These are my (slightly reorganized) live-TeXed notes for the course Math 223b: Algebraic Number Theory taught by Joe Rabinoff at Harvard, Spring 2012. This course is a continuation of Math 223a. Please let me know if you notice any errors or have any comments!

01/28/2013

TopOverview

Suppose $K$ is a global field or a local field. Denote $A_K=\mathbb{A}_K^\times/K^\times$ or $A_K=K^\times$ respectively. Recall that the main theorems of class field theory states the following.

Theorem 1 There exists a canonical continuous homomorphism with dense image $\Psi_K: A_K\rightarrow G_K^\mathrm{ab}$, called the Artin map, such that
  1. (norm and verlagerung functoriality) For any finite separable extension $L/K$, the following two diagrams commute $$\xymatrix{A_L\ar[r] \ar[d]^{N_{L/K}}& G_L^\mathrm{ab}\ar[d]\\ A_K \ar[r]& G_K^\mathrm{ab},}\quad \xymatrix{A_L\ar[r] & G_L^\mathrm{ab}\\ A_K \ar[r]\ar[u] & G_K^\mathrm{ab}\ar[u]^{\Ver}.}$$
  2. (existence) Every finite index open subgroup of $A_K$ arises as the kernel of $\Psi_{L/K}: A_K\rightarrow G_K^\mathrm{ab}\rightarrow\Gal(L/K)$ for a unique finite abelian extension $L/K$ (so $\ker \Psi_{L/K}=N_{L/K}(A_L)$).
  3. When $K$ is a number field, $\Psi_K$ is surjective and $D_K$ consists of divisible elements. In all other cases, $\Psi_K$ is injective. Write $A_K^1=(\mathbb{A}_K^\times)^1$ (global) or $\mathcal{O}_K^\times$ (local) respectively. Then $\Psi_K$ induces an isomorphism between $A_K$ and the Weil group $W_K^\mathrm{ab}$, between $A_K^1$ and the inertia group $I$.
Theorem 2 (local-global compatibility) Suppose $v$ is a place of a global field $K$, then the following diagram commutes $$\xymatrix{K_v^\times \ar[r]^-{\mathrm{incl.}} \ar[d]_{\Psi_{K_v}} & \mathbb{A}_K^\times/K^\times \ar[d]^{\Psi_K}\\ G_{K_v}^\mathrm{ab} \ar[r] & G_K^\mathrm{ab}.}$$ In particular, if $L/K$ is finite abelian , then $\Psi_K$ kills $\mathcal{O}_v^\times$ if and only if $v$ is unramified, in which case it sends $\pi_v$ to $\Frob_v$. This determines $\Psi_K$ by continuity.

We will prove the main theorems of local and global class field theory in the first part of this semester (as sketched at the end of last semester). In the remaining part, our additional topics may include central simple algebras and quaternion algebras, Lubin-Tate formal groups (explicit local class field theory), CM elliptic curves (explicit class field theory for imaginary quadratic fields), (possibly) Drinfeld modules (explicit class field theory for global function fields), local and global Tate duality theorems (c.f., Serre's Galois cohomology [1]) and (possibly) Langlands classification of representations of tori.

01/30/2013

Top$G$-modules

We will start with basics on group cohomology (c.f., Serre's local fields [2]).

Definition 1 Let $G$ be any group. A left (right) $G$-module is an abelian group $A$ equipped with a left (right) action of $G$, i.e., a homomorphism $G\rightarrow\Aut(A)$.
Example 1 If $G=\Gal(L/K)$ is a Galois group of the field extension $L/K$, then $(L,+)$ and $(L^\times, \times)$ are both $G$-modules.
Example 2 Denote the free abelian group on the elements of $G$ by $\mathbb{Z}[G]$ (called the group ring). Under the left (right) action of $G$ on $\mathbb{Z}[G]$ by $h\sum a_g(g)=\sum a_g (hg)$ ($\sum a_g (g) h=\sum a_g(gh)$), $\mathbb{Z}[G]$ becomes a left (right) $G$-module.
Remark 1 A $G$-module is the same thing as a $\mathbb{Z}[G]$-module (in the usual sense when viewing $\mathbb{Z}[G]$ as a ring).
Remark 2 For any commutative ring $R$, one similarly defines the group ring $R[G]=R\otimes_\mathbb{Z} \mathbb{Z}[G]$. Then an $R[G]$-module is nothing but an $(R,G)$-module, i.e., an $R$-modules $A$ equipped with a $G$-action $G\rightarrow \Aut_R(A)$).
Definition 2 A $G$-module homomorphism is a $\mathbb{Z}[G]$-module homomorphism. The category of $G$-modules, denoted by $\mathbf{Mod}_G$, is an abelian category. For $A, A'\in\mathbf{Mod}_G$, the set of $G$-module homomorphisms between them is denoted by $\Hom_G(A,A')$.
Definition 3 A $G$-module $A$ is called injective (projective) if the functor $\Hom_G(-,A)$ ($\Hom_G(A,-)$) is exact.
Remark 3 $\mathbf{Mod}_G$ has enough injectives (in fact, it holds over any ring, c.f. Lang's algebra), i.e., every $G$-module can be embedded into an injective module (using the fact that any divisible module is injective). Any free $\mathbb{Z}[G]$-module is projective and $\mathbf{Mod}_G$ has enough projective since any $G$-module is the quotient of some free $\mathbb{Z}[G]$-module.
Definition 4 For $A, A'\in \mathbf{Mod}_G$, we endow the $G$-module structure on $A\otimes_\mathbb{Z}A'$ by $g(a\otimes a')=(ga)\otimes (ga')$ and the $G$-module structure on $\Hom_\mathbb{Z}(A,A')$ by $g(\phi)(a)=g.\phi(g^{-1}a)$ (c.f., Remark 18).
Definition 5 Let $\phi: H\rightarrow G$ be a group homomorphism. This induces a ring homomorphism $\phi: \mathbb{Z}[H]\rightarrow \mathbb{Z}[G]$. We define three functors
  1. restriction $\Res_H^G: \mathbf{Mod}_G\rightarrow \mathbf{Mod}_H$, viewing a $G$-module as a $H$-module via $\phi$;
  2. (compact) induction $\ind_H^G: \mathbf{Mod}_H\rightarrow \mathbf{Mod}_G$, $A\mapsto \mathbb{Z}[G]\otimes_{\mathbb{Z}[H]} A$;
  3. (co-)induction $\Ind_H^G: \mathbf{Mod}_H\rightarrow \mathbf{Mod}_G$, $A\mapsto \Hom_{\mathbb{Z}[H]}(\mathbb{Z}[G],A)=\{f: G\rightarrow A: f(gh)=h^{-1}f(g)\}$.
Remark 4 Suppose $A$ is right $R$-module and $B$ is left $R$-module, then $A\otimes_RB$ is an abelian group. So $\ind_H^GA=\mathbb{Z}[G]\otimes_{\mathbb{Z}[H]}A$ is an abelian group when viewing $\mathbb{Z}[G]$ as a right $\mathbb{Z}[H]$-module. It further becomes a $G$-module when $G$ acts on $\mathbb{Z}[G]$ from the left, i.e., $g(x\otimes y)=(gx)\otimes y$.
Remark 5 The $G$-action on $\Ind_H^GA$ is given by $(gf)(x)=f(xg)$.
Remark 6 Suppose $H$ is a subgroup of $G$ and $\{g_i\}$ is a set of coset representatives of $H\backslash G$. Then $\mathbb{Z}[G]\cong \bigoplus g_i^{-1}\mathbb{Z}[H]$ and $\ind_H^GA\cong \bigoplus g_i^{-1}A$ and the $G$-action is given by $g(\sum g_i^{-1}a)=\sum g_j^{-1}(g_j g g_i^{-1}) a_i$, where $gg_i^{-1}\in g_j^{-1}H$. Under $f\mapsto \sum g_i^{-1}\otimes f(g_i)$, $\ind_H^G(A)$ can also be identified with the set of maps $f: H\backslash G\rightarrow A$ where $f$ vanishes almost everywhere: this explains the name "compact induction".
Remark 7 Suppose $H$ is a subgroup of $G$ and $\{g_i\}$ a set of coset representatives of $H\backslash G$. Then $f\mapsto f(g_i)\cdot g_i$ gives $\Ind_H^G=\prod A\cdot g_i$ and the $G$-action is given by $g(a_i\cdot g_i)=((g_jgg_i^{-1})a_i\cdot g_i)$, where $g_jg\in H g_i$. $\Ind_H^GA$ can be identified with all maps $f: H\backslash G\rightarrow A$ via $(f: \mathbb{Z}[G]\rightarrow A)\mapsto (g_i\mapsto f(g_i))$.
Remark 8 Suppose $H$ is a subgroup of $G$, we see that there is an $G$-module inclusion $\ind_H^GA\hookrightarrow \Ind_H^GA$. When $[G:H]<\infty$, $\ind_H^GA=\Ind_H^GA$ since the direct sum is the same as the direct product.
Definition 6 For $H=1$ and $A$ any abelian group, we have $\ind_H^GA=\mathbb{Z}[G]\otimes_\mathbb{Z} A=A[G]$. We say $A'$ is induced if $A'$ is isomorphic to $A[G]$ as a $G$-module. A direct summand of an induced module is called relatively projective.
Definition 7 Similarly, $\Ind_1^GA=\Hom(G,A)$ is called co-induced. A direct summand of a co-induced module is called relatively injective.
Proposition 1 (Frobenius reciprocity) $A\in \mathbf{Mod}_H$ and $A'\in \mathbf{Mod}_G$. Then $$\Hom_H(A,\Res_H^GA')=\Hom_G(\ind_H^GA, A').$$
Remark 9 This is the usual adjointness between the restriction of scalars and the extension of scalars of modules.
Remark 10 For $A=\Res_H^GA'$, $\Id\in \Hom_H(A,\Res_H^GA')$ corresponds to the surjective map $\ind_H^G \Res_H^G A'\rightarrow A'$, $g\otimes a'\mapsto ga'$. So any $G$-module $A$ is canonically a quotient of the induced module $A_*=A[G]$.
Remark 11 For $A'{}=\ind_H^GA$, $\Id\in \Hom_G(\ind_H^GA,A')$ corresponds to the map $A\rightarrow \Res_H^G\ind_H^G A$, $a\mapsto 1\otimes a$.
Remark 12 Notice that $\mathbb{Z}[G]\otimes_\mathbb{Z}A\cong \mathbb{Z}[G]\otimes_\mathbb{Z} \Res_1^G A$ (as $G$-modules) given by $g\otimes ga\mapsto g\otimes a$. So there is no ambiguity to view $\mathbb{Z}[G]\otimes_\mathbb{Z}A$ as the tensor product as $G$-modules or as abelian groups.
Proposition 2 (Frobenius reciprocity) Suppose $A\in \mathbf{Mod}_H$ and $A'\in \mathbf{Mod}_G$. Then $$\Hom_H(\Res_H^GA',A)=\Hom_G(A',\Ind_H^GA).$$
Remark 13 Similarly, we have a canonical injection $A\hookrightarrow \Ind_H^G\Res_H^GA$ via $a\mapsto (x\mapsto xa)$. So any $G$-module $A$ is canonically injects into the co-induced module $A^*=\Hom_{\mathbb{Z}}(\mathbb{Z}[G], A)$. There is also a canonical map $\Res_H^G\Ind_H^G A'\rightarrow A'$ given $f\mapsto f(1)$.
Remark 14 Under Frobenius reciprocity, the inclusion $\ind_H^G A\hookrightarrow \Ind_H^G A$ comes from the canonical map $A\rightarrow \Res_H^G \Ind_H^G A$ given by $$a\mapsto \quad g\mapsto
\begin{cases}
  ga, & g\in H, \\
  0, & g\not\in H.
\end{cases}$$ or the canonical map $\Res_H^G \ind_H^G A\rightarrow A$ given by $$g\otimes a\mapsto
\begin{cases}
  ga, & g\in H, \\
  0, & g\not\in H.
\end{cases}$$ So explicitly $\ind_H^G A\hookrightarrow \Ind_H^G A$ is given by $$x\otimes a\mapsto \quad gx^{-1}\mapsto
\begin{cases}
  ha, & g\in H, \\
  0, & g\not\in H.
\end{cases}$$
Remark 15 When $[G:H]<\infty$ (e.g., $G$ is finite), $\Ind, \Res$ are both left and right adjoint.

02/01/2013

TopGroup cohomology and homology

Definition 8 Let $A$ be a $G$-module. We define the invariants $A^G=\{a\in A: ga=a, \forall g\in G\}$.
Remark 16 Notice $A^G=\Hom(\mathbb{Z}, A)$ (where $G$ acts on $\mathbb{Z}$ trivially). So any short exact sequence of $G$-modules $0\rightarrow A\rightarrow B\rightarrow C\rightarrow0$ gives a left exact sequence of invariants $$0\rightarrow A^G\rightarrow B^G\rightarrow C^G.$$
Definition 9 Let $A$ be a $G$-module. The co-invariants $A_G$ is defined to be the largest quotient on which $G$ acts trivially, i.e, $A_G=A/\langle ga-a, a\in A, g\in G\rangle$.
Remark 17 Let $I_G=\ker (\mathbb{Z}[G]\xrightarrow{\deg} \mathbb{Z})$ be the argumentation ideal, then $I_G=\langle g-e,g\in G\rangle$. Tensoring with $A$ gives a right exact sequence $I_G\otimes A\rightarrow A\rightarrow \mathbb{Z}\otimes_GA\rightarrow0$, where the image is of the first map is the same as $\langle ga-a, a\in A, g\in G\rangle$. So $A_G= \mathbb{Z}\otimes_G A$ and any short exact sequence of $G$-modules $0\rightarrow A\rightarrow B\rightarrow C\rightarrow0$ gives a right exact sequence of co-invariants $$A_G\rightarrow B_G\rightarrow C_G\rightarrow 0.$$
Remark 18 One can check directly (Definition 4 ) that $$A\otimes_G A=(A\otimes A')_G,\quad \Hom_G(A,A')=\Hom(A,A')^G.$$

The group cohomology (resp. homology) of $G$ measures the failure of the right (resp. left) exactness of taking $G$-invariants (resp. co-invariants).

Definition 10 The group cohomology functor $H^i(G,-): \mathbf{Mod}_G\rightarrow \mathbf{Ab}$ is defined to be the right derived functor of $(-)^G$. The group homology functor $H_i(G,-): \mathbf{Mod}_G\rightarrow \mathbf{Ab}$ is defined to be the left derived functor of $(-)_G$.
Remark 19 Recall that the right derived functor $H^i(G,-)$ can be computed using injective resolutions. Let $0\rightarrow A\rightarrow I^0\rightarrow I^1\rightarrow\cdots $ be an injective resolution $G$-modules of $A$, then $H^i(G,A)$ is the same as the cohomology of the complex $(I^\bullet)^G$. Similarly, $H_i(G,A)$ can computed using a projective resolution $P^\bullet\rightarrow A$ and taking the cohomology of the complex of co-invariants $(P^\bullet)_G$.

The group cohomology (homology) is a cohomological (homological) functor satisfying the following basic properties.

Proposition 3
  1. $H^i(G,A)$ and $H_i(G,A)$ are independent of the choice of the resolutions (up to canonical isomorphisms).
  2. $H^i(G,A)$ and $H_i(G,A)$ are covariant functors in $A$.
  3. $H^0(G,A)=A^G$ and $H_0(G,A)=A_G$.
  4. Any short exact sequence $0\rightarrow A\rightarrow B\rightarrow C\rightarrow0$ of $G$-modules induces long exact sequences in group cohomology and homology. The connecting maps are natural.
Remark 20 If $A$ is injective, then $0\rightarrow A\rightarrow A\rightarrow0$ is an injective resolution of $A$, so $H^i(G,A)=0$ for $i\ge1$. Similarly, if $A$ is projective, then $H_i(G,A)=0$ for $i\ge1$.
Remark 21 The product of two injective modules are injective. So the product of injective resolutions is an injective resolution of the product. Consequently, we have $H^i(G,\prod A_j)=\prod H^i(G,A_j)$. Similarly, $H_i(G, \bigoplus A_j)=\bigoplus H_i(G,A_j)$.

TopBehavior under induction

Remark 22 Suppose $H\rightarrow G$ is a group homomorphism. If $I\in \mathbf{Mod}_H$ is injective and $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ is an exact sequence in $\mathbf{Mod}_G$, then taking $\Res_H^G$ (which is exact) gives a exact sequence $$0\rightarrow \Hom(\Res_H^G C, I)\rightarrow \Hom(\Res_H^G B, I)\rightarrow \Hom(\Res_H^G A, I)\rightarrow 0.$$ Hence $\Ind_H^G I$ is injective in $\mathbf{Mod}_G$. (in general, a functor with exact left adjoint preserves injectives).
Remark 23 Suppose $H$ is a subgroup of $G$, then $\Ind_H^G A=\prod_{H\backslash G} A g_i$, so $\Ind_H^G$ is evidently exact (however, this is not true for an arbitrary homomorphism $H\rightarrow G$, e.g., for $G=1$, $\Ind_H^1 A=\Hom_H(\mathbb{Z},A)=A^H$, which is only left exact).
Remark 24 Using Frobenius reciprocity, we know that $$(\Ind_H^G A)^G=\Hom_G( \mathbb{Z},\Ind_H^GA)=\Hom_H(\mathbb{Z},A)=A^H.$$ Explicitly, $a\in A^H$ is identified with the constant function $(g\mapsto a)\in (\Ind_H^G A)^G$. When $H$ is a subgroup of $G$, $\Ind_H^G$ is exact and so $\Ind_H^GI^\bullet$ is an injective resolution of $\Ind_H^G A$. We can calculate the cohomology $$H^i(G, \Ind_H^G A)=H^i((\Ind_H^G I^\bullet)^G)=H^i((I^\bullet)^H)=H^i(H,A).$$ In other words, $\Ind_H^G$ preserves group cohomology. This result is usually known as Shapiro's Lemma.
Remark 25 Similarly, $\ind_H^G$ has an exact right adjoint, so it preserves projectives. When $H$ is a subgroup of $G$, $\ind_H^GA$ is exact and $H_i(G, \ind_H^G A)=H_i(H, A)$. In other words, $\ind_H^G$ preserves group homology.
Corollary 1
  1. If $A$ is relatively injective, then $H^i(G, A)=0$ for $i\ge1$.
  2. If $A$ is relatively projective, then $H_i(G, A)=0$ for $i\ge1$.
Proof Suppose $A$ is relatively injective, then $A\oplus B=\Ind_1^G C$ for some $B, C$, where $C$ is injective. Thus $$0=H^i(1, C)=H^i(1, \Ind_1^GC)=H^i(G,A)\oplus H^i(G,B)$$ for $i\ge1$. Similarly for the other part.
Remark 26 Notice that we can also compute the group cohomology and homology using acyclic resolutions. By the previous proposition, there is canonical acyclic resolution $0\rightarrow A\rightarrow \Ind_1^G A\rightarrow \Ind_1^G(\Ind_1^GA)\rightarrow\cdots$ to compute the group cohomology (similarly for group homology), which will be useful in the future.

TopCochains and chains

Notice that $(-)^G=\Hom_G(\mathbb{Z},-)$, so $H^i(G,-)$ is simply another name for $\Ext^i_{\mathbb{Z}[G]}(\mathbb{Z},-)$. Similarly, $H_i(G,-)$ is simply the other name for $\Tor^i_{\mathbb{Z}[G]}(\mathbb{Z},-)$. It is a general principle that one can compute these groups via resolutions in both variables. In our cases, we can start with a projective resolution of the trivial $\mathbb{Z}[G]$-module $\mathbb{Z}$. Finding such an explicit resolution will give us an concrete description of cohomology and homology groups in terms of cochains and chains.

Definition 11 Let $P_i=\mathbb{Z}[G^i]$. Then the diagonal morphism makes $P_i$ a $\mathbb{Z}[G]$-module. Define the boundary map $\delta: P_{i+1}\rightarrow P_i$ by $$\delta(g_0,\ldots, g_{i+1})=\sum_{j=0}^{i+1} (-1)^j (g_0,\ldots, \hat{g_j}, \ldots, g_{i+1})$$ and $\delta: P_0=\mathbb{Z}[G]\rightarrow \mathbb{Z}$ be the degree map. The complex $\cdots P_2\rightarrow P_1\rightarrow P_0\rightarrow \mathbb{Z}$ is exact and $P_i$ is free (hence projective). This projective resolution of $\mathbb{Z}$ is called the standard resolution.

Thus $H^i(G,A)=H^i(\Hom_G(P_\bullet, A))$. Concretely, we have $$\Hom_G(P_i,A)=\{f: G^{i+1}\rightarrow A: f(gg_0, \ldots, gg_i)=g\cdot f(g_0,\ldots ,g_i)=\Ind_G^{G^{i+1}}(A).$$ Such an element $f$ is called a homogeneous cochain. The boundary map is concretely $$(\delta f)(g_0,\ldots, g_{i+1})=\sum_{j=0}^{i+1}(-1)^{j} f(g_0,\ldots,\hat{g_j},\ldots g_{i+1}).$$

Definition 12 A homogeneous cochain $f$ is called a cocycle if $\delta f=0$ and a coboundary if $f=\delta f_0$ for some $f_0$. Then $H^i(G,A)$ is the quotient group of the $i $-cocycles by the $i $-coboundaries.

02/04/2013

Notice that $f$ is uniquely determined on its values at $(1, g_1,\ldots, g_i)$. But inhomogeneous cochains are defined slightly differently as follows to make the boundary maps look nicer.

Definition 13 We define the an inhomogeneous cochain to be $$\tilde f(g_1,\ldots, g_i)=f(1,g_1,g_1g_2,\ldots, g_1\cdots g_i).$$ Then $$f(g_0,\ldots, g_i)=g_0 \tilde f(g_0^{-1}g_1,\ldots , g_{i-1}^{-1}g_i).$$ So $\Hom_G(P_i,A)$ can be identified with inhomogeneous cochains, denoted by $C^i(G,A)$.
Remark 27 Notice that $C^\bullet(G,A)$ forms a cochain complex and its cohomology computes $H^i(G,A)$. The boundary map can be computed as 
\begin{align*}
  \delta\tilde f(g_1,\ldots,g_{i+1})&=f(g_1,g_1g_2,\ldots g_1\cdots g_{i+1}) \\&\quad+\sum_{j=1}^{i+1}(-1)^j f(1,g_1,g_1g_2,\ldots \widehat{g_1\cdots g_j},\ldots g_1\cdots g_{i+1})\\
  &=g_1\tilde f(g_2,g_3,\ldots g_{i+1})+\sum_{j=1}^{i} (-1)^j\tilde f(g_1, g_2, \ldots, g_jg_{j+1},\ldots, g_{i+1})\\&\quad+(-1)^{i+1}\tilde f(g_1,\ldots, g_i).
\end{align*}
Example 3 A 0-coboundary is of the form $\delta(a)(g)=ga-a$. So it verifies that $H^0(G,A)=Z^0(G,A)$.
Example 4 A 1-coboundary is of the form $(\delta f)(g_1,g_2)=g_1f(g_2)-f(g_1g_2)+f(g_1)$. A function $f:G\rightarrow A$ such that $\delta f=0$ (i.e. $f(g_1g_2)=f(g_1)+g_1f(g_2)$) is called a twisted homomorphism. So we can identify set of 1-coboundaries as the set of twisted homomorphisms. In particular, when $G$ acts on $A$ trivially, $H^1(G,A)=\Hom(G,A)$ (there is no "twisting").

Analogously, homology can be computed using homogeneous and inhomogeneous chains. We start with a right $\mathbb{Z}[G]$-module projective resolution $P_\bullet\rightarrow \mathbb{Z}$ of $\mathbb{Z}$, where $P_i=\mathbb{Z}[G^{i+1}]$ with the right $G$-action $$(g_1,\ldots, g_i).g=(g^{-1}g_0,\ldots, g^{-1}g_i)$$ Choose the coset representatives $\{x_j\}$ of $G\backslash G^{i+1}$, Then $P_i=\bigoplus x_j \mathbb{Z}[G]$ and $P_i\otimes_GA=\bigoplus x_jA$ can be identified with functions $f: G^{i+1}\rightarrow A$ such that $f(hx)=hf(x), \forall h\in G$, and $f$ has finite support modulo $G$ (which is the same as $\ind_G^{G^{i+1}A}$). Such an element $f$ is called a homogeneous chain. Similarly we can define inhomogeneous chains and the chain complex $C_\bullet(G,A)$ of inhomogeneous chains computes $H_i(G,A)$. We have a similar formula for boundary maps (adding one index instead of deleting): 
\begin{align*}
  df(g_1,\ldots,g_{i-1})&=\sum_{g\in G}g^{-1}f(g, g_1,\ldots,g_{i-1})\\&\quad+\sum_{j=1}^{i-1}(-1)^j\sum_{g\in G}f(g_1,\ldots, g_jg,g^{-1},g_{j+1},\ldots, g_{i-1})\\&\quad+(-1)^i\sum_{g\in G}f(g_1,\ldots,g_{i-1},g).
\end{align*}

Example 5 Suppose $f\in C_1(G,A)$ is a 1-chain, then $$d f=\sum_{g\in G}(g^{-1}f(g)-f(g)).$$ Hence $B_0(G,A)=\langle ga-a,a\in A,g\in G\rangle$ and it verifies that $H_0(G,A)=C_0(G,A)/B_0(G,A)=A_G$.
Proposition 4 $H_1(G,\mathbb{Z})\cong G^\mathrm{ab}$.
Remark 28 The group cohomology of $G$ can be viewed as the cohomology of the classifying space $BG=K(G,1)$ (with coefficient in a local system). The above proposition simply reflects the fact that the first homology group (with $\mathbb{Z}$-coefficient) is the abelianization of the fundamental group of a space. (In fact, since $BG=EG/G$, we have $C^\bullet(BG, \mathbb{Z})=C^\bullet(EG,\mathbb{Z})^G$. Notice $EG$ is contractible, $\mathbb{Z}\rightarrow C^\bullet(EG,\mathbb{Z})$ forms a free resolution of $\mathbb{Z}$. Hence $H^i(BG,\mathbb{Z})=H^i(C^\bullet(EG,\mathbb{Z})^G)=H^i(G,\mathbb{Z})$).
Proof Let $I_G$ be the argumentation ideal of $G$. Then $A_G=A/I_GA$. The long exact sequence of group homology implies that $$H_1(G, \mathbb{Z}[G])\rightarrow H_1(G, \mathbb{Z})\rightarrow H_0(G, I_G)\rightarrow H_0(G, \mathbb{Z}[G])\rightarrow H_0(G, \mathbb{Z})$$ is exact. Notice that $H_1(G,\mathbb{Z}[G])=0$ (as $\mathbb{Z}[G]$ is free) and $H_0(G, \mathbb{Z}[G])\cong \mathbb{Z}\cong H_0(G,\mathbb{Z})$. Hence $H_1(G,\mathbb{Z})\cong H_0(G,I_G)=I_G/I_G^2$. On the other hand, $$G^\mathrm{ab}\cong I_G/I_G^2,\quad g\mapsto g-1.$$ In fact, $G\rightarrow I_G/I_G^2, g\mapsto g-1$ is easily seen to be a group homomorphism, hence factors through $G^\mathrm{ab}$. Conversely, $I_G$ is a $\mathbb{Z}$-module freely generated on $\{g-1,g\in G\}$. The map $I_G\rightarrow G^\mathrm{ab}, g-1\mapsto g$ and $I_G^2\mapsto 0$ gives the inverse.

TopChange of groups

TopRestriction and corestriction

Definition 14 Let $\phi: H\rightarrow G$ be a homomorphism. Then $A\mapsto H^i(H, \Res_H^G A)$ is also a $\delta$-functor (i.e. a short exact sequence gives a long exact sequence). Since $H^i(G,-)$ is universal repelling (derived functors are universal $\delta$-functors), we obtain a morphism of $\delta$-functors $$\phi^*: H^i(G,-)\rightarrow H^i(H,\Res_H^G-).$$ Alternatively, we even have a morphism on the level of cochains $C^\bullet (G,A)\rightarrow C^\bullet (H, \Res_H^G A)$ via composing with $\phi$.
Remark 29 When $H$ is a subgroup of $G$, there is a natural map $H^0(G,A)=A^G\rightarrow A^H=H^0(H,\Res_H^G A)$. For higher cohomology, we can define it by "dimension-shifting". Namely, let $0\rightarrow A\rightarrow I\rightarrow B\rightarrow0$ an exact sequence of $G$-modules where $I$ is injective, then we can define $H^i(G,A)\rightarrow H^i(H,\Res_H^G A)$ inductively using the following diagram
$$\xymatrix{H^{i-1}(G,B) \ar[r]^\cong \ar[d]&H^i(G,A) \ar@{-->}[d]\\ H^{i-1}(H, \Res_H^G B) \ar[r]^\cong & H^i(H, \Res_H^G A).}$$

02/06/2013

Definition 15 Here is a slightly more general construction. Suppose $A\in \mathbf{Mod}_G$ and $A'\in \mathbf{Mod}_H$. We say $\psi: A\rightarrow A'$ is a homomorphism compatible with $\phi$ if $\psi(\phi(h)a)=h\psi(a)$ for any $h\in H$, equivalently, $\psi: \Res_H^G A\rightarrow A'$ is a $H$-module homomorphism. Composing $H^i(G,A)\rightarrow H^i(H, \Res_H^G H)$ with $\psi$ gives a functor $\phi^*: H^i(G,A)\rightarrow H^i(H,A')$. When $H$ is a subgroup of $G$, we call this functor the restriction functor, denoted by $$\Res: H^i(G,A)\rightarrow H^i(H,A').$$
Definition 16 Similarly, since homology is a universal attracting $\delta$-functor, $\phi: H\rightarrow G$ induces a functor $$\phi_*: H_i(H,\Res_H^G A)\rightarrow H_i(G,A).$$ On the level of chain complex, this is given by $C_\bullet(H,\Res_H^G A)\rightarrow C_\bullet(G,A): f\mapsto (\vec g\mapsto \sum_{\phi(\vec h)=\vec g}f(\vec h))$.
Exercise 1 Prove the above map is a map of chain complexes and it induces $\phi_*: H_i(H,A)\rightarrow H_i(G,A)$.
Remark 30 When $H$ is a subgroup of $G$, we have a natural map $H_0(H,\Res_H^G A)=A_H\rightarrow A_G=H_0(G,A)$. For higher homology, the corresponding maps again can be defined by "dimension-shifting".
Definition 17 Similarly, suppose $\psi: A\rightarrow A'$ is a homomorphism compatible with $\phi$, then composing with $\psi$ gives a functor $\phi_*: H_i(H, A')\rightarrow H_i(G,A)$. When $H$ is a subgroup of $G$, we call this functor the corestriction functor, denoted by $$\Cor: H_i(H, A')\rightarrow H_i(G,A).$$
Exercise 2 Use Shapiro's Lemma (Remark 24) and the natural map $A\rightarrow \Ind_H^G \Res_H^G A$ (Remark 13) to construct the restriction functor $\Res_H^G : H^i(G,A)\rightarrow H^i(H, \Res_H^GA)$. Do the same for the corestriction functor using the natural map $\ind_H^G \Res_H^G A\rightarrow A$ (Remark 10).
Remark 31 If $H$ is a finite index subgroup of $G$. In this case $\ind_H^G A=\Ind_H^G A$. We further have corestriction on cohomology $$\Cor: H^i(G,A)\cong H^i(H,\Ind_H^G \Res_H^G A)\cong H^i(H,\ind_H^G \Res_H^G A)\rightarrow H^i(H,A)$$ and similarly restriction on homology $$\Res :H_i(G,A)\rightarrow H_i(H, A).$$
Remark 32 Unwrapping the definition, we can compute the effect of $\Cor$ on degree 0: $$A^H=(\Ind_H^G \Res_H^G A)^G\cong(\ind_H^G \Res_H^G A)^G\rightarrow A^G,$$ where $a\in A^H$ is identified with $\sum_{g\in G/H}g\otimes a\in (\ind_H^G \Res_H^G A)^G$, which gets sent to $\sum_{g\in G/H}ga\in A^G$ (c.f., Remark 24 and Remark 10).
Definition 18 We define the norm $N_{G/H}(a)=\sum_{g\in G/H}ga$.
Remark 33 By the dimension shifting argument, we know that $\Cor$ on cohomology is the unique extension of the norm $N_{G/H}$ to higher degrees.

Similarly, we can compute the effect of $\Res$ on degree 0 : $$A_G\rightarrow(\Ind_H^G \Res_H^G A)_G\cong(\ind_H^G \Res_H^G A)_G=A_H,$$ where $[a]\in A_G$ is sent to $g\mapsto ga$, which is identified with $\sum_{g\in H\backslash G} 1\otimes ga=[\sum_{g\in H\backslash G} ga]\in A_H$.

Definition 19 We define the conorm $N_{G/H}': A_G\rightarrow A_H$, $N_{G/H}'([a])=[\sum_{g\in H\backslash G} ga]$. Then $\Res$ on homology is the unique extension of the norm $N_{G/H}$ to higher degrees.
Exercise 3 The restriction on homology $H_1(G,\mathbb{Z})$ is compatible with the isomorphism $H_1(G,\mathbb{Z})\cong G^\mathrm{ab}$.

TopInflation and coinflation

Definition 20 Suppose $H$ is a normal subgroup of $G$. Then $A^H$ is again a $G$-module and $H$ acts trivially on it, hence $A^H$ is a $G/H$-module. Suppose $\psi: A^\#\rightarrow A$ is compatible with the quotient map $\phi: G\rightarrow G/H$. Then $\phi$ and $\psi$ induce the inflation functor $$\Inf: H^i(G/H, A^\#)\rightarrow H^i(G, A)$$ and coinflation functor $$\Conf: H_i(G, A)\rightarrow H_i(G/H, A^\#).$$
Example 6 Suppose $H=G$ and $\phi: G\rightarrow G, \quad g\mapsto hgh^{-1}$, for some $h\in G$. Then $\psi: A\rightarrow A,\quad a\mapsto h^{-1}a$ is compatible with $\phi$ and they induce an automorphism $H^i(G,A)\rightarrow H^i(G,A)$. We claim that this map is the identity. In fact, since this is a morphism of $\delta$-functors, we only need to check it on the degree 0 part ("dimension-shifting" argument), in which case $\psi: A^G\rightarrow A^G$ is the identity.
Remark 34 Suppose $H$ is a normal subgroup of $G$, $\phi_g: H\rightarrow H$ is the conjugation by $g\in G$ and $\psi: A\rightarrow A,\quad a\mapsto g^{-1}a$. Similarly we obtain an automorphism $(\phi_g)_*: H^i(H,A)\rightarrow H^i(H,A)$, which is trivial when $g\in H$ by the previous example. It follows that $G/H$ acts on $H^i(H,A)$ and there will be a Hochschild-Serre spectral sequence $H^i(G/H, H^j(H,A))\Rightarrow H^{i+j}(G,A)$ (more on this later).

Top(Co)inflation-(co)restriction exact sequence

Suppose $H$ is a normal subgroup of $G$ and $A\in \mathbf{Mod}_G$.

Theorem 3 (Inflation-restriction exact sequence)
  1. The sequence $$0\rightarrow H^1(G/H,A^H)\xrightarrow{\Inf} H^1(G,A)\xrightarrow{\Res} H^1(H, A)$$ is exact.
  2. If $H^i(H,A)=0$ for $1\le i\le q-1$, then $$0\rightarrow H^q(G/H,A^H)\xrightarrow{\Inf} H^q(G,A)\xrightarrow{\Res} H^q(H, A)$$ is exact. Moreover, $$\Inf : H^i(G/H,A^H)\rightarrow H^i(G,A)$$ is an isomorphism for $1\le i\le q-1$.
Remark 35 This can be viewed as part of the spectral sequence mentioned in the previous remark. Nevertheless our direct proof with be fun with computation of cocycles.
Proof
  1. First we show the injectivity. Say $f:G/H\rightarrow A^H$ is a cocycles such that $$\tilde f=\Inf(f):G\rightarrow G/H\xrightarrow{f} A^H\rightarrow A$$ is a coboundary. We need to show that $f$ it self is a coboundary. Suppose $\tilde f(g)=ga-a$. Then $f(\bar g)=ga-a$. But $f(\bar g \bar h)=f(\bar g)$, we find that $gha= ga$, hence $ha=a$ for any $h\in H$, i.e. $a\in A^H$. It follows that $f$ is a coboundary. Next let us show that composition is zero. Suppose $f:G/H\rightarrow A^H$ is a cocycle. Then $$\tilde f=\Res\Inf(f): H\rightarrow G\rightarrow G/H\rightarrow A^H\rightarrow A,$$ which is clearly zero. Finally let us show the exactness in the middle. Suppose $f:G\rightarrow A$ is a coboundary in $H^1(H,A)$. Then there exists $a\in A$ such that $f(h)=ha-a$ for any $h\in H$. Subtracting $f$ by the coboundary $ga-a$, we may assume $f|H=0$. Now for any $g\in G, h\in H$, we know that $f(gh)=f(g)+gf(h)=f(g)$, i.e. $f$ factors through $G\rightarrow G/H$. On the other hand, since $H$ is normal, $hg=gh'$ for some $h'\in H$, hence $hf(g)=f(h)+hf(g)=f(hg)=f(gh')=f(g)$, namely, $f(g)\in A^H$. Thus $f$ factors through $G/H\rightarrow A^H$.
  2. Induction on $q $. Suppose $q\ge2$. Choose $I$ injective such that $0\rightarrow A\rightarrow I\rightarrow B\rightarrow 0$. Notice $$\Hom_{G/H}(X,A^H)=\Hom_G(\Res_G^{G/H} X, A)\xeq{\text{Frob}}\Hom_{G/H}(X,\Ind_G^{G/H}A),$$ so $A^H\cong \Ind_G^{G/H}A$. It follows that $I^H$ is an injective $G/H$-module as $\Ind$ preserves injectives (Remark 22). Since we assume $H^1(H,A)=0$, we obtain a short exact sequence $$0\rightarrow A^H\rightarrow I^H\rightarrow B^H\rightarrow 0.$$ By the assumption $H^i(H, B)\cong H^{i+1}(H,A)=0$ for $1\le i\le q-2$. We obtain the following diagram $$\xymatrix{H^{q-1}(G/H,B^H) \ar[r]^{\Inf} \ar[d]^\delta_\cong& H^{q-1}(G,B) \ar[r]^{\Res} \ar[d]^\delta_\cong& H^{q-1}(H,B) \ar[d]^\delta_\cong \\ H^{q}(G/H,A^H) \ar[r]^{\Inf}& H^{q}(G,A) \ar[r]^{\Res} & H^q(H,A). }$$ The induction hypothesis implies the exactness of the first row, hence the second row is exact as desired.
Exercise 4 Prove the analogous theorem on homology. For example, the sequence $$H_1(H,A)\xrightarrow{\Cor}H_1(G,A)\xrightarrow{\Conf}H_1(G/H,A_H)\rightarrow 0$$ is exact.

02/11/2013

Proposition 5 $\Cor\circ\Res=[G:H]$ on $H^i(G,A)$ and $H_i(G,A)$. In particular, if $H^i(H,A)=0$, then $[G:H]$ kills $H^i(G,A)$.
Proof By dimension shifting, it suffices to check on degree zero, e.g., for cohomology, the composition is given by $A^G\rightarrow A^H\rightarrow A^G: a\mapsto \sum_{g\in H\backslash G} ga\mapsto [G:H]a$.

TopTate cohomology

From now on we assume that $G$ is a finite group. A phenomenon unique to finite groups is that the group homology can be also understood as group cohomology — the Tate cohomology as we shall define.

Definition 21 Define $N=\sum_{g\in G}g\in \mathbb{Z}[G]$ to be the absolute norm.

Let $A\in\mathbf{Mod}_G$. Notice that $N: A\rightarrow A$ descends to a map $N^*: A_G\rightarrow A^G$. Hence we obtain a map $N^*: H_0(G,A)\rightarrow H^0(G,A)$ which is functorial in $A$.

Definition 22 We define $$\hat H_0(G,A):=\ker N^*=\ker(N)/I_GA$$ and $$\hat H^0(G,A)=\coker(N^*)=A^G/\im(N).$$
Remark 36 These two groups can be thought of as "reduced" cohomology and homology group, which makes the trivial group have all its cohomology and homology groups zero.

One can check directly that Shapiro's lemma (Remark 24) also holds for these two groups:

Proposition 6 Let $H$ be a subgroup of $G$ and $A\in\mathbf{Mod}_H$, Then $$\hat H^0(G,\Ind_H^G A)=\hat H^0(H, A)$$ and $$\hat H_0(G, \ind_H^G A)=\hat H_0(H, A).$$
Proof Let us prove the first identity (similarly for the second). By Shapiro's lemma (Remark 24), there is an isomorphism $(\Ind_H^G A)^G\cong A^H$ induced by $f\mapsto f(1)$. By the definition of $\hat H^0$, it remains to show that $N_G(\Ind_H^G A)\cong N_HA$. On the one hand, for $f\in \Ind_H^G A$, we have $$(N_Gf)(1)=N_H\sum_{g\in H\backslash G}f(g)\in N_HA.$$ On the other hand, for any $a\in A$ , if we let $$f(g)=
\begin{cases}
   ga, & g\in H \\
   0, & g\not\in H,
\end{cases}$$ then $$(N_Gf)(1)=\sum_{h\in H}ha=N_H(a).$$ Thus $N_G(\Ind_H^G A)\cong N_HA$ as desired.
Corollary 2 If $A$ is relatively injective (equivalently, relatively projective since $G$ is finite), then $\hat H^0(G,A)=\hat H_0(G,A)=0$.
Proof We may assume $A$ is (co-)induced since $\hat H^0$ and $\hat H_0$ commutes with direct sum. Then the result follows from the previous proposition (c.f., Corollary 1).
Remark 37 Given a short exact sequence of $G$-modules $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$, $N^*$ fits into the following commutative diagram $$\xymatrix{ H_1(G,C) \ar[r] & A_G \ar[r] \ar[d]^{N^*} & B_G \ar[r] \ar[d]^{N^*}  & C_G \ar[r] \ar[d]^{N^*} & 0\\ 0 \ar[r] & A^G \ar[r] & B^G \ar[r] & C^G \ar[r] & H^1(G,A).}$$ Thus the snake lemma implies the following long exact sequence $$\scriptstyle\cdots\rightarrow H_1(G,C)\rightarrow \hat H_0(G,A)\rightarrow \hat H_0(G,B)\rightarrow \hat H_0(G,C)\rightarrow\hat H^0(G,A)\rightarrow\hat H^0(G,B)\rightarrow \hat H^0(G,C)\rightarrow H^1(G,A)\rightarrow\cdots,$$ which "glues" the two long exact sequences of homology and cohomology together.
Definition 23 We define the Tate cohomology group $\hat H^n(G,A):=H^n(G,A)$, for $n\ge1$, $\hat H^{-1}(G,A):=\hat H_0(G,A)$ and $\hat H^{-n}(G,A):=H_{n-1}(G,A)$ for $n\ge2$. So $\hat H^n(G,-)$ ($n\in \mathbb{Z}$) form a $\delta$-functor (infinite in both directions).
Remark 38 The functors $\Cor$ and $\Res$ on cohomology and homology extend to Tate cohomology: it suffices to check they extend to $\hat H^0$ and $\hat H^{-1}$. For example, if $N_Ha=0$ then $$N_G(a)=\sum_{g\in G}ga=\sum_{g\in G/H} g N_H(a)=0,$$ hence $\Cor: H_0(H,A)\rightarrow H_0(G,A)$ descends to $\Cor: \hat H^{-1}(H,A)\rightarrow \hat H^{-1}(G,A)$. Similarly for the $\Cor$ on homology and $\Res$ on cohomology and homology. One can also check directly by diagram chasing that they induce morphisms of $\delta$-functors.
Remark 39 Suppose $A$ is induced, then $\hat H^n(G,A)=0$ for all $n\in \mathbb{Z}$. Similarly for co-induced $A$. This fact makes the dimension shifting argument even easier: we can start anywhere and the induction will extend in both direction (c.f., Remark 10 and Remark 13).

02/13/2013

Proposition 7 $\Cor\circ\Res=[G:H]: \hat H^i(G,A)\rightarrow \hat H^i(G,A)$. In particular, $\hat H^i(G,A)$ is killed by $\#G$ (take $H=\{1\}$).
Proof The same argument as in Proposition 5.
Corollary 3 If $A$ is a finitely generated abelian group, then $\hat H^i(G,A)$ is finite.
Proof Notice that if $A$ is finitely generated and $G$ is finite, then $Z^i(G,A)$ is a finitely generated abelian group, hence $H^i(G,A)$ is finitely generated. But $\hat H^i(G,A)$ is also torsion by the previous proposition.

TopTate cohomology via complete resolution

A more conceptual way to understand the Tate cohomology is via complete resolution. For a finite group $G$, the existence of a complete resolution boils down to the nice duality properties of $G$-modules (c.f., Cassels-Frohlich [3] and Brown [4]).

Let us start with a discussion of the linear duality of $G$-modules. Let $G$ be a finite group and $P $ be a finitely generated left $\mathbb{Z}[G]$-module.

Definition 24 We define $P^*:=\Hom_\mathbb{Z}(P, \mathbb{Z})$ with the left $G$-module structure $(g\phi)(x)=\phi(g^{-1}x)$ (c.f., Definition 4).
Example 7 Taking $P=\mathbb{Z}[G]$, then $P^*=\Hom_\mathbb{Z}(\mathbb{Z}[G],\mathbb{Z})$ has a dual basis $\{g^*: g\in G\}$, where $g^*(h)=\delta_{g,h}$. The action of $G$ on $P^*$ is given by $$h.g^*(x)=g^*(h^{-1}x)=\delta_{hg,x}=(hg)^*.$$ Hence $\mathbb{Z}[G]^*\cong \mathbb{Z}[G]$ as left $G$-modules. Consequently, if $P $ is a free $\mathbb{Z}[G]$-module, then so is $P^*$.
Remark 40 If $P $ is furthermore a finitely generated free abelian group. Then for any $G$-module $A$, we have $$P\otimes A\cong \Hom(P^*, A),\quad x\otimes a\mapsto(\phi\mapsto(\phi(x)a))$$ as left $G$-modules.
Remark 41 Suppose $P\cong \mathbb{Z}[G]^n$ is free as a $\mathbb{Z}[G]$-module. Then $$P\otimes A\cong \mathbb{Z}[G]\otimes A^n\cong \ind_1^G\Res_1^G A^n$$ as left $G$-modules. So $P\otimes A$ is induced and all its Tate cohomology groups are zero. In particular $$N^*: P\otimes_GA\cong (P\otimes A)_G\rightarrow (P\otimes A)^G\cong \Hom_G(P^*,A)$$ is an isomorphism (c.f., Remark 18).

Now we are in position to construct the complete resolution (i.e. a $G$-module resolution of $\mathbb{Z}$ which extends in both directions). Let $P_\bullet\rightarrow \mathbb{Z}$ be a resolution by finite free $\mathbb{Z}[G]$-modules (e.g. the standard resolution). Taking dual gives another resolution by finite free $\mathbb{Z}[G]$-modules $$0\rightarrow \mathbb{Z}\xrightarrow{\epsilon^*}P_0^*\rightarrow P_1^*\rightarrow P_2^*\rightarrow\cdots.$$ Write $P_{-n}=P_{n-1}^*$ for $n\ge1$, then gluing the above two resolutions together gives a complete resolution $$P_\bullet: \xymatrix{\cdots \ar[r] & P_2 \ar[r] & P_1 \ar[r] & P_0 \ar[rr] \ar[rd]^{\epsilon}&& P_{-1}\ar[r] & P_{-2} \ar[r] & \cdots\\ &&&& \mathbb{Z} \ar[ru]^{\epsilon^*}&}$$

As we expected, the complete resolution computes all Tate cohomology groups.

Proposition 8 $\hat H^i(G,A)\cong H^i(\Hom_G(P_\bullet,A))$ is an isomorphism of $\delta$-functors.
Proof For $n\ge1$, this is true by definition. For $n\le-2$, this is true since $P_n\otimes_G A=\Hom_G(P_n^*,A)$ by Remark 41. The remaining cases $n=-1,0$ can be checked by hand.
Remark 42 For $i\ge1$, $d: P_{-i}\rightarrow P_{-(i+1)}$ is explicitly given by 
\begin{multline*}
  d(x_1^*,\ldots x_{i}^*)(y_1,\ldots,y_{i+1})=(x_1^*,\ldots, x_i^*)(d(y_1,\ldots, y_{i+1}))\\
  =\sum_{s\in G}\sum_{j=0}^i (-1)^j (x_1^*,\ldots, x_j^*, s^*, x_{j+1}^*,\ldots, x_i^*)(y_1,,\ldots, y_{i+1}).
\end{multline*}
and $d: P_0\rightarrow P_{-1}$ is given by $\epsilon^*\circ \epsilon=N\epsilon$. so the notion of a homogeneous cochain in negative degree $-n$ coincides with that of in nonnegative degree $n-1$ (c.f., Example 7).

TopCup products

We first give an axiomatic description of cup products, its resemblance to the usual cup product (e.g., in singular cohomology) should not surprise you too much.

Theorem 4 Suppose $G$ is a finite group. Then there exists a system of homomorphisms $$\smile: \hat H^p(G,A)\otimes \hat H^q(G,A)\rightarrow \hat H^{p+q}(G,A\otimes_GB)$$ for all $p,q\in \mathbb{Z}$ and $G$-modules $A$, $B$, satisfying:
  1. These homomorphisms are functorial in $A$ and $B$.
  2. In degree 0, the homomorphism is induced by $A^G\otimes B^G\rightarrow (A\otimes B)^G$ and passing the quotient.
  3. Suppose $0\rightarrow A'\rightarrow A\rightarrow A''\rightarrow0$ is exact and $0\rightarrow A'\otimes B\rightarrow A\otimes B\rightarrow A''\otimes B\rightarrow 0$ is also exact. Then $$(\delta a'')\smile b=\delta(a''\smile b),$$ for $a''\in \hat H^p(G,A'')$ and $b\in \hat H^q(G,B)$, where $\delta$ is the connecting homomorphism.
  4. Suppose $0\rightarrow B'\rightarrow B\rightarrow B''\rightarrow0$ is exact and $0\rightarrow A\otimes B'\rightarrow A\otimes B\rightarrow A\otimes B''\rightarrow 0$ is also exact. Then $$a\smile (\delta b'')=(-1)^p\delta(a\smile b''),$$ for $a\in \hat H^p(G,A)$ and $b\in \hat H^q(G,B'')$.
Remark 43 The uniqueness immediately follows from these four axioms using dimension shifting. The only trick here is the following. Embedding $A$ into the co-induced $G$-module $\Ind_1^GA$ gives $0\rightarrow A\rightarrow \Ind_1^GA\rightarrow Q\rightarrow0$ an exact sequence of $\mathbb{Z}[G]$-modules. Notice there is always a splitting $\Ind_1^GA\rightarrow A$ as abelian groups sending $f$ to $f(1)$. Hence this sequence remains exact when $\otimes B$. Furthermore $\Ind_1^GA\otimes B\cong\Ind_1^G(A\otimes B)$ is again co-induced. Thus the dimension shifting argument applies using (c) in the first variable and (d) in the second variable.

Some basic properties of cup products are in order before we prove the existence.

Proposition 9
  1. $(a\smile b)\smile c=a\smile(b\smile c)$.
  2. $a\smile b=(-1)^{\dim a\dim b}b\smile a$.
  3. Suppose $H$ is a subgroup of $G$, then $\Res_H^G(a\smile b)=\Res_H^G(a)\smile \Res_H^G(b)$.
  4. Suppose $H$ is a subgroup of $G$, $\Cor_H^G(a\smile \Res_H^G b)=\Cor_H^G(a)\smile b$.
Proof It suffices to check on degree 0 by dimension shifting. (a), (b) and (c) then follow immediately. For (d), it reduces to the fact that $N_{G/H}(a\otimes b)=(N_{G/H}a)\otimes b$, for $a\in A^H/N_HA$ and $b\in A^G/N_GA$.
Remark 44 Viewing $\Cor$ as the pushforward and $\Res$ as the pullback induced by the map $BH\rightarrow BG$, (d) is analogous to the projection formula (c.f., Remark 28).

02/15/2013

To prove the existence of cup products, we will construct a family of $G$-module homomorphisms $\phi_{p,q}: P_{p+q}\rightarrow P_p\otimes P_q$ for a complete resolution $P_\bullet$ of $\mathbb{Z}$ as in the following proposition.

Proposition 10 Suppose there is a family of $G$-module homomorphisms $\phi_{p,q}: P_{p+q}\rightarrow P_p\otimes P_q$ satisfying:
  1. $\varepsilon\otimes \varepsilon\circ\phi_{0,0}=\varepsilon$.
  2. $\phi_{p,q}\circ d=(d\otimes 1)\circ\phi_{p+1,q}+(-1)^p(1\otimes d)\circ \phi_{p,q+1}$ (this gives a chain complex map $\phi: P_\bullet\rightarrow (P_\bullet\hat\otimes P_\bullet)$).

Then such a family of $G$-module homomorphisms is enough to construct the cup product in Theorem 4.

Proof First, suppose $f\in \Hom_G(P_p,A)$, $g\in\Hom_G(P_q, B)$. We define $$f\smile g:=(f\otimes g)\circ\phi_{p,q}\in \Hom_G(P_{p+q}, A\otimes B)$$ Then one can check that $$\delta(f\smile g)=(\delta f\smile g)+(-1)^p(f\smile \delta g).$$ It follows easily that the $\smile$ thus defined indeed descends to the level of Tate cohomology.

Second, we need to check this construction actually satisfies the axioms in Theorem 4. This construct is certainly functorial in $A$ and $B$ and Axiom (b) follows from Requirement (a) in Proposition 10. For Axiom (c), since $P_p$'s are free, the diagram $$\xymatrix@C=.7em{0 \ar[r] &\Hom_G(P_p, A') \ar[r] \ar[d]^{\smile\beta} &\Hom_G(P_p, A) \ar[r] \ar[d]^{\smile\beta} & \Hom_G(P_p, A'') \ar[r] \ar[d]^{\smile\beta} & 0\\ 0 \ar[r] &\Hom_G(P_{p+q}, A'\otimes B) \ar[r] & \Hom_G(P_{p+q}, A\otimes B) \ar[r] & \Hom_G(P_{p+q}, A''\otimes B) \ar[r] &0}$$ has exact rows. Now one can check that $\delta \alpha''\smile \beta=\delta(\alpha''\smile \beta)$ by diagram-chasing the connecting homomorphisms. Similarly for Axiom (d).

Now we use the standard complete resolution $P_\bullet$ to construct a family of $\phi_{p,q}$ (see the previous section). There are six cases depending on the signs of the degrees.

Definition 25
  1. For $p,q\ge0$, we define $\phi_{p,q}(x_0,\ldots, x_{p+q})=(x_0,\ldots, x_p)\otimes (x_{p},\ldots, x_{p+q})$.
  2. For $p,q\ge1$, we define $\phi_{-p,-q}(x_1^*,\ldots, x_{p+q}^*)=(x_1^*,\ldots, x_p^*)\otimes (x_{p+1}^*,\ldots, x_{p+q}^*)$.
  3. For $p\ge0, q\ge1$, we define 
\begin{align*}
  \phi_{p,-p-q}(x_1^*,\ldots, x_q^*)&=\sum_{s\in G^p}(x_1, s_1,\ldots, s_p)\otimes (s_p^*,\ldots s_1^*, x_1^*,\ldots x_q^*)\\
  \phi_{-p-q,p}(x_1^*,\ldots, x_q^*)&=\sum_{s\in G^p}(x_1^*,\ldots, x_q^*,s_1^*,\ldots, s_p^*)\otimes (s_p,\ldots s_1,x_q),\\
  \phi_{p+q,-q}(x_0,\ldots, x_p)&=\sum_{s\in G^q}(x_0,\ldots, x_p,s_1,\ldots s_q)\otimes (s_q^*,\ldots, s_1^*), \\
  \phi_{-q,p+q}(x_0,\ldots, x_p)&=\sum_{s\in G^q}(s_1^*,\ldots,s_q^*)\otimes (s_q,\ldots,s_1,x_0,\ldots,x_p)
\end{align*}
Remark 45 One motivation here is to mimic the construction of the diagonal map $C_*(X)\rightarrow C_*(X)\otimes C_*(X)$ as in singular cohomology. It is straightforward but (extremely) tedious to verify $\phi_{p,q}$ thus constructed satisfy the requirements in Proposition 10, we shall not do the verification. Notice there are choices made in this construction and such a family $\phi_{p,q}$ is certainly not unique. You can construct your favorite one too.
Remark 46 Let $f$, $g$ be a $p $-cochain and $q $-cochain. Let $f^*$, $g^*$ be a $p $-chain and $q $-chain. Using the constructed $\phi_{p,q}$, one can compute explicitly the cup products in all six cases. For example, in the first two cases. $$(f\smile g)(x_0,\ldots, x_{p+q})=f(x_0,\ldots, x_p)\otimes g(x_{p},\ldots x_{p+q}),$$ and $$(f^*\smile g^*)(x_1^*,\ldots, x_{p+q}^*)=f^*(x_1^*,\ldots, x_p^*)\otimes g(x_{p+1}^*,\ldots x_{p+q}^*).$$ Notice the latter does not make sense for infinite groups $G$: if you have two functions with compact support modulo the $G$-action, the tensor product may not have compact support modulo the diagonal $G$-action. This again explains that the there is a natural cup product structure $\smile$ on group cohomology but not on group homology (as in topology). In contrast, when $q\ge p$, $$(f\smile g^*)(x_1^*,\ldots,x_{q-p}^*)=\sum_{s\in G^p}f(x_1,s_1,\ldots, s_{p})\otimes g^*(s^*_p,\ldots,s^*_1,x_1^*, \ldots,x_{q-p}^*)$$ is always compact modulo the $G$-action for an infinite group $G$. In fact there is always a cap product $$\frown: H^p(G,A)\otimes H_q(G,B)\rightarrow  H_{q-p}(G,A \otimes B).$$ In the case of finite groups, the notion of cup product and cap product coincide due to the nice linear duality of $G$-modules.

02/20/2013

TopCohomology of finite cyclic groups

The general strategy of computing cohomology of a finite group is by reducing to computing the cohomology of its Sylow $p $-groups. By filtering these $p $-groups, the problem reduces to the case of finite cyclic $p $-groups. In this section we shall discuss the general theory of cohomology of finite cyclic groups.

Suppose $G$ is a finite cyclic group of order $n$. Then its argumentation ideal $I_G$ is generated by $D=s-1$ as a $\mathbb{Z}[G]$-module, where $s $ is any generator of $G$.

Remark 47 Let $A$ be a $G$-module. Notice the action of $G$ on $A$ is determined by the action of $s $, hence $A^G=\ker (D: A\rightarrow A)$ and $I_GA=\im (D: A\rightarrow A)$. So we have a nice symmetric expression (Definition 22) $$\hat H^0(G,A)=\ker (D)/\im(N),\quad \hat H_0(G,A)=\ker (N)/\im(D).$$ In particular, if $A$ is induced, then $\ker(N)=\im(D)$ and $\ker(D)=\im(N)$. Applying this to the free $G$-module $\mathbb{Z}[G]$, we obtain a complete resolution of $\mathbb{Z}$, $$L_\bullet: \xymatrix{\cdots \ar[r] &\mathbb{Z}[G] \ar[r]^{D} & \mathbb{Z}[G] \ar[rr]^{N} \ar[rd]^{\epsilon} &&\mathbb{Z}[G] \ar[r]^{D} & \mathbb{Z}[G] \ar[r] & \cdots\\ & & & \mathbb{Z} \ar[ru]^{N} & & & &}$$ Thus $\hat H^i(G,A)\cong H^i(\Hom(L_\bullet,A))$ and in particular, $\hat H^i(G,A)\cong \hat H^{i+2}(G,A)$ since $L_\bullet$ is of period 2. Explicitly, $$\hat H^i(G,A)=
\begin{cases}
  A^G/NA, & i \text{ even}, \\
  \ker(N)/DA, & i\text{ odd}.
\end{cases}$$

The long exact sequence of Tate cohomology thus retracts to an exact hexagon due to the above periodicity.

Corollary 4 A short exact sequence of $G$-module $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ gives an exact diagram $$\xymatrix@=1em{ & \hat H^0(G,B) \ar[r] & \hat H^0(G,C) \ar[rd]  & \\ \hat H^0(G, A) \ar[ru]  & &  & \hat H^1(G,A) \ar[ld] \\ & \hat H^1(G, C) \ar[lu] & \hat H^1(G, B) \ar[l] & }$$
Exercise 5 Define $\theta_s\in \Hom(G, \mathbb{Q}/\mathbb{Z}),\quad s\mapsto1/n$. The exact sequence $$0\rightarrow \mathbb{Z} \rightarrow \mathbb{Q} \rightarrow \mathbb{Q}/\mathbb{Z} \rightarrow0$$ gives a connecting homomorphism $$\Hom(G,\mathbb{Q}/\mathbb{Z})=H^1(G,\mathbb{Q}/\mathbb{Z})\rightarrow H^2(G,\mathbb{Z}).$$ Denote $\chi_s$ the image of $\theta_s$ under this homomorphism.
  1. Prove that $\smile\chi_s: \hat H^i(G,A)\rightarrow \hat H^{i+2}(G,A)$ is the isomorphism above.
  2. Prove that $\smile \chi_s: \hat H^0(G,A)\rightarrow \hat H^2(G,A)$ depends on the choice of the generator $s $ (it should not surprise you since the complete resolution $L_\bullet$ depends on a choice of $s $).
Definition 26 Suppose $G$ is a finite cyclic group and $A$ is a $G$-module, we define the Herbrand quotient of $A$ to be $h(A)=\#  \hat H^0(G,A)/\# \hat H^1(G,A)$ if it makes sense.
Remark 48 The Herbrand quotient should be thought of as (a multiplicative version of) the Euler characteristic of $A$. Euler characteristic is additive in short exact sequences. Analogous Herbrand quotient is multiplicative in short exact sequences.
Proposition 11 Suppose $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ is a short exact sequence of $G$-modules. If two of $h(A)$, $h(B)$ and $h(C)$ exist, then the third also exists and $h(B)=h(A)h(C)$.
Proof It follows easily from the exact hexagon (Corollary 4).
Proposition 12 If the $G$-module $A$ is finite (as a set), then $h(A)=1$.
Proof Since the number of elements is always multiplicative in exact sequence, looking at the exact sequence $$0\rightarrow A^G\rightarrow A\xrightarrow{D} A\rightarrow A_G\rightarrow 0,$$ we know that $\# A^G=\# A_G$. The same trick applies to the exact sequence $$0\rightarrow \hat H^1(G,A)\rightarrow A_G\xrightarrow{N^*} A^G\rightarrow \hat H^0(G,A)\rightarrow 0$$ and gives the desired result.
Proposition 13 Suppose there is a $G$-module homomorphism $f: A\rightarrow B$ with finite kernel and cokernel. Then $h(A)=h(B)$ (meaning that if one of them exists, then the other also exists).
Proof Applying the previous two propositions to the two short exact sequences $$0\rightarrow \ker f\rightarrow A\rightarrow \im f\rightarrow0,\quad 0\rightarrow\im f\rightarrow B\rightarrow\coker f\rightarrow0$$ implies what we want.

Now suppose $G$ is cyclic of prime order $p $. The structure theory of $G$-modules in this case is not so complicated and we are going to classify them completely.

Definition 27 The trivial Herbrand quotient of $A$ is defined to be the Herbrand quotient of $A$ regarded trivial $G$-action, i.e., $\phi(A)=\#A/pA/\#A[p]$ if it exists (notice if $G$ acts trivially on $A$, then $\hat H^0 A=A/pA$ and $\hat H_0A=A[p]$).

The goal in remaining of the this section is to prove the following formula of the Herbrand quotient.

Proposition 14 Let $G$ be a cyclic group of prime order $p $ and $A$ be a $G$-module. Suppose $\phi(A)$ exists, then $\phi(A^G)$, $\phi(A_G)$, $h(A)$ all exist and $$h(A)^{p-1}=\phi(A^G)^p/\phi(A)=\phi(A_G)^p/\phi(A).$$
Remark 49 The second equality should not surprise you since $\phi(A^G)=\phi(A_G)$ by multiplicativity.
Lemma 1 Suppose $0\rightarrow A'\rightarrow A\rightarrow A''\rightarrow 0$ is exact and $\phi(A'), \phi(A'')$ are defined. If Proposition 14 is true for $A', A''$, then it is also true for $A$.
Proof Notice $\phi(A)=\phi(A')\phi(A'')$ and $h(A)=h(A')h(A'')$ are both defined. It remains to show that $\phi(A^G)$ and $\phi(A_G)$ are defined. The long exact sequence in group cohomology gives $$0\rightarrow (A')^G\rightarrow A^G\rightarrow (A'')^G\rightarrow B\rightarrow 0,$$ where $B $ is the image of $(A'')^G$ in $\hat H^1(G,A')$. By the assumption that $h(A')$ is defined, we know $\hat H^1(G,A')$ is finite, so $B$ is also finite and $\phi(B)=1$ by Proposition 12. Hence $\phi(A^G)=\phi((A')^G)\phi((A'')^G)$ by multiplicativity. The same argument for $A_G$.

To prove Proposition 14, we can filter the $G$-module $A$ and apply the above lemma.

Lemma 2 If $\phi(A)$ is defined, then there exists a short exact sequence of $G$-modules $$0\rightarrow A'\rightarrow A\rightarrow A''\rightarrow 0$$ such that
  1. $A'$ is finitely generated over $\mathbb{Z}$.
  2. $A''$ is $p $-divisible (i.e., $pA''{}=A''$).

Furthermore, $\phi(A')$ and $\phi(A'')$ are both defined.

Proof Since $A/pA$ is finite, there exists $A'\subseteq A$ finitely generated over $\mathbb{Z}$ such that $A'\twoheadrightarrow A/pA$. Replacing $A'$ by $\mathbb{Z}[G]A'$ (still finitely generated over $\mathbb{Z}$ since $G$ is finite), we obtain a $G$-submodule $A'$ such that $A'\twoheadrightarrow A/pA$. Write $A''{}=A/A'$. Applying the snake lemma to multiplication by $p $ gives a short exact sequences $$A'/pA'\rightarrow A/pA\rightarrow A''/pA''\rightarrow 0.$$ Since $A'/pA'\rightarrow A/pA$ is surjective by construction, we know that $A''/pA''{}=0$, i.e., $A''$ is $p $-divisible. Since $A'$ is finitely generated over $\mathbb{Z}$, we know that $A'/pA'$ and $A'[p]$ are both finite and $\phi(A')$ is defined, so $\phi(A'')$ is also defined.
Proof (Proof of Proposition 14) By the previous two lemmas, we may assume $A$ is finitely generated over $\mathbb{Z}$ or $A$ is $p $-divisible.

First assume $A$ is finitely generated over $\mathbb{Z}$. Suppose $A_1$ and $A_2$ are two $G$-modules finitely generated over $\mathbb{Z}$ and $A_1\otimes \mathbb{Q} \cong A_2\otimes \mathbb{Q}\cong V$ as $\mathbb{Q}[G]$-modules. Let $A_3=A_1+A_2\subseteq V$. Then $A_3$ is a $G$-stable lattice in $V$. For $i=1,2$, the inclusion $A_i\rightarrow A_3$ has finite cokernel. Hence $h(A_i)=h(A_3)$ and $\phi(A_i)=\phi(A_3)$ for $i=1,2$. In other words, we have proved that if $V$ is a finite dimensional $\mathbb{Q}[G]$-module, and $A\subseteq V$ is a $G$-stable lattice, then $h(A)$ and $\phi(A)$ do not depend on the choice of $A$. By Proposition 12, $\phi(A^G)$ and $\phi(A_G)$ do not depend on the choice of $A$ either. So we reduce to the case of finite dimensional $G$-representations by replacing $A$ with $A\otimes \mathbb{Q}$.

02/22/2013

By filtering the finite dimensional $G$-representation, it suffices to prove the formula when $V=A\otimes \mathbb{Q}$ is simple. As a ring $\mathbb{Z}[G]=\mathbb{Z}[s]/(s^p-1)$ and $\mathbb{Q}[G]=\mathbb{Q}[s]/(s^p-1)\cong \mathbb{Q} \times \mathbb{Q}(\zeta_p)$. So any $\mathbb{Q}[G]$-module is a direct product of a $\mathbb{Q}$-module and a $\mathbb{Q}(\zeta_p)$-module. In particular, $\mathbb{Q}[G]$ has exactly two simple modules: $\mathbb{Q}$ (the trivial representation) and $\mathbb{Q}(\zeta_p)$ (the representation of dimension $p-1 $). For the first case $\mathbb{Q}$ (with $\mathbb{Z}$ a lattice inside it), then formula is obvious. For the second case, we identify $\mathbb{Q}(\zeta_p)=\mathbb{Q}[G]/\mathbb{Q}$, where $\mathbb{Q}\hookrightarrow \mathbb{Q}[G]$ via $1\mapsto N$. Let $A=\mathbb{Z}[G]/\mathbb{Z}$ be a $G$-stable lattice inside $\mathbb{Q}[G]/\mathbb{Q}$. Then $$h(A)=1/h(\mathbb{Z})=1/p,\quad \phi(A)=\# (A/p)=p^{p-1},\quad \phi(A^G)=1.$$ The shape of the desired formula essentially boils down to this computation.

It remains to treat the case where $A$ is $p $-divisible. From the exact sequence $$0\rightarrow A[p^\infty]\rightarrow A\rightarrow A'\rightarrow0,$$ and the fact $A[p^\infty][p]=A[p]$ and $pA[p^\infty]=A[p^\infty]$, the snake lemma tells us that $A'[p]=0$ and $A/pA=A'/pA'{}=0$. In particular, $p: A'\rightarrow A'$ is an isomorphism and $p: \hat H^i(G, A')\rightarrow\hat H^i(G, A')$ is an isomorphism by functoriality. But $p=\# G$ kills $\hat H^i(G,A')$ by Proposition 7, hence $\hat H^i(G,A')=0$ and $\phi(A')=h(A')=1$. We now have reduced to the case $A=A[p^\infty]$ with $\#A[p]<\infty$. Recall that there is a duality between such discrete torsion abelian groups and finitely generated free $\mathbb{Z}_p$-modules given by taking the Pontryagin dual $A\mapsto \hat A=\Hom_\mathbb{Z}(A, \mathbb{Q}_p/\mathbb{Z}_p)$. This duality further hold on the $G$-module level. One can check that $$\hat A^G=\widehat{A_G},\quad \hat A_G=\widehat{A^G},$$ and $N^*: A_G\rightarrow A^G$ dualizes to $\hat N^*=N^*: \hat A_G\rightarrow \hat A^G$. Moreover, $$(\hat H^1A)^\wedge=\hat H^0 \hat A,\quad (\hat H^0A)^\wedge=\hat H^1\hat A.$$ In particular, $$h(\hat A)=1/h(A),\quad \phi(A)=1/\phi(\hat A),\quad \phi(A^G)=1/\phi(\hat A_G).$$ So it remains to prove the formula for $A$ a finitely generated free $\mathbb{Z}_p$-module. The exactly same argument for finitely generated $\mathbb{Z}$-modules works using the fact that $\mathbb{Q}_p[G]\cong \mathbb{Q}_p\times \mathbb{Q}_p(\zeta_p)$, where $\mathbb{Q}_p(\zeta_p)$ has degree $p-1 $ over $\mathbb{Q}_p$.

03/15/2013

The notes are incomplete at this point because I was out of town for AWS 2013. Meanwhile we did two essential inputs for the proof of class field theory: the study of cohomology of finite groups which leads to the proof of Tate-Nakayama (c.f., Chap VII-IX in [1]), and a detailed analysis of ramification and norm which allows one show that $K^\mathrm{ur}$, the maximal unramified extension of a complete discretely valued field $K$ with perfect residue field, has universally trivial Brauer group (c.f., Chap IV-V, X in [1]). We summarize these two main results:

Theorem 5 (Tate-Nakayama) Let $G$ be a finite group and $A$ be $G$-module. $a\in \hat H^2(G,A)$. Suppose $G_p$ is a $p $-Sylow subgroup of $G$ for any $p $, and
  1. $\hat H^1(G_p,A)=0$ (think: Hilbert 90).
  2. $\hat H^2(G_p,A)\cong \mathbb{Z}/\# G_p \mathbb{Z}=\langle a_{G_p}\rangle$, where the subscript denotes the restriction of $a$ to $G_p$ (think: Brauer group).

Then for any $G$-module $D$ such that $\Tor(A,D)=0$, $$a_H\smile (\cdot): \hat H^n(H,D)\rightarrow \hat H^{n+2}(H, A \otimes D)$$ is an isomorphism for any $n\in \mathbb{Z}$ and any subgroup $H$ of $G$ (think: the inverse of the Artin map).

Theorem 6 Suppose $K$ is a complete discretely valued field with perfect residue field. Then $\Br(K^\mathrm{ur})=0$.

TopClass formations

The Tate-Nakayama lemma provides cohomological testing hypotheses for class field theory. These cohomological data can be formalized as class formation. We will first state in full generality purely in group cohomological terms, then specialize to Galois cohomology.

Let $G$ be any group and $\{G_E\}_{E\in X}$ be a nonempty collection of finite index subgroups of $G$. Assume that $G_E=G_{E'}$ if and only if $E=E'$. We make the following hypotheses (formal properties that an infinite Galois group must satisfy)

  1. For any $F_1,\ldots, F_n\in X$, there exists $G_F=\bigcup G_{F_i}$.
  2. If $G_F$ is contained in a subgroup $G'$ of $G$, then there exists $F'$ such that $G'{}=G_{F'}$.
  3. For any $s\in G$, $F\in X$, there exists $F'\in X$ such that $s G_F s^{-1}=G_{F'}$.
Example 8 Let $K$ be a field and $\Omega$ be any Galois extension of $K$. Then $G=\Gal(\Omega/K)$ and $X=\{E\subseteq \Omega,\text{ finite over }K\}$ and $G_E=\Gal(\Omega/E)$ satisfy the previous hypotheses. $\Omega=K^\mathrm{sep}$ is the most interesting case for class field theory.
Remark 50 We borrow the terminology from Galois theory in an obvious fashion. For instance, we call $F\in X$ fields. We say $F'/F$ Galois if and only and only if $G_F\lhd G_{F'}$ and write $G_{F'/F}=G_{F'}/G_F$. We say $F=\prod F_i$ is the composite of $F_i$ if and only if $G_F=\bigcup G_{F_i}$, etc.
Definition 28 A formation is the above data $(G, \{G_F\}_{F\in X})$ along with $A\in \mathbf{Mod}_G$ such that $A=\bigcup_{F\in X} A^{G_F}$. Notice that for any $a\in A$, the the stabilizer of $a$ is $G_{F}$ for some $F$.
Example 9 For $K$ a local field, we are interested in the formation coming from $A=(K^\mathrm{sep})^\times$. Similarly, for $K$ a number field, we are interested in the class formation coming from $A=\bigcup_{F/K\text{ finite}} \mathbb{A}_F^\times/F^\times$.
Remark 51
  1. We write $A_E:=A^{G_E}$ for short.
  2. For $F/E$ Galois, $G_{F/E}$ acts on $A_F$ and $H^0(G_{F/E},A_F)=A_E$.
  3. We write $H^i(E/F):=H^i(G_{F/E},A_F)$, $H_i(E/F)=H_i(G_{F/E},A_F)$ and likewise $\hat H^i(E/F)=\hat H^i(G_{F/E},A_F)$. Given Galois extensions $F/E$, $F'/E'$ with $E\subseteq E'$ and $F\subseteq F'$. Then
    1. We have a canonical map $G_{F'/E'}\rightarrow G_{F/E}$ compatible with the inclusion $A_F\rightarrow A_F'$.
    2. When $E=E'$, this induces the inflation $$\Inf: H^i(F/E)\rightarrow H^i(F'/E).$$ Notice the inflation only makes sense for the cohomology (which does not extend to Tate cohomology in negative degrees).
    3. When $F=F'$, this induces the restriction $$\Res: \hat H^i(F/E)\rightarrow \hat H^i(F/E'),$$ and the corestriction $$\Cor: \hat H^i(F/E')\rightarrow \hat H^i(F/E).$$
    4. For $s\in G$, we have isomorphisms $t\mapsto s^{-1}ts: G_{sF/sE}\cong G_{F/E}$ and $s: a\mapsto: s^{-1}a, A_F\cong A_{sF}$, which induces $$s^*: \hat H^i(F/E)\cong \hat H^i(sF/sE).$$ If furthermore $s\in G_E$, then $s^*=\Id$.
Definition 29 A class formation is a formation $(G, \{G_E\}_{E\in X},A)$ along with a homomorphism $$\inv_E: H^2(E)\rightarrow \mathbb{Q}/\mathbb{Z}$$ (called the invariant) for any $E\in X$ satisfying the following 2 axioms:

Axiom I: For any $F/E$ Galois, $H^1(F/E)=0$.

This is Hilbert 90 for $A=\mathbb{G}_m$. Since $H^n(G,A)=0$ if and only $H^n(G_p,A)=0$ for all $p$'s, Axiom I equivalent to say that for any $F/E$ cyclic of prime degree, $H^1(F/E)=0$.

Axiom II:

  1. For any $F/E$ Galois, $$\inv_{F/E}: H^2(F/E)\xrightarrow{\Inf} H^2(E)\xrightarrow{\inv_E} \mathbb{Q} /\mathbb{Z}$$ is injective and maps $H^2(F/E)$ isomorphically to $1/[F:E]\mathbb{Z} /\mathbb{Z}$. We denote the generator $u_{F/E}\in H^2(F/E)$ such that $\inv_{F/E}(u_{F/E})=1/[F:E]$ and call it the fundamental class.
  2. For any $E'/E$, $\inv_{E'}\circ\Res_{E/E'}=[E': E] \inv_{E}$.
Proposition 15 Let $E'/E$ be any extension and $F/E$ be a Galois extension containing $E'$. Then
  1. IF $E'/E$ is Galois, then $\Inf_{F/E}(u_{E'/E})=[F:E']u_{F/E}$.
  2. $\Res_{E/E'}(u_{F/E})=u_{F/E'}$.
  3. $\Cor_{E'/E}(u_{F/E'})=[E': E]u_{F/E}$.
  4. $u_{sF/sE}=s^*(u_{F/E})$ for any $s\in G$.
Proof To check two things in $H^2$ to be equal, only need to check after composing with $\inv$ by the injectivity.
Remark 52 Here is my own way to remember the proposition. The elements of the Brauer group of a field $E$ correspond to equivalence classes of central simple algebras $B$ over $E$. Under this identification, $\Inf_{F/E}$ and $\Cor_{E'/E}$ do not change $B$ and keep the same invariant, hence there must be a degree factor to balance the change of the fundamental class. $\Res_{E/E'}$ is the operation $B\otimes_E E'$ and multiply the invariant by the degree, hence there is no more degree factor before the fundamental class.
Remark 53 In the sequel we will prove that the formation coming from $A=\mathbb{G}_m$ is indeed a class formation for any complete discretely valued field with quasi-finite residue field, and thus construct the Artin map using the Tate-Nakayama theorem. However, the existence theorem does not follow automatically. In fact, we need the residue field to be furthermore finite in order to prove the existence theorem (in other words, the locally-compactness is the extra input needed).

03/25/2013

Take $G=G_{F/E}$, $A=A_F$, $D=\mathbb{Z}$ and $a=u_{F/E}$ in the Tate-Nakayama Theorem 5, we obtain

Theorem 7 For any $F/E$ Galois and $q\in \mathbb{Z}$, $$\theta_{F/E}^q:= u_{F/E}\smile(\cdot): \hat H^q(G_{F/E},\mathbb{Z})\rightarrow \hat H^{q+2}(F/E)$$ is an isomorphism.
Example 10 When $q=1$, then $\hat H^1(G_{F/E},\mathbb{Z})=\Hom(G_{F/E},\mathbb{Z})=0$ since $G_{F/E}$ is finite. Therefore $\hat H^3(F/E)=0$.
Example 11 When $q=2$, using the exact sequence $$0\rightarrow \mathbb{Z} \rightarrow \mathbb{Q} \rightarrow \mathbb{Q}/\mathbb{Z} \rightarrow0$$ and the fact $\mathbb{Q}$ is cosmologically trivial (it is divisible), we compute that $\hat H^2(G_{F/E}, \mathbb{Z})\cong \hat H^1(G_{F/E},\mathbb{Q}/\mathbb{Z})=G_{F/E}^\vee$. Hence $\hat H^4(E/F)=G_{F/E}^\vee$.
Example 12 When $q=-2$, $\hat H^{-2}(G_{F/E},\mathbb{Z})=H_1(G_{F/E},\mathbb{Z})=G_{F/E}^\mathrm{ab}$ and $\hat H^0(G_{F/E},A_F)=A_E/N_{F/E}A_F$. So $G^\mathrm{ab}_{F/E}\cong A_E/N_{F/E}A_F$. This is the expected (inverse of) the Artin map.
Exercise 6 Let $u: G_{F/E}^2\rightarrow A_F$ be the cocycle representing $u_{F/E}\in H^2(F/E)$. Then $\theta(s)=\sum_{t\in G_{F/E}}u(t,s)\mod{N_{F/E}A_F}$.
Definition 30 The inverse to the isomorphism $\theta$, called the (Artin) reciprocity homomorphism, is indeed more useful than $\theta$ itself. We denote it by $a\mapsto (a,F/E)\in G_{F/E}^\mathrm{ab}$.
Proposition 16 Let $\chi\in \Hom(G_{F/E},\mathbb{Q}/\mathbb{Z})=G_{F/E}^\vee$. For $s\in G_{F/E}^\mathrm{ab}$, set $\langle\chi,s\rangle=\chi(s)\in \mathbb{Q}/\mathbb{Z}$. Let $d\chi\in \hat H^2(G_{F/E},\mathbb{Z})$ be the coboundary associated to the exact sequence $0\rightarrow \mathbb{Z} \rightarrow \mathbb{Q} \rightarrow \mathbb{Q}/\mathbb{Z}\rightarrow 0$. Then $$\langle\chi, (a,F/E)\rangle=\inv_{F/E}(a\smile d\chi),\quad a\in \hat H^0(F/E).$$
Remark 54 It characterizes $(a,F/E)$ uniquely by duality. This is more convenient than the previous exercise since $a$ is in degree 0.
Proof Write $s_a=(a,F/E)$ for short. By definition, $a=\theta(s_a)=u_{F/E}\smile s_a$. So $$a\smile d\chi=u_{F/E}\smile s_a\smile d\chi=u_{F/E}\smile d(s_a\smile \chi)$$ since $ds_a=0$. Notice that $s_a\smile \chi\in \hat H^{-1}(G_{F/E},\mathbb{Q}/\mathbb{Z})\cong1/n \mathbb{Z} /\mathbb{Z}$, where $n=[F:E]$.

We claim that for any $s\in G_{F/E}$, $s\smile \chi=\langle\chi, s\rangle\in 1/n\mathbb{Z} /\mathbb{Z}$. In fact, first we can check that $s\in H_1(G_{F/E},\mathbb{Z})$ is represented by the homogeneous 1-cycle which maps $(t,ts^{-1})\mapsto 1$ and maps other things to 0. Then one can compute explicitly that $$s\smile \chi(t)=\sum_{u\in G_{F/E}}s(u,t)\chi(u,t)=\chi(t,ts)=\chi(1,s)=\chi(s),$$ where the last equality identifies the homogeneous cycle with the inhomogeneous cycle.

Suppose $s_a\smile \chi=r/n$, then we compute $d(s_a\smile \chi)=r\in  \mathbb{Z}/n \mathbb{Z}\cong \hat H^0(G_{F/E},\mathbb{Z})$, then $$\inv_{F/E}(a\smile d\chi)=\inv_{F/E}(u_{F/E}\smile d(s_a\smile\chi))=r/n\in \mathbb{Q} /\mathbb{Z},$$ as desired.

Proposition 17 (Functoriality) Let $F'\supseteq F\supseteq E'\supseteq E$, where $F', F/E$ are Galois. Then we have norm and verlagerung functoriality for the reciprocity homomorphism $$\xymatrix{A_{E'} \ar[d] \ar[r]^{N_{F/E}}& A_E \ar[d] \\  G_{F/E'}^\mathrm{ab} \ar[r] &G_{F/E}^\mathrm{ab},}\quad \xymatrix{A_E \ar[d] \ar@{^(->}[r] & A_{E'} \ar[d] \\  G_{F/E}^\mathrm{ab} \ar[r]^-{\mathrm{Ver}} &G_{F/E'}^\mathrm{ab}.} $$ For $s\in G$, we have the follow commutative diagram $$\xymatrix{A_E \ar[d] \ar[r]^{s^*} &  A_{sE} \ar[d] \\ G_{F/E}^\mathrm{ab} \ar[r]^{s^*} & G_{sF/sE}^\mathrm{ab}.}$$ Also the compatibility with respect to field extensions $F'/F$, $$\xymatrix{A_E \ar@{=}[r] \ar[d] & A_E \ar[d] \\ G_{F'/E}^\mathrm{ab} \ar@{->>}[r] & G_{F/E}^\mathrm{ab}.}$$
Proof The norm functoriality follows from that $$\Cor(\Res(u_{F/E'}\smile x))=u_{F/E}\smile\Cor(x).$$ The verlagerung functoriality follows from that $$\Res(u_{F/E}\smile x)=\Res(u_{F/E})\smile \Res(x)=u_{F/E'}\cup\Res(x).$$ The third diagram follows from $u_{sF/sE}=s^*u_{F/E}$ (c.f., Proposition 15). For the last diagram, we use the previous proposition. Let $\chi\in G_{F/E}^\vee$ and $a\in A_E$. Then by the previous proposition 
\begin{align*}
  \langle\chi,(a,F/E)\rangle&=\inv_{F/E}(a\smile d\chi)=\inv_{F/E'}\Inf(a\smile d\chi)\\
  &=\inv_{F'/E}(\Inf(a)\smile d\Inf(\chi))=\langle\Inf(\chi),(a,F'/E)\rangle,
\end{align*}
which finishes the proof.
Remark 55 Set $G_E^\mathrm{ab}=\varprojlim G_{F/E}^\mathrm{ab}$. Then we can pass to the inverse limit using the last diagram and obtain a reciprocity homomorphism $(\cdot, {*}/E): A_E\rightarrow G_E^\mathrm{ab}$. It again satisfies the first three functoriality (in particular applies to the Artin maps in class field theory).
Remark 56 One can also formulate the existence theorem using class formation. We won't do this here: instead we will discuss Lubin-Tate theory which not only proves the existence theorem but also provide an explicit construction of the local class fields.

TopBrauer group of a complete discretely valued field

Let $K$ be a complete discretely valued with perfect residue field $k $ (not necessarily characteristic $p $). We proved that $\Br(K^\mathrm{ur})=0$ (Theorem 6). It follows that $\Br(K^\mathrm{ur}/K)\cong\Br(K)$. So every element of $\Br(K)$ is split by a finite unramified Galois extension $L/K$ (the Galois condition is needed when $k $ is not finite). We would like to relate $\Br(K)$ and $\Br(k)$. Let $l$ be the residue field of $L$ and write $\mathfrak{g}=\Gal(L/K)\cong\Gal(l/k)$ and write $\mathfrak{g}=\Gal(L/K)=\Gal(l/k)$. Then $$0\rightarrow \mathcal{O}_L^\times\rightarrow L^\times\rightarrow \mathbb{Z} \rightarrow0$$ is an exact sequence of $\mathfrak{g}$-modules, split by a choice of uniformizer $\pi_L\in \mathcal{O}_K$ (this exists because $L/K$ is unramified!). Hence for any $q $, $$0\rightarrow H^q(\mathfrak{g} ,\mathcal{O}_L^\times)\rightarrow H^q(L/K)\rightarrow H^q(\mathfrak{g} ,\mathbb{Z})\rightarrow 0$$ is a split exact sequence.

Lemma 3 For any $q\ge1$, $H^q(\mathfrak{g} ,1+\mathfrak{p}_L)=0$.
Proof $1+\mathfrak{p}_L$ is filtered by $1+\mathfrak{p}_L^n$, with successive quotients $\cong l$. Since $H^q(\mathfrak{g} ,l)=0$ (normal basis theorem), it remains to prove the following lemma.
Lemma 4 Let $M$ be a $\mathfrak{g}$-module filtered by $(M_n)_{n\ge1}$. Suppose $M$ is complete and Hausdorff with respect to the topology defined by $(M_n)$ and $H^q(\mathfrak{g}, M_n/M_{n-1})=0$ for $n\ge1$. Then $H^q(\mathfrak{g} ,M)=0$.
Proof We omit the proof. The idea is that using completeness we can add up coboundaries valued in the filtration $M_n$ to get a coboundary valued in $M$.
Corollary 5 For any $q\ge1$, $H^q(\mathfrak{g},\mathcal{O}_L^\times)\cong H^q(\mathfrak{g} ,l^\times)=H^q(l/k)$. So $$0\rightarrow H^q(l/k)\rightarrow H^q(L/K)\rightarrow H^q(\mathfrak{g} ,\mathbb{Z})\rightarrow0$$ is a split exact sequence. Taking direct limits with respect to $L$ implies that $$0\rightarrow H^q(k)\rightarrow H^q(K^\mathrm{ur}/K)\rightarrow H^q(G_k,\mathbb{Z})\rightarrow0$$ is a split exact sequence.
Remark 57 Taking $q=2$ gives an split exact sequence $0\rightarrow\Br(k)\rightarrow\Br(K)\rightarrow H^2(G_k,\mathbb{Z})\rightarrow 0$. When $k $ is finite, $\Br(k)=0$ (by Hilbert 90 and Herbrand quotient). So $\Br(K)\cong H^2(G_k,\mathbb{Z})\cong H^1(\hat{\mathbb{Z}},\mathbb{Q}/\mathbb{Z})\cong\mathbb{Q}/\mathbb{Z}$ (Corollary 6). These hold for a general quasi-finite $k $: this is actually a purely group cohomological consequence of $G_k\cong \hat{\mathbb{Z}}$ as we shall see next time (Proposition 21).

03/27/2013

TopCohomology of $\hat{\mathbb{Z}}$

Recall that every open subgroup of $\hat{\mathbb{Z}}$ is of the form $n \hat{\mathbb{Z}}$ for some $n\in \mathbb{Z}$. A discrete $\hat{\mathbb{Z}}$-module is none other than an abelian group $A$ equipped with an automorphism $F:A\rightarrow A$ (the action of $1\in \hat{\mathbb{Z}}$) such that $A=\bigcup_n A^{F^n}$, i.e., every element of $A$ is fixed by some power of $F $.

Proposition 18 Let $A'{}=\{a\in A: N_na:=(1+F+F^2+\cdots F^{n-1}) a=0 \text{ for some } n\}$. Then $H^1(\hat{\mathbb{Z}}, A)=A'/(F-1)A$.
Remark 58 This can be viewed as a "pro-cyclic" version of $\hat H^1\cong\hat H^{-1}$.
Proof It follows from $H^1(\mathbb{Z}/n \mathbb{Z}, A)=\hat H^{-1}(\mathbb{Z}/n \mathbb{Z},A)=\ker (N)/\im (F-1)$ and then passing to the inverse limit.
Remark 59 The above isomorphism takes a cocycle $\phi: \hat{\mathbb{Z}} \rightarrow A$ to $\phi(1)\mod{(F-1)A}$. When $\hat{\mathbb{Z}}$ acts trivially on $A$, we know that $A'{}=A_\mathrm{tor}$, $(F-1)A=0$ and hence $\Hom_\mathrm{cont}(\hat{\mathbb{Z}},A)\cong A_\mathrm{tor}$.
Remark 60 $A_\mathrm{tor}\subseteq A'$.
Corollary 6 $\Hom_\mathrm{cont}(\hat{\mathbb{Z}}, \mathbb{Q}/\mathbb{Z})=\mathbb{Q}/\mathbb{Z}$.
Proposition 19 Let $A$ be a discrete $\hat{\mathbb{Z}}$-module. If $A$ is divisible or torsion , then $H^2(\hat{\mathbb{Z}},A)=0$.
Proof First suppose $A$ is finite. Then $H^2(\mathbb{Z}/n \mathbb{Z},A)=\hat H^0(\mathbb{Z}/n \mathbb{Z},A)=A^F/N_nA$. For $m\ge1$, we claim that $$\Inf: H^2(\mathbb{Z}/n \mathbb{Z}, A^{F^n})\rightarrow H^2(\mathbb{Z}/nm \mathbb{Z}, A^{F^{nm}})$$ is induced by multiplication by $m $, $$A^F/N_n A^{F^n}\xrightarrow{\times m} A^F/N_{nm}A^{F^nm}.$$ Recall the isomorphism between degree 0 and degree 2 is by cup product with $\theta_n$, where $\theta_n\in H^2(\mathbb{Z}/n \mathbb{Z},\mathbb{Z})$ is the coboundary of $\chi_n:1\mapsto 1/n\in H^1(\mathbb{Z}/n \mathbb{Z} ,\mathbb{Q}/\mathbb{Z})$ (Exercise 5). So 
\begin{align*}
  \Inf(x\smile\theta_n)&=\Inf(a)\smile\Inf(\theta_n)=\Inf(x)\smile d(\chi_n)\\
  &=x\smile m d\chi_{nm}=m(x\smile \theta_{nm}),
\end{align*}
because $x\in A^F$, $\Inf(x)=x$ and $\Inf(\chi_n)=m\chi_{nm}$. Our claim follows. So taking $m=\#A$ implies that $\Inf(x)=0$ for any $x $, hence $H^2(\hat{\mathbb{Z}},A)=\varinjlim A^F/N_nA^{F^n}=0$.

When $A$ is torsion, we can write $A$ as the union of finite $\mathfrak{g}$-submodules. Then we are done by taking the direct limit.

When $A$ is divisible, we have $A[n]=0$, the taking the cohomology of the exact sequence $0\rightarrow A[n]\rightarrow A\xrightarrow{\times n}A\rightarrow0$ shows that $n$ is injective on $H^2(\hat{\mathbb{Z}}, A)$ for any $n$. But any cohomology group is torsion and we are done again.

Our next goal is to show that any quasi-finite field has trivial Brauer group.

Definition 31 Let $k $ be a field and $F\in G_k$. We say $(k,F)$ is quasi-finite if
  1. $k $ is perfect,
  2. the map $\hat{\mathbb{Z}} \rightarrow G_k: 1\mapsto F$ is an isomorphism.
Remark 61 Any finite extension of a quasi-finite field $k$ is cyclic of the form $k_n:=\bar k^{F^n}$. Moreover, $(k_n,F^n)$ is also quasi-finite.
Example 13
  1. Every finite field $k $ is quasi-finite with $F $ being the (arithmetic or geometric) Frobenius.
  2. Let $C$ be an algebraically closed field of characteristic 0 and $k=C((T))$. Then $\bar k=C\{\{T\}\}=\bigcup C((T^{1/n}))$ is the field of Puiseux series. Choose a compatible system $(\zeta_n)_{n\ge1}$ of primitive $n$-th root of unity and let $F\in G_k$ such that $F(T^{1/n})=\zeta_n T^{1/n}$. Then $\Gal(C((T^{1/n}))/C((T))=\langle F\rangle\cong \mathbb{Z}/n \mathbb{Z}$. Hence $(k,F)$ is quasi-finite.
Proposition 20 Let $(k, F)$ be a quasi-finite field.
  1. If $w\in \bar k^\times$ is a root of unity, then there exists $y\in \bar k^\times$ such that $w=y^{F-1}:=F(y)/y$.
  2. If $\Char(k)>0$, then $F-1: \bar k\rightarrow \bar k$ is surjective.
Proof
  1. Let $A=\bar k^\times$. Then $0=H^1(G_k,A)=A'/(F-1)A$. But $A'$ contains all the roots of unity.
  2. The same argument with $A=\bar k$ using additive Hilbert 90.
Proposition 21 If $k $ is quasi-finite. Then $\Br(k)=0$.
Proof Since $\bar k^\times$ is divisible, it follows from Proposition 19that $\Br(k)=H^2(\hat{\mathbb{Z}}, \bar k^\times)=0$
Corollary 7 If $k'/k$ is a finite extension, then $N(k')^\times=k^\times$.
Proof Write $k'{}=\bar k^{F^n}$. Then $k^\times/ N(k')^\times=\hat H^0(\Gal(k'/k), (k')^\times)=\hat H^2(\Gal(k'/k), (k')^\times)\hookrightarrow \Br(k)=0$.

TopLocal class field theory

We can now finally construct the local class formation for a complete discretely valued field $K$ with quasi-finite residue field $k $. Write $\mathfrak{g}=\Gal(K^\mathrm{ur}/K)\cong G_k\cong \hat{\mathbb{Z}}$. Notice that any finite extension of $K$ also satisfies the same hypothesis. Let $\inv_k: \Br(k)\cong \mathbb{Q}/\mathbb{Z}$ be as in Remark 57. We would like to check the two axioms of the class formation is satisfied. Axiom I is simply Hilbert 90. Axiom II is the following

Theorem 8
  1. For $L/K$ finite Galois, $\inv_{L/K}: H^2(L/K)\rightarrow H^2(K)\xrightarrow{\inv_K}\mathbb{Q}/\mathbb{Z}$ induces $H^2(L/K)\cong 1/[L:K] \mathbb{Z} /\mathbb{Z}$.
  2. For $L/K$ finite separable. $\inv_L\circ\Res_{L/K}=[L:K]\inv_K$.
Remark 62 Part (a) follows formally from (b) since $\inv_L$ is an isomorphism. The real content lies in (b).
Proof (Part (a) of Theorem 8) Write $n=[L:K]$.

When $L/K$ is unramified, $L^\mathrm{ur}=K^\mathrm{ur}$ and $\mathfrak{g}_L\cong\Gal(K^\mathrm{ur}/L)\cong n \mathfrak{g}_K$. The result follows from the following commutative diagram $$\xymatrix@C=.8em{\Br(K) \ar[d]^{\Res} & \ar[l] H^2(\mathfrak{g}_K, (K^\mathrm{ur})^\times) \ar[r]^-{\val_*} \ar[d]^{\Res}& H^2(\hat{\mathbb{Z}}, \mathbb{Z}) \ar[r]^-d \ar[d]^{\Res}& H^1(\hat{\mathbb{Z}},\mathbb{Q}/\mathbb{Z}) \ar[r] \ar[d]^{\Res}& \mathbb{Q}/\mathbb{Z}  \ar[d]^{\times n} \\ \Br(L)  & \ar[l] H^2(\mathfrak{g}_L, (K^\mathrm{ur})^\times) \ar[r]^-{\val_*} & H^2(n \hat{\mathbb{Z}},\mathbb{Z}) \ar[r]^-d  & H^1(n \hat{\mathbb{Z}} , \mathbb{Q}/\mathbb{Z}) \ar[r] & \mathbb{Q}/\mathbb{Z}.}$$

When $L/K$ is totally ramified, $L^\mathrm{ur}=LK^\mathrm{ur}$ and $\mathfrak{g}_L\cong \mathfrak{g}_K$. The result follows from tracing the same diagram by noticing that $\val_L=n\val_K$: $$\xymatrix@C=.8em{\Br(K) \ar[d]^{\Res} & \ar[l] H^2(\mathfrak{g}_K, (K^\mathrm{ur})^\times) \ar[r]^-{\val_*} \ar[d]^{i_*}& H^2(\hat{\mathbb{Z}}, \mathbb{Z}) \ar[r]^-d \ar[d]^{\times n}& H^1(\hat{\mathbb{Z}},\mathbb{Q}/\mathbb{Z}) \ar[r] \ar[d]^{\times n}& \mathbb{Q}/\mathbb{Z}  \ar[d]^{\times n} \\ \Br(L)  & \ar[l] H^2(\mathfrak{g}_L, (L^\mathrm{ur})^\times) \ar[r]^-{\val_*} & H^2(\hat{\mathbb{Z}},\mathbb{Z}) \ar[r]^-d  & H^1(\hat{\mathbb{Z}} , \mathbb{Q}/\mathbb{Z}) \ar[r] & \mathbb{Q}/\mathbb{Z}.}$$

For a general $L/K$, we can take $L'$ be the maximal unramified extension of $K$ in $L$ and reduce to the previous two cases by looking at $L/L'$ (totally ramified) and $L'/K$ (unramified).

03/29/2013

Corollary 8 $a\in \Br(K)$ is split by a finite extension $L/K$ if and only if $[L:K]a=0$.
Proof By definition, $a$ is split by $L/K$ if and only if $\Res_{K/L}(a)=0$. Since $\inv_L$ is an isomorphism, this is is equivalent to $0=\inv_{L}\cdot\Res_{K/K}(a)=[L:K]\inv_K(a)$ by Part (b) of Theorem 8, if and only if $[L:K]a=0$ since $\inv_K$ is an isomorphism.
Proof (Part (b) of Theorem 8) By the previous corollary, $$H^2(L/K)\cong\{x\in \mathbb{Q}/\mathbb{Z} : [L:K]x=0\}=1/[L:K]\mathbb{Z}/\mathbb{Z}.$$
Remark 63 We have verified all the axioms in the class formation. So we get the Artin maps (Definition 30) isomorphism $$\Psi_{L/K}: K^\times/N L^\times\cong \Gal(L/K)^\mathrm{ab},\quad a\mapsto (a,L/K)$$ for any $L/K$ finite Galois. Taking the inverse limit of $\Psi_{L/K}$ we obtain $$\Psi_K: K^\times\rightarrow G_K^\mathrm{ab},\quad a\mapsto (a, {*}/K).$$ It follows that $\Psi_K$ is continuous and has dense image.
Remark 64 These Artin maps satisfy norm and verlagerung functoriality (Remark 55).

In particular, the local norm index equality (which we treated as a black box last semester) follows:

Theorem 9 (local norm index inequality) Suppose $L/K$ is finite Galois, then $NL^\times$ is a finite index subgroup of $K^\times$ and $[K^\times: NL^\times]\mid [L:K]$. This becomes an equality if and only if $L/K$ is abelian.
Proposition 22 If $L/K$ is an unramified finite extension. Then $(x, L/K)=F^\mathrm{\val(x)}\in \Gal(l/k)\cong\Gal(L/K)$.
Proof By Proposition 16, it suffices to compute $\langle\chi,(a,L/K)\rangle=\inv_{L/K}(a\smile d\chi)$. Write $\mathfrak{g}=\Gal(l/k)\cong\Gal(L/K)$. Tracing the definition of $\inv_{L/K}$ $$\inv_{L/K}:\xymatrix{ H^2(\mathfrak{g}, L^\times) \ar[r]^{\val_*} & H^2(\mathfrak{g}, \mathbb{Z}) \ar[r]^-d & H^1(\mathfrak{g} ,\mathbb{Q}/\mathbb{Z}) \ar[r]^-{F\mapsto \frac{1}{[L:K]}} & \mathbb{Q}/\mathbb{Z}}, $$ we find that $\inv_{L/K}(a\smile d\chi)=\val(a)\chi(F)=\chi(F^{\val(a)})$ as wanted.
Remark 65 In particular, when $L/K$ is unramified, then Artin map kills $\mathcal{O}_K^\times$. Notice the unramified condition is crucial, otherwise we do not have an identification between $H^2(L/K)$ and $H^2(\mathfrak{g},L^\times)$!
Corollary 9 Suppose $L/K$ is finite abelian with $G=\Gal(L/K)$. Let $I=G^0$ be the inertia group. Then $\Psi_{L/K}(\mathcal{O}_K^\times)=I$.
Proof Let $L'{}=L^I$ be the maximal unramified subextension of $L/K$. So $\Psi_{L'/K}(\mathcal{O}_K^\times)=\{1\}$ by the previous proposition. Hence $\Psi_{L/K}(\mathcal{O}_K^\times)\subseteq I$.

Conversely, let $\sigma\in I$ and choose $a\in K^\times$ such that $(a,L/K)=\sigma$. Since $\sigma|_{L'}=1$. we know that $1=(a,L'/K)=F^{\val_K(a)}$ by the previous proposition. Hence $[L':K]\mid \val_K(a)$. On the other hand, for $b\in L^\times$, $$\val_K(N_{L/K}b)=1/e\val_L(N_{L/K}b)=[L:K]/e\val_L(b)=[L':K]\val_L(b).$$ So we can choose $b\in L^\times$ such that $\val_K(N_{L/K}b)=\val_K(a)$. Let $a'{}=a/N_{L/K}(b)$, then $a'\in \mathcal{O}_K^\times$ and $(a',L/K)=(a,L/K)=\sigma$ since the Artin map kills the norm.

Remark 66 To summarize, we have proved that
  1. The Artin $\Psi_K: K^\times\rightarrow G_K^\mathrm{ab}$ is continuous and has dense image and $\Psi_{L/K}: K^\times/NL^\times\cong\Gal(L/K)$ for $L/K$ finite abelian (Remark 63).
  2. $\Psi_K$ satisfies the norm and verlagerung functoriality (Remark 64).
  3. $\Psi_K$ restricts to $\mathcal{O}_K^\times\rightarrow I_K$ (Corollary 9) with dense image. In other words, we have the following commutative diagram $$\xymatrix{0 \ar[r]  & \mathcal{O}_K^\times \ar[r] \ar[d]& K^\times \ar[r]^{\val} \ar[d]^{\Psi_K} & \mathbb{Z} \ar[r] \ar[d]& 0\\ 0 \ar[r] & I_K\ar[r] & W \ar[r]& \mathbb{Z} \ar[r]& 0.}$$ Moreover, when $\mathcal{O}_K^\times$ is compact (e.g, $K$ has finite residue field), $\mathcal{O}_K^\times\rightarrow I_K$ is actually surjective.

We are still missing

  1. $\Psi_{L/K}(1+\mathfrak{p}_K^n)=\Gal(L/K)^n$. This part can be done directly, but we shall postpone the discussion due to the time.
  2. The existence theorem. Notice this is equivalent to $\mathcal{O}_K^\times\rightarrow I_K$ being injective. Because then $K^\times\cong W$ (the abelianized Weil group) and thus the open subgroups of $K^\times$ of finite index correspond to the finite abelian extensions of $K$ (Theorem 20). Lubin-Tate theory will construct the inverse of $\mathcal{O}_K^\times\rightarrow I_K$ and the injectivity follows. Indeed Lubin-Tate theory does more than that by constructing explicit splitting polynomials for the abelian extensions of $K$.
Remark 67 We really need the fact the residue field is finite. When $k $ is not finite, $\mathcal{O}_K^\times\rightarrow I_K$ can be neither injective nor surjective.

04/01/2013

TopLubin-Tate theory

Definition 32 Let $R$ be a ring. A formal group law over $R$ is a power series $F(X,Y)\in R\llbracket X,Y\rrbracket$ such that
  1. $F(X,Y)=X+Y+O(2)$ (meaning that $F(X,Y)\equiv X+Y\pmod{ (X,Y)^2}$).
  2. $F(X,F(Y,Z))=F(F(X,Y),Z)$ in $R\llbracket X,Y,Z\rrbracket$.
  3. There exists $\iota(X)\in R\llbracket X\rrbracket$ such that $F(X,\iota(X))=F(\iota(X),X)=0$.
  4. $F(X,0)=F(0,X)=X$.

We say $F $ is commutative if in addition $F(X,Y)=F(Y,X)$.

Remark 68 (a), (b), (d) implies (c) and the inverse in uniquely determined. So we do not need to worry about the inverse when constructing formal group laws. Notice condition (d) implies that $F(X,Y)$ is of the form $X+Y+XY G(X,Y)$, i.e., all the higher order terms are crossed terms.
Definition 33 A homomorphism $F_1(X,Y)\rightarrow F_2(X,Y)$ between two formal group laws is given by a power series $\phi(X)\in R\llbracket X\rrbracket$ such that
  1. $\phi(X)\in XR\llbracket X\rrbracket$.
  2. $F_2(\phi(X),\phi(Y))=\phi(F_1(X,Y))$.
Example 14 Let $R=\mathcal{O}_K$ be the ring of integers of a local field $K$. Suppose $\mathfrak{m}=\pi \mathcal{O}_K$. If $F(X,Y)$ is a formal group law over $\mathcal{O}_K$, then it gives $\mathfrak{m} $ a structure of a group by simply evaluating $x+_F+y=F(x,y)$ for $x,y\in \mathfrak{m}$.
Example 15 The formal additive group $\hat{\mathbb{G}}_a$ is defined to be $F(X,Y)=X+Y$. Given $x,y\in \mathfrak{m}$, $x+_F y=F(x,y)=x+y$.
Example 16 The formal multiplicative group $\hat{\mathbb{G}}_m$ is defined to be $F(X,Y)=(X+1)(Y+1)-1=X+Y+XY$. Then $x\mapsto 1+x$ gives an isomorphism $(\mathfrak{m}, +_F)\cong(1+\mathfrak{m}, \times)$.

The more interesting formal groups are those not so easily described in terms of explicit power series.

From now on let $K$ be a local field with uniformizer $\pi$ and $k=\mathcal{O}_K/(\pi)$ with $q=p^f$ elements.

Definition 34 Let $\mathcal{F}_\pi=\{f(T)\in \mathcal{O}_K\llbracket T\rrbracket: f(T)\equiv \pi T\pmod{T^2}, f(T)\equiv T^q\pmod{\pi}\}$.
Lemma 5 For any $f,g\in \mathcal{F}_\pi$ and $a_1,\ldots,a_n\in \mathcal{O}_K$. Then there exists a unique $F(X_1,\ldots,X_n)\in \mathcal{O}_K\llbracket X_1,\ldots,X_n\rrbracket$ such that
  1. $F(X_1,\ldots,X_n)=a_1X_1+\ldots a_nX_n+O(2)$
  2. $f(F(X_1,\ldots,X_n))=F(g(X_1),\ldots,g(X_n))$.
Proof The proof is bases on the successive approximations of the desired result, i.e., the induction on the degree of $F $. Write $\underline{X}=(X_1,\ldots,X_n)$ and $g(\underline{X})=(g(X_1),\ldots,g(X_n))$ for short. We inductively show that the system of equations
  1. $F_r(\underline{X})=a_1X_1+\ldots a_nX_n+O(2)$,
  2. $f(F_r(\underline{X}))=F_r(g(\underline{X}))+O(r+1)$,

has a unique solution $F_r$ in $\mathcal{O}_K\llbracket \underline{X}\rrbracket/(X_1,\ldots,X_n)^r$.

When $r=1$, we can (and are forced to) take $F_1(\underline{X})=a_1X_1+\ldots a_nX_n$. Suppose we are given $F_r(\underline{X})$. Set $F_{r+1}(\underline{X})=F_r(\underline{X})+\Delta_r$, where $\Delta_r\in (X_1,\ldots,X_n)^{r+1}$. Since 
\begin{align*}
  f(F_{r+1}(X))&=f(F_r(X))+\pi \Delta_r(X)+O(r+2),\\
  F_{r+1}(g(X))&=F_r(g(X))+\pi^{r+1}\Delta_r(X)+O(r+2),
\end{align*}
to fulfill condition (b) we are forced to take $$\Delta_r(\underline{X})=\frac{f(F_r(\underline{X}))-F_r(g(\underline{X}))}{\pi^{r+1}-\pi}+O(r+2).$$ Since $\pi^{r+1}-\pi=\pi(\pi^r-1)$ and $\pi^r-1$ is a unit, to show that $\Delta_r\in \mathcal{O}_K\llbracket \underline{X}\rrbracket$ it suffices to show $\pi$ divides the numerator. It is true because $$f(F_r(\underline{X}))-F_r(g(\underline{X}))\equiv F_r(\underline{X})^q-F_r(\underline{X}^q)\equiv0\pmod{(\pi)}.$$ Now taking $r\rightarrow\infty$ gives the desired $F(\underline{X})$. It is furthermore uniquely determined by this process.

Definition 35 Given $f\in \mathcal{F}_\pi$. We define $F_f(X,Y)\in\mathcal{O}_K\llbracket X,Y\rrbracket$ by
  1. $F_f(X,Y)=X+Y+O(2)$.
  2. $f(F_f(X,Y))=F_f(f(X),f(Y))$.

Such a power series exists and is unique by Lemma 5.

Remark 69 We will see soon that $F_f$ is indeed a formal group.
Definition 36 Given $a\in \mathcal{O}_K$ and $f,g\in\mathcal{F}_\pi$. We define $[a]_{f,g}\in\mathcal{O}_K\llbracket T\rrbracket$ such that
  1. $[a]_{f,g}=aT+O(2)$.
  2. $f([a]_{f,g}(T))=[a]_{f,g}(g(T))$.

Again such a power series exists and is unique by Lemma 5.

Theorem 10 The following identities hold:
  1. $F_f(X,Y)=F_f(Y,X)$.
  2. $F_f(F_f(X,Y),Z)=F_f(X,F_f(Y,Z))$.
  3. $F_f([a]_{f,g}(X),[a]_{f,g}(Y))=[a]_{f,g}(F_g(X,Y))$.
  4. $[a]_{f,g}([b]_{f,g}(T))=[ab]_{f,g}(T)$.
  5. $[a+b]_{f,g}(T)=F_f([a]_{f,g}(T),[b]_{f,g}(T))$.
  6. $[\pi]_{f,f}(T)=f(T)$.
  7. $[1{}]_{f,f}(T)=T$.
Proof Apply Lemma 5 repeatedly.

The importance of this theorem lies in the following definition (intuitively, a formal group with "extra endomorphism").

Definition 37 A formal $\mathcal{O}_K$-module is a (commutative) formal group law $F(X,Y)\in \mathcal{O}_K\llbracket X,Y\rrbracket$ together with a homomorphism $\mathcal{O}_K\rightarrow\End_{\mathcal{O}_K}(F)$, denoted by $a\mapsto [a](T)$ such that $[a](T)= aT+O(2)$.
Corollary 10 Given $f\in\mathcal{F}_\pi$. There exists a unique formal $\mathcal{O}_K$-module $(F,[\cdot])$ such that $f(T)\in\End_{\mathcal{O}_K}(F)$ and $[\pi]=f$.
Proof $(F_f,[\cdot]_f:=[\cdot]_{f,f})$ works by the previous theorem. The only thing needs to check is that $F_f(X,0)=F_f(0,X)=X$. In fact, it again follows from the uniqueness of Lemma 5 applied to $F(X):=F_f(X,0)$ and $f=g\in\mathcal{F}_\pi$.
Remark 70 Moreover, for $f,g\in\mathcal{F}_\pi$, there exists a canonical isomorphism of formal $\mathcal{O}_K$-modules $(F_f,[\cdot]_f)\cong(F_g,[\cdot]_g)$ given by $[1{}]_{f,g}$.
Example 17 When $K=\mathbb{Q}_p$, $\pi=p$, we can take $f(T)=(T+1)^p-1$. Then $F_f=\hat{\mathbb{G}}_m$ and $[p]=(T+1)^p-1$. The $p^n$-torsion of $F_f$ is exactly the $p^n$-th roots of unity and adjoining all of them gives us a maximal totally ramified extension of $\mathbb{Q}_p$! This picture will generalize to any local field and is the main content of Lubin-Tate theory.
Definition 38 Now fix a $f\in \mathcal{F}_\pi$. For a valued field extension $L/K$, we define $$F_f(L):=\{x\in L: |X|<1\}$$ endowed with the structure of a $\mathcal{O}_K$-module: for $x,y\in F_f(L)$, $a\in \mathcal{O}_K$, we set $x+_F y=F_f(x,y)$ and $ax=[a]_f(x)$. It is easy to check that when $L/K$ is a finite Galois extension, $F_f(L)$ is an $\mathcal{O}_K$-module on which $\Gal(L/K)$ acts by $\mathcal{O}_K$-linear maps. Moreover, $G_K=\Gal(K^s/K)$ acts on $F_f(K^s)$ by $\mathcal{O}_K$-linear maps.

04/03/2013

Definition 39 For $m\ge1$, we define $$F_f(K^s)[\pi^m]=\{x\in F_f(K^s): [\pi^m]_f(x)=0\}.$$ Define $L_{\pi,m}=K(F_f(K^s)[\pi^m])$ and $G_{\pi,m}=\Gal(L_{\pi,m}/K)$. These only depend on the choice of $\pi$ and $m  $ and do not depend on the choice of $f\in\mathcal{F}_\pi$. In fact, since $F_f$ and $F_g$ are isomorphic as $\mathcal{O}_K$-module by $[1{}]_{f,g}$ (Remark 70) and thus there is an induced isomorphism $F_f(K^s)\cong F_g(K^s)$ as both $\mathcal{O}_K$ and $G_K$-modules. Also let $L_\pi=\bigcup L_{\pi,m}$ and $G_\pi=\varprojlim_m G_{\pi,m}$.

The main theorem for today is the following.

Theorem 11
  1. $[\pi]_f: F_f(K^s)\rightarrow F_f(K^s)$ is surjective.
  2. $F_f(K^s)[\pi^m]\cong \pi^{-m}\mathcal{O}_K/\mathcal{O}_K\subseteq K/\mathcal{O}_K$.
  3. $F_f(K^s)[\pi^\infty]\cong K/\mathcal{O}_K$.
  4. For any $\tau\in G_\pi$, there exists a unique $u\in \mathcal{O}_K^\times$ such that for any $x\in F_f(K^s)[\pi^\infty]$, $\tau(x)=[u]_f(x)$.
  5. $\tau\mapsto u$ induces an isomorphism $G_\pi\cong \mathcal{O}_K^\times$ and $G_{\pi,m}\cong (\mathcal{O}_K/\pi^m)^\times$.
  6. $\pi\in N_{L_{\pi,m}/K}(L_{\pi,m}^\times)$.
Remark 71 Though the objects $F_f(K^s)[\pi^m]$ are rather complicated, the proof is actually very elementary (using counting argument).
Proof We may assume $f(T)=\pi T+T^q$ (Remark 70).
  1. For any $x\in F_f(K^s)$. By definition, $[\pi]_f (y)=x$ is the same as $y$ being a root of $f(T)-x$. We must show that $f(T)-x$ has a root in $\{y\in K^s: |y|<1\}$. In fact, the Newton polygon of $f(T)-x$ is strictly above the $x$-axis, hence all the roots of $f(T)-x$ has positive valuation.
  2. When $m=1  $, $F_f(K^s)[\pi]$ is a $\mathcal{O}_K/\pi$ is a $k $-vector space. But $\{x\in K^s:|x|<1: f(x)=0\}$ has exactly $q$ elements, so $F_f(K^s)[\pi]$ is 1-dimensional $k $-vector space. In general, we assume by induction that there are elements $x_j\in F_f(K^s)[\pi^j]$, $j=1,2,\ldots,m-1$ such that $a\mapsto [a]_f(x_j)$ induces an isomorphism $\mathcal{O}_K/\pi^j\cong F_f(k^s)[\pi^j]$ and furthermore $[\pi]_f(x_j)=x_{j-1}$ for $j=2,\ldots,m-1$. Now look at the sequence $$0\rightarrow F_f(k^s)[\pi]\rightarrow F_f(k^s)[\pi^m]\xrightarrow{[\pi]_f} F_f(k^s)[\pi^{m-1}]\rightarrow0.$$ It is exact: the injectivity and the exactness in the middle are obvious; the surjectivity follows from the (a). Now we are done by induction if we choose any preimage $x_m$ of $x_{m-1}$ under $[\pi]_f$. In fact, $a\mapsto[a]_f(x_m)$ induces an injection $\mathcal{O}_K/\pi^m\hookrightarrow F_f(K^s)[\pi^m]$ by construction and it must be an isomorphism by counting.
  3. It follows from (b) and $\bigcup \pi^{-m}\mathcal{O}_K/\mathcal{O}_K=K/\mathcal{O}_K$.
  4. The isomorphism in (c) induces an injection $G_\pi\hookrightarrow \Aut_{\mathcal{O}_k}(F_f(K^s)[\pi^\infty])\cong\Aut_{\mathcal{O}_K}(K/\mathcal{O}_K)\cong\mathcal{O}_K^\times$. Similarly, $G_{\pi,m}\hookrightarrow \Aut_{\mathcal{O}_K}(F_f(K^s)[\pi^m])\cong(\mathcal{O}_K/\pi^m)^\times$.
  5. It suffices to show the surjectivity of the maps in (d). By injectivity, we only need to check that $|G_{\pi,m}|\ge(q-1)q^{m-1}$. By definition $L_{\pi,m}$ is the splitting field of the polynomial $[\pi^m]_f=f(f(\cdots f(T))=f^{(m)}(T)$, which a polynomial of degree $q^m$. Notice that $f^{(m-1)}(T)\mid f^{(m)}(T)$. We define $\Phi_m(T)=f^{(m)}(T)/f^{(m-1)}(T)$. Then $\Phi_m(T)=f^{(m-1)}(T)^{q-1}+\pi$. Now $\Phi_m(T)\equiv T^{q^{m-1}(q-1)}\pmod{\pi}$ and has constant coefficient $\pi$, thus is an Eisenstein polynomial. It follows that $\Phi_m(T)$ is irreducible and in particular $[L_{\pi,m}:K]\ge q^{m-1}(q-1)$ as wanted. As a byproduct, we have also shown that $L_{\pi,m}/K$ is totally ramified.
  6. It is immediate since $\pi$ is the constant term of $\Phi_m(T)$.
Remark 72 If $L/K$ is any finite abelian totally ramified extension, the Artin map $\Psi_{L/K}: K^\times\twoheadrightarrow\Gal(L/K)$ maps $\mathcal{O}_K^\times\twoheadrightarrow I_{L/K}=\Gal(L/K)$. So for any $\pi\in K^\times$, there exists $u\in \mathcal{O}_K^\times$ such that $\Psi_{L/K}(\pi)=\Psi_{L/K}(u)$. Namely, $\Psi_{L/K}(\pi/u)=1$. Define $\pi'{}=\pi/u$. Then $L\subseteq (K^\mathrm{ab})^{\Psi_K(\pi')=1}$. Conversely, any subextension of $(K^\mathrm{ab})^{\Psi_K(\pi')=1}$ is totally ramified. In fact we will see that $(K^\mathrm{ab})^{\Psi_K(\pi')=1}=L_{\pi'}$ constructed by Lubin-Tate theory and every maximal totally ramified abelian extension of $K$ has this form. (Warning: there is no unique maximal totally ramified extension since the compositum of two totally ramified extensions are not necessarily totally ramified!)

04/05/2013

Today we are going to remove the choice of the uniformizer $\pi$ from the whole picture, by adding the maximal unramified extension $K^\mathrm{ur}$.

Definition 40 Define $r_\pi: K^\times\rightarrow\Gal(L_\pi K^\mathrm{ur}/K)\cong \Gal(L_\pi/K)\times \Gal(K^\mathtt{ur}/K)$ by sending $\pi^m\cdot u$ to $(\phi_\pi^{-1}(u^{-1}), \Frob^m)$, where $\phi_\pi:G_\pi\xrightarrow{\sim} \mathcal{O}_K^\times$ is the isomorphism in Theorem 11 (d). Namely, $(u^{-1},m)\in \mathcal{O}_K^\times\times \hat{\mathbb{Z}}$ under the identification $\Gal(L_\pi/K)\times \Gal(K^\mathtt{ur}/K)\cong \mathcal{O}_K^\times\times \hat{\mathbb{Z}}$. We will soon see this naturally defined homomorphism $r_\pi$ agrees with the Artin map $\Psi_K$.
Theorem 12 The compositum field $L_\pi L^\mathrm{ur}$ and the isomorphism $r_\pi$ are independent of the choice of $\pi$.
Proof Choose another uniformizer $\pi'{}=u\pi$, where $u\in\mathcal{O}_K^\times$. Choose $f\in F_\pi$ and $g\in F_{\pi'}$. Write $\mathcal{O}_{K^\mathrm{ur}}$ be the ring of integers of $K^\mathrm{ur}$ and $\hat{\mathcal{O}}_K^\mathrm{ur}$ be the ring of integers of the completion $\hat K^\mathrm{ur}$. The following lemma is crucial:
Lemma 6 There exists $\theta(X)\in \hat{\mathcal{O}}_{K^\mathrm{ur}}\llbracket X \rrbracket$ such that $\theta(x)=\varepsilon X+O(2)$, where $\varepsilon\in \hat{\mathcal{O}}_{K^\mathrm{ur}}^\times$ such that
  1. $\theta(F_f(X,Y))=F_g(\theta(X),\theta(Y))$.
  2. $\theta([a]_f(X))=[a]_g(\theta(X))$.
  3. $\Frob(\theta)(X)=\theta([u]_f(X))$, where $\Frob$ acts on $\theta$ coefficient-wisely.
Remark 73 The lemma tells us that $\theta$ is an isomorphism of the formal $\mathcal{O}_K$-modules $(F_f,[\cdot]_f)$ and $(F_g,[\cdot]_g)$ but only defined over the larger coefficient ring $\hat{\mathcal{O}}_{K^\mathrm{ur}}$. That explains why $L_\pi K^\mathrm{ur}$ is independent of $\pi$ (though $L_\pi$ is not).

Assuming the lemma 6, we know that $\theta$ gives an isomorphism of $\mathcal{O}_K$-modules $F_f(K^s)[\pi^\infty]\cong F_g(K^s)[\pi^\infty]$. This isomorphism does not respect the action of $G_K$, but it does respect the action of $\Gal(K^s/K^\mathrm{ur}$. Hence $L_\pi L^\mathrm{ur}= L_{\pi'}K^\mathrm{ur})$. To show that $r_\pi$ and $r_{\pi'}$ are equal, it suffices to show that $r_\pi(\pi')=r_{\pi'}(\pi')$, since the left hand side does not depend on $\pi$ and all such $\pi'{}=u\pi$ generate $K^\times$ when $u$ runs over $\mathcal{O}_K^\times$. Notice that $r_\pi(\pi')|_{K^\mathrm{ur}}=\Frob=r_{\pi'}(\pi')|_{K^\mathrm{ur}}$, so it suffices to show that $r_\pi(\pi')|_{L_{\pi'}}=r_{\pi'}(\pi)|_{L_{\pi'}}=\Id_{L_{\pi'}}$, i.e., $r_\pi(\pi')$ acts trivially on $$L_{\pi'}=K(F_g(K^s)[\pi^\infty])=K(\theta(F_f(K^s)[\pi^\infty])).$$ Let $x\in F_f(K^s)[\pi^\infty]$. Then $r_\pi(\pi')=r_\pi(\pi) r_\pi(u)$ where $$r_\pi(\pi)|_{L_\pi}=\Id,\quad r_\pi(\pi)|_{K^\mathrm{ur}}=\Frob,\quad r_\pi(u)|_{L_\pi}=[u]^{-1}_f,\quad r_\pi(u)|_{K^\mathrm{ur}}=\Id.$$ Using Lemma 6 , we can now compute that 
\begin{align*}
  \theta(x)^{r_\pi(\pi')}&=\theta(x)^{r_\pi(\pi)r_\pi(u)}=(\Frob(\theta)(x))^{r_\pi(u)}\\
  &=\Frob(\theta)([u^{-1}]_f(x))=\theta([u]_f([u^{-1}]_f(x))=\theta(x),
\end{align*}
as desired.

Corollary 11 $L_\pi K^\mathrm{ur}=K^\mathrm{ab}$ and $r_\pi=\Psi_K$.
Proof Write $E=L_\pi K^\mathrm{ur}$. Then $E\subseteq K^\mathrm{ab}$ and we have the Artin map $\Psi_{E/K}: K^\times\rightarrow \Gal(E/K)$. By Theorem 11 (f), for any $\pi'{}=u\pi$ and $m\ge1$, $\pi'$ is a norm from $L_{\pi',m}$. So $\Psi_{E/K}(\pi')$ acts as the identity on $L_{\pi',m}$. It follows that $\Psi_{E/K}(\pi')|_{L_{\pi'}}=\Id$ and $\Psi_{E/K}(\pi')|_{K^\mathrm{ur}}=\Frob$. Therefore $\Psi_{E/K}(\pi')=r_\pi(\pi')$ and thus $\Psi_{E/K}=r_\pi$.

If $E\ne K^\mathrm{ab}$, then $K^\mathrm{ab}/E$ is totally ramified and hence $\Gal(K^\mathrm{ab}/E)\subseteq I_K$, where $I_K$ is the inertia group of $K^\mathrm{ab}/K$. But $I_K=\Psi_K(\mathcal{O}_K^\times)$, so we can choose $1\ne a\in \mathcal{O}_K^\times$ such that $\Psi_K(a)|_E=\Id$ and $\Psi_K(a)\ne\Id_{K^\mathrm{ab}}$. But then $\Psi_K(a)|_E=\Psi_{E/K}(a)=r_\pi(a)\ne1$ since $r_\pi|_{\mathcal{O}_K^\times}$ is an isomorphism, a contradiction.

As Remark 66, the existence theorem then follows from the isomorphism $r_{\pi}|_{\mathcal{O}_K^\times}=\Psi_K|_{\mathcal{O}_K^\times}:\mathcal{O}_K^\times\cong I_K$.

Corollary 12 (Existence theorem)
  1. $\Psi_K$ induces an isomorphism $K^\times\cong W_K^\mathrm{ab}$.
  2. Any open finite index subgroup of $K^\times$ has the form $N_{E/K}(E^\times)$ for some finite abelian extension $E/K$.

References

[1]Jean-Pierre Serre, Galois Cohomology, Springer, 2001.

[2]Jean-Pierre Serre, Local Fields (Graduate Texts in Mathematics), Springer, 1980.

[3]John William Scott Cassels and Albrecht Frohlich, Algebraic Number Theory, London Mathematical Society, 2010.

[4]Kenneth S. Brown, Cohomology of Groups (Graduate Texts in Mathematics, No. 87), Springer, 1982.