (\mathbb{A}_K^\times/K^\times)/H$ using the splitting $\mathbb{A}_K^\times/K^\times\cong (\mathbb{A}_K^\times)^1/K^\times\times \mathbb{R}_{>0}$. But $(\mathbb{A}_K^\times)^1/K^\times$ is compact, hence any open subgroup of $\mathbb{A}_K^\times/K^\times$ is of finite index. This is false for $K$ a global function field, e.g., $(\mathbb{A}_K^\times)^1/K^\times$ is open and has infinite quotient is $q^\mathbb{Z} $.

Now suppose $K$ is a global function field. We have an isomorphism of topological groups $\mathbb{A}_K^\times/K^\times\cong q^\mathbb{Z}\times (\mathbb{A}_K^\times)^1/K^\times$, a product of a discrete group and a profinite group. Then the intersection of all open subgroup $H\subseteq \mathbb{A}_K^\times/K^\times$ is trivial, hence $\Psi_K$ is injective.

But $\Psi_K$ is not surjective, indeed we claim that if $||x||_K=q^{-n}$, then $\Psi_K(x)|_{\bar k}=\Frob^{n}$, an integral power of the Frobenius. In fact, for a finite extension $l/k$, then $L=K\otimes_kl$ is finite abelian and unramified everywhere and $\Gal(L/K)=\Gal(l/k)=\langle\Frob\rangle$ (Theorem 13, or the base change of an etale map is etale). It suffices to show that $\Psi_K(x)|_l=\Frob^n$. For $v\in V_K$, $\Frob_v\in \Gal(L/K)$ lifts $x\mapsto x^{\# k_v}=(x^q)^{[k_v: k]}$, By definition, $\Psi_{L/K}(x)=\prod_v\Frob_v^{\ord_v(x)}=\Frob^n$ as $L/K$ is unramified everywhere and $n=\sum_v \ord_v(x)$. In other words, we have proved the following diagram commutes $$\xymatrix{0\ar[r] & (\mathbb{A}_K^\times)^1/K^\times \ar[r] \ar[d] & \mathbb{A}_K^\times/K^\times \ar[r] \ar[d]^{\Psi_K} & q^\mathbb{Z} \ar[r] \ar[d] & 0\\ 0 \ar[r] & I_K \ar[r] & G_K^\mathrm{ab} \ar[r] & G_{\bar k}\cong \hat{\mathbb{Z}} \ar[r] & 0.}$$

It follows that $\Psi_K((\mathbb{A}_K^\times)^1/K^\times)\subseteq I_K$. We claim that $\Psi_K: (\mathbb{A}_K^\times)^1/K^\times\rightarrow I_K$ is actually an isomorphism. We already know it is injective, so it suffices to show the surjectivity. Choose $z\in \mathbb{A}_K^\times/K^\times$ such that $||z||_K=q^{-1}$. Then there exists a unique continuous homomorphism $G_K\cong \hat{\mathbb{Z}}\rightarrow G_K^\mathrm{ab}$ sending 1 to $\Psi_K(z)$ by the universal property of profinite groups. Hence $G_K^\mathrm{ab}\cong I_K\times \hat{\mathbb{Z}}$ as topological groups. Under this identification, $\Psi_K$ is simply $(\mathbb{A}_K^\times)^1/K^\times\times \mathbb{Z}\xrightarrow{(\Psi_K,\Id)} I_K\times \hat{\mathbb{Z}}$. Let $H=\Psi_K((\mathbb{A}_K^\times)^1/K^\times)\subseteq I_K$. Then $H $ is a closed subgroup and $\Psi_K(\mathbb{A}_K^\times/K^\times)= H\times \mathbb{Z}$, hence the closure of $\Psi_K(\mathbb{A}_K^\times/K^\times)$ in $G_K^\mathrm{ab}$ is $H\times \hat{\mathbb{Z}}$. Since the image of $\Psi_K$ is dense, it follows that $H\times \hat{\mathbb{Z}}=I_K\times \hat{\mathbb{Z}}$, which proves the surjectivity.

We have proved

Proposition 7 $\Psi_K$ induces vertical isomorphisms of topological groups $$\xymatrix{0 \ar[r] & (\mathbb{A}_K^\times)^1/K^\times \ar[r] \ar[d]^{\cong} & \mathbb{A}_K^\times/K^\times \ar[r] \ar[d]^{\cong}& q^\mathbb{Z} \ar[r] \ar[d]^{\cong} & 0\\0\ar[r] & I_K \ar[r] & W \ar[r] & \mathbb{Z} \ar[r] &0.}$$ So $\mathbb{A}_K^\times/K^\times$ can viewed as the Weil group $W$ and $G_K^\mathrm{ab}$ is the profinite completion of $W$.
Corollary 5 $L\mapsto \ker(\Psi_{L/K})$ gives a bijection between all abelian extensions $L/K$ such that the constant field extension $l/k$ is either finite or equal to $\bar k$, and closed subgroups of $\mathbb{A}_K^\times/K^\times$.
Proof By Exercise 11.

10/12/2012

TopNorm and Verlagerung functoriality

Let $K$ be a global field and $L/K$ be any finite separable extension. $L^\mathrm{sep}\cong K^\mathrm{sep}$ gives us a canonical map $G_L^\mathrm{ab}\rightarrow G_K^\mathrm{ab}$. We also have the compatibility of local and global norms: $$\xymatrix{\mathbb{A}_L^\times\ar[r]^-{N_{L/K}} &\mathbb{A}_K^\times\\   L_w^\times \ar[r]^-{N_{L_w/K_v}} \ar[u] & K_v^\times. \ar[u]}$$

Proposition 8 (Norm functoriality) The follows diagram commutes: $$\xymatrix{\mathbb{A}_L^\times/L^\times \ar[r]^-{\Psi_L} \ar[d]^{N_{L/K}}& G_L^\mathrm{ab}\ar[d]\\ \mathbb{A}_K^\times/K^\times \ar[r]^-{\Psi_K} & G_K^\mathrm{ab}.}$$
Proof We only need to show the diagram for $K'/K$ is finite abelian and $L'=LK'$ (which finite abelian over $L$): $$\xymatrix{\mathbb{A}_L^\times/L^\times \ar[r]^-{\Psi_{L'/L}} \ar[d]^{N_{L/K}}& \Gal(L'/L) \ar[d]^\phi\\ \mathbb{A}_K^\times/K^\times \ar[r]^-{\Psi_{K'/K}} & \Gal(K'/K).}$$ Suppose $S\subseteq V_L$ is a finite set of places containing:
  • $v$ such that $w=v|_K$ is ramified in $L/K$
  • $v$ such that $w=v|_K$ is ramified in $K'/K$
  • $v$ ramified in $L'/L$
  • $v|\infty$.

Since $L^\times\{x\in \mathbb{A}_L^\times: x_v=1, \forall v\in S\}$ is dense in $\mathbb{A}_L^\times$ by weak approximation. It suffices to show that $$\phi\circ\Psi_{L'/L}(\pi_v)=\Psi_{K'/K}N_{L/K}(\pi_v),\quad 1=\Psi_{K'/K}N_{L/K}(\mathcal{O}_v^\times)$$ for any $v\not\in S$. Let $v'|v$ and $w'{}=v'|_{K'}$. For the latter equality, notice that $N_{L/K}(\mathcal{O}_v^\times)=\mathcal{O}_v^\times\subseteq \mathcal{O}_w^\times$ and $\Psi_{K'/K}(\mathcal{O}_w^\times)=1$ since $w$ is unramified. It remains to prove the first equality. Let $\pi_w\in K_w^\times$ be a uniformizer. Then $\pi_v=\pi_w$ since $L_v/K_w$ is unramified. We compute $$\Psi_{K'/K}N_{K_v/K_w}(\pi_w)=\Psi_{K'/K}(\pi_v^{[L_w:K_v]})=\Frob_w^{f(v/w)}.$$ On the other hand, $\phi\circ\Psi_{L'/L}(\pi_v)=\Frob_v|_{K'}$. Since both side of the first equality are the unique element of $\Gal(K'/K)$ which fixes $v$ and reduces to $x\mapsto x^{\# l_v}$, they are actually equal.

Corollary 6 Suppose $L/K$ is finite abelian, then $\Psi_{L/K}N_{L/K}=1$. In other words,$N_{L/K}(\mathbb{A}_L^\times/L^\times)\subseteq\ker(\Psi_{L/K})$.
Proof Take $K'{}=L$ in the previous theorem.
Theorem 24 (Global norm index inequality) Suppose $K$ is a global field and $L/K$ is a finite and separable extension. Then $[\mathbb{A}_K^\times/K^\times: N_{L/K}(\mathbb{A}_L^\times/L^\times)]\le[L:K]$.
Proof We will give an easy analytic proof later (cf. Exercise 17). In fact, the cohomological proof will even give a division relation.
Remark 60 If particular, we conclude that when $L/K$ is finite abelian we actually have an equality, as the $\Psi_K$ maps surjectively onto $\Gal(L/K)$ in this case.
Theorem 25 (Existence theorem III) Suppose $K$ is a global field and $L/K$ is a finite and separable extension. Then $[\mathbb{A}_K^\times/K^\times: N_{L/K}(\mathbb{A}_L^\times/L^\times)]=[L:K]$ if and only if $L/K$ is abelian, in which case $\ker(\Psi_{L/K})=N_{L/K}(\mathbb{A}_L^\times/L^\times)$.
Proof It remains to prove the "only if" direction, which is left as an exercise (hint: the left-hand-side is always $[L^\mathrm{ab}:K]$.

Now let us briefly turn to the verlagerung functoriality of the Artin map.

Definition 52 Let $G$ be a group and $S\subseteq G$ be a subgroup of finite index. For any $g,\sigma\in G$, let $f_\sigma$ be the smallest $f\ge1$ such that $\sigma g^f \sigma^{-1}\in S$. This only depends the choice of $\sigma$ in $S\setminus G/ \langle g\rangle$, a finite double coset. Write $\{\sigma_1,\ldots,\sigma_r\}$ be the coset representatives. We define the verlagerung (or transfer) $\Ver: g\mapsto \prod_{i=1}^r\sigma_i g^{f_{\sigma_i}}\sigma_i^{-1}\in S$. Then $\Ver: G^\mathrm{ab}\rightarrow S^\mathrm{ab}$ is a group homomorphism. In terms of group cohomology, this is the restriction map $\Res: H_1(G, \mathbb{Z})\rightarrow H_1(S, \mathbb{Z})$ and functorial in $(G,S)$.

Suppose $K$ is a field and $L/K$ is finite separable. Then $G_L\subseteq G_K$ is of finite index (depending on the choice of an isomorphisms $L^\mathrm{sep}\cong K^\mathrm{sep}$). We then obtain the verlagerung $\Ver: G_K^\mathrm{ab}\rightarrow G_L^\mathrm{ab}$, a continuous map of the topological abelianization of $G_K$ and $G_L$, which does not depend on the choice of $L^\mathrm{sep}\cong K^\mathrm{sep}$.

Proposition 9 Suppose $K$ is a global field and $L/K$ is a finite separable extension. The following diagram commutes: $$\xymatrix{\mathbb{A}_K^\times/K^\times \ar[r]^-{\Psi_K} \ar[d] & G_K^\mathrm{ab} \ar[d]^{\Ver }\\ \mathbb{A}_L^\times/L^\times \ar[r]^-{\Psi_L} & G_L^\mathrm{ab}.}$$
Proof Suppose $K'/K$ is finite Galois and $L=K'L$. It suffices to show at the finite level. The remaining check will be an easy calculation which we leave as an exercise.
Remark 61 Notice the nice interchange of position between $\Ver$ and $N_{L/K}$ in the previous two propositions.

TopStatement of local class field theory

Theorem 26 (Local class field theory) Let $K$ be a nonarchimedean local field. There exists a canonical injective homomorphism $\Psi_K: K^\times\rightarrow G_K^\mathrm{ab}$ satisfying:
  1. $$\xymatrix{0 \ar[r]  & \mathcal{O}_K^\times \ar[r] \ar[d]^{\cong} & K^\times \ar[r]^{\ord} \ar[d]_{\Psi_K}^{\cong} & \mathbb{Z} \ar[r] \ar[d]& 0\\ 0 \ar[r] & I_K\ar[r] & W \ar[r]& \mathbb{Z} \ar[r]& 0}$$ Namely, the Artin map identifies $K^\times$ with the Weil group $W$
  2. If $L/K$ is finite separable, then we have norm functoriality and verlagerung functoriality for $\Psi_L$ and $\Psi_K$: $$\xymatrix{L^\times\ar[r] \ar[d]^{N_{L/K}}& G_L^\mathrm{ab}\ar[d]\\ K^\times \ar[r]& G_K^\mathrm{ab},}\quad \xymatrix{L^\times\ar[r] & G_L^\mathrm{ab}\\ K^\times \ar[r]\ar[u] & G_K^\mathrm{ab}\ar[u]^{\Ver}.}$$
  3. If $L/K$ is finite abelian, then $\Psi_{L/K}: K^\times\rightarrow G_K^\mathrm{ab}\rightarrow\Gal(L/K)$ is surjective with $\ker(\Psi_{L/K})=N_{L/K}(L^\times)$, $\Psi_{L/K}(1+\mathfrak{m}_K^i)=G^i$ (the $i$-th ramification group in the upper numbering, $i\ge1$) and $\Psi_{L/K}(\mathcal{O}_K^\times)=G^0$ (the inertia group).
  4. (Existence theorem) Every open subgroup of finite index of $K^\times$ is of the form $N_{L/K}(L^\times)$ for some $L/K$ finite abelian.

10/15/2012

Remark 62 The statements in the previous theorem are not independent. For example, given the local norm index equality, the norm functoriality implies the first part of (c). Also (a) implies the second part of (c) concerning $\mathcal{O}_K^\times$. Moreover, (a) and (c) implies (d) by Theorem 20.
Remark 63 This theorem does not (at least, not obviously) uniquely characterize $\Psi_K$. But if $L/K$ is unramified, then $\Psi_{L/K}$ is uniquely determined by (a).
Remark 64 Of course, there are obvious Artin maps for archimedean local fields: $$\Psi_\mathbb{R}: \mathbb{R}^\times/\mathbb{R}_{>0}\xrightarrow{\cong}\Gal(\mathbb{C} /\mathbb{R}),$$ and $$\Psi_\mathbb{C}: \mathbb{C}^\times/\mathbb{C}^\times\xymatrix{\cong}\Gal(\mathbb{C} /\mathbb{C}).$$

Analogously, the local existence theorem will follow from the local norm index inequality.

Theorem 27 (Local norm index inequality) Suppose $K$ is a local field and $L/K$ is a finite and separable extension. Then $[K^\times: N_{L/K}(L^\times)]\le[L:K]$.
Theorem 28 (Existence Theorem) Suppose $K$ is a local field and $L/K$ is a finite and separable extension. Then $[K^\times: N_{L/K}(L^\times)]=[L:K]$ if and only if $L/K$ is abelian, in which case, $N_{L/K}(L^\times)\cong\ker(\Psi_{L/K})$.

The local-global compatibility will follow from defining the global Artin map via "gluing" local Artin maps.

Theorem 29 (Local-global compatibility) Suppose $K$ is a global field and $v\in V_K$. Then we have a commutative diagram $$\xymatrix{ K_v^\times \ar[d]  \ar[r]^-{\Psi_{K_v}} & G_{K_v}^\mathrm{ab}\ar[d] \\ \mathbb{A}_K^\times/K^\times \ar[r]^-{\Psi_K}  & G_K^\mathrm{ab} .}$$ If $L/K$ is finite abelian and $w|v$, then $L_w/K_v$ is also finite abelian and we have a corresponding commutative diagram at finite level.

We can then derive part of (a) in global class field theory (Theorem 21) using the local-global compatibility at archimedean places.

Corollary 7 Suppose $K$ is a number field and $v\in V_K$ is a real place. If $L/K$ is finite abelian, then $v$ is unramified in $L$ if and only if $\Psi_{L/k}$ kills all $K_v^\times$. Otherwise $\Psi_{L/K}$ kills $(K_v^\times)_{>0}$ and $\Psi_{L/K}(-1)$ is the complex conjugation.

TopRay class fields and conductors

Now let us discuss the classical formulation of class field theory in terms of ideal classes.

Definition 53 Let $K$ be a global field. A modulus of $K$ is a formal product $\mathfrak{m}=\prod_{v\in V_K} \mathfrak{p}_v^{e_v}$, where $e_v\ge0$ and $e_v=0$ for almost all $v$; for $v$ a real place, $e_v=0,1$; $v$ a complex place, $e_v=0$. We denote by $\mathfrak{m}_v=\mathfrak{p}_v^{e_v}$ the $v$-component of $\mathfrak{m}$. We write $v|\mathfrak{m}$ or $\mathfrak{p}_v|\mathfrak{m}$ if $e_v>0$. The modulus is used to keep track of the ramification of the places of $K$ in some sense.
Definition 54 Let $x_v\in K_v^\times$, we write $x_v\equiv1\mod{\mathfrak{m}_v}$ if
  • $v$ is nonarchimedean and $e_v>0$, then $x_v\in 1+\mathfrak{p}_v^{e_v}$;
  • $v$ is nonarchimedean and $e_v=0$, then $x_v\in \mathcal{O}_v^\times$;
  • $v$ is archimedean and $e_v=1$, then $x_v>0$;
  • no requirement for $v$ archimedean and $e_v=0$.

In other words, $x_v$ should go to $G^{e_v}$ under the Artin map.

Definition 55 Let $x\in \mathbb{A}_K^\times$, we say $x\equiv1 \mod{\mathfrak{m}}$ if and only if $x_v\equiv1\mod{\mathfrak{m}_v}$ for any $v$. This is a multiplicative condition, so it makes sense to say that $x\equiv y\mod{\mathfrak{m}}$ whenever $x_vy_v^{-1}\equiv 1\mod{\mathfrak{m}_v}$.
Example 21 For $K=\mathbb{Q}$, each modulus is of the form $\mathfrak{m}=m$ or $\mathfrak{m}=m\infty$, where $m\in \mathbb{Z}_{>0}$. It follows easily that an idele $x=(x_f,x_\infty)\equiv 1\mod{m}$ if and only the finite part $x_f\in \ker(\hat{\mathbb{Z}}^\times\rightarrow(\mathbb{Z}/m \mathbb{Z})^\times)$ (which can be thought of the usual congruence relation $x\equiv 1\mod{m}$), and $x\equiv1\mod{m\infty}$ if we further have $x_\infty>0$.
Definition 56 Suppose $\mathfrak{m}$ is a modulus, we define $U_\mathfrak{m}=\{x\in \mathbb{A}_K^\times: x\equiv 1\mod{\mathfrak{m}} \}$, namely $$U_\mathfrak{m}=\prod_{v|\infty\atop e_v=0}K_v^\times\prod_{v|\infty\atop e_v=1}(K_v^\times)_{>0}\prod_{v\nmid\infty\atop e_v=0}\mathcal{O}_v^\times\prod_{v\nmid\infty\atop e_v>0}(1+\mathfrak{p}_v^{e_v}).$$ It is an open subgroup of $\mathbb{A}_K^\times$ and $U_{\mathfrak{m}'}\subseteq U_{\mathfrak{m}}$ if and only if $\mathfrak{m}|\mathfrak{m}'$. We know that $\{U_\mathfrak{m}\}$ forms a cofinal filtered system of open subgroups in $\mathbb{A}_K^\times$. This generalizes Proposition 4.
Definition 57 Define
  • $I_K^{(\mathfrak{m})}$ be the free abelian group of fractional ideals of $K$ generated by $\{\mathfrak{p}_v: \mathfrak{p}_v\nmid \mathfrak{m}\}$.
  • $K_{\mathfrak{m}}^\times=\{x\in K^\times: x\equiv 1\mod{\mathfrak{m}_v}, \forall v|\mathfrak{m}\}$
  • $P_\mathfrak{m}$ be the image of $K_{\mathfrak{m}}^\times$ under the natural map $K_{\mathfrak{m}}^\times\rightarrow I_K^{(\mathfrak{m})}$.
  • the ray class group $\Cl_\mathfrak{m}(K)=I_K^{(\mathfrak{m})}/P_{\mathfrak{m}}$.
Remark 65 When $K$ is a number field and $\mathfrak{m}=(1)$, we recover that $I_K^{(\mathfrak{m})}=I_K$, $P_\mathfrak{m}=P_K$ and $\Cl_\mathfrak{m}(K)=\Cl(K)$. If $\mathfrak{m}$ is the product of all real places, then $I_{K}^{(\mathfrak{m})}=I_K$, $P_{\mathfrak{m}}=\{(x): x_v>0, \forall v\text{ real}\}$ is generated by totally positive elements and $\Cl_\mathfrak{m}(K)$ is called the narrow class group.
Example 22 For $K=\mathbb{Q}$ and $\mathfrak{m}=m\infty$, we have $I_K^{(\mathfrak{m})}=\{a/b \in \mathbb{Q}^\times_{>0}: (a,m)=(b,m)=1\}$. We have a surjection $I_K^{(\mathfrak{m})}\rightarrow (\mathbb{Z}/m \mathbb{Z})^\times$ and the kernel is exactly $P_{\mathfrak{m}}=\{x\in \mathbb{Q}_{>0}: x\equiv 1\mod{m}\}$. Therefore $\Cl_{\mathfrak{m}}(\mathbb{Q})\cong (\mathbb{Z}/m \mathbb{Z})^\times$. Similarly, when $\mathfrak{m}=m$, we find that $\Cl_\mathfrak{m}(\mathbb{Q})= (\mathbb{Z}/m \mathbb{Z})^\times/\pm1$.

10/17/2012

Proposition 10 There are isomorphisms $$(\mathbb{A}_K^\times/K^\times )/(K^\times U_\mathfrak{m})\cong\mathbb{A}_K^\times/K^\times U_\mathfrak{m}\cong \Cl_\mathfrak{m}(K).$$
Proof We define $\mathbb{A}_{K,\mathfrak{m}}^\times=\{ x\in \mathbb{A}_K^\times : x\equiv1 \mod{\mathfrak{m}}\}$ (it properly contains $U_\mathfrak{m}$) and $K_{\mathfrak{m}}^\times=K^\times\cap \mathbb{A}_{K,\mathfrak{m}}^\times$. Then we have an injection $$\mathbb{A}_{K,\mathfrak{m}}^\times/K^\times_{\mathfrak{m}}U_\mathfrak{m}\hookrightarrow\mathbb{A}_K^\times/K^\times U_\mathfrak{m}.$$ It is actually an isomorphism by weak approximation. Now define the homomorphism $$\mathbb{A}_{K,\mathfrak{m}}^\times\rightarrow I_K^{(\mathfrak{m})},\quad x\mapsto \prod_{v\nmid\infty} \mathfrak{p}_v^{\ord_v(x_v)},$$ which is obviously a surjection with kernel $U_\mathfrak{m}$. Taking quotient by the image of $K_\mathfrak{m}^\times$, we obtain that $\Cl_\mathfrak{m}(K)\cong \mathbb{A}_{K,\mathfrak{m}}^\times/K_{\mathfrak{m}}^\times U_\mathfrak{m}$.
Remark 66 The second isomorphism is a bit hard to write down explicitly since the weak approximation is involved in the proof.
Remark 67 When $K$ is a number field, $\Cl_\mathfrak{m}(K)$ is finite since any open subgroup of $\mathbb{A}_K^\times/K^\times$ is of finite index. But when $K$ is a global function field $\Cl_\mathfrak{m}(K)$ is infinite, since $K^\times U_\mathfrak{m}\subseteq (\mathbb{A}_K^\times)^1$ and the quotient contains at least $q^\mathbb{Z} $.
Definition 58 The ray class field of the modulus $\mathfrak{m}$ is the class field $K_\mathfrak{m}$ corresponding to $K^\times U_\mathfrak{m}/K^\times\subseteq \mathbb{A}_K^\times/K^\times$. When $K$ is a number field, $K_\mathfrak{m}/K$ is finite abelian. When $K$ is a global function field, $K_\mathfrak{m}$ contains $K\otimes_k\bar k$ since the image of $K^\times U_\mathfrak{m}$ has trivial image when projected to $q^\mathbb{Z} $.
Remark 68 When $K$ is a number field, the Artin map induces an isomorphism $\Cl_\mathfrak{m}(K)\cong\Gal(K_\mathfrak{m}/K)$. When $K$ is a global function field, the Artin map induces a map $\Cl_\mathfrak{m}(K)\rightarrow\Gal(K_\mathfrak{m}/K)$ with dense image. In either case, $\Psi_{K_\mathfrak{m}/K}(\pi_v)=\Frob_v$ for any $v\nmid \infty$, $v\nmid \mathfrak{m}$.
Exercise 12 When $K=\mathbb{Q}$, we have $\mathbb{Q}_{m\infty}=\mathbb{Q}(\zeta_m)$ and $\mathbb{Q}_{m}=\mathbb{Q}(\zeta_m+\bar \zeta_m)$ and the following diagram commutes $$\xymatrix{\Cl_{m\infty}(\mathbb{Q})\ar[r]^-{\Psi_K}_-\cong \ar[d]^{\cong}&\Gal(\mathbb{Q}_{m\infty}/\mathbb{Q})\ar[d]^{=}\\ (\mathbb{Z}/m \mathbb{Z})^\times & \Gal(\mathbb{Q}(\zeta_m)/\mathbb{Q}). \ar[l]^-{\alpha_m}_-\cong}$$
Remark 69 Notice that $K^\times U_\mathfrak{m}$ forms a filtered cofinal system of open subgroups of $\mathbb{A}_K^\times/K^\times$. Hence $\{K_\mathfrak{m}\}$ forms a cofinal system of abelian extensions of $K$, in which every finite abelian extension of $K$ is contained in some $K_\mathfrak{m}$. In particular, the classical Kronecker-Weber theorem for $\mathbb{Q}$ follows from the previous exercise.

We summarize the easy properties of ray class fields as follows.

Proposition 11
  1. $K_\mathfrak{m}$ is unramified at all $v\nmid \mathfrak{m}$.
  2. $K_\mathfrak{m} \cap K_{\mathfrak{m}'}=K_{\gcd(\mathfrak{m} ,\mathfrak{m}')}$
  3. If $\mathfrak{m}|\mathfrak{m}'$, then $K_\mathfrak{m}\subseteq K_{\mathfrak{m}'}$.
Definition 59 Suppose $L/K$ is a finite abelian extension. We say a modulus $\mathfrak{m}$ of $K$ is admissible for $L/K$ if $L\subseteq K_\mathfrak{m}$. The gcd of all admissible modulus of $L/K$ is called the conductor $\mathfrak{f}_{L/K}$ of $L/K$. It follows that $L\subseteq K_{\mathfrak{f}_{L/K}}$.
Example 23 When $K=\mathbb{Q}$. $\mathfrak{f}_{L/K}$ is essentially the same as Proposition 3, except that $\infty\nmid \mathfrak{f}$ if and only if $L$ is totally real.
Lemma 15 $\mathfrak{m}$ is admissible for $L/K$ if and only if $N_{L/K}(\mathbb{A}_L^\times)\supseteq U_\mathfrak{m}$.
Proof The "if" direction follows from the fact that $K^\times U_\mathfrak{m}/K^\times\subseteq N_{L/K}(\mathbb{A}_L^\times/L^\times)$. For the "only if" direction, suppose $L\subseteq K_\mathfrak{m}$, then the local-global compatibility implies that $\Psi_{L_w/K_v}(1+\mathfrak{p}_v^{e_v})=1$ for $v$ nonarchimedean (and a similar thing for archimedean places), hence by the definition of idelic norm, $N_{L/K}(\mathbb{A}_L^\times)\supseteq N_{L_w/K_v}(L_w^\times)\supseteq 1+\mathfrak{p}_v^{e_v}$. Therefore $U_\mathfrak{m}\subseteq N_{L/K}(\mathbb{A}_L^\times)$.
Remark 70 In some literature, the notion of admissible modulus is defined via this lemma.
Proposition 12 $v$ is ramified in $L$ if and only if $v| \mathfrak{f}_{L/K}$.
Proof Let $N=N_{L/K}(\mathbb{A}_L^\times/L^\times)\supseteq K^\times U_\mathfrak{m}/K^\times$, where $\mathfrak{m}=\mathfrak{f}_{L/K}$. If $v$ is ramified, the $\mathcal{O}_v^\times\not\subseteq N$, hence $\mathcal{O}_v^\times\not\subseteq K^\times U_\mathfrak{m}/K^\times$, hence $\mathcal{O}_v^\times\not\subseteq U_\mathfrak{m}$, which implies that $v|\mathfrak{m}$. Now suppose $v$ is unramified, then for any $w|v$, $N_{L_w/K_v}(L_w^\times)\supseteq\mathcal{O}_v^\times$. Then we can find a modulus $\mathfrak{m}$ such that $v\nmid \mathfrak{m}$ and $N_{L/K}(\mathbb{A}_L^\times)\supseteq U_\mathfrak{m}$. It follows that $v|\mathfrak{f}_{L/K}$ by the previous lemma and the definition of $\mathfrak{f}_{L/K}$.

TopIdeal-theoretic formulation of global class field theory

Definition 60 Let $L/K$ be a finite abelian extension and $\mathfrak{m}$ be a modulus which is divisible by all ramified places. We define the Artin symbol $$\Art_{L/K}: I_K^{(\mathfrak{m})}\rightarrow \Gal(L/K),\quad \mathfrak{p}\mapsto (\mathfrak{p}, L/K)=\Frob_\mathfrak{p}.$$
Exercise 13 $\mathfrak{m}$ is admissible if and only if $P_\mathfrak{m}\subseteq \ker \Art_{L/K}$. (This is the way Artin originally introduced the notion of admissible moduli. The existence of admissible moduli is truly surprising and is the key difficulty of class field theory!)

10/19/2012

Definition 61 Suppose $L/K$ is a finite extension and $\mathfrak{m}$ is a modulus of $K$. We denote by $I_L^{(\mathfrak{m})}$ the free abelian group generated by prime ideals of $L$ whose restriction to $K$ are coprime to $\mathfrak{m}$. The usual ideal norm restricts to a group homomorphism $N_{L/K}: I_L^{(\mathfrak{m})}\rightarrow I_K^{(\mathfrak{m})}$. We denote the image of $\mathfrak{m}$ by $N_{L/K}(\mathfrak{m})$.
Exercise 14 Let $L/K$ be a finite extension. Suppose $K'/K$ is finite abelian and $L'{}=K'L$. Then we have a commutative diagram $$\xymatrix{I_L^{(\mathfrak{n})} \ar[r] \ar[d]^{N_{L/K}}& \Gal(L'/L)\ar[d] \\ I_K^{(\mathfrak{m})} \ar[r] & \Gal(K'/K),}$$ where $\mathfrak{m}$ is a modulus of $K$ divisible by the primes of ramified in $K'$, $\mathfrak{n}$ is a modulus of $L$ divisible by primes ramified in $L'$ or restrict to primes of $K$ ramified in $K'$.
Remark 71 Suppose $L/K$ is finite abelian and $\mathfrak{m}$ is divisible by primes ramified primes. Then it follows from the previous exercise (norm functoriality) that $\Art_{L/K}$ kills $N_{L/K}(\mathfrak{m})$. $\Art_{L/K}$ is surjective by weak approximation and surjectivity of $\Psi_{L/K}$ (which is completely nontrivial without class field theory!). Suppose further that $\mathfrak{m}$ is admissible for $L/K$, then $\Art_{L/K}$ also kills $P_\mathfrak{m}$ by Exercise 13.
Theorem 30 (Class field theory in ideal-theoretic language) Let $K$ be a number field and $L/K$ be a finite abelian extension.
  1. (Reciprocity law) There exists an admissible modulus $\mathfrak{m}$ for $L/K$ and $\Art_{L/K}$ induces an isomorphism $I_K^{(\mathfrak{m})}/P_\mathfrak{m}N_{L/K}(\mathfrak{m})\cong\Gal(L/K)$.
  2. (Existence theorem) For any modulus $\mathfrak{m}$, there exists a unique finite abelian extension $K_\mathfrak{m}/K$ for which $\mathfrak{m}$ is admissible and $N_{K_\mathfrak{m}/K}(\mathfrak{m})\subseteq P_\mathfrak{m}$ (i.e., $\Cl_\mathfrak{m}(K)\cong \Gal(K_\mathfrak{m}/K)$).
Remark 72 One can show that this classical version will imply the idele-theoretic version of global class field theory.
Remark 73 An inconvenience in idele-theoretic language is that one has to constantly change the moduli when dealing with several field extensions.

TopWeber L-functions

Before stepping into the cohomological proof of class field theory, we will discuss various applications of class field theory in the following several weeks. From now on we will assume $K$ is a number field for simplicity (though some results are also valid for function fields). Write the degree $N=[K:\mathbb{Q}]$.

Definition 62 We defined the Dedekind zeta function $$\zeta_K(s)=\prod_{\mathfrak{p}}\frac{1}{1- (N \mathfrak{p})^{-s}}=\sum_{\mathfrak{a}}\frac{1}{(N\mathfrak{a})^s},$$ where $N\mathfrak{a}=\#(\mathcal{O}_K/\mathfrak{a})$ is the absolute norm.

We omit the proof of the simple analytic property.

Proposition 13 $\zeta_K(s)$ is analytic on $\Re(s)>1-1/N$ except a simple pole at $s=1$.
Definition 63 Let $\mathfrak{m}$ be a modulus of $K$ and $\mathcal{R}\in \Cl_\mathfrak{m}(K)$ be an ideal class. We define $$\zeta_K(s,\mathcal{R})=\sum_{\mathfrak{a}\in \mathcal{R}}\frac{1}{(N \mathfrak{a})^s}.$$

A similar analytic property holds for $\zeta_K(s,\mathcal{R})$ too.

Theorem 31 $\zeta_K(s,\mathcal{R})$ is analytic on $\Re(s)>1-1/N$ except a simple pole at $s=1$. The residue at $s=1$ depends on the modulus $\mathfrak{m}$ but not on ideal class $\mathcal{R}$.

Recall that for $G$ a finite abelian group, we have the notion of Pontryakin dual $\hat G=\{\chi: G\rightarrow \mathbb{C}^\times\}$ consisting of characters of $G$, and the following elementary properties holds:

Proposition 14
  1. $\hat{\hat G}\cong G$ canonically.
  2. If $\chi\in\hat G$, then $\sum_{g\in G}\chi(g)=
\begin{cases}
  \# G, & \chi=1,\\
  0, & \chi\ne1.
\end{cases}$
  3. If $g\in G$, then $\sum_{\chi\in \hat G}\chi(g)=
\begin{cases}
  \# G, & g=1,\\
  0, & g\ne1.
\end{cases}$
Definition 64 The Weber $L$-function for a character $\chi: \Cl_\mathfrak{m}(K)\rightarrow \mathbb{C}^\times$ is defined to be $$L_{K,\mathfrak{m}}(s,\chi)=\prod_{\mathfrak{p}\nmid \mathfrak{m}}\frac{1}{1-\chi(\mathfrak{p})(N \mathfrak{p})^{-s}}=\sum_{(\mathfrak{m},\mathfrak{a})=1}\frac{\chi(\mathfrak{a})}{(N \mathfrak{a})^s}=\sum_{\mathcal{R}\in \Cl_\mathfrak{m}(K)}\chi(\mathcal{R})\zeta_K(s,\mathcal{R}).$$ It follows from the previous theorem that $L_{K,\mathfrak{m}}(s,\chi)$ is analytic on $\Re(s)>1-1/N$ except a possible simple pole at $s=1$. When $\chi\ne1$, it is actually analytic at $s=1$ by the previous proposition, since all the residues of $\zeta_K(s,\mathcal{R})$ at $s=1$ are the same.
Example 24 Consider $K=\mathbb{Q}$ and $\mathfrak{m}=m\infty$. Then $$\zeta_\mathbb{Q}(s)=\sum_{n\ge1}\frac{1}{n^s}$$ is simply the Riemann zeta function. A character $\chi: \Cl_\mathfrak{m}(\mathbb{Q})\rightarrow \mathbb{C}$ is the same thing as a Dirichlet character $(\mathbb{Z}/m \mathbb{Z})^\times\rightarrow \mathbb{C}^\times$. Then $$L_{\mathbb{Q},m\infty}=\sum_{n\ge1}\frac{\chi(n)}{n^s}$$ is simply a Dirichlet $L$-function, where we extend $\chi$ on $\mathbb{Z}$ by letting $\chi(x)=0$ whenever $(x,m)\ne1$.

Now class field theory easily imply the following result on special values of Weber $L$-functions.

Theorem 32 $L_{K,\mathfrak{m}}(1,\chi)\ne0$ if $\chi\ne1$.
Proof By global class field theory, there exists a class field $K_\mathfrak{m}$ such that $\Cl_\mathfrak{m}(K)\cong\Gal(K_\mathfrak{m}/K)$. So $\chi$ can be viewed as a character of $G_K$. Let $\mathfrak{n}$ be a modulus of $K_\mathfrak{m}$ divisible exactly by the primes of $K_\mathfrak{m}$ restricting to $\mathfrak{p}|\mathfrak{m}$. Then as a special case of the lemma below ($L=K_\mathfrak{m}$) $$L_{K_\mathfrak{m},\mathfrak{n}}(s,1)=\prod_{\chi}L_{K,\mathfrak{m}}(s,\chi).$$ The result then follows since each of the $L$-functions $L_{K_\mathfrak{m},\mathfrak{n}}(s,1)$ and $L_{K,\mathfrak{m}}(s,1)$ has a simple pole at $s=1$.

10/22/2012

Lemma 16 Let $L/K$ be a finite abelian extension. Suppose $\mathfrak{m}$ is an admissible modulus for $L/K$ and $\mathfrak{n}$ is exactly divisible by primes of $L$ restricting to primes dividing $\mathfrak{m}$. Then $$L_{L,\mathfrak{n}}(s,1)=\prod_\chi L_{K, \mathfrak{m}}(\chi),$$ where $\chi$ runs over all characters $\Cl_\mathfrak{m}(K)\twoheadrightarrow\Gal(L/K)\xrightarrow{\chi}\mathbb{C}^\times$.
Proof This equality actually holds at the level of local Euler factors. Say $\mathfrak{p}\nmid \mathfrak{m}$, then $\mathfrak{p}$ is unramified and $\mathfrak{p} \mathcal{O}_L=\prod_i \mathfrak{q}_i$, we claim that $$\prod_i \frac{1}{1-(N \mathfrak{q}_i)^{-s}}=\prod_\chi \frac{1}{1-\chi(\mathfrak{p})(N \mathfrak{p})^{-s}}.$$ Write $n=[L:K]$, then $N \mathfrak{q}_{i=1}^{n/f}=N \mathfrak{p}^f$. The right hand side becomes $$\frac{1}{(1-(N \mathfrak{p})^{-sf})^{n/f}}.$$ Write $y=(N \mathfrak{p})^{-s}$ for short. Taking the logarithms of both sides, it reduces to show that $$\frac{n}{f}\sum_{m\ge1} \frac{y^{fm}}{m}=\sum_\chi\sum_{m\ge1}\frac{\chi(\mathfrak{p})^my^m}{m}.$$ Notice that $\chi(\mathfrak{p})=\chi(\Frob_\mathfrak{p})$ and $\Frob_\mathfrak{p}$ has order $f$ in $\Gal(L/K)$. Then the character $\chi\mapsto \chi(\mathfrak{p})^m$ of the Pontryagin dual of $\Gal(L/K)$ is nontrivial if and only if $f\nmid m$. Therefore the right-hand-side is simply $$\sum_\chi\sum_{f\nmid m}\frac{y^m}{m}=n\sum_{m\ge1}\frac{y^{fm}}{fm}=\frac{n}{f}\sum_{m\ge1}\frac{y^{\mathfrak{m}}}{m},$$ which coincides with the left-hand-side.
Remark 74 The above lemma can be viewed as a special case in the more general setting of Galois representations. Consider a Galois representation $\rho: G_K\rightarrow GL(V)$, where $V$ is a $n$-dimensional complex vector space. Since $GL(V)$ has no small subgroup, we know that $\rho$ factors through a finite extension $\Gal(L/K)$. We define the Artin $L$-function of $\rho$ to be $$L(s,\rho)=\prod_\mathfrak{p}\det (1- (N \mathfrak{p})^{-s}\rho(\Frob_\mathfrak{p})|V^{I_\mathfrak{p}})^{-1}.$$ It converges on $\Re(s)>1$ and satisfy the following nice (and easy) properties:
  1. $L(s, \rho_1\oplus \rho_2)=L(s,\rho_1)L(s,\rho_2)$.
  2. $L(s,\Ind_{G_L}^{G_K}\tilde\rho)=L(s,\tilde \rho)$ for $L/K$ a finite extension and $\tilde \rho: G_L\rightarrow GL(V)$.
Example 25 Consider a one-dimensional Galois representation $\rho: G_K\rightarrow \mathbb{C}^\times$. Then $\rho$ factors through a finite cyclic extension $L/K$, where $L=(\bar K)^{\ker\rho}$. Let $\mathfrak{m}$ be the conductor of $L/K$, then $\rho$ induces a character $\Cl_\mathfrak{m}(K)\twoheadrightarrow\Gal(L/K)\xrightarrow{\chi} \mathbb{C}^\times$. By definition, $$L_{K,\mathfrak{m}}(s,\chi)=L(s,\rho),$$ since $\mathbb{C}^{I_\mathfrak{p}}\ne0$ if and only if $\mathfrak{p}$ is unramified in $L$. In other words, Weber $L$-functions can be viewed as the same thing as one-dimensional Artin $L$-functions via class field theory. Notice Weber $L$-functions involves geometry of numbers (ideal classes, etc.), which are crucial for establishing the analytic properties. On the other hand, the functoriality properties of Artin $L$-functions give handy ways to establish non-vanishing results of special values. This picture motivates the Langlands program of the study of general Artin $L$-functions via relating Galois representations and automorphic representations and class field theory can be viewed as the Langlands program for $GL_1$.
Example 26 Suppose $L/K$ is a finite abelian extension with $G=\Gal(L/K)$. Then $\Ind_{G_L}^{G_K}(\mathbf{1}_L)\cong\bigoplus_\chi \chi$ (notice the left-hand-side is the regular representation of $G$). Thus we have $$\zeta_L(s)=L(s,\mathbf{1}_L)=L(s,\Ind_{G_L}^{G_K}\mathbf{1}_L)=L(s,\bigoplus_\chi \chi)=\prod_\chi L(s,\chi).$$ This is analogous to the previous lemma except that more local Euler factors are involved here.

General Artin $L$-functions can be reduced to one-dimensional Artin $L$-functions via Brauer's induction theorem.

Theorem 33 (Brauer) Suppose $G$ is a finite group and $\rho: G\rightarrow GL(V)\cong GL_n(\mathbb{C})$. Then there exists $n_1,\ldots, n_r\in \mathbb{Z}$, $H_1,\ldots, H_i$ subgroups of $G$ and characters $\chi_i: H_i\rightarrow \mathbb{C}^\times$ such that $$\rho\cong\sum_{i=1}^r n_i\Ind_{H_i}^G\chi_i,$$ as virtual representations.
Corollary 8 With the same notation, $$L(s,\rho)=\prod_{i=1}^r L(s,\chi_i)^{n_i}.$$
Remark 75 In particular, we obtain a meromorphic continuation for Artin $L$-functions from the meromorphic continuation of Weber $L$-functions (but the Artin conjecture for holomorphy when $V^G=0$ is still open in general!).

TopChebotarev's density theorem

Definition 65 Let $K$ be a number field of degree $N $. Let $P $ be a set of finite primes of $K$. We define the natural density of $P $ to be the limit (if it exists) $$\mu(P)=\lim_{x\rightarrow\infty}\frac{\# \{\mathfrak{p} \in P: N \mathfrak{p}<x\}}{\#\{\mathfrak{p}: N \mathfrak{p}<x\}}.$$ We define the Dirichlet density to be $$\delta(P)=\lim_{s\rightarrow 1^+,s\in \mathbb{R}}\frac{\sum_{\mathfrak{p}\in P} (N \mathfrak{p})^{-s}}{\log (s-1)^{-1}}.$$
Remark 76 The natural density is the "strongest notion" of density in the following sense: when $\mu(P)$ exists, $\delta(P)$ also exists and $\delta(P)=\mu(P)$. We will mainly talk about the Dirichlet density since they are more convenient for analytic arguments.

10/24/2012

Remark 77 For $f,g$ holomorphic in a neighborhood of 1 except possibly at 1, we write $f\sim_1 g$ if $f-g$ is holomorphic at 1. It may be helpful to notice that $\sum_\mathfrak{p} (N \mathfrak{p})^{-s}\sim_1\log (s-1)^{-1}$.Then $$\log \zeta_K(s)=-\sum_{\mathfrak{p}}\log(1-(N \mathfrak{p})^{-s})=\sum_\mathfrak{p}\sum_{m\ge1} \frac{1}{m (N \mathfrak{p})^{ms}}.$$ Since $\sum_\mathfrak{p}\sum_{m\ge2}\frac{1}{(N \mathfrak{p})^{ms}}$ is analytic for $\Re(s)>1/2$. We see that $$\log \zeta_K(s)\sim_1\sum_\mathfrak{p}\frac{1}{(N \mathfrak{p})^s}.$$ Also observe that $\zeta_K(s)$ has a simple at $s=1$, so $(s-1)\zeta_K(s)$ is analytic and non-vanishing at 1. Hence $$\log\zeta_K(s)\sim_1\log(s-1)^{-1}.$$ Because $\sim_1$ is an equivalence relation, it follows that $\sum_\mathfrak{p} (N \mathfrak{p})^{-s}\sim_1\log (s-1)^{-1}$ and consequently $$\lim_{s\rightarrow 1^+}\frac{\sum_\mathfrak{p}(N \mathfrak{p})^{-s}}{\log (s-1)^{-1}}=1.$$

The following basic properties of the (Dirichlet) density $\delta$ follows easily from definition.

Proposition 15
  1. If $P $ has a density, then $\delta(P)\in [0,1]$.
  2. If $P $ is finite, then $\delta(P)=0$.
  3. If $P $ is the disjoint union of $P_1$ and $P_2$ and two of $P $, $P_1$, $P_2$ have density, then so does the third one and $\delta(P)=\delta(P_1)+\delta(P_2)$.
  4. If $P\subseteq P'$ and both have density, then $\delta(P)\le\delta(P')$.
  5. If $P $ has a density and $\delta(P')=1$,then $\delta(P)=\delta(P\cap P')$.
  6. If $P $ has density and $P'$ is the complement of $P $, then $P $ has density and $\delta(P)+\delta(P')=1$.
Exercise 15 Suppose $K$ is a number field. Denote the degree of a prime $\mathfrak{p}$ by $\deg \mathfrak{p}=[\mathcal{O}_K/\mathfrak{p}: \mathbb{F}_p]$. Prove that $\{\mathfrak{p} : \deg \mathfrak{p}=1\}$ has density 1. (Not to be confused with the density of of split primes of $\mathbb{Q}$ in $K$, which will be shown to be $1/N $).
Proposition 16 Let $\mathfrak{m}$ be a modulus and $\mathcal{R}_0\in \Cl_\mathfrak{m}(K)$. Then $\delta(\{\mathfrak{p}\in \mathcal{R}_0\})=1/\#\Cl_\mathfrak{m}(K)$.
Proof Notice that $$\log L_{K,\mathfrak{m}}(s,\chi)\sim_1\sum_{\mathfrak{p}\nmid \mathfrak{m}}\frac{\chi(\mathfrak{p})}{(N \mathfrak{p})^{s}}=\sum_{\mathcal{R}\in \Cl_\mathfrak{m}(K)}\chi(\mathcal{R})\sum_{\mathfrak{p}\in \mathcal{R}}\frac{1}{(N \mathfrak{p})^s}.$$ On the other hand, since $L_{K,\mathfrak{m}}(s,\chi)$ is analytic at 1 whenever $\chi\ne1$, we know that $$\log(s-1)^{-1}\sim_1\log\zeta_K(s)\sim_1\log L_{K,\mathfrak{m}}(s,1)\sim_1\sum_\chi \chi(\mathcal{R}_0)^{-1}\log L_{K,\mathfrak{m}}(s,\chi).$$ Expanding the last sum gives $$\sum_{\chi,\mathcal{R}}\chi(\mathcal{R}/\mathcal{R}_0)\sum_{\mathfrak{p}\in \mathcal{R}}\frac{1}{(N \mathfrak{p})^s}=\#\Cl_\mathfrak{m}(K)\sum_{\mathfrak{p}\in \mathcal{R}_0}\frac{1}{(N \mathfrak{p})^s}.$$ The desired result then follows.
Remark 78 The non-vanishing result of $L_{K,\mathfrak{m}}(1,\chi)$ can be proved without class field theory, which we leave as an exercise (cf. Exercise 17).
Example 27 When $K=\mathbb{Q}$, $\mathfrak{m}=m\cdot\infty$, we have $\Cl_\mathfrak{m}(\mathbb{Q})\cong (\mathbb{Z}/m \mathbb{Z})^\times$ and an ideal class is given by simply given by a residue class modulo $m$. From the previous proposition we recover the classical theorem of Dirichlet on primes in progressions: $\delta(\{p:p\equiv n\mod{m}\})=1/\varphi(m)$. Namely, the primes are equidistributed modulo $m$.
Theorem 34 (Chebotarev's density) Suppose $L/K$ is a Galois (but not necessarily abelian) extension of global fields with $G=\Gal(L/K)$ and $N=[L:K]$. Let $\sigma\in G$ and $C $ be the conjugacy class of $\sigma$ in $G$ (with $c=\#C$). Then $S_C=\{\mathfrak{p}: \mathfrak{p} \text{ unramified in } L, \Frob_\mathfrak{p}\in C\}$ has density $\delta(S_C)=c/N$. In other words, the Frobenius conjugacy classes are equidistributed in $\Gal(L/K)$.
Proof Let us first show the case that $L/K$ is abelian. Let $\mathfrak{m}$ be the conductor of $L/K$. Let $H $ be the kernel of $\Art_{L/K}: I_K^{(\mathfrak{m})}\rightarrow G$. Then $C=\{\sigma\}$ and $S_C=\{\mathfrak{p}\nmid \mathfrak{m}: \Frob_\mathfrak{p}=\sigma\}=\{\mathfrak{p} \nmid \mathfrak{m}: \mathfrak{p}\in \mathfrak{a}+H\}$, where $\mathfrak{a}$ is any preimage of $\sigma$. Since $(\mathfrak{a}+H)/P_\mathfrak{m}\subseteq \Cl_\mathfrak{m}(K)$ contains exactly $[H: P_\mathfrak{m}]$ ideal classes, we know that $$\delta(S_C)=\frac{[H: P_\mathfrak{m}]}{\#\Cl_\mathfrak{m}(K)}=\frac{[H: P_\mathfrak{m}]}{[I_K^{(\mathfrak{m})}: P_\mathfrak{m}]}=\frac{1}{[I_K^{(\mathfrak{m})}:H]}=\frac{1}{N}$$ as needed. This is the only step where the usage of class field theory is crucial.

10/26/2012

For the general case, we let $K'$ be the fixed field of $L$ under the cyclic group generated by $\sigma$. Then $L/K'$ is a cyclic extension with Galois group $\langle\sigma\rangle\cong \mathbb{Z}/f \mathbb{Z}$ and we can reduce the previous case as follows. Let $S'$ be the primes $\mathfrak{p}'$ of $K'$ unramified in $L$ and $(\mathfrak{p}', L/K')=\sigma$. By the abelian case, we know that $\delta(S')=1/f$. Also, let $S''$ be the primes $\mathfrak{p}'$ in $S'$ such that $f(\mathfrak{p}'/K)=1$, then $\delta(S'')=1/f$ by Exercise 15. We claim that for any $\mathfrak{p}\in S_C$, $\#\{\mathfrak{p}'\in S'': \mathfrak{p}'|_K=\mathfrak{p}\}=N/cf$. Assuming this claim, we know that $$\sum_{\mathfrak{p} \in S_C}\frac{1}{(N \mathfrak{p})^s}=\frac{cf}{N}\sum_{\mathfrak{p}'\in S''}\frac{1}{(N \mathfrak{p}')^s}\sim_1\frac{cf}{N}\cdot\delta(S'')\log\frac{1}{s-1}=\frac{c}{N}\log \frac{1}{s-1},$$ as desired.

It remains to show the claim. Let $T$ be the set of primes $\mathfrak{q}$ of $L$ such that $\mathfrak{q}$ is unramified in $L/K$ and $\Frob_{\mathfrak{q}/K}=\sigma$. For $\mathfrak{q}\in T$, write $\mathfrak{p}'{}=\mathfrak{q}|_{K'}$ and $\mathfrak{p}=\mathfrak{q}|_K$. Since $\sigma$ acts trivially on $K'$, it also acts trivially on $\mathcal{O}_{K'}/\mathfrak{p}'$. Therefore $f(\mathfrak{p}'/\mathfrak{p})=1$, i.e. $\mathfrak{p}'\in S''$. Since $[L:K']=f=f(\mathfrak{q}/\mathfrak{p}')$, we know that $\mathfrak{q}$ is the unique prime of $L$ over $\mathfrak{p}'$. On the other hand, given $\mathfrak{p}'\in S''$, let $\mathfrak{p}=\mathfrak{p}'|K$ and $\mathfrak{q}$ be a prime of $L$ over $\mathfrak{p}'$. Since $\sigma=(\mathfrak{p}', L/K')$ has order $f$, we know that $f(\mathfrak{q}/\mathfrak{p}')=f$. Hence $\mathfrak{q}$ is the unique prime of $L$ over $\mathfrak{p}'$ and $\Frob_{\mathfrak{q}/K}=\sigma$. Thus $\mathfrak{q}\in T$. In this way we have exhibited a bijection between $T$ and $S''$.

Now let $\mathfrak{p}\in S_C$ and $\mathfrak{q}$ a prime of $L$ over $\mathfrak{p}$. We can choose $\mathfrak{q}\in T$. Then $\{\mathfrak{q}\in T: \mathfrak{q}|_K=\mathfrak{p}\}$ is the orbit of $\mathfrak{q} $ under the centralizer $G_\sigma$ of $\sigma$. So $$\#\{\mathfrak{q}\in T: \mathfrak{q}|_K=\mathfrak{p}\}=\frac{\# G_\sigma}{\# D(\mathfrak{q}/K)}=\frac{\# G_\sigma}{f}=\frac{\# G}{[G: G_\sigma]f}=\frac{N}{cf},$$ which proves the claim.

Remark 79 The corresponding Chebotarev's density theorem for global function fields can be proved using Riemann hypothesis for curves.

TopSplit primes

Definition 66 Let $K$ be a global field and $L/K$ be a finite separable extension. We define $$\Spl(L/K)=\{v\in V_K: v\text{ splits completely in } L\}.$$ For $S\subseteq V_K$, we also define $\Spl_S(L/K)=\Spl(L/K)\setminus S$.
Corollary 9 Suppose $L/K$ is Galois and $\delta(S)=0$. Then $\Spl(L/K)$ and $\Spl_S(L/K)$ have density $1/[L:K]$.
Proof Use Chebotarev's Density Theorem 34 for $\sigma=1$.
Remark 80 Notice this special case doesn't use class field theory since it only involves the second part of the proof of Chebotarev's density theorem: $$\sum_{\mathfrak{p}\in \Spl(L/K)}\frac{1}{(N \mathfrak{p})^s}\sim_1\frac{1}{N}\sum_{f(\mathfrak{q})=1}\frac{1}{(N \mathfrak{p})^s}\sim_1\frac{1}{N}\sum_{\mathfrak{q}}\frac{1}{(N \mathfrak{q})^s}\sim_1\frac{1}{N}\log(s-1)^{-1}.$$
Exercise 16
  1. Suppose $L,L'\subseteq K^\mathrm{sep}$ are finite separable. Then $\Spl(LL'/K)=\Spl(L/K)\cap\Spl(L'/K)$.
  2. Suppose $L/K$ is finite separable and $L'$ is its Galois closure. Then $\Spl(L/K)=\Spl(L'/K)$. In particular, $\delta(\Spl(L/K))=1/[L:K]$ if and only if $L/K$ is Galois.
Corollary 10 If $\Spl(L/K)$ has density 1, then $L=K$.
Proof Apply the previous exercise to the Galois closure of $L$.
Remark 81 We used this corollary in Theorem 13. But as remarked above, it doesn't depend on class field theory and there is no danger of circular reasoning.

We can now prove the following results without using class field theory.

Exercise 17
  1. $L_{K,\mathfrak{m}}(1,\chi)\ne0$ for $\chi\ne1$.
  2. (Global norm index inequality) $[\mathbb{A}_K^\times: K^\times N_{L/K}(\mathbb{A}_L^\times)]\le[L:K]$ for $L/K$ a finite extension of number fields.
Theorem 35 Suppose $L,L'/K$ are finite Galois, $\delta(S)=0$. Then the following are equivalent:
  1. $L\subseteq L'$.
  2. $\Spl_S(L'/K)\subseteq\Spl_S(L/K)$.
  3. $\Spl(L'/K)\subseteq\Spl(L/K)$.
Proof (a) implies (c) and (c) implies (b) are obvious. For (b) implies (a), notice that $\Spl_S(LL'/K)=\Spl_S(L)\cap\Spl_S(L')=\Spl_S(L'/K)$. Therefore $LL'{}=L'$ and $L\subseteq L'$.
Remark 82 The statement is obviously false without the Galois assumption.
Corollary 11 Suppose $L/K$ and $L'/K$ are finite Galois, $\delta(S)=0$. Then $L=L'$ if and only if $\Spl_S(L/K)=\Spl_S(L'/K)$. In other words, a Galois extension is determined by the set of split primes.
Remark 83 Which subsets of $L/K$ are of the form $\Spl(L/K)$ for $L/K$ Galois? The answer for general case is still unknown. But when $L/K$ is abelian, we can answer this question using class field theory as follows. Suppose $\mathfrak{m}$ is the conductor of $L/K$. Let $H\subseteq I_K^{(\mathfrak{m})}$ be the kernel of $\Art_{L/K}$. Then $\mathfrak{p}$ splits completely if and only if $\mathfrak{p}\in H$, if and only if $\mathfrak{p}$ is an element of the subgroup $\bar H=H/P_\mathfrak{m}\subseteq \Cl_\mathfrak{m}(K)$.
Remark 84 The last condition can be thought as a congruence condition. For example, $p$ splits completely in $\mathbb{Q}(\zeta_m)$ if and only if $p\equiv1\mod{m}$. If $L\subseteq \mathbb{Q}(\zeta_m)$ corresponds to the subgroup $\bar H\subseteq(\mathbb{Z}/m \mathbb{Z})^\times$, then $p$ splits completely in $L$ if and only if $p\mod{m}\in \bar H$.

TopHilbert class fields

Definition 67 Let $K$ be a number field. The Hilbert class field $H $ of $K$ is defined to be the ray class field $K_\mathfrak{m}$ for $\mathfrak{m}=1$. Then $\Gal(H/K)=\Cl(K)$ and $H/K$ is unramified at every place (including archimedean places). The narrow Hilbert class field $H^+$ of $K$ is defined to be the ray class field $K_\mathfrak{m}$, where $\mathfrak{m}$ is the product of all real places of $K$. Then $\Gal(H^+/K)\cong\Cl^+(K)$ (the narrow class group) and $H^+/K$ is unramified at all finite places. Notice that $H\subseteq H^+$ and $\Cl^+(K)$ surjectis onto $\Cl(K)$.
Lemma 17 $H $ (resp. $H^+$) is the maximal abelian extension of $K$ which is unramified everywhere (resp. at all finite places).
Proof If $L/K$ is unramified everywhere, then $\mathfrak{f}_{L/K}=1$ by Proposition 12. Hence $L\subseteq K_{1}=H$. Similarly for $H^+$.

10/31/2012

Remark 85 Denote by $K_{>0}$ the totally possibly elements of $K^\times$. By weak approximation, $K^\times/K_{>0}\cong\prod_{v \text{ real}} K_v^\times/(K_v^\times)_{>0}\cong (\mathbb{Z}/2 \mathbb{Z})^{r_1}$. So $(\mathbb{Z}/2 \mathbb{Z})^{r_1}$ surjects onto the kernel of quotient map $\Cl^+(K)\twoheadrightarrow \Cl(K)$.
Proposition 17 Suppose $L/K$ is finite extension of number fields and is totally ramified at some $v\in V_K$. Then $h_K\mid h_L$.
Proof Since $L H_K/L$ is abelian and unramified everywhere, we know that $L H_K\subseteq H_L$. So it suffices to show that $L$ and $H_L$ are linearly disjoint over $K$ (i.e. $L\otimes_K H_K\cong LH_K$. Suppose $H_K=K(\alpha)$ with $f\in K[x]$ the minimal polynomial of $\alpha$. Let $g\in L[x]$ be a monic polynomial dividing $f$ over $L$. Since $H_K/K$ is Galois, $f$ splits into linear factors over $H_K$, which implies $g$ also splits over $H_K$. Hence $g\in H_K[x]\cap L[x]$, which is equal to $K[x]$ by the assumption that $L/K$ is totally ramified at some place. Hence $g\in K[x]$ and $f$ is irreducible over $L$ as needed.
Example 28 Consider $K=\mathbb{Q}(\zeta_m+\overline{\zeta_m})$, $L=\mathbb{Q}(\zeta_m)$. We know that $h_K\mid h_L$ using the previous proposition. When $m=p$ is a prime, the famous criterion of Kummer asserts that $p$ divides $h_L/h_K$ (indeed equivalent to that $p$ divides $h_L$) if and only if $p$ divides the numerator of some Bernoulli number $B_2,B_4,\cdots, B_{p-3}$. Such a prime $p$ is called irregular.

TopArtin's principal ideal theorem

Theorem 36 (Principal Ideal Theorem) Let $K$ be a number field and $H $ be its Hilbert class field. Then any ideal $\mathfrak{a}\subseteq \mathcal{O}_K$ becomes principal in $\mathcal{O}_H$.

Using class field theory, it will reduce to the following purely group-theoretic theorem (we omit the proof).

Theorem 37 (Furtwangler) Let $G$ be a finite group and $H=[G,G]$. Then $\Ver: G^\mathrm{ab}\rightarrow H^\mathrm{ab}$ is trivial.
Proof (Proof of the Principal Ideal Theorem) Let $H'$ be the Hilbert class field of $H=H_K$. Then $H'/K$ is Galois since $H' $ is intrinsic to $K$. Notice that $H $ is the maximal abelian subextension of $H' $. Therefore $\Gal(H'/H)=[G,G]$ and $\Gal(H/K)=G^\mathrm{ab}$, where $G=\Gal(H'/K)$. We have the following commutative diagram $$\xymatrix{I_K \ar[r]^-{\Art_{H/K}} \ar[d] & \Gal(H/K)=\Gal(H'/K)^\mathrm{ab} \ar@<6ex>[d]^{\Ver} \\ I_H \ar[r]^-{\Art_{H'/H}} & \Gal(H'/H)=\Gal(H'/H)^\mathrm{ab}}$$ By the previous theorem, we know that the $\Art_{H'/H}(\mathfrak{a}\mathcal{O}_H)=0$, hence $\mathfrak{a}\mathcal{O}_H$ is principal.

TopClass field towers

The principal ideal theorem motivates the following construction. Let $K=K_0$ be a number field and $K_{i+1}$ be the Hilbert class field of $K_i$. The we obtain the Golod-Shafarevich tower $$K=K_0\subseteq K_1\subseteq\cdots K_\infty=\bigcup K_i.$$ Does $\{K_i\}$ stabilize (i.e., $[K_\infty:K]<\infty$)? The answer in general is no.

Exercise 18 The following conditions are equivalent:
  1. $[K_\infty: K]<0$.
  2. $\Cl(K_i)=0$ for some $K_i$.
  3. There exists an number field $L$ such that $\Cl(L)=0$ and $K$ embeds into $L$.
Remark 86 Clearly $K_\infty\subseteq K^\mathrm{ur}$ (the maximal unramified extension of $K$). But this is usually not an equality (e.g., $K_\infty$ is always solvable).

The analysis on this tower may be easier if we consider a single prime $p$ at one time. Let $K_{i+1}^{(p)}$ be the maximal unramified abelian $p$-extension of $K_i$. We obtain a tower $$K_0^{(p)}\subseteq K_1^{(p)}\subseteq \cdots K_\infty^{(p)}=\bigcup K_i^{(p)}\subseteq K_\infty.$$ We now state a theorem of Golod-Shafarevich (for the proofs, cf. Cassels-Frolich Ch. IX).

Theorem 38 (Golod-Shafarevich) Let $K$ be a number field of degree $N $. If $[K_\infty^{(p)}:K]<\infty$, then $$d^{(p)}\Cl(K)<2+2\sqrt{N+1},$$ where $d^{(p)}(G):=\dim_{\mathbb{F}_p}G/pG$ is the $p$-rank of a finite group of $G$.

On the other hand, Brumer's theorem provides a lower bound on $d^{(p)}\Cl(K)$.

Theorem 39 (Brumer) Suppose $K/\mathbb{Q}$ is Galois of degree $N $. Let $t_k^{(p)}$ be the number of primes $q $ such that $p\mid e(\mathfrak{q}/q)$ for any $\mathfrak{q}$ above $q $. Then $$d^{(p)}\Cl(K)\ge t_K^{(p)}-2(N-1).$$
Corollary 12 Suppose $K/\mathbb{Q}$ is Galois of degree $N $. If $t_K^{(p)}\ge2N+2\sqrt{N+1}$, then $[K_\infty^{(p)}: K]=\infty$.
Example 29 When $N=p=2$, it follows from the previous corollary that a quadratic field has infinite Golod-Shafarevich tower whenever the number of ramified primes is at least 8, e.g., $\mathbb{Q}(\sqrt{d})$ where $d$ has at least 8 different prime factors. In particular, there are infinite such quadratic fields.
Remark 87 In general, for any $N>1$, there exists infinitely many number fields of degree $N $ with infinite Golod-Shafarevich towers. For this, one needs the non-Galois version of Brumer's theorem.

TopHilbert class fields of global function fields

Now suppose $K$ is a global function field. Then $K_{(1)}$ is the maximal abelian unramified extension of $K$. In particular, $K\otimes_k \bar k\subseteq K_{(1)}$ and $K_{(1)}$ is infinite. To get better situation, we may ask what is the maximal unramified extension of $K$ with constant field $k$. As class fields with constant fields $k$ corresponds to subgroups $H\subseteq \mathbb{A}_K^\times/K^\times$ that surjects onto $q^\mathbb{Z} $ under $||\cdot||_K$, we know that a maximal unramified extension with constant field $k$ corresponds to a minimal subgroup $H\supseteq K^\times U_{(1)}$ such that there exists an element $\beta\in H$ with $||\beta||_K=q$. However, there may be more than one such minimal subgroup. The set of such subgroups $H=\langle K^\times U_{(1)},\beta\rangle$ forms a principal homogeneous space under $(\mathbb{A}_K^\times)^1/K^\times U_{(1)}\cong\Pic^0(X)$ (in geometrical terms, it is simply $\Pic^1(X)$). So there are $h_K=\#\Pic^0(X)$ such groups $H $ and hence there are $h_K$ maximal abelian unramified extension of $K$ with constant field $k$.

11/02/2012

Nevertheless, the following construction gives a maximal unramified extension which is canonical in some sense. Choose $\beta\in \mathbb{A}_K^\times$ with $||\beta||_K=q$. Then $\beta^{h_K}$ modulo $K^\times U_{(1)}$ is independent on the choice of $\beta$ and the subgroup $U= \langle\beta^{h_K}, K^\times U_{(1)}\rangle/K^\times\subseteq \mathbb{A}_K^\times/K^\times$ is canonically defined. Let $H $ be the class field of $U$. We then have an exact sequence $$0\rightarrow\Pic^0(X)\rightarrow\Gal(H/K)\rightarrow \mathbb{Z}/h_K \mathbb{Z}\rightarrow0.$$ So $H/K$ is everywhere unramified of degree $h_K^2$ with constant field the degree $h_K$ extension of $k$.

We end the discussion by applying a similar idea to prove a useful proposition concerning $\ell$-adic characters of the Weil group of a global function field.

Definition 68 Let $G$ be a Hausdorff group. A continuous character $\chi: W_K\rightarrow G$ is called unramified if the class field corresponding to $\ker\chi$ is unramified at $v$.
Proposition 18 Suppose $K$ is a global function field of characteristic $p$ and $E $ is a finite extension of $\mathbb{Q}_\ell$ for some $\ell\ne p$. Let $\chi: W_K\rightarrow\mathcal{O}_E^\times$ be a continuous homomorphism unramified outside a finite set $S$. Then there exists $c\in \mathcal{O}_E^\times$ and $\chi_1: W_K\rightarrow\mathcal{O}_E^\times$ continuous of finite order such that $\chi=\chi_1\cdot c^{\deg(\cdot)}$.
Proof Since $\mathcal{O}_E^\times$ is abelian, $\chi$ factors through $W_K^\mathrm{ab}$. Pick $\sigma\in W_K^\mathrm{ab}$ such that $\deg \sigma=1$. Let $c=\chi(\sigma)$ and $\chi_1=\chi\cdot c^{-\deg(\cdot)}$. Notice that $\chi_1(W_K)=\chi_1(I)$ by construction. But $I\cong (\mathbb{A}_K^\times)^1/K^\times$, it suffices to show that $\chi(K^\times U_{(1)}/K^\times)$ is finite as $\Pic^0(X)$ is finite. But $K^\times U_{(1)}/K^\times\cong U_{(1)}/K^\times\cap U_{(1)}\cong U_{(1)}/k^\times$ and by assumption $\chi(\mathcal{O}_v^\times)=1$ for $v\not\in S$, it suffices to show that $\chi(\prod_{v\in S}\mathcal{O}_v^\times)$ is finite. But $\mathcal{O}_E^\times$ has a finite index pro-$\ell$ group and $\prod_{v\in S}\mathcal{O}_v^\times$ has a finite index pro-$p$ group. Because there is no nontrivial map from a pro-$p$ group to a pro-$\ell$ group, the image must be finite.

TopGrunwald-Wang theorem

Proposition 19 Let $K$ be a global field. Suppose $\Char(K)\nmid n$ and $K^\times$ contains the $n$-th roots of unity. If $x\in K$ is an $n$-th power in $K_v$ for almost all $v$, then $x$ itself is an $n$-th power in $K$.
Proof Let $\alpha\in K^\mathrm{sep}$ be an $n$-th root of $x$ and $f(t)\in K[t]$ be its minimal polynomial. So $f(t)\mid t^n-x$. If $x$ is an $n$-th power in $K_v$, then $f(t)$ splits completely in $K_v$ as $K^\times$ contains the $n$-th roots of unity. Since $f(t)$ is separable, it follows that $v$ splits completely in $K_v(\alpha)$. Now the split primes $\Spl(K(\alpha)/K)$ has density 1, hence $K(\alpha)=K$.
Exercise 19
  1. 16 is an 8-th power in $\mathbb{R}$ and in $\mathbb{Q}_p$ for $p\ne2$, but not in $\mathbb{Q}_2$ (hence not in $\mathbb{Q}$, which is obvious).
  2. Let $K=\mathbb{Q}(\sqrt{7})$, then 16 is an 8-th power in $K_v$ for any place $v$ but not an 8-th power in $K$.
Theorem 40 (Grunwald-Wang) Let $K$ be a global field. If $x\in K$ is a $n$-th power in $K_v$ for almost all $v$, then $x$ is an $n$-th power in $K$, except potentially if $K$ is a number field and $K(\zeta_{2^t})/K$ is not cyclic, where $n=2^t\cdot m$ with $m$ odd.
Remark 88 $\Gal(\mathbb{Q}(\zeta_8)/\mathbb{Q})\cong(\mathbb{Z}/8 \mathbb{Z})^\times\cong (\mathbb{Z}/2 \mathbb{Z})^2$ is not cyclic, which explains the previous exercise.
Remark 89 Grunwald proved and published the theorem without noticing the exceptional cases (and even a second proof due to Whaples was published and reviewed by Chevalley later!). Wang found a counter-example and proved the right version of the theorem in his thesis. The mistake occurred in one of Grunwald's lemmas where the case $p=2$ fails and is essentially due to the sloppy usage of the notation $K(\sqrt[n]{x})$ whose meaning depends on a choice of $n$-th root of unity. For this reason Artin-Tate strongly urged that the notation $K(\sqrt[n]{x})$ should never be used unless $K$ contains the $n$-th roots of unity!

The proof only uses the result on split primes (Corollary 10), so it is not really an application of class field theory.

11/05/2012

Proof It suffices to treat the case that $n=p^a$ is a power of a prime. In fact, suppose $n=n_1n_2$ with $n_1, n_2$ coprime, then by the Euclidean algorithm we can write $a_1n_1+a_2n_2=1$. Suppose $x=y_1^{n_1}=y_2^{n_2}$, then $x=x^{a_1n_1+a_2n_2}=y_2^{a_1 n} y_1^{a_2 n}=(y_1^{a_1}y_2^{a_2})^n$. We can further assume that $p\nmid \Char(K)$. If $n=p^a$ and $y\in K_v$ such that $y^n=x$. Then $K(y)/K$ is purely inseparable over $K$. But $K_v/K$ is separable for any $v$, which implies that $K(y)=K$ and $x\in (K^\times)^n$.

First consider the case that $K(\zeta_{p^a})/K$ is cyclic of $p$-power order. We know that $t^n-x$ splits into linear factors over $K(\zeta_{p^a})$ by the previous proposition. Now look at the factorization $t^n-x=\prod f_i(t)$ over $K$ and choose a root $y_i$ of $f_i$ in $K(\zeta_{p^a})$ for each $i$. If $x\in (K_v^\times)^n$ for some $v$, then $y_i\in K_v$ for some $i$ and $v$ splits in $K(y_i)$(notice that $K(y_i)/K$ is always abelian). Since $K(\zeta_{p^a})/K$ is a cyclic of $p$-power order, its subfields are totally ordered. Hence there exists an $i_0$ such that $K(y_{i_0})\subseteq K(y_i)$ for any $i$. Therefore $v$ splits in $K(y_{i_0})$ for almost all $v$. By Corollary 10, we know that $K(y_{i_0})=K$ and $t^n-x$ has a linear factor $t-y_{i_0}$ over $K$, i.e., $x\in (K^\times)^n$.

The case $p=2$ and $p\nmid \Char(K)$ the follows since $K(\zeta_{2^a})/K$ is always cyclic of 2-power order, due to the assumption in the number field case and the fact that every finite extension of the constant field is cyclic extension in the global function field case. It remains to prove the case $p$ odd and $p\nmid \Char(K)$. Notice that the extension $K(\zeta_{p^a})/K(\zeta_p)$ is always cyclic of $p$-power order and we can apply the first case to find $y\in K(\zeta_p)$ such that $x=y^n$. Let $z=N_{K(\zeta_p)/K}(y)$, then $x^d=z^n$ is an $n$-th power, where $d\nmid p-1$ is the degree of $K(\zeta_p)/K$. Because $d$ is coprime to $p$, we know that $x$ itself is an $n$-th power.

Let us analyze the exceptional case in more detail. It is remarkable that we can write down the exceptional cases completely. In general when the local-global principle fails, it is quite rare that the failure can be completely classified.

Suppose $K(\zeta_{2^t})/K$ is not cyclic. We choose a $2^r$-th root of unity $\zeta_r\in K^\mathrm{sep}$ such that $\zeta_1=-1$, $\zeta_2=i$ and $\zeta_{r+1}^2=\zeta_r$. Write $\eta_r=\zeta_r+\zeta_r^{-1}$.

Lemma 18
  1. $K(\eta_{r+1})\supseteq K(\eta_r)$.
  2. $K(i, \eta_{r+1})=K(\zeta_r,\eta_{r+1})=K(\zeta_{r+1})$.
  3. $K(\eta_r)/K$ is cyclic of 2-power order.
Proof Corresponding to the decomposition of $\Gal(\mathbb{Q}(\zeta_r)/\mathbb{Q})\cong \mathbb{Z} /2 \mathbb{Z}\times \mathbb{Z}/2 ^{r-2} \mathbb{Z}$, we have the decomposition of $\mathbb{Q}(\zeta_r)=\mathbb{Q}(i)\cdot \mathbb{Q}(\eta_r)$. The lemma follows immediately from this decomposition.

Let $s\ge2$ be the unique integer such that $\eta_s\in K$ and $\eta_{s+1}\not\in K$.

Lemma 19 $K(\zeta_r)/K $ is cyclic for any $r\ge1$ if and only if $i\in K(\eta_{s+1})$.
Proof If $K(\zeta_r)/K $ is cyclic, then it contains a unique quadratic extension $K(\eta_{s+1})$ as $\eta_{s+1}^2=\eta_s+2$. We find that $K(i)\subseteq K(\eta_{s+1})$, hence $i\in K(\eta_{s+1})$. If $i\in K(\eta_{s+1})$, then $K(\eta_r)=K(\zeta_r)$ is cyclic for all $r\ge s+1$ by the previous lemma. For $r<s+1$, $K(\zeta_r)$ is also cyclic because $K(\zeta_r)\subseteq K(\zeta_{r+1})$.
Remark 90 Suppose we are in the exceptional case, then the previous lemma implies that $i\not\in K(\eta_{s+1})$. Then $K(\zeta_{s+1})=K(\eta_{s+1},i)$ is a degree 4 abelian extension of $K$, which is the compositum of the three quadratic extensions $K(\eta_{s+1})$, $K(i)$ and $K(i\eta_{s+1})$. In particular, $\eta_{s+1}^2=\eta_{s}+2$, $-1$ and $-(\eta_{s}+2)$ are all non-squares in $K$.
Definition 69 Let $S\subseteq V_K$ be a finite set. We define $P(m, S)=\{x\in K^\times: x\in (K_v^\times)^m, \forall v\not\in S\}$.
Remark 91 Write $m=2^t\cdot m'$. Suppose $x\in P(m, S)$ and $x\not\in (K^\times)^m$, then we know that $x\in (K^\times)^{m'}$ by the Grunwald-Wang theorem, hence $x\in P(2^t, S)$ but $x\not\in (K^\times)^{2^t}$ by the Euclidean algorithm. Moreover, because $K(\zeta_s)/K $ is cyclic, the Grunwald-Wang theorem implies that $t>s\ge 2$.

The following key lemma characterizes all the counter examples.

Lemma 20 If $x\in P(m, S)$ and $x\not\in (K^\times)^m$, then $x\in x_0 (K^\times)^m$ , where $x_0=\eta_{s+1}^m\in K^\times$.
Proof $K(\zeta_r)/K(i)$ is always cyclic, so by the Grunwald-Wang theorem there exists $y\in K(i)^\times$ such that $y^{2^t}=x$. Let $\sigma\in \Gal(\mathbb{Q}(i)/\mathbb{Q})$ be the complex conjugation. We compute $$\sigma \zeta_s=\zeta_s^{-1},\quad \sigma \eta_s=\eta_s,\quad (y^{1-\sigma})^{2^t}=x^{1-\sigma}=1.$$ So $y^{1-\sigma}$ is a $2^t$-th root of unity. But $\bbmu_{2^t}\cap K(i)= \bbmu_{2^s}$, hence $y^{1-\sigma}$ is a $2^s$-root of unity. So we can write $y^{1-\sigma}=\zeta_s^\mu$ for some $\mu$. Define an element $y_1=y\zeta_s^\lambda$ for some $\lambda$ to be determined. Then $$y_1^{1-\sigma}=y^{1-\sigma}(\zeta_s^\lambda)^{1-\sigma}=\zeta_s^{\mu+2\lambda}.$$ If $\mu$ is even, then we choose $\lambda$ so that $y_1^{1-\sigma}=1$. Then $y_1\in K^\times$ and $y_1^{2^t}=x$ (as $t>s$), which contradicts our assumption. If $\mu$ is odd, we choose $\lambda$ so that $y_1^{1-\sigma}=\zeta_s^{m'}$. Replacing $y$ by $y_1$, we may assume that $y^{1-\sigma}=\zeta_s^{m'}$. Notice that $\zeta_s=\frac{1+\zeta_s}{1+\zeta_s^{-1}}$, we compute $$y^{1-\sigma}=\zeta_s^{m'}=\left(\frac{1+\zeta_s}{1+\zeta_s^{-1}}\right)^{m'}=(1+\zeta_s)^{m'(1-\sigma)}.$$ If $z=y(1+\zeta_s)^{-m'}$, then $z^{1-\sigma}=1$, hence $z\in K^\times$. So $x=y^{2^t}=z^{2^t}(1+\zeta_s)^m=x_0z^{2^t}$, where $$x_0=(1+\zeta_s)^m=(\zeta_{s+1}\eta_{s+1})^m=\eta_{s+1}^m.$$ Notice that $x_0=(\eta_s+2)^{2^{t-1}m'}\in (K^\times)^{m'}$. Hence $x/x_0\in (K^\times)^{m'}$. Also $x/x_0=z^{2^t}\in (K^\times)^{2^t}$, hence the lemma follows.

11/07/2012

We can actually show that $x_0\not\in (K^\times)^m$. If not, then $x_0\in (K^\times)^{2^t}$. Thus $\zeta (\eta_s+2)\in (K^\times)^2$ for some $2^{t-1}$-th root of unity $\zeta$. But $\zeta\in K^\times\cap\bbmu_2=\{\pm1\}$, which contradicts our assumption. Therefore if $P(m,S)\ne (K^\times)^m$ then $P(m,s)$ is the disjoint union of $(K^\times)^m$ and $x_0(K^\times)^m$.

We have shown that in order to ensure $K(\zeta_{2^t})/K$ to be non-cyclic, we necessarily have $t>s$, $-1, \eta_s+2,-(\eta_s+2)$ are non-squares in $K^\times$. We shall show this obstruction always fails with the further condition (d) as in the following strong version of Grunwald-Wang theorem.

Theorem 41 (Strong Grunwald-Wang) $P(m,S)=(K^\times)^m$ except the following condition holds:
  1. $K$ is a number field.
  2. $-1, \pm(\eta_s+2)\not\in (K^\times)^2$.
  3. $m=2^tm'$, where $t>s$.
  4. $S$ contains the set $S_0$ of places over 2 such that $-1,\pm(\eta_s+2)$ are non-squares in $(K_v^\times)^2$. In the exceptional case, $P(m,S)=(K^\times)^m\coprod x_0(K^\times)^m$.
Proof Suppose $K$ is a number field and $P(m,s)\ne (K^\times)^m$, then (a)-(c) holds and $P(m,S)=(K^\times)^m\coprod x_0(K^\times)^m$. We have to show that (d) holds, i.e., for $v\in S_0$, then $x_0\not\in (K_v^\times)^m$. If $x_0=\eta_{s+1}^m\in (K_v^\times)^m$. Then we deduce that $(\eta_s+2)^{2^{t-1}}\in (K_v^\times)^{2^t}$ by Euclidean algorithm. Hence there exits $\zeta\in \bbmu_{2^{t-1}}$ and $\zeta(\eta_s+2)\in (K_v^\times)^2$. If $-1\not\in (K_v^\times)^2$, then $\bbmu_{2^{t-1}}\cap K_v^\times=\{\pm1\}$, we know $\pm(\eta_s+2)\in (K_v^\times)^2$. Otherwise, $-1\in(K_v^\times)^2$, hence $v\not\in S_0$.

For the other direction, we must show that if (a)-(d) hold, then $x_0\in P(m,S)$. It is equivalent to showing that $x_0\in (K_v)^\times$ for any $v\not\in S$. Let $F=K(i,\eta_{s+1})=K(\zeta_{s+1})$. Condition (b) implies that $F/K$ is a degree four extension with Galois group $\mathbb{Z}/2 \mathbb{Z}/2$. Moreover, $F$ is unramified away from 2 by construction. If $w$ is a palce of above $v\nmid2$. Then $F_w/K_v$ is quadratic. Consequently one of $-1,\pm(\eta_s+2)$ is a square in $K_v^\times$. If $v\nmid2$ and $v\not\in S$, the one of $-1,\pm(\eta_s+2)$ is a square in $K_v^\times$ by the assumption (d). In either case,

  1. if $\eta_s+2$ or $-)(\eta_s+2)$ is a square in $K_v^\times$. Then $\eta_{s+1}, i\eta_{s+1}\in K_v^\times$. Therefore $x_0=(\eta_{s+1})^m=(i\eta_{s+1})^m\in (K_v^\times)^m$.
  2. if $i\in K_v^\times$, then $\zeta_s\in K_v(i)^\times=K_v^\times$, hence $x_0=(1+\zeta_s)^m\in (K_v^\times)^m$. We conclude that $x_0\in P(m,S)$.

11/30/2012

TopOutline of the proof

TopGroup cohomology

Suppose $G$ is any group. From an exact sequence of $G$-modules, $$0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0,$$ taking $G$-invariants gives an exact sequence $$0\rightarrow A^G\rightarrow B^G\rightarrow C^G.$$ But the last map is not necessarily surjective, e.g., consider $G=G_K$ and the Kummer sequence $$0\rightarrow \mu_n\rightarrow \mathbb{G}_m\rightarrow \mathbb{G}_m\rightarrow 0.$$ The group cohomology functor $A\mapsto H^q(G,A)$, $q\ge0$ extends the above left exact sequence to the expected long exact sequence.

Dually, we can also take the $G$-coinvariants $A_G=A/\langle ga-a,a\in A\rangle$ (the largest quotient on which $G$ acts trivially) which does not preserve injectivity and the group homology functor $A\mapsto H_q(G,A)$ fills in the corresponding long exact sequences.

When $G$ is a finite group, Tate defines the norm map $N: A\rightarrow A, \quad a\mapsto\sum_{g\in G} ga$. It maps into $A^G$ and factors through $A_G$. Then we can connect both long exact sequence in cohomology and homology as $$\xymatrix{ H_1(G,C) \ar[r] & A_G \ar[r] \ar[d] & B_G \ar[r] \ar[d]  & C_G \ar[r] \ar[d] & 0\\ 0 \ar[r] & A^G \ar[r] & B^G \ar[r] & C^G \ar[r] & H^1(G,A).}$$ Write $\hat H_0(G,A)=\ker (N: A_G\rightarrow A^G)$, $\hat H^0(G,A)=\mathrm{coker}(N: A_G\rightarrow A^G)$ and $\hat H^{-q}(G,A)=H_q(G,A)$. Then the five lemma gives a long exact sequence in Tate group cohomology $\hat H^n(G,A)$ in both directions.

When $G$ is cyclic, we further have $\hat H^q(G,A)\cong\hat H^{q-2}(G,A)$, i.e., the Tate cohomology groups are periodic of period 2.

Example 30 Let $G=\Gal(L/K)$ be finite. Then the Hilbert 90 theorem asserts that $H^1(G,L^\times)=0$. In particular, when $G$ is cyclic, $\ker(N)/(\sigma-1)L^\times=\hat H_0(G,L^\times)=\hat H^{-1}(G, L^\times)=0$, which implies that $N(x)=1$ if and only if $x=\sigma(y)/y$ (the classical formulation of Hilbert 90).
Example 31 $\hat H^0(G,L^\times)=K^\times/N(L^\times)$.
Example 32 $\hat H^{-2}(G,\mathbb{Z})=H_1(G,\mathbb{Z})=G^\mathrm{ab}$ and dually $H^1(G,\mathbb{Z})=\Hom(G,\mathbb{Z})$, where $\mathbb{Z}$ is endowed with the trivial $G$-action.

TopClass formations

Class formations tell you all the group cohomology input in order to derive all the statements in class field theory (e.g., Artin maps).

Let $K$ be a local or global field. Suppose $G=G_K$ and $A $ is a $G$-module. We say $A $ is a continuous $G$-module if $A=\bigcup_{[L:K]<\infty} (A_L=A^{G_L})$ (e.g., $A=(K^\mathrm{sep})^\times$ and $A=\bigcup_{[L:K]<\infty}\mathbb{A}_L^\times/L^\times$. Fix such an $A $, for any Galois extension $L/K$, we define $H^q(L/K):=H^q(\Gal(L/K), A_L)$ and $\hat H^q(L/K):=\hat H^q(\Gal(L/K), A_L)$.

Suppose $L'/K$ is Galois and $L\subseteq L'$, we have an inflation map $\Inf_{L'/L}: H^q(L/K)\rightarrow H^q(L'/K)$ given by the precompostion by the natural surjection on cocycles.

Suppose $K'\subseteq L$, we have a restriction map $\Res_{K'/K}: H^2(L/K)\rightarrow H^q(L/K')$ given by the precompostion by the natural inclusion on cocycles.

Definition 70 A class formation is the data of $(G=G_K,A)$ plus an invariant map $\inv_{L/K'}: H^q(L/K')\rightarrow \mathbb{Q}/\mathbb{Z}$ for any $K'/K$ finite and $L/K'$ finite Galois satisfying
  1. $H^1(L/K)=0$ for any $L/K$ finite Galois.
  2. $\inv_L$ is injective and induces an isomorphism $H^2(L/K)\cong 1/[L:K]\mathbb{Z}/\mathbb{Z}$. Moreover, $\inv_{L/K'}\circ \Res_{K'/K}=[K':K]\inv_{L/K}$. Equivalently, there exists a generator (called the fundamental class) $u_{L/K'}\in H^2(L/K')$ of order $[L:K']$ such that $\Res(u_{L/K})=u_{L/K'}$ (which formally implies that $\Inf(u_{L/K})=[L':L]u_{L'/K}$).

Using pure group cohomology, we will prove the following main theorem.

Theorem 42 Cup product with $u_{L/K}$ gives isomorphisms $\hat H^q(\Gal(L/K),\mathbb{Z})\cong \hat H^{q+2}(\Gal(L/K), A_L)=\hat H^{q+2}(L/K)$. In particular, when $q=-2$ and $L/K$ is abelian, we obtain that $\Gal(L/K)\cong A_K/N(A_L)$. We define the Artin map $\Psi_{L/K}$ to be the inverse of this isomorphism.

TopLocal class field theory

The main reference will be Serre's local fields and Galois cohomology. We set $A=(K^\mathrm{sep})^\times$, then $A_L=L^\times$. The first axiom in class formation is simply Hilbert 90 and the second axiom amounts to proving that $H^2(G_K,(K^\mathrm{sep})^\times)\cong \mathbb{Q}/\mathbb{Z}$. Let $G_\mathrm{ur}=\Gal(K^\mathrm{ur}/K)$ (for local fields $\cong\hat{\mathbb{Z}}$, but this als0 works for general complete discretely valued fields with quasi-finite residue fields, however, the existence does not work, the local compactness is still needed). For any field $E $, we write $\Br_E=H^2(G_E, (E^\mathrm{sep})^\times)$.

Theorem 43 We have an exact sequence $$0\rightarrow \Br_k\rightarrow \Br_K\rightarrow\Hom_\mathrm{cont}(G_\mathrm{ur},\mathbb{Q}/\mathbb{Z})\rightarrow 0,$$ where $k$ is the residue field of $K$.

The only non-formal part is the following

Proposition 20 $\Br_K=H^2(G_\mathrm{ur},(K^\mathrm{sep})^\times)$ (in other words, any central simple algebra over $K$ is split after an unramified base change.
Theorem 44 When $k$ is finite, $\Br_k=0$ (in other words, there are no noncommutative central division algebra over a finite field).
Corollary 13 When $K$ is local, $\Br_K\cong \Hom_\mathrm{cont}(\hat{\mathbb{Z}}, \mathbb{Q}/\mathbb{Z})\cong \mathbb{Q}/\mathbb{Z}$.

The class formation machinery then proves local class field theory except the existence theorem. Proving the latter boils down to constructing enough abelian extensions using cyclotomic, Kummer and Artin-Schrier extensions (together with pure topological arguments).

TopGlobal class field theory

The main reference will be Cassels-Frolich and Artin-Tate. In global class field theory, the idele class group $A_L=\mathbb{A}_L^\times/L^\times$ plays the role of $L^\times$ in local class field theory. The first axiom in class formation is again Hilbert 90. For the second axiom, we will prove $\Psi_{L/K}$ is the the product of local Artin maps and it kills all the global elements. By the exact sequence $0\rightarrow L^\times\rightarrow \mathbb{A}_L^\times\rightarrow \mathbb{A}_L^\times/L^\times\rightarrow 0$, the Brauer group of a global field $K$ fits into $$0\rightarrow\Br_K\rightarrow\bigoplus\Br_{K_v} \rightarrow \mathbb{Q}/\mathbb{Z}\rightarrow 0.$$ As we will see, the proof of the existence theorem (norm index inequality) interplays with the proof of Artin maps (class formation): first proved for cyclotomic extensions and then in general.