Sharedwww / tables / Notes / tate-pcmi.texOpen in CoCalc
Author: William A. Stein
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% Galois Cohomology: tate.tex
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\begin{document}
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\part*{Galois Cohomology}
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\pauth{John Tate}
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\setcounter{tocdepth}{2} % all the way to subsections
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\tableofcontents
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\mainmatter
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\setcounter{page}{1}
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\LogoOn
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\lectureseries[Galois Cohomology]{Galois Cohomology}
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% The ^{1} is a ``fake'' footnote indicator. It's there because
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\auth[J. Tate]{John Tate$\mbox{}^{1}$}
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\address{Department of Mathematics; Austin, TX}
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\email{tate@math.utexas.edu}
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%The following items will become first page footnotes; they are optional.
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\subjclass{11}
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\keywords{Galois cohomology, Elliptic curves}
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%\date{August 22, 1999}
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\setaddress
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%\lecture{Galois cohomology}
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I thank Helena Verrill and William Stein for their help in
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getting this account of my talks at Park City into print. After
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Helena typed up her original notes of the talks, William was a
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great help with the editing, and put them in the canonical format
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for this volume.
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The somewhat inefficient organization of this account is
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mainly a result of the fact that, after the first talk had been
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given with the idea that it was to be the only one, a second was
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later scheduled, and these are the notes of the material in the
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two talks in the order it was presented.
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The bible for this subject is Serre~\cite{serre:gc}, in
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conjunction with~\cite{serre:local} or \cite{cassels-frohlich}.
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Haberland~\cite{haberland} is also an excellent reference.
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\section{Group modules}
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Consider a group~$G$ and an abelian group~$A$
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equipped with a map
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$$G\times A\rightarrow A,$$
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$$(\sigma,a)\mapsto \sigma a.$$
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We use notation $\sigma,\,\tau,\,\rho,\dots$
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for elements of~$G$, and $a,\,b,\,a',\,b',\dots$ for elements of~$A$.
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To say that~$A$ is a \emp{$G$-set} means that
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$$\tau(\sigma a) = (\tau\sigma)a \quad\text{and}\quad 1a=a,$$
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for all $\sigma, \tau\in G$ and $a\in A$,
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where~$1$ is the identity in~$G$.
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To say that~$A$ is a \emp{$G$-module} means that, in addition, we
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have
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$$\sigma(a+b)=\sigma a + \sigma b,$$
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for all $\sigma\in G$ and $a,b\in A$.
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This is all equivalent to giving~$A$ the structure
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of ${\Z}[G]$-module.
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Given a $G$-module~$A$ as above, the subgroup of fixed elements of~$A$ is
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$$A^G:=\left\{ a\in A \mid \sigma a = a \text{ for all } \sigma \in G\right\}.$$
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We say~$G$ \emp{acts trivially} on~$A$ if $\sigma a=a$
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for all $a\in A$; thus
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$A^G=A$ if and only if the action is trivial.
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When ${\Z}$, ${\Q}$, ${\Q}/{\Z}$ are
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considered as $G$-modules, this is with the trivial action,
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unless stated otherwise.
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If we take $G=\Gal(K/k)$, with~$K$ a Galois
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extension of~$k$ of possibly infinite degree,
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then we have the following examples of fixed subgroups of
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$G$-modules:
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$$\begin{array}{l|l}
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A & A^G\\
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\hline
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K^+ \text{ as an additive group}&k^+\\
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K^{*} \text{ as a multiplicative group}&k^*\\
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E(K), \text{ where } E/k \text{ is an elliptic curve}&E(k).\\
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\end{array}
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$$
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The action on $E(K)$ above is given by
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$\sigma(x,y)=(\sigma x,\sigma y)$ for a point $P=(x,y)$,
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if~$E$ is given as a plane cubic.
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In general, if~$C$ is a commutative algebraic group over~$K$,
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we can take $A=C(K)$, and then $A^G=C(k)$.
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\section{Cohomology}
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We now define the cohomology groups
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$H^r(G,A)$,
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for $r\in\Z$.
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Abstractly, these are the right derived functors of the left exact functor
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$$\left\{G\text{-modules}\right\}\rightarrow\left\{\text{abelian groups}\right\}$$
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that sends $A\mapsto A^G$.
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Since $A^G=\Hom_{{\Z}[G]}({\Z},A)$,
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we have a canonical isomorphism
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$$H^r(G,A)=\Ext^r_{{\Z}[G]}({\Z},A).$$
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More concretely, the cohomology groups $H^r(G,A)$ can be computed
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using the ``standard cochain complex'' (see,
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e.g., \cite[pg.~96]{cassels-frohlich}). Let
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$$C^r(G,A):=\Maps(G^r,A);$$
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an element of $C^r(G,A)$ is a function~$f$ of~$r$ variables in~$G$,
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$$f(\sigma_1,\dots,\sigma_r)\in A,$$
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and is called an \emp{$r$-cochain}.
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(If, in addition,~$A$ and~$G$ have a topological structure, then we
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instead consider continuous cochains.)
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There is a sequence
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$$\cdots \ra 0 \ra 0\rightarrow C^0(G,A)\stackrel{\delta}{\rightarrow}
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C^1(G,A)\stackrel{\delta}{\rightarrow}
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C^2(G,A)\stackrel{\delta}{\rightarrow}\cdots$$
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Here $C^0(G,A)=A$, since an element~$f$ of $C^0(G,A)$ is given by
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the single element $f_0(\bullet)\in A$,
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its value at the unique element $\bullet\in G^0$.
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The maps~$\delta$ are defined by
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$$\begin{array}{lll}
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(\delta f_0)(\sigma)&
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=&\sigma f_0(\bullet) - f_0(\bullet),\\
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(\delta f_1)(\sigma,\tau)&=&
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\sigma f_1(\tau)-f_1(\sigma\tau)+f_1(\sigma),\\
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(\delta f_2)(\sigma,\tau,\rho)&=&
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\sigma f_2(\tau,\rho)-f_2(\sigma\tau,\rho)
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+f_2(\sigma,\tau\rho)-f_2(\sigma,\tau),\\
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\end{array}
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$$
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and so on. Note that $\delta\circ\delta=0$.
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The cohomology groups are given by
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$$H^r(G,A)=\ker\delta/\im\delta\subset C^r(G,A)/\im\delta.$$
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Cocycles are elements of the kernel of~$\delta$, and coboundaries are elements
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of the image of~$\delta$.
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We have
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$$\begin{array}{lll}
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H^0(G,A)&=&A^G,\\
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H^1(G,A)&=&\frac{\text{crossed-homomorphisms}}
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{\text{principal crossed-homomorphisms}}\\
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&=&\Hom(G,A), \text{ if action is trivial},\\
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H^2(G,A)&=&\text{classes of ``factor sets''}.\\
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\end{array}
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$$
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The groups $H^2(G,A)$ and $H^1(G,A)$ arise in many situations.
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Perhaps the simplest is their connection with group extensions
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and their automorphisms.
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Given a $G$-module~$A$, suppose~${\mathcal G}$
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is a group extension of $G$ by $A$, that is,
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$\mathcal G$ is a group which contains~$A$ as
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a normal subgroup such that ${\mathcal G}/A\cong G$,
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where the given action of~$G$ on~$A$ is the same as the
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conjugation action induced by this isomorphism.
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Construct a $2$-cocycle $a_{\sigma,\tau}$ as follows.
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For each element $\sigma\in G$, let
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$u_\sigma\in{\mathcal G}$ be a coset representative
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corresponding to~$\sigma$.
267
Then
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${\mathcal G} = \coprod_{\sigma} Au_\sigma,$
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i.e.,
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every element of $\mathcal G$
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is uniquely of the form $au_{\sigma}$.
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Thus
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$$u_\sigma u_\tau = a_{\sigma,\tau}u_{\sigma\tau}$$
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for some $a_{\sigma,\tau}\in A$. The
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map $(\sigma,\tau)\mapsto a_{\sigma,\tau}$
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is a $2$-cocycle.
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\begin{exercise}
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Using the associative law, check that $a_{\sigma,\tau}$ is a
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$2$-cocycle, and if~${\mathcal G}'$ is another extension of~$G$ by~$A$,
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then there is an isomorphism ${\mathcal G}'\cong{\mathcal G}$ that
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induces the identity on~$A$ and~$G$ if and only if the corresponding
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$2$-cocycles differ by a coboundary.
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\end{exercise}
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\begin{exercise}
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Conversely, show that every 2-cocycle arises in this way.
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For example, in the trival case, if $a_{\sigma,\tau}=1$
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for every~$\sigma$ and~$\tau$,
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then we can take~${\mathcal G}$ to be the semidirect product
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$G\ltimes A$.
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\end{exercise}
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Therefore we may view $H^2(G,A)$ as the group of isomorphism classes of
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extensions of~$G$ by~$A$ with a given action of~$G$ on~$A$.
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% (Note our terminology: we call~$\G$ an
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% extension of~$G$ by~$A$ rather than an extension of~$A$
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% by~$G$.)
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\begin{exercise}
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Show that an automorphism of~$\G$ that induces the identity on~$A$
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and on $G=\G/A$ is of the form $au_\sigma\mapsto ab_\sigma u_\sigma$
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with $\sigma\mapsto b_\sigma$ a $1$-cocycle, and it is an inner
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automorphism induced by an element of~$A$ if and only
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if $\sigma\mapsto b_\sigma$ is a coboundary.
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\end{exercise}
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\subsection{Examples}
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\label{brau}
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Given a finite Galois extension $K/k$, and
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a commutative algebraic group~$C$ over~$k$, the
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following notation is frequently used:
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$$H^r(K/k,C):=H^r(\Gal(K/k),C(K)).$$
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We have $H^0(K/k,C)=C(k)$ as above, and
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\begin{align*}
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H^1(K/k,{\bG}_m)&=H^1(K/k,K^*)=0,\\
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H^2(K/k,{\bG}_m)&=\Br(K/k)\subset \Br(k)
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\end{align*}
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The first equality is Hilbert's Theorem 90.
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In the second equality $\Br(k)$ is the Brauer group of~$k$;
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this is the
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group of equivalence classes of central simple algebras with
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center~$k$ that are finite dimensional over~$k$;
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two such algebras are equivalent if they are matrix
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algebras over $k$-isomorphic division algebras.
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The map from $H^2(K/k,{\bG}_m)$ to
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$\Br(K/k)$ is defined as follows.
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Given a $2$-cocycle $a_{\sigma,\tau}$,
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define a central simple algebra over~$k$ by
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${\mathcal A}=\oplus K {u_\sigma}$, which is a
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vector spaces over~$K$ with a basis
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$\{u_\sigma\}$ indexed by the elements $\sigma\in{}G$.
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Multiplication is defined by the same rules
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as for group extensions above (with $A=K^*$),
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extended linearly.
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\subsection{Characterization of $H^r(G,-)$}
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\label{subsec:char}
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For fixed $G$ and varying $A$ the groups $H^r(G,A)$ have the
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following fundamental properties:
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\begin{enumerate}
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\item $H^0(G,A)=A^G$.
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\item $H^r(G,-)$ is a functor
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$$\left\{G\text{-modules}\right\}
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\rightarrow \left\{\text{abelian groups}\right\}.$$
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\item Each short exact sequence
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$$0\rightarrow A' \rightarrow A\rightarrow A''\rightarrow 0$$
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gives rise to connecting homomorphisms (see below)
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$$\delta:H^r(G,A'')\rightarrow H^{r+1}(G,A')$$
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from which we get a long exact sequence of cohomology groups,
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functorial in short exact sequences in the natural sense.
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\item If~$A$ is ``induced'' or ``injective'', then
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$H^r(G,A)=0$ for all $r\not=0$.
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\end{enumerate}
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These properties characterize the sequence of functors
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$H^i$ equipped with the
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$\delta$'s uniquely, up to unique isomorphism.
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For $c\in H^r(G,A'')$, define $\delta(c)$ as follows.
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Let $c_1:G^r\rightarrow A''$ be a cocycle representing~$c$.
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Lift~$c_1$ to any map (cochain)
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$c_2:G^r\rightarrow A$.
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Since $\delta(c_1)=0$, the map $\delta(c_2):G^{r+1}\rightarrow A$ has
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image in $A'$, so defines a map
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$\delta(c_2):G^{r+1}\rightarrow A'$,
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and thus represents
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a class $\delta(c)\in{}H^{r+1}(G,A')$.
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For an infinite Galois extension, one uses cocycles that come by
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inflation from finite Galois subextensions.
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This amounts to using continuous cochains, where continuous means
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with respect to the Krull topology on~$G$
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and the discrete topology on~$A$.
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Abstracting this situation leads to the notion of the cohomology
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of a profinite group~$G$ (i.e., a projective limit, in the category of
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topological groups, of finite groups~$G_i$) operating continuously on a
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discrete module~$A$. Without loss of generality the~$G_i$ can be taken to
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be the quotients~$G/U$ of~$G$ by its open normal subgroups~$U$, and
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then~$A$ is the union of its subgroups $A^U$. The cohomology groups
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$H^r(G,A)$ computed with continuous cochains are direct limits, relative
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to the inflation maps (see Section~\ref{sec:infres}),
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of the cohomology groups $H^r(G/U,A^U)$
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of the finite quotients, because the continuous cochain complex
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$C^*(G,A)$ is the direct limit of the complexes $C^*(G/U,A^U)$. Also, it is
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easy to see that the groups $\{H^r(G,-)\}_r$ are characterized by
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$\delta$-functoriality on the category of {\em discrete} $G$-modules.
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\section{Kummer theory}
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Let $k^{\sep}$ be a separable closure of a field~$k$,
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and put $G_k=\Gal(k^{\sep}/k)$.
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Let $m\ge 1$ be an integer, and assume that the image
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of~$m$ in~$k$ is nonzero.
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Associated to the exact sequence
398
$$0\longrightarrow \mu_m\longrightarrow (k^{\sep})^{*}\stackrel{m}{\longrightarrow}
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(k^{\sep})^{*}\longrightarrow 0,$$
400
we have a long exact sequence
401
$$\xymatrix{
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0\ar[r]
403
&{\mu_m\cap k}
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\ar[r]
405
& k^{*}
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\ar[r]^m
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&k^{*}
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\ar`r[d]`d[]`[dlll]`d[][dll]\\
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&H^1(G_k,\mu_m)\ar[r]
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&H^1(G_k,(k^{\sep})^{*})=0,&{}\\
411
}
412
$$
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where the last equality is
414
by Hilbert's Theorem 90.
415
Thus $H^1(G_k,\mu_m)\isom k^*/(k^*)^m.$
416
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Now assume that the group of $m$th roots of unity $\mu_m$ is contained
418
in~$k$. Then
419
\begin{align*}
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H^1(G_k,\mu_m)&=\Hom_{\cont}(G_k,\mu_m),\\
421
\intertext{so}
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k^{*}/(k^{*})^m&\cong \Hom_{\cont}(G_k,\mu_m).
423
\end{align*}
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Using duality, this isomorphism describes the finite abelian
425
extensions of~$k$ whose Galois group is killed by~$m$.
426
For example, consider a Galois extension $K/k$ such that
427
$G=\Gal(K/k)$
428
is a finite abelian group that is killed by~$m$.
429
Since~$G$ is a quotient of $G_k=\Gal(k^{\sep}/k)$,
430
we have a diagram
431
$$\xymatrix{
432
k^*/(k^{*})^m
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\ar[rr]^{\cong}&&\Hom_{\cont}(G_k,\mu_m)\\
434
B\ar[rr]^{\cong} \ar[u]
435
&&{\widehat{G}:=\Hom(G,\mu_m),} \ar[u]\\
436
}$$
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where~$B$ is the subgroup of $k^*/(k^{*})^m$
438
corresponding to $\widehat{G}$ under the isomorphism.
439
\begin{exercise}
440
Show that
441
$$K=k(\sqrt[m]{B})=k(\{\sqrt[m]{b} \mid b\in B\}),$$
442
and $[K:k]=\# B$.
443
\end{exercise}
444
The case when~$G$ cyclic is the crucial step in showing that
445
a polynomial with solvable Galois group can be solved by
446
radicals.
447
448
For the rest of this section, we assume that~$k$ is a number field
449
and continue to assume that~$k$ contains~$\mu_m$.
450
Let~$S$ be a finite set of primes of~$k$ including all divisors
451
of~$m$ and large enough so that the ring $\O_S$ of $S$-integers
452
of~$k$ is a principal ideal ring.
453
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\begin{exercise}
455
Show that the extension
456
$K(\sqrt[m]{B})$ above is unramified outside~$S$ if and
457
only if $B\subset U_S {k^*}^m/{k^*}^m \isom U_S/U_S^m$,
458
where $U_S=\O_S^*$ is the
459
group of $S$-units of~$k$.
460
\end{exercise}
461
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\begin{exercise}\label{ex:finite}
463
Let $k_S$ be the maximal extension of~$k$ which is unramified
464
outside~$S$, and let $G_S=\Gal(k_S/k)$. Then
465
$\Hom_{\cont}(G_S,\mu_m)=U_S/U_S^m$.
466
It follows that $\Hom_{\cont}(G_S,\mu_m)$ is finite,
467
because $U_S$ is finitely generated.
468
\end{exercise}
469
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Now let~$E$ be an elliptic curve over~$k$.
471
The $m$-torsion points of~$E$ over~$\overline{k}$ form a group
472
$E_m=E_m(\overline{k})\ncisom (\Z/m\Z)^2.$ Suppose
473
that, in addition to the conditions above,~$S$ also contain
474
the places at which~$E$ has bad reduction. Then it is
475
a fact that $E(k_S)$ is divisible by~$m$, so we have an exact
476
sequence
477
$$0 \ra E_m \ra E(k_S) \xrightarrow{m} E(k_S)\ra 0.$$
478
Taking cohomology we obtain an exact sequence
479
$$0 \ra E(k)/m E(k) \ra H^1(k_S/k,E_m)
480
\ra H^1(k_S/k,E)_m\ra0,$$
481
where the subscript~$m$ means elements killed by~$m$.
482
Thus, to prove that $E(k)/m E(k)$ is finite (the
483
``weak Mordell-Weil theorem''), it suffices to show that
484
$H^1(k_S/k,E_m)$ is finite. Let $k'=k(E_m)$ be the extension of~$k$
485
obtained by adjoining the coordinates of the points of order~$m$. Then
486
$k'/k$ is finite and unramified outside~$S$.
487
Hence $H^1(k'/k,E_m)$ is finite, and the exact
488
inflation-restriction sequence (see Section~\ref{sec:infres})
489
$$0 \ra H^1(k'/k,E_m)\ra H^1(k_S/k,E_m)
490
\ra H^1(k_S/k',E_m)$$
491
shows that it suffices
492
to prove $H^1(k_S/k,E_m)$ is
493
finite when $k=k'$. But then
494
$$H^1(k_S/k,E_m) \isom \Hom_{\cont}(G_S,E_m)
495
\isom \Hom_{\cont}(G_S,\mu_m)^2$$
496
is finite by Exercise~\ref{ex:finite}.
497
498
\begin{exercise}
499
Take $k=\Q$ and let~$E$ be the elliptic curve $y^2=x^3-x$.
500
Let $m=2$ and $S=\{2\}$, $U_S=\langle -1, 2\rangle$, and
501
show $(E(\Q):2 E(\Q))\leq 16$. (In fact, $E(\Q)=E_2$ is of order~$4$,
502
killed by~$2$, but to show that we need to
503
examine what happens over~$\R$ and over~$\Q_2$, not just use the
504
lack of ramification at the other places.)
505
\end{exercise}
506
507
\begin{exercise}
508
Suppose $S'=S\union\{P_1,P_2,\ldots,P_t\}$ is obtained
509
by adding~$t$ new primes to~$S$. Then
510
$U_{S'}\isom U_S\cross\Z^t$. Hence
511
$H^1(k_{S'}/k,E)_m\isom H^1(k_S/k,E)\cross(\Z/m\Z)^{2t}$.
512
Hence $H^1(k,E)$ contains an infinite number of independent elements of
513
order~$m$. Hilbert Theorem 90 is far from true for~$E$.
514
\end{exercise}
515
516
\section{Functor of pairs $(G,A)$}
517
A {\em morphism of pairs}
518
$(G,A)\mapsto(G',A')$
519
is given by a pair of maps~$\phi$ and~$f$,
520
$$\xymatrix{
521
G&G'\ar[l]_{\phi}}\>\>\>\text{and}\>\>\>
522
\xymatrix{
523
A_\phi\ar[r]^{f}&A'},
524
$$
525
where~$\phi$ is a group homomorphism, and~$f$ is a homomorphism of
526
$G'$-modules, and
527
$A_\phi$ means~$A$ with the~$G'$ action induced by~$\phi$.
528
A morphism of pairs induces a map
529
$$H^r(G,A)\ra H^r(G',A')$$
530
got by composing the map
531
$H^r(G,A)\rightarrow H^r(G',A_{\phi})$
532
induced by~$\phi$
533
with the map
534
$H^r(G',A_\phi)\ra H^r(G',A')$
535
induced by~$f$.
536
We thus consider $H^r(G,A)$ as a functor of pairs $(G,A)$.
537
538
If~$G'$ is a subgroup of~$G$ then there are maps
539
$$\xymatrix{
540
H^r(G,A)\[email protected]/^/[rrr]^{\text{restriction}}
541
&&&H^r(G',A).\[email protected]/^/[lll]^{\text{corestriction}}
542
}$$
543
Here the corestriction map (also called the ``transfer map'')
544
is defined only if the index $[G:G']$ is finite.
545
546
When $r=0$ the corestriction map is the trace or norm:
547
$$\xymatrix{
548
K\[email protected]{-}[dr]^{G'}\ar[dd]^G\\
549
&K'\[email protected]{-}[dl]\\
550
k}\>\>\phantom{bigspace}\>\>
551
\xymatrix{
552
{A^G}\ar[r]^{\res} & {A^{G'}}\[email protected]/^/[l]^{\cores}\\
553
{\displaystyle{\sum_{g\in\{\text{coset reps for $G/G'$}\}} ga}}
554
&{\begin{array}{c}a.\\ {}\\ \end{array}}\[email protected]{|->}@<-1ex>[l]\\
555
}$$
556
557
\begin{corollary}
558
If~$G$ is of finite cardinality~$m$, then
559
$$mH^r(G,A)=0 \text{ for } r\not=0.$$
560
\end{corollary}
561
\begin{proof}
562
Letting $G'=\{1\}$,
563
we have
564
$$\text{(corestriction)}\circ\text{(restriction)} = [G:G'] = [G:\{1\}]=m.$$
565
Since
566
$H^r(\{1\},A)=0$ for $r\neq 0$,
567
this composition is~$0$, as claimed.
568
\end{proof}
569
570
\begin{exercise}
571
Restriction to a $p$-Sylow subgroup is injective
572
on the $p$-primary component of $H^r(G,A)$.
573
\end{exercise}
574
575
\section{The Shafarevich group}
576
Let~$k$ be a number field,~$\nu$ a place of~$k$,
577
and $k_\nu$ the completion of~$k$ at $\nu$.
578
Let $\knubar$ be an algebraic closure of $k_\nu$ and
579
let~$\kbar$ be the algebraic closure of~$k$ in~$\knubar$.
580
These four fields are illustrated in the following diagram.
581
$$\xymatrix{
582
&{\overline{k}_\nu}\[email protected]{-}[dr]^{G_\nu}\[email protected]{-}[dl]\\
583
{\overline{k}} && {k_\nu}\\
584
&{k}\[email protected]{-}[ul]^{G_k}\[email protected]{-}[ur]\\
585
}$$
586
Let $E$ be an elliptic curve over $k$. We have
587
natural morphisms of pairs
588
$$(G_k,E(\kbar))\ra(G_\nu,E(\knubar)),$$
589
for each place~$\nu$,
590
hence a homomorphism
591
$$H^1(k,E)\rightarrow \prod_\nu H^1(k_\nu, E),$$
592
where the product is taken over all places of~$k$.
593
The kernel of this map is the Shafarevich group
594
$\Sha(k,E)$, which is conjectured
595
to be finite.
596
597
If you can prove that~$\Sha$ is finite, then you will be famous,
598
and you will have shown that the descent algorithm to compute
599
the Mordell-Weil group, which seems to work in practice, will always work.
600
Until 1986, there was no single instance where it was known that~$\Sha$
601
was finite! Now much is known for $k=\Q$ if the
602
rank of $E(\Q)$ is~$0$ or~$1$; see ~\cite{kolyvagin} and~\cite{rubin} for
603
results in this direction.
604
Almost nothing is known for higher ranks.
605
606
\section{The inflation-restriction sequence}
607
\label{sec:infres}
608
Recall that a morphism of pairs
609
$$(G,A)\rightarrow (G',A')$$
610
is a map $G'\rightarrow G$
611
and a $G'$-homomorphism $A\rightarrow A'$,
612
where $G'$ acts on~$A$ via $G'\rightarrow G$.
613
In particular, we can take~$G'$ to be a subgroup~$H$ of~$G$.
614
Here are three special instances of the above map:
615
$$\begin{array}{lll}
616
1)&\text{restriction}& H^r(G,A)\rightarrow H^r(H,A)\\
617
2)&\text{inflation}& H^r(G/H,A^H)\rightarrow H^r(G,A)\\
618
&&(\text{for } H\triangleleft G,\> G\rightarrow G/H,\> A^H\subset A) \\
619
3)&\text{conjugation}& H^r(H,A)\stackrel{\tilde{\sigma}}{\rightarrow}
620
H^r(\sigma H\sigma^{-1},A), \>\sigma\in G\\
621
&&(\text{for } \sigma h\sigma^{-1}\mapsto h\text{ and }
622
a \mapsto \sigma a)\\
623
\end{array}$$
624
625
\begin{theorem}
626
\label{conjth}
627
If $\sigma\in H$, then the conjugation map $\tilde{\sigma}$ is
628
the identity.
629
\end{theorem}
630
631
\begin{exercise}
632
Given a commutative algebraic group $C$ defined over $k$ one sometimes
633
uses the notation $H^r(k,C):=H^r(k^{\sep}/k,C)$, where $k$
634
is a separable algebraic closure of $k$. Show that this makes
635
sense, in the sense that if $k_1^s$ and $k_2^s$ are two separable closures of $k$,
636
then the isomorphism
637
$H^r(\Gal(k_1^s/k),C(k_1^s))\isom
638
H^r(\Gal(k_2^s/k),C(k_2^s))$
639
induced by a $k$-isomorphism $\vphi:k_1^s\ra k_2^s$
640
is independent of the choice of $\vphi$.
641
\end{exercise}
642
643
\begin{theorem}
644
If~$H$ is a normal subgroup of~$G$, then
645
there is a ``Hochschild-Serre'' spectral sequence
646
$$E_2^{rs}=H^r(G/H,H^s(H,A))\Rightarrow H^{r+s}(G,A)$$
647
\end{theorem}
648
By Theorem~\ref{conjth},~$G$ acts on $H^r(H,A)$ and~$H$
649
acts trivially, so this makes sense.
650
(The profinite case follows immediately from the
651
finite one by direct limit; cf. the end of Section~\ref{subsec:char}.)
652
The low dimensional corner of the spectral sequence
653
can be pictured as follows.
654
$$\xymatrix{
655
&{E^{02}}\[email protected]{-}[d]\[email protected]{-}[dr]\\
656
&E^{01}\[email protected]{-}[d]\[email protected]{-}[r]\[email protected]{-}[dr]&E^{11}\[email protected]{-}[d]\[email protected]{-}[dr]&{}\\
657
{}&E^{00}\[email protected]{-}[r]&E^{10}\[email protected]{-}[r]&{E^{20}}\\
658
}$$
659
Inflation and restriction are ``edge homomorphisms'' in the
660
spectral sequence.
661
The lower left corner pictured above gives the obvious isomorphism
662
$A^G\isom (A^H)^{G/H}$, and the exact sequence
663
$$\xymatrix{
664
0\ar[r]
665
&H^1(G/H,A^H)\ar[r]^{\infff}
666
&H^1(G,A)\ar[r]^{\res}
667
&H^1(H,A)^{G/H}\ar`r[d]`d[]`[dlll]_d`d[][dll]\\
668
&H^2(G/H,A^H)\ar[r]^{\inf}
669
&H^2(G,A).&{}\\
670
}
671
$$
672
The map~$d$ is the ``transgression'' and is induced by
673
$d_2:E_2^{01}\ra E_2^{20}$.
674
675
\begin{exercise}\mbox{}\vspace{-3ex}\newline
676
\begin{enumerate}
677
\item Show that this last sequence, or at least the first line,
678
is exact by using standard $1$-cocycles.
679
680
\item If $H^1(H,A)=0$, so that $E_2^{r1}=0$ for all~$r$, then
681
the sequence obtained by increasing the superscripts on the
682
$H$'s by~$1$ is exact.
683
\end{enumerate}
684
\end{exercise}
685
686
Consider a subfield~$K$ of $k^{\sep}$ that is Galois over~$k$, and let~$C$
687
be a commutative algebraic group over~$k$.
688
$$\xymatrix{
689
&{k^{\sep}}\[email protected]{-}[d]\[email protected]{-}@/^1pc/[d]^{G_K}\[email protected]{-}@/_1pc/[dd]_{G_k}\\
690
&{K}\[email protected]{-}[d]\\
691
{C}\[email protected]{-}[r]&{k.}\\
692
}
693
$$
694
The inflation-restriction sequence is
695
$$\xymatrix{
696
0\ar[r]
697
&H^1(K/k,C(K))\ar[r]
698
&H^1(k,C(k^{\sep}))\ar[r]
699
&H^1(K,C(k^{\sep}))^{\Gal(K/k)}\ar`r[d]`d[]`[dlll]`d[][dll]\\
700
&H^2(K/k,C(K))\ar[r]&H^2(k,C(k^{\sep})).
701
&{}\\
702
}
703
$$
704
If $C={\bG}_m$, then
705
$H^1(K,C(k^{\sep}))=0$, and there is an inflation-restriction
706
sequence with $(1,2)$ replaced by $(2,3)$:
707
$$\xymatrix{
708
0\ar[r]
709
&H^2(K/k,K^*)\ar[r]
710
&H^2(k,(k^{\sep})^*))\ar[r]
711
&H^2(K,(k^{\sep})^*))^{\Gal(K/k)}\ar`r[d]`d[]`[dlll]`d[][dll]\\
712
&H^3(K/k,K^*)\ar[r] &{H^3(k,(k^{\sep})^*)}.&{}\\
713
}
714
$$
715
An element $\alpha\in H^2(K,(k^{\sep})^*)^{\Gal(K/k)}$
716
represents a central simple algebra~$A$ over~$K$
717
which is isomorphic to all of its conjugates by $\Gal(K/k)$. As
718
the diagram indicates, the image $\alpha$ in $H^3(K/k,K^*)$ is the
719
``obstruction'' whose vanishing is the necessary and sufficient
720
condition for such an algebra~$A$ to come by base extension from an
721
algebra over~$k$.
722
723
\section{Cup products}
724
725
\subsection{$G$-pairing}
726
727
If $A$, $A'$, and~$B$ are $G$-modules, then
728
$$A\times A'\stackrel{b}{\rightarrow}B$$
729
is a \emp{$G$-pairing}
730
if it is bi-additive, and respects the action of $G$:
731
$$b(\sigma a,\sigma a')=\sigma b(a,a').$$
732
Such a pairing induces a map $\tilde{b}$
733
$$\cup:H^r(G,A)\times H^s(G,A')\stackrel{\tilde{b}}{\rightarrow}
734
H^{r+s}(G,B),$$
735
as follows:
736
given cochains~$f$ and~$f'$, one defines
737
(for a given $b$) a cochain $f \cup f'$ by
738
$$(f\cup f')(\sigma_1,\dots,\sigma_{r+s})=
739
b(f(\sigma_1,\dots,\sigma_r),\sigma_1\dots\sigma_r
740
f'(\sigma_{r+1},\dots,\sigma_{r+s})),$$
741
and checks the rule
742
$$\delta(f\cup f')=\delta f\cup f' + (-1)^rf\cup \delta f'.$$
743
If $\delta f=\delta f'=0$, then also
744
$\delta(f\cup f')=0$; i.e., if~$f$ and~$f'$ are cocycles, so
745
is $f\cup f'$.
746
Similarly one checks that the cohomology class of $f\cup f'$
747
depends only on the classes of~$f$ and~$f'$.
748
Thus we obtain the desired pairing $\tilde{b}$.
749
750
If $r=0$ and $a\in A^G$ is fixed, then $a'\mapsto b(a,a')$
751
defines a $G$-homomorphism $\vphi_a:A'\ra B$,
752
and $\alpha'\mapsto a\cup \alpha'$ is the map
753
$H^r(G,A')\ra H^r(G,B)$ induced by $\vphi_a$.
754
755
If~$H$ is a subgroup of~$G$, and
756
$\alpha\in H^r(G,A)$ and $\beta\in H^s(H,A')$,
757
then we can form
758
$$\res (\alpha) \cup \beta\in H^{r+s}(H,B).$$
759
Suppose that the index of~$H$ in~$G$ is finite, so that
760
corestriction is defined; then one can show that
761
$$\cores(\res (\alpha) \cup \beta)=
762
\alpha \cup \cores (\beta)\in H^{r+s}(G,B).$$
763
764
\subsection{Duality for finite modules}
765
If~$A$ and~$B$ are $G$-modules, we make the group
766
$\Hom_\Z(A,B)$ into a $G$-module by
767
defining $(\sigma f)(a)=\sigma(f(\sigma^{-1}a)).$
768
Note then that $\Hom_G(A,B)=(\Hom_\Z(A,B))^G$. Also,
769
the obvious pairing $A\cross \Hom_\Z(A,B)\ra B$
770
is a $G$-pairing. The canonical map
771
% I CAN'T REMEMBER THE RIGHT WAY TO DO THIS!
772
$$(*)\qquad\qquad\qquad\qquad\qquad
773
A \ra \Hom_\Z(\Hom_\Z(A,B))\hspace{1.6in}$$
774
is a $G$-homomorphism. In case $A$ is finite, killed by~$m$,
775
and~$B$ has a unique cyclic subgroup of order~$m$,
776
the map ($*$) is an isomorphism; one can thus recover~$A$
777
from its ``dual'' $\Hom_\Z(A,B)$ which has the same
778
order as~$A$.
779
There are two especially important such duals for finite~$A$.
780
\begin{itemize}
781
\item The \emp{Pontrjagin Dual} of~$A$ is
782
$\Hom_\Z(A,\Q/\Z)$; this equals $\Hom_\Z(A,\Z/m\Z)$ if $mA=0$.
783
\item The \emp{Cartier Dual} of~$A$ is
784
$\Hom_\Z(A,\mu(k^{\sep}))$; this equals $\Hom_\Z(A,\mu_m(k^{\sep}))$
785
if $mA=0$.
786
\end{itemize}
787
788
In the Pontrjagin case,~$G$ is an arbitrary profinite group
789
and acts trivially on $\Q/\Z$. Taking limits, this duality extends to a
790
perfect duality (i.e., an anti-equivalence of categories) between
791
discrete abelian torsion groups and profinite abelian groups.
792
793
In the Cartier case, $G=\Gal(k^{\sep}/k)$ or some quotient thereof,
794
and $m\neq 0$ in~$k$. (The Cartier dual of a $p$-group in characteristic~$p$
795
is a \emp{group scheme}, not just a Galois module.)
796
If~$E$ is an elliptic curve over~$k$ and the image of~$m$ in~$k$ is nonzero, the
797
\emp{Weil pairing} $E_m(k^{\sep})\cross E_m(k^{\sep})\ra\mu_m$
798
identifies $E_m$ with its Cartier dual.
799
800
801
\section{Local fields}
802
803
Let~$k$ be a local field, i.e., the field of fractions
804
of a complete discrete valuation ring
805
with finite residue field~$F$. Let~$K$
806
be a finite extension of~$k$.
807
808
Fundamental facts:
809
\begin{align*}
810
H^1(K/k,K^{*})&=0 \qquad \text{(Hilbert's Theorem 90)}\\
811
H^2(K/k,K^{*})&={\Z}/[K:k]{\Z}\\
812
H^2(k,{\bG}_m)&=\Br(k)={\Q}/{\Z} \\
813
\end{align*}
814
815
The equality $\Br(k)=\Q/{\Z}$
816
is given canonically, by the Hasse invariant, as follows:
817
The group $\Br(k)$ is the Brauer group, defined in \S\ref{brau}.
818
Consider the inflation-restriction sequence for
819
$H^2(-,\Gm)$ in the tower of fields
820
$$\[email protected]=1.2pc{{\kbar}\[email protected]{-}[d]\\
821
{k^{\ur}}\[email protected]{-}[d]\\
822
{k}}$$
823
where $k^{\ur}$ is the maximal unramified extension of~$k$.
824
Since every central division algebra over a local field
825
has an unramified splitting field, we have
826
$\Br(k^{\ur})=0$, and hence an isomorphism
827
$$\Br(k) \isom H^2(k^{\ur}/k,\Gm)
828
=H^2(\Frob^{\widehat{\Z}},(k^{\ur})^{*}).$$
829
Using the exact sequence
830
$$\xymatrix{
831
0\ar[r]
832
& U(k^{\ur})\ar[r]
833
&(k^{\ur})^*\ar[rr]^{\text{valuation}}&{}
834
&{\Z}\ar[r] &0}$$
835
and the fact that the unit group of an unramified extension has
836
trivial cohomology in dimension $\neq 0$,
837
we find that we can replace
838
$(k^{\ur})^*$ by $\Z$, and hence
839
$$\Br(k)\isom H^2(\hat{\Z},\Z)
840
=H^1(\hat{\Z},\Q/\Z)=\Q/\Z;$$
841
the middle equality comes from the short exact sequence
842
$$ 0 \ra \Z \ra \Q \ra \Q/\Z\ra 0$$
843
and the fact that~$\Q$, being uniquely divisible,
844
has trivial cohomology in nonzero dimensions.
845
The resulting map
846
$$\Br(k)\rightarrow{\Q}/{\Z}$$
847
is the called the Hasse invariant.
848
849
\begin{theorem}\label{thmdual}
850
Let~$A$ be a finite $G_k$-module of
851
order prime to the characteristic of~$k$. Let
852
$$A^*=\Hom(A,{\bG}_m)=\Hom(A,\mu(\overline{k}))$$
853
be the Cartier dual of~$A$.
854
Then the $G$-pairing
855
$$A\times A^*\rightarrow \overline{k}^{*}$$
856
induces a pairing
857
$$H^r(k,A)\times
858
H^{2-r}(k,A^*)\rightarrow
859
H^2(G_k,(k^{\sep})^{*})=\Br(k)={\Q}/{\Z}.$$
860
This is a perfect pairing of finite groups, for
861
all $r\in\Z$. It is nontrivial only if $r=0,1,2$, since
862
for $r\ge 3$,
863
$$H^r(k,A)=0\>\>\text{ for all }A,$$
864
i.e., ``the cohomological dimension of a
865
non-archimedean local field is~$2$.''
866
\end{theorem}
867
868
\emp{Example.} By Kummer theory, we have
869
$$k^{*}/(k^{*})^m=H^1(k,\mu_m(\overline{k})).$$
870
Thus there is a perfect pairing
871
$$\[email protected]=.3pc{
872
H^1(k,{\Z}/m{\Z})
873
&{\times }
874
&H^1(k,\mu_m({\overline{k}}))\ar[rrr]
875
&&&{\Q}/{\Z}\\
876
\\
877
\\
878
{\Hom(G_k,{\Z}/m{\Z})}\[email protected]{=}[uuu]
879
&{\times} & {k^{*}/(k^{*})^m}\[email protected]{=}[uuu]\\
880
}$$
881
The left hand equality is because the action is trivial.
882
Conclusion:
883
$$G_k^{\ab}/(G_k^{\ab})^m \isom k^*/(k^*)^m.$$
884
885
886
Taking the limit gives
887
\emp{Artin reciprocity}:
888
$$\xymatrix{
889
k^{*}\[email protected]{^{(}->}[r]&G_k^{ab}
890
};
891
$$
892
the image is dense.
893
894
Let $E/k$ be an elliptic curve.
895
In some sense,
896
$$E=\Ext^1(E,{\bG}_m)$$
897
in the category of algebraic groups.
898
There is a pairing
899
$$H^r(k,E)\times H^s(k,E)\rightarrow H^{r+s+1}(k,{\bG}_m).$$
900
For example, taking $r=0$ and $s=1$, we have the following theorem.
901
\begin{theorem}\label{thmpontrjagin}
902
Let~$E$ be an elliptic curve over a non-archimedean local field~$k$,
903
then we have the following
904
perfect pairing between Pontrjagin duals.
905
$$\[email protected]=.3pc{
906
H^0(k,E)&\times &H^1(k,E)\ar[rrr] &&&{H^2(k,\bG_m)={\Q}/{\Z}}\\
907
\\
908
\\
909
{\begin{array}{c}E(k)\\ \text{profinite}\end{array}}\[email protected]{=}[uuu]
910
&{\begin{array}{c} \times \\ \end{array}}
911
&{\begin{array}{c} H^1(k,E) \\
912
\text{discrete, torsion}\end{array}}\[email protected]{=}[uuu]
913
}
914
$$
915
\end{theorem}
916
\begin{proof}[Sketch of Proof]
917
We use the Weil pairing.
918
Letting~$D$ denote ``Pontrjagin dual'',
919
we have a diagram
920
$$\xymatrix{
921
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\\
922
0\ar[r]&H^1(k,E)_m^D\ar[r]&H^1(k,E_m)^D\ar[r]
923
&(E(k)/mE(k))^D\ar[r]&0}$$
924
The rows are exact. The top one from the Kummer sequence, and the
925
bottom is the dual of the top one. The middle vertical
926
arrow is an isomorphism by Theorem~\ref{thmdual}.
927
The outside vertical arrows are induced by the pairing
928
of Theorem~\ref{thmpontrjagin}.
929
The diagram commutes, so they are also isomorphisms,
930
and Theorem~\ref{thmpontrjagin} follows by passage
931
to the limit with more and more divisible~$m$.
932
\end{proof}
933
934
It was in trying to prove Theorem~\ref{thmpontrjagin} that I was
935
led to Theorem~\ref{thmdual} in the late 1950's. Of course
936
the ``fundamental facts'' and the Artin isomorphism are a
937
much older story.
938
939
\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
940
\begin{thebibliography}{1}
941
942
\bibitem{cassels-frohlich}
943
J.\thinspace{}W.\thinspace{}S. Cassels and A.~\protect{Fr{\"o}hlich} (eds.),
944
\emph{Algebraic number theory}, London, Academic Press Inc. [Harcourt Brace
945
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