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\part*{Galois Cohomology}
\pauth{John Tate}

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\mainmatter
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\LogoOn

\lectureseries[Galois Cohomology]{Galois Cohomology}

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\auth[J. Tate]{John Tate$\mbox{}^{1}$} 

\address{Department of Mathematics; Austin, TX}
\email{tate@math.utexas.edu}

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\subjclass{11}
\keywords{Galois cohomology, Elliptic curves}
%\date{August 22, 1999}
 \setaddress

%\lecture{Galois cohomology}

        I thank Helena Verrill and William Stein for their help in
    getting this account of my talks at Park City into print. After
    Helena typed up her original notes of the talks, William was a
    great help with the editing, and put them in the canonical format
    for this volume.
    
        The somewhat inefficient organization of this account is
    mainly a result of the fact that, after the first talk had been
    given with the idea that it was to be the only one, a second was
    later scheduled, and these are the notes of the material in the
    two talks in the order it was presented.


The bible for this subject is Serre~\cite{serre:gc}, in
conjunction with~\cite{serre:local} or \cite{cassels-frohlich}.  
Haberland~\cite{haberland}  is also an excellent reference. 

\section{Group modules}
Consider a group~$G$ and an abelian group~$A$ 
equipped with a map 
$$G\times A\rightarrow A,$$
$$(\sigma,a)\mapsto \sigma a.$$
We use notation $\sigma,\,\tau,\,\rho,\dots$ 
for elements of~$G$, and $a,\,b,\,a',\,b',\dots$ for elements of~$A$.
To say that~$A$ is a \emp{$G$-set} means that
$$\tau(\sigma a) = (\tau\sigma)a \quad\text{and}\quad 1a=a,$$
for all $\sigma, \tau\in G$ and $a\in A$,
where~$1$ is the identity in~$G$. 
To say that~$A$ is a \emp{$G$-module} means that, in addition, we
have
$$\sigma(a+b)=\sigma a + \sigma b,$$
for all $\sigma\in G$ and $a,b\in A$.
This is all equivalent to giving~$A$ the structure
of ${\Z}[G]$-module.

Given a $G$-module~$A$ as above, the subgroup of fixed elements of~$A$ is
  $$A^G:=\left\{ a\in A \mid \sigma a = a \text{ for all } \sigma \in G\right\}.$$
We say~$G$ \emp{acts trivially} on~$A$ if $\sigma a=a$
for all $a\in A$; thus
$A^G=A$ if and only if the action is trivial.
When ${\Z}$, ${\Q}$, ${\Q}/{\Z}$ are
considered as $G$-modules, this is with the trivial action, 
unless stated otherwise.

If we take $G=\Gal(K/k)$, with~$K$ a Galois 
extension of~$k$ of possibly infinite degree,
then we have the following examples of fixed subgroups of
$G$-modules:
$$\begin{array}{l|l}
A & A^G\\
\hline
K^+ \text{ as an additive group}&k^+\\
K^{*} \text{ as a multiplicative group}&k^*\\
E(K), \text{ where } E/k \text{ is an elliptic curve}&E(k).\\
\end{array}
$$
The action on $E(K)$ above is given by
$\sigma(x,y)=(\sigma x,\sigma y)$ for a point $P=(x,y)$,
if~$E$ is given as a plane cubic.
In general, if~$C$ is a commutative algebraic group over~$K$,
we can take $A=C(K)$, and then $A^G=C(k)$.

\section{Cohomology}
We now define the cohomology groups 
   $H^r(G,A)$,
for $r\in\Z$.
Abstractly, these are the right derived functors of the left exact functor
$$\left\{G\text{-modules}\right\}\rightarrow\left\{\text{abelian groups}\right\}$$
that sends $A\mapsto A^G$.
Since $A^G=\Hom_{{\Z}[G]}({\Z},A)$,
we have a canonical isomorphism
 $$H^r(G,A)=\Ext^r_{{\Z}[G]}({\Z},A).$$

More concretely, the cohomology groups $H^r(G,A)$ can be computed
using the ``standard cochain complex'' (see,
e.g., \cite[pg.~96]{cassels-frohlich}). Let
      $$C^r(G,A):=\Maps(G^r,A);$$
an element of $C^r(G,A)$ is a function~$f$ of~$r$ variables in~$G$,
    $$f(\sigma_1,\dots,\sigma_r)\in A,$$
and is called an \emp{$r$-cochain}.
(If, in addition,~$A$ and~$G$ have a topological structure, then we 
instead consider continuous cochains.)
There is a sequence
$$\cdots \ra 0 \ra 0\rightarrow C^0(G,A)\stackrel{\delta}{\rightarrow}
C^1(G,A)\stackrel{\delta}{\rightarrow}
C^2(G,A)\stackrel{\delta}{\rightarrow}\cdots$$
Here $C^0(G,A)=A$, since an element~$f$ of $C^0(G,A)$ is given by
the single element $f_0(\bullet)\in A$,
its value at the unique element $\bullet\in G^0$. 
The maps~$\delta$ are defined by
$$\begin{array}{lll}
(\delta f_0)(\sigma)&
=&\sigma f_0(\bullet) - f_0(\bullet),\\
(\delta f_1)(\sigma,\tau)&=&
\sigma f_1(\tau)-f_1(\sigma\tau)+f_1(\sigma),\\
(\delta f_2)(\sigma,\tau,\rho)&=&
\sigma f_2(\tau,\rho)-f_2(\sigma\tau,\rho)
+f_2(\sigma,\tau\rho)-f_2(\sigma,\tau),\\
\end{array}
$$
and so on.  Note that $\delta\circ\delta=0$.
The cohomology groups are given by
  $$H^r(G,A)=\ker\delta/\im\delta\subset C^r(G,A)/\im\delta.$$
Cocycles are elements of the kernel of~$\delta$, and coboundaries are elements
 of the image of~$\delta$.
We have
$$\begin{array}{lll}
H^0(G,A)&=&A^G,\\
H^1(G,A)&=&\frac{\text{crossed-homomorphisms}}
{\text{principal crossed-homomorphisms}}\\
&=&\Hom(G,A), \text{ if action is trivial},\\
H^2(G,A)&=&\text{classes of ``factor sets''}.\\
\end{array}
$$

The groups $H^2(G,A)$ and $H^1(G,A)$ arise in many situations.
Perhaps the simplest is their connection with group extensions
and their automorphisms. 
Given a $G$-module~$A$, suppose~${\mathcal G}$ 
is a group extension of $G$ by $A$, that is,
$\mathcal G$ is a group which contains~$A$ as 
a normal subgroup such that ${\mathcal G}/A\cong G$,
where the given action of~$G$ on~$A$ is the same as the
conjugation action induced by this isomorphism.
Construct a $2$-cocycle  $a_{\sigma,\tau}$ as follows.
For each element $\sigma\in G$, let
$u_\sigma\in{\mathcal G}$ be a coset representative 
corresponding to~$\sigma$.
Then
${\mathcal G} = \coprod_{\sigma} Au_\sigma,$
i.e., 
every element of $\mathcal G$
is uniquely of the form $au_{\sigma}$.
Thus
$$u_\sigma u_\tau = a_{\sigma,\tau}u_{\sigma\tau}$$
for some $a_{\sigma,\tau}\in A$.  The
map $(\sigma,\tau)\mapsto a_{\sigma,\tau}$
is a $2$-cocycle.


\begin{exercise} 
Using the associative law, check that $a_{\sigma,\tau}$ is a
$2$-cocycle, and if~${\mathcal G}'$ is another extension of~$G$ by~$A$,
then there is an isomorphism ${\mathcal G}'\cong{\mathcal G}$ that
induces the identity on~$A$ and~$G$ if and only if the corresponding
$2$-cocycles differ by a coboundary.
\end{exercise}

\begin{exercise} 
Conversely, show that every 2-cocycle arises in this way.
For example, in the trival case, if $a_{\sigma,\tau}=1$
for every~$\sigma$ and~$\tau$,
then we can take~${\mathcal G}$ to be the semidirect product
$G\ltimes A$.
\end{exercise}

Therefore we may view $H^2(G,A)$ as the group of isomorphism classes of 
extensions of~$G$ by~$A$ with a given action of~$G$ on~$A$.
% (Note our terminology: we call~$\G$ an
% extension of~$G$ by~$A$ rather than an extension of~$A$ 
% by~$G$.)

\begin{exercise}
Show that an automorphism of~$\G$ that induces the identity on~$A$
and on $G=\G/A$ is of the form $au_\sigma\mapsto ab_\sigma u_\sigma$
with $\sigma\mapsto b_\sigma$ a $1$-cocycle, and it is an inner
automorphism induced by an element of~$A$ if and only 
if $\sigma\mapsto b_\sigma$ is a coboundary.
\end{exercise}

\subsection{Examples}
\label{brau}
Given a finite Galois extension $K/k$, and
a commutative algebraic group~$C$ over~$k$, the
following notation is frequently used:
     $$H^r(K/k,C):=H^r(\Gal(K/k),C(K)).$$
We have $H^0(K/k,C)=C(k)$ as above, and
\begin{align*}
H^1(K/k,{\bG}_m)&=H^1(K/k,K^*)=0,\\
H^2(K/k,{\bG}_m)&=\Br(K/k)\subset \Br(k)
\end{align*}
The first equality is Hilbert's Theorem 90.
In the second equality  $\Br(k)$ is the Brauer group of~$k$;
this is the 
group of equivalence classes of central simple algebras with
center~$k$ that are finite dimensional over~$k$;
two such algebras are equivalent if they are matrix
algebras over $k$-isomorphic division algebras.

The map from $H^2(K/k,{\bG}_m)$ to 
$\Br(K/k)$ is defined as follows.  
Given a $2$-cocycle $a_{\sigma,\tau}$, 
define a central simple algebra over~$k$ by
${\mathcal A}=\oplus K {u_\sigma}$, which is a 
vector spaces over~$K$ with a basis 
$\{u_\sigma\}$ indexed by the elements $\sigma\in{}G$.  
Multiplication is defined by the same rules
as for group extensions above (with $A=K^*$), 
extended linearly.

\subsection{Characterization of $H^r(G,-)$}
\label{subsec:char}
For fixed $G$ and varying $A$ the groups $H^r(G,A)$ have the
following fundamental properties:
\begin{enumerate}
\item $H^0(G,A)=A^G$.
\item $H^r(G,-)$ is a functor 
     $$\left\{G\text{-modules}\right\} 
           \rightarrow \left\{\text{abelian groups}\right\}.$$
\item Each short exact sequence
   $$0\rightarrow A' \rightarrow A\rightarrow A''\rightarrow 0$$
   gives rise to connecting homomorphisms (see below)
    $$\delta:H^r(G,A'')\rightarrow H^{r+1}(G,A')$$
   from which we get a long exact sequence of cohomology groups,
   functorial in short exact sequences in the natural sense. 
\item If~$A$ is ``induced'' or ``injective'', then  
      $H^r(G,A)=0$ for all $r\not=0$. 
\end{enumerate}
These properties characterize the sequence of functors
$H^i$ equipped with the
$\delta$'s uniquely, up to unique isomorphism.

For $c\in H^r(G,A'')$, define $\delta(c)$ as follows.
Let $c_1:G^r\rightarrow A''$ be a cocycle representing~$c$.
Lift~$c_1$ to any map (cochain) 
$c_2:G^r\rightarrow A$.
Since $\delta(c_1)=0$, the map $\delta(c_2):G^{r+1}\rightarrow A$ has
image in $A'$, so defines a map
$\delta(c_2):G^{r+1}\rightarrow A'$,
and thus represents 
a class $\delta(c)\in{}H^{r+1}(G,A')$.
 
For an infinite Galois extension, one uses cocycles that come by
inflation from finite Galois subextensions.
This amounts to using continuous cochains, where continuous means
with respect to  the Krull topology on~$G$ 
and the discrete topology on~$A$.

Abstracting this situation leads to the notion of the cohomology
of a profinite group~$G$ (i.e., a projective limit, in the category of
topological groups, of finite groups~$G_i$) operating continuously on a
discrete module~$A$. Without loss of generality the~$G_i$ can be taken to
be the quotients~$G/U$ of~$G$ by its open normal subgroups~$U$, and 
then~$A$ is the union of its subgroups $A^U$. The cohomology groups 
$H^r(G,A)$ computed with continuous cochains are direct limits, relative 
to the inflation maps (see Section~\ref{sec:infres}), 
of the cohomology groups $H^r(G/U,A^U)$
of the finite quotients, because the continuous cochain complex
$C^*(G,A)$ is the direct limit of the complexes $C^*(G/U,A^U)$. Also, it is
easy to see that the groups $\{H^r(G,-)\}_r$ are characterized by
$\delta$-functoriality on the category of {\em discrete} $G$-modules.


\section{Kummer theory}
Let $k^{\sep}$ be a separable closure of a field~$k$,
and put $G_k=\Gal(k^{\sep}/k)$.
Let $m\ge 1$ be an integer, and assume that the image
of~$m$ in~$k$ is nonzero.
Associated to the exact sequence
$$0\longrightarrow \mu_m\longrightarrow (k^{\sep})^{*}\stackrel{m}{\longrightarrow}
(k^{\sep})^{*}\longrightarrow 0,$$
we have a long exact sequence
$$\xymatrix{
0\ar[r]
&{\mu_m\cap k}
\ar[r]
& k^{*}
\ar[r]^m
&k^{*}
\ar`r[d]`d[]`[dlll]`d[][dll]\\
&H^1(G_k,\mu_m)\ar[r]
&H^1(G_k,(k^{\sep})^{*})=0,&{}\\
}
$$
where the last equality is
by Hilbert's Theorem 90. 
Thus $H^1(G_k,\mu_m)\isom k^*/(k^*)^m.$

Now assume that the group of $m$th roots of unity $\mu_m$ is contained
in~$k$. Then
\begin{align*}
H^1(G_k,\mu_m)&=\Hom_{\cont}(G_k,\mu_m),\\
\intertext{so}
k^{*}/(k^{*})^m&\cong \Hom_{\cont}(G_k,\mu_m).
\end{align*}
Using duality, this isomorphism describes the finite abelian 
extensions of~$k$ whose Galois group is killed by~$m$.
For example, consider a Galois extension $K/k$ such that 
$G=\Gal(K/k)$ 
is a finite abelian group that is killed by~$m$.  
Since~$G$ is a quotient of $G_k=\Gal(k^{\sep}/k)$,
we have a diagram
$$\xymatrix{
k^*/(k^{*})^m 
\ar[rr]^{\cong}&&\Hom_{\cont}(G_k,\mu_m)\\
B\ar[rr]^{\cong} \ar[u]
  &&{\widehat{G}:=\Hom(G,\mu_m),} \ar[u]\\
}$$
where~$B$ is the subgroup of $k^*/(k^{*})^m$
corresponding to $\widehat{G}$ under the isomorphism.
\begin{exercise}
Show that
$$K=k(\sqrt[m]{B})=k(\{\sqrt[m]{b} \mid b\in B\}),$$
and $[K:k]=\# B$.
\end{exercise}
The case when~$G$ cyclic is the crucial step in showing that
a polynomial with solvable Galois group can be solved by 
radicals.

For the rest of this section, we assume that~$k$ is a number field
and continue to assume that~$k$ contains~$\mu_m$.
Let~$S$ be a finite set of primes of~$k$ including all divisors
of~$m$ and large enough so that the ring $\O_S$ of $S$-integers
of~$k$ is a principal ideal ring.  

\begin{exercise}
Show that the extension
$K(\sqrt[m]{B})$ above is unramified outside~$S$ if and
only if $B\subset U_S {k^*}^m/{k^*}^m \isom U_S/U_S^m$,
where $U_S=\O_S^*$ is the
group of $S$-units of~$k$. 
\end{exercise}

\begin{exercise}\label{ex:finite}
Let $k_S$ be the maximal extension of~$k$ which is unramified
outside~$S$, and let $G_S=\Gal(k_S/k)$. Then 
$\Hom_{\cont}(G_S,\mu_m)=U_S/U_S^m$. 
It follows that $\Hom_{\cont}(G_S,\mu_m)$ is finite,
because $U_S$ is finitely generated. 
\end{exercise}

Now let~$E$ be an elliptic curve over~$k$.
The $m$-torsion points of~$E$ over~$\overline{k}$ form a group
$E_m=E_m(\overline{k})\ncisom (\Z/m\Z)^2.$  Suppose
that, in addition to the conditions above,~$S$ also contain
the places at which~$E$ has bad reduction.  Then it is
a fact that $E(k_S)$ is divisible by~$m$, so we have an exact 
sequence
 $$0 \ra E_m \ra E(k_S) \xrightarrow{m} E(k_S)\ra 0.$$
 Taking cohomology we obtain an exact sequence
 $$0 \ra E(k)/m E(k) \ra H^1(k_S/k,E_m)
    \ra H^1(k_S/k,E)_m\ra0,$$
where the subscript~$m$ means elements killed by~$m$. 
Thus, to prove that $E(k)/m E(k)$ is finite (the 
``weak Mordell-Weil theorem''), it suffices to show that 
$H^1(k_S/k,E_m)$ is finite.  Let $k'=k(E_m)$ be the extension of~$k$
obtained by adjoining the coordinates of the points of order~$m$.  Then
$k'/k$ is finite and unramified outside~$S$.  
Hence $H^1(k'/k,E_m)$ is finite, and the exact
inflation-restriction sequence (see Section~\ref{sec:infres})
$$0 \ra H^1(k'/k,E_m)\ra H^1(k_S/k,E_m)
         \ra H^1(k_S/k',E_m)$$
shows that it suffices
to prove $H^1(k_S/k,E_m)$ is
finite when $k=k'$.  But then
$$H^1(k_S/k,E_m) \isom \Hom_{\cont}(G_S,E_m)
                 \isom \Hom_{\cont}(G_S,\mu_m)^2$$
is finite by Exercise~\ref{ex:finite}.

\begin{exercise}
Take $k=\Q$ and let~$E$ be the elliptic curve $y^2=x^3-x$.
Let $m=2$ and $S=\{2\}$, $U_S=\langle -1, 2\rangle$, and
show $(E(\Q):2 E(\Q))\leq 16$. (In fact, $E(\Q)=E_2$ is of order~$4$, 
killed by~$2$, but to show that we need to
examine what happens over~$\R$ and over~$\Q_2$, not just use the
lack of ramification at the other places.)
\end{exercise}

\begin{exercise}
Suppose $S'=S\union\{P_1,P_2,\ldots,P_t\}$ is obtained
by adding~$t$ new primes to~$S$.  Then
$U_{S'}\isom U_S\cross\Z^t$.  Hence
$H^1(k_{S'}/k,E)_m\isom H^1(k_S/k,E)\cross(\Z/m\Z)^{2t}$.
Hence $H^1(k,E)$ contains an infinite number of independent elements of
order~$m$.  Hilbert Theorem 90 is far from true for~$E$.
\end{exercise}

\section{Functor of pairs $(G,A)$}
A {\em morphism of pairs}
$(G,A)\mapsto(G',A')$
is given by a pair of maps~$\phi$ and~$f$,
$$\xymatrix{
G&G'\ar[l]_{\phi}}\>\>\>\text{and}\>\>\>
\xymatrix{
A_\phi\ar[r]^{f}&A'},
$$
where~$\phi$ is a group homomorphism, and~$f$ is a homomorphism of
$G'$-modules, and
$A_\phi$ means~$A$ with the~$G'$ action induced by~$\phi$.
A morphism of pairs induces a map
$$H^r(G,A)\ra H^r(G',A')$$
got by composing the map
$H^r(G,A)\rightarrow H^r(G',A_{\phi})$
induced by~$\phi$
with the map
$H^r(G',A_\phi)\ra H^r(G',A')$
induced by~$f$.
We thus consider $H^r(G,A)$ as a functor of pairs $(G,A)$.

If~$G'$ is a subgroup of~$G$ then there are maps
$$\xymatrix{
H^r(G,A)\ar@/^/[rrr]^{\text{restriction}}
&&&H^r(G',A).\ar@/^/[lll]^{\text{corestriction}}
}$$
Here the corestriction map (also called the ``transfer map'')
is defined only if the index $[G:G']$ is finite.

When $r=0$ the corestriction map is the trace or norm:
$$\xymatrix{
K\ar@{-}[dr]^{G'}\ar[dd]^G\\
&K'\ar@{-}[dl]\\
k}\>\>\phantom{bigspace}\>\>
\xymatrix{ 
{A^G}\ar[r]^{\res} & {A^{G'}}\ar@/^/[l]^{\cores}\\
{\displaystyle{\sum_{g\in\{\text{coset reps for $G/G'$}\}} ga}} 
     &{\begin{array}{c}a.\\ {}\\ \end{array}}\ar@{|->}@<-1ex>[l]\\
}$$

\begin{corollary}  
If~$G$ is of finite cardinality~$m$, then
$$mH^r(G,A)=0 \text{ for } r\not=0.$$
\end{corollary}
\begin{proof}
Letting $G'=\{1\}$, 
we have
$$\text{(corestriction)}\circ\text{(restriction)} = [G:G'] = [G:\{1\}]=m.$$
Since
$H^r(\{1\},A)=0$ for $r\neq 0$, 
this composition is~$0$, as claimed.
\end{proof}

\begin{exercise}
Restriction to a $p$-Sylow subgroup is injective
on the $p$-primary component of $H^r(G,A)$.
\end{exercise}

\section{The Shafarevich group}
Let~$k$ be a number field,~$\nu$ a place of~$k$, 
and $k_\nu$ the completion of~$k$ at $\nu$. 
Let $\knubar$ be an algebraic closure of $k_\nu$ and
let~$\kbar$ be the algebraic closure of~$k$ in~$\knubar$.
These four fields are illustrated in the following diagram.
$$\xymatrix{
&{\overline{k}_\nu}\ar@{-}[dr]^{G_\nu}\ar@{-}[dl]\\
{\overline{k}} && {k_\nu}\\
&{k}\ar@{-}[ul]^{G_k}\ar@{-}[ur]\\
}$$
Let $E$ be an elliptic curve over $k$.  We have
natural morphisms of pairs
$$(G_k,E(\kbar))\ra(G_\nu,E(\knubar)),$$ 
for each place~$\nu$,
hence a homomorphism
$$H^1(k,E)\rightarrow \prod_\nu H^1(k_\nu, E),$$
where the product is taken over all places of~$k$.
The kernel of this map is the Shafarevich group
$\Sha(k,E)$, which is conjectured
to be finite.

If you can prove that~$\Sha$ is finite, then you will be famous,
and you will have shown that the descent algorithm to compute
the Mordell-Weil group, which seems to work in practice, will always work. 
Until 1986, there was no single instance where it was known that~$\Sha$ 
was finite!  Now much is known for $k=\Q$ if the
rank of $E(\Q)$ is~$0$ or~$1$; see ~\cite{kolyvagin} and~\cite{rubin} for 
results in this direction.
Almost nothing is known for higher ranks. 

\section{The inflation-restriction sequence}
\label{sec:infres}
Recall that a morphism of pairs 
$$(G,A)\rightarrow (G',A')$$
is a map $G'\rightarrow G$
and a $G'$-homomorphism $A\rightarrow A'$,
where $G'$ acts on~$A$ via $G'\rightarrow G$.
In particular, we can take~$G'$ to be a subgroup~$H$ of~$G$.
Here are three special instances of the above map:
$$\begin{array}{lll}
1)&\text{restriction}& H^r(G,A)\rightarrow H^r(H,A)\\
2)&\text{inflation}& H^r(G/H,A^H)\rightarrow H^r(G,A)\\
  &&(\text{for } H\triangleleft G,\> G\rightarrow G/H,\> A^H\subset A) \\
3)&\text{conjugation}&  H^r(H,A)\stackrel{\tilde{\sigma}}{\rightarrow}
 H^r(\sigma H\sigma^{-1},A), \>\sigma\in G\\
  &&(\text{for } \sigma h\sigma^{-1}\mapsto h\text{ and }
          a \mapsto \sigma a)\\
\end{array}$$

\begin{theorem}
\label{conjth}
If $\sigma\in H$, then the conjugation map $\tilde{\sigma}$ is 
the identity.
\end{theorem}

\begin{exercise}
Given a commutative algebraic group $C$ defined over $k$ one sometimes
uses the notation $H^r(k,C):=H^r(k^{\sep}/k,C)$, where $k$
is a separable algebraic closure of $k$.  Show that this makes
sense, in the sense that if $k_1^s$ and $k_2^s$ are two separable closures of $k$,
then the isomorphism 
   $H^r(\Gal(k_1^s/k),C(k_1^s))\isom 
       H^r(\Gal(k_2^s/k),C(k_2^s))$
induced by a $k$-isomorphism $\vphi:k_1^s\ra k_2^s$
is independent of the choice of $\vphi$. 
\end{exercise}

\begin{theorem}
If~$H$ is a normal subgroup of~$G$, then
there is a ``Hochschild-Serre'' spectral sequence
    $$E_2^{rs}=H^r(G/H,H^s(H,A))\Rightarrow H^{r+s}(G,A)$$
\end{theorem}
By Theorem~\ref{conjth},~$G$ acts on $H^r(H,A)$ and~$H$ 
acts trivially, so this makes sense.
(The profinite case follows immediately from the
 finite one by direct limit; cf. the end of Section~\ref{subsec:char}.)
The low dimensional corner of the spectral sequence 
can be pictured as follows.
$$\xymatrix{
&{E^{02}}\ar@{-}[d]\ar@{-}[dr]\\
&E^{01}\ar@{-}[d]\ar@{-}[r]\ar@{-}[dr]&E^{11}\ar@{-}[d]\ar@{-}[dr]&{}\\
{}&E^{00}\ar@{-}[r]&E^{10}\ar@{-}[r]&{E^{20}}\\
}$$
Inflation and restriction are ``edge homomorphisms'' in the
spectral sequence.
The lower left corner pictured above gives the obvious isomorphism 
$A^G\isom (A^H)^{G/H}$, and the exact sequence
$$\xymatrix{
0\ar[r]
&H^1(G/H,A^H)\ar[r]^{\infff}
&H^1(G,A)\ar[r]^{\res}
&H^1(H,A)^{G/H}\ar`r[d]`d[]`[dlll]_d`d[][dll]\\
&H^2(G/H,A^H)\ar[r]^{\inf}
&H^2(G,A).&{}\\
}
$$
The map~$d$ is the ``transgression'' and is induced by
$d_2:E_2^{01}\ra E_2^{20}$.

\begin{exercise}\mbox{}\vspace{-3ex}\newline
\begin{enumerate}
\item Show that this last sequence, or at least the first line,
is exact by using standard $1$-cocycles.  

\item If $H^1(H,A)=0$, so that $E_2^{r1}=0$ for all~$r$, then
the sequence obtained by increasing the superscripts on the
$H$'s by~$1$ is exact.
\end{enumerate}
\end{exercise}

Consider a subfield~$K$ of $k^{\sep}$ that is Galois over~$k$, and let~$C$ 
be a commutative algebraic group over~$k$.
$$\xymatrix{        
&{k^{\sep}}\ar@{-}[d]\ar@{-}@/^1pc/[d]^{G_K}\ar@{-}@/_1pc/[dd]_{G_k}\\
&{K}\ar@{-}[d]\\
{C}\ar@{-}[r]&{k.}\\
}
$$
The inflation-restriction sequence is
$$\xymatrix{
0\ar[r]
&H^1(K/k,C(K))\ar[r]
&H^1(k,C(k^{\sep}))\ar[r]
&H^1(K,C(k^{\sep}))^{\Gal(K/k)}\ar`r[d]`d[]`[dlll]`d[][dll]\\
&H^2(K/k,C(K))\ar[r]&H^2(k,C(k^{\sep})).
&{}\\
}
$$
If $C={\bG}_m$, then 
$H^1(K,C(k^{\sep}))=0$, and there is an inflation-restriction
sequence with $(1,2)$ replaced by $(2,3)$:
$$\xymatrix{
0\ar[r]
&H^2(K/k,K^*)\ar[r]
&H^2(k,(k^{\sep})^*))\ar[r]
&H^2(K,(k^{\sep})^*))^{\Gal(K/k)}\ar`r[d]`d[]`[dlll]`d[][dll]\\
&H^3(K/k,K^*)\ar[r] &{H^3(k,(k^{\sep})^*)}.&{}\\
}
$$
An element $\alpha\in H^2(K,(k^{\sep})^*)^{\Gal(K/k)}$ 
represents a central simple algebra~$A$ over~$K$ 
which is isomorphic to all of its conjugates by $\Gal(K/k)$.  As
the diagram indicates, the image $\alpha$ in $H^3(K/k,K^*)$ is the
``obstruction'' whose vanishing is the necessary and sufficient
condition for such an algebra~$A$ to come by base extension from an
algebra over~$k$.

\section{Cup products}

\subsection{$G$-pairing}

If $A$, $A'$, and~$B$ are $G$-modules, then
$$A\times A'\stackrel{b}{\rightarrow}B$$
is a \emp{$G$-pairing} 
if it is bi-additive, and respects the action of $G$:
$$b(\sigma a,\sigma a')=\sigma b(a,a').$$
Such a pairing induces a map $\tilde{b}$
$$\cup:H^r(G,A)\times H^s(G,A')\stackrel{\tilde{b}}{\rightarrow}
H^{r+s}(G,B),$$
as follows:
given cochains~$f$ and~$f'$, one defines 
(for a given $b$) a cochain $f \cup f'$ by
$$(f\cup f')(\sigma_1,\dots,\sigma_{r+s})=
b(f(\sigma_1,\dots,\sigma_r),\sigma_1\dots\sigma_r
    f'(\sigma_{r+1},\dots,\sigma_{r+s})),$$
and checks the rule
$$\delta(f\cup f')=\delta f\cup f' + (-1)^rf\cup \delta f'.$$
If $\delta f=\delta f'=0$, then also 
$\delta(f\cup f')=0$; i.e., if~$f$ and~$f'$ are cocycles, so 
is $f\cup f'$.  
Similarly one checks that the cohomology class of $f\cup f'$
depends only on the classes of~$f$ and~$f'$.  
Thus we obtain the desired pairing $\tilde{b}$.

If $r=0$ and $a\in A^G$ is fixed, then $a'\mapsto b(a,a')$
defines a $G$-homomorphism $\vphi_a:A'\ra B$,
and  $\alpha'\mapsto a\cup \alpha'$ is the map
$H^r(G,A')\ra H^r(G,B)$ induced by $\vphi_a$. 

If~$H$ is a subgroup of~$G$, and
$\alpha\in H^r(G,A)$ and $\beta\in H^s(H,A')$, 
then we can form
   $$\res (\alpha) \cup \beta\in H^{r+s}(H,B).$$
Suppose that the index of~$H$ in~$G$ is finite, so that 
corestriction is defined; then one can show that
$$\cores(\res (\alpha) \cup \beta)=
\alpha \cup \cores (\beta)\in H^{r+s}(G,B).$$

\subsection{Duality for finite modules}
If~$A$ and~$B$ are $G$-modules, we make the group
$\Hom_\Z(A,B)$ into a $G$-module by
defining $(\sigma f)(a)=\sigma(f(\sigma^{-1}a)).$
Note then that $\Hom_G(A,B)=(\Hom_\Z(A,B))^G$. Also,
the obvious pairing $A\cross \Hom_\Z(A,B)\ra B$
is a $G$-pairing.  The canonical map
% I CAN'T REMEMBER THE RIGHT WAY TO DO THIS!
 $$(*)\qquad\qquad\qquad\qquad\qquad    
       A \ra \Hom_\Z(\Hom_\Z(A,B))\hspace{1.6in}$$
is a $G$-homomorphism.  In case $A$ is finite, killed by~$m$,
and~$B$ has a unique cyclic subgroup of order~$m$,
the map ($*$) is an isomorphism; one can thus recover~$A$ 
from its ``dual'' $\Hom_\Z(A,B)$ which has the same
order as~$A$.
There are two especially important such duals for finite~$A$.
\begin{itemize}
\item The \emp{Pontrjagin Dual} of~$A$ is
     $\Hom_\Z(A,\Q/\Z)$; this equals $\Hom_\Z(A,\Z/m\Z)$ if $mA=0$.
\item The \emp{Cartier Dual} of~$A$ is 
    $\Hom_\Z(A,\mu(k^{\sep}))$; this equals $\Hom_\Z(A,\mu_m(k^{\sep}))$
    if $mA=0$.
\end{itemize}

In the Pontrjagin case,~$G$ is an arbitrary profinite group
and acts trivially on $\Q/\Z$.  Taking limits, this duality extends to a 
perfect duality (i.e., an anti-equivalence of categories) between 
discrete abelian torsion groups and profinite abelian groups.

In the Cartier case, $G=\Gal(k^{\sep}/k)$ or some quotient thereof,
and $m\neq 0$ in~$k$. (The Cartier dual of a $p$-group in characteristic~$p$ 
is a \emp{group scheme}, not just a Galois module.)
If~$E$ is an elliptic curve over~$k$ and the image of~$m$ in~$k$ is nonzero, the
\emp{Weil pairing} $E_m(k^{\sep})\cross E_m(k^{\sep})\ra\mu_m$
identifies $E_m$ with its Cartier dual. 


\section{Local fields}

Let~$k$ be a local field, i.e., the field of fractions
of a complete discrete valuation ring
with finite residue field~$F$.  Let~$K$
be a finite extension of~$k$. 

Fundamental facts:
\begin{align*}
H^1(K/k,K^{*})&=0 \qquad \text{(Hilbert's Theorem 90)}\\
H^2(K/k,K^{*})&={\Z}/[K:k]{\Z}\\
H^2(k,{\bG}_m)&=\Br(k)={\Q}/{\Z} \\
\end{align*}

The equality $\Br(k)=\Q/{\Z}$
is given canonically, by the Hasse invariant, as follows:
The group $\Br(k)$ is the Brauer group, defined in \S\ref{brau}.
Consider the inflation-restriction sequence for 
$H^2(-,\Gm)$ in the tower of fields
$$\xymatrix@=1.2pc{{\kbar}\ar@{-}[d]\\
            {k^{\ur}}\ar@{-}[d]\\
            {k}}$$
where $k^{\ur}$ is the maximal unramified extension of~$k$.
Since every central division algebra over a local field
has an unramified splitting field, we have
$\Br(k^{\ur})=0$, and hence an isomorphism
$$\Br(k) \isom H^2(k^{\ur}/k,\Gm)
    =H^2(\Frob^{\widehat{\Z}},(k^{\ur})^{*}).$$
Using the exact sequence
$$\xymatrix{
0\ar[r]
& U(k^{\ur})\ar[r]
&(k^{\ur})^*\ar[rr]^{\text{valuation}}&{}
&{\Z}\ar[r] &0}$$
and the fact that the unit group of an unramified extension has
trivial cohomology in dimension $\neq 0$,
we find that we can replace
$(k^{\ur})^*$ by $\Z$, and hence
$$\Br(k)\isom H^2(\hat{\Z},\Z)
    =H^1(\hat{\Z},\Q/\Z)=\Q/\Z;$$
the middle equality comes from the short exact sequence
  $$ 0 \ra \Z \ra \Q \ra \Q/\Z\ra 0$$
and the fact that~$\Q$, being uniquely divisible,
has trivial cohomology in nonzero dimensions.
The resulting map 
    $$\Br(k)\rightarrow{\Q}/{\Z}$$
is the called the Hasse invariant.

\begin{theorem}\label{thmdual}
Let~$A$ be a finite $G_k$-module of
order prime to the characteristic of~$k$.  Let
$$A^*=\Hom(A,{\bG}_m)=\Hom(A,\mu(\overline{k}))$$
be the Cartier dual of~$A$. 
Then the $G$-pairing
$$A\times A^*\rightarrow \overline{k}^{*}$$
induces a pairing
$$H^r(k,A)\times
H^{2-r}(k,A^*)\rightarrow
H^2(G_k,(k^{\sep})^{*})=\Br(k)={\Q}/{\Z}.$$
This is a perfect pairing of finite groups, for
all $r\in\Z$.  It is nontrivial only if $r=0,1,2$, since
for $r\ge 3$,
$$H^r(k,A)=0\>\>\text{ for all }A,$$
i.e., ``the cohomological dimension of a 
non-archimedean local field is~$2$.''
\end{theorem}

\emp{Example.} By Kummer theory, we have
$$k^{*}/(k^{*})^m=H^1(k,\mu_m(\overline{k})).$$
Thus there is a perfect pairing
$$\xymatrix@=.3pc{
H^1(k,{\Z}/m{\Z})
  &{\times }
  &H^1(k,\mu_m({\overline{k}}))\ar[rrr]
                &&&{\Q}/{\Z}\\
\\
\\
{\Hom(G_k,{\Z}/m{\Z})}\ar@{=}[uuu]
    &{\times} & {k^{*}/(k^{*})^m}\ar@{=}[uuu]\\
}$$
The left hand equality is because the action is trivial.
Conclusion:  
$$G_k^{\ab}/(G_k^{\ab})^m \isom k^*/(k^*)^m.$$


Taking the limit gives 
\emp{Artin reciprocity}:
$$\xymatrix{
k^{*}\ar@{^{(}->}[r]&G_k^{ab}
};
$$
the image is dense.

Let $E/k$ be an elliptic curve.
In some sense, 
$$E=\Ext^1(E,{\bG}_m)$$
in the category of algebraic groups.
There is a pairing 
$$H^r(k,E)\times H^s(k,E)\rightarrow H^{r+s+1}(k,{\bG}_m).$$
For example, taking $r=0$ and $s=1$, we have the following theorem.
\begin{theorem}\label{thmpontrjagin}
Let~$E$ be an elliptic curve over a non-archimedean local field~$k$,
then we have the following
perfect pairing between Pontrjagin duals.
$$\xymatrix@=.3pc{
H^0(k,E)&\times &H^1(k,E)\ar[rrr] &&&{H^2(k,\bG_m)={\Q}/{\Z}}\\
\\
\\
         {\begin{array}{c}E(k)\\ \text{profinite}\end{array}}\ar@{=}[uuu]    
        &{\begin{array}{c} \times \\ \end{array}}
&{\begin{array}{c} H^1(k,E) \\ 
 \text{discrete, torsion}\end{array}}\ar@{=}[uuu]
}
$$
\end{theorem}
\begin{proof}[Sketch of Proof]
We use the Weil pairing.
Letting~$D$ denote ``Pontrjagin dual'',
we have a diagram
$$\xymatrix{
0\ar[r]&E(k)/mE(k)\ar[r]\ar[d]&H^1(k,E_m)\ar[r]\ar[d]&H^1(k,E)_m\ar[r]\ar[d]&0\\
0\ar[r]&H^1(k,E)_m^D\ar[r]&H^1(k,E_m)^D\ar[r]
 &(E(k)/mE(k))^D\ar[r]&0}$$
The rows are exact.  The top one from the Kummer sequence, and the
bottom is the dual of the top one.  The middle vertical
arrow is an isomorphism by Theorem~\ref{thmdual}.
The outside vertical arrows are induced by the pairing
of Theorem~\ref{thmpontrjagin}. 
The diagram commutes, so they are also isomorphisms, 
and Theorem~\ref{thmpontrjagin} follows by passage
to the limit with more and more divisible~$m$.
\end{proof}

It was in trying to prove Theorem~\ref{thmpontrjagin} that I was
led to Theorem~\ref{thmdual} in the late 1950's.  Of course
the ``fundamental facts'' and the Artin isomorphism are a 
much older story.

\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\begin{thebibliography}{1}

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\bibitem{serre:gc}
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\end{thebibliography}

\end{document}