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\begin{document}
\vskip -.75in \centerline{\bf Math 252: Modular Abelian Varieties,
Fall 2003, Prof. William Stein}
\bigskip
\medskip
\medskip
\bigskip
\title{N\'{e}ron models and the Shafarevich-Tate group}
\maketitle { \centerline{Tseno V. Tselkov}
\centerline{\texttt{tselkov@fas.harvard.edu}}}
\pagestyle{headings} \pagenumbering{arabic}
%\renewcommand{\thepage}{\centerline{--\ \arabic{page}\ --}}
\bigskip
\begin{abstract}
This is a review paper, which establishes the existence of
N\'{e}ron models for elliptic curves and proves a theorem of Mazur
relating the Shafarevich-Tate group of an abelian variety to the
\'{e}tale cohomology of its N\'{e}ron model.
\end{abstract}
\section{Introduction}
The organization of this paper is the following.
In Section 2, we define the Shafarevich-Tate and the Selmer groups
and give some motivation together with the most important facts
about them.
Section 3 deals with the basic facts about reductions of elliptic
curves.
In Section 4, we define arithmetic surfaces. Then we state a few
results about them, which will be key in the sequel. We then move
on to define N\'{e}ron models of elliptic curves. We establish the
uniqueness of the N\'{e}ron model and prove that it behaves well
under unramified base extensions.
Section 5 contains a complete proof of the existence of N\'{e}ron
models for elliptic curves. We first show that N\'{e}ron models
exist for elliptic curve with good reduction. Then we show that
N\'{e}ron models also exist in the case when we work over a
strictly Henselian discrete valuation ring. Finally, we use
faithfully flat descend and gluing to show that N\'{e}ron models
does exist in the general case, when we work over a Dedekind
domain.
In Section 6, we give a proof of a theorem of Mazur relating the
Shafarevich-Tate group of an abelian variety with the \'{e}tale
cohomology of its N\'{e}ron model.
\section{The Shafarevich-Tate and Selmer Groups}
Let $E/K$ and $E'/K$ be elliptic curves defined over a number
field $K$. Suppose we are given a non-zero isogeny $\phi: E \to
E'$ defined over $K$. (The classic example is when $E' = E$ and
$\phi = [m]$.) Then we have a natural exact sequence of Galois
$G_{\bar{K}/K}$-modules,
{
\[
\begin{CD}
0 @> >> E[\phi] @> >> E @> \phi >> E' @> >> 0,\\
\end{CD}
\]
} where $E[\phi]$ denotes the kernel of $\phi$. Taking its Galois
cohomology we obtain a long exact sequence
{
\[
\begin{CD}
0 @> >> E(K)[\phi] @> >> E(K) @> \phi >> E'(K) @> \delta >>
\end{CD}
\]
} {
\[
\begin{CD}
@> \delta >> H^1(G_{\bar{K}/K}, E[\phi]) @> >> H^1(G_{\bar{K}/K},
E) @> >> H^1(G_{\bar{K}/K}, E') @> >> 0
\end{CD}
\]
} From it we directly obtain the fundamental short exact sequence
{\small
\[
\begin{CD}
0 @> >> E'(K)/\phi(E(K)) @> \delta >> H^1(G_{\bar{K}/K}, E[\phi]) @> >> H^1(G_{\bar{K}/K}, E)[\phi] @> >> 0.\\
\end{CD}
\]
}
For each place $v$, let us fix an extension of $v$ to $\bar{K}$,
which gives us an embedding $\bar{K} \subset \bar{K}_v$ and a
decomposition group $G_v \subset G_{\bar{K}/K}$. Now $G_v$ clearly
acts on $E(\bar{K}_v)$ and $E'(\bar{K}_v)$, so repeating the
argument above gives us exact sequences {\small
\[
\begin{CD}
0 @> >> E'(K_v)/\phi(E(K_v)) @> \delta >> H^1(G_v, E[\phi]) @> >> H^1(G_v, E)[\phi] @> >> 0.\\
\end{CD}
\]
}
We have the natural inclusions $G_v \subset G_{\bar{K}/K}$ and
$E(\bar{K}) \subset E(\bar{K}_v)$ and they give us restriction
maps on cohomology as usual. Thus we obtain the following
commutative diagram {\small
\[
\begin{CD}
0 @> >> E'(K)/\phi(E(K)) @> \delta >> H^1(G_{\bar{K}/K}, E[\phi]) @> >> H^1(G_{\bar{K}/K}, E)[\phi] @> >> 0.\\
@. @VV V @VV V @VV V \\
0 @> >> \prod_{v} E'(K_v)/\phi(E(K_v)) @> \delta >> \prod_v H^1(G_v, E[\phi]) @> >> \prod_v H^1(G_v, E)[\phi] @> >> 0.\\
\end{CD}
\]
}
Now the following definitions come naturally
\begin{definition}
Let $\phi: E/K \to E'/K$ be an isogeny as above. The
\textit{Shafarevich-Tate} group of $E/K$ is the subgroup of
$H^1(G_v, E)$ defined by \beqa \Sha(E/K) = \ker
\left(H^1(G_{\bar{K}/K}, E) \to \prod_v H^1(G_v, E)\right).\eeqa
The \textit{$\phi$-Selmer group $E/K$} is the subgroup of
$H^1(G_{\bar{K}/K}, E[\phi])$ defined by \beqa S^{(\phi)} = \ker
\left(H^1(G_{\bar{K}/K}, E[\phi]) \to \prod_v H^1(G_v,
E)\right).\eeqa
\end{definition}
\begin{remark}
The groups $\Sha(E/K)$ and $S^{(\phi)}(E/K)$ depend only on $E$
and $K$, and not on the extensions of $v$ to $\bar{K}$.
\end{remark}
The commutative diagram above and the definitions directly give us
the following fact.
\begin{prop}
Let $\phi: E/K \to E'/K$ be an isogeny of elliptic curves defined
over $K$. Then there is an exact sequence \beqa 0 \to
E'(K)/\phi(E(K)) \to S^{(\phi)}(E/K) \to \Sha(E/K)[\phi] \to
0.\eeqa
\end{prop}
\begin{remark}
It is well-known that $H^1(G_{\bar{K}/K})$ can be identified with
the Weil-Ch\^{a}telet group $WC(E/K)$ of equivalence classes of
homogeneous spaces (or torsors) for $E/K$. Then we can view the
Shafarevich-Tate group as the group of homogeneous spaces which
have a $K_v$-rational point for every place $v$, i.e. $\Sha(E/K)$
is the group of homogeneous spaces which are everywhere locally
trivial modulo equivalence.
\end{remark}
The following theorem asserts the finiteness of the Selmer group,
which is not hard to establish. For a complete proof we refer the
reader to [6], Theorem 4.2.
\begin{thm}
Let $\phi: E/K \to E'/K$ be an isogeny of elliptic curves defined
over $K$. Then the Selmer group $S^{(\phi)}(E/K)$ is finite.
\end{thm}
On the contrary, the finiteness of the Shafarevich-Tate group is
among the open problems in mathematics. The famous conjecture due
to Birch and Swinnerton-Dyer asserts that it is also finite.
\begin{conjecture}
Let $E/K$ be an elliptic curve. The the Shafarevich-Tate group
$\Sha(E/K)$ is finite.
\end{conjecture}
\section{Reduction of Elliptic Curves}
Let $K$ be a local field, $R$ - its ring of integers, $\pi$ - an
uniformizer for $R$, and $k$ - the residue field of $R$.
Suppose we are given an elliptic curve $E/K$ with Weierstrass
equation \beqa y^2 + a_1xy + a_3y = x^3 + a_2 x^2 + a_4x +
a_6.\eeqa
\begin{definition}
Let $E/K$ be an elliptic curve. A Weierstrass equation as above is
called a minimal Weierstrass equation for $E$ if $v(\Delta)$ is
minimal subject to $a_1, a_2, a_3, a_4, a_6 \in R$.
\end{definition}
It is quite easy to show that every elliptic curve has a minimal
Weierstrass equation.
\begin{prop}
Every elliptic curve has a minimal Weierstrass equation.
\end{prop}
\begin{proof}
We can certainly find a Weierstrass equation with all $a_i \in R$.
Among those there is a minimal $v(\Delta)$ since $v$ is discrete.
\end{proof}
Now let us consider the natural reduction map $R \to R/\pi R$
denoted by $t \mapsto \tilde{t}$. Having chosen a minimal
Weierstrass equation for $E/K$, we can reduce its coefficients
modulo $\pi$ to obtain a curve over $k$, namely \beqa \tilde{E}:
y^2 + \tilde{a_1}xy + \tilde{a_3}y = x^3 + \tilde{a_2} x^2 +
\tilde{a_4}x + \tilde{a_6}.\eeqa
\begin{definition}
The curve $\tilde{E}/k$ is called the reduction of $E$ modulo
$\pi$.
\end{definition}
\begin{remark}
It is not hard to show that the minimality condition on the
equation for $E$ implies that the equation for $\tilde{E}$ is
unique up to the standard change of coordinates.
\end{remark}
Now let $P \in E(K)$. We can find homogeneous coordinates $P =
[x_0, y_0, z_0]$ with $x_0, y_0, z_0 \in R$ and at least one of
them in $R^\times$. Then the reduced point $\tilde{P} =
[\tilde{x_0}, \tilde{y_0}, \tilde{z_0}]$ clearly is in
$\tilde{E}$(k). Thus we obtain a reduction map \beqa E(K) \to
\tilde{E}(k), & P \mapsto \tilde{P}.\eeqa
The reduced curve $\tilde{E}/k$ may or may not be singular. In
fact it might be one of three types and we classify $E$ according
to these possibilities.
\begin{definition}
Let $E/K$ be an elliptic curve and let $\tilde{E}$ be the reduced
curve for a minimal Weierstrass equation.\\(a) $E$ has good (or
stable) reduction over $K$ if $\tilde{E}$ is non-singular.\\(b)
$E$ has multiplicative (or semi-stable) reduction over $K$ if
$\tilde{E}$ has a node.\\(c) $E$ has additive (or unstable)
reduction over $K$ if $\tilde{E}$ has a cusp.\\ In cases (b) and
(c) $E$ is said to have bad reduction.
\end{definition}
\begin{remark}
It is a standard fact how the reduction type can be read off from
a minimal Weierstrass equation. For a precise explanation the
reader can consult [6], Proposition 5.1.
\end{remark}
\section{Arithmetic Surfaces and Properties of N\'{e}ron Models}
To even define N\'{e}ron models for elliptic curves we shall need
some preliminaries on arithmetic surfaces. To simplify the
discussion we shall make the following convention: All Dedekind
domains and all discrete valuation rings have perfect residue
fields. Let $R$ be a Dedekind domain. Intuitively, an arithmetic
surface is an $R$-scheme $\mathcal{C} \to \spec(R)$ whose fibers
are curves.
\begin{definition}
Let $R$ be a Dedekind domain with fraction field $K$. An
arithmetic surface (over $R$) is an integral, normal, excellent
scheme $\mathcal{C}$, which is flat and of finite type over $R$,
and whose generic fiber is a non-singular connected projective
curve $C/K$ and whose special fibers are unions of curves over the
appropriate residue fields.
\end{definition}
\begin{remark}
The special fibers may be reducible or singular or even
non-reduced.
\end{remark}
In the sequel, we shall use the following key results about
arithmetic surfaces, the proofs of which can be found in [5],
Chapter 4, Section 4.
\begin{prop}
Let $R$ be a Dedekind domain with fraction field $K$, let
$\mathcal{C}/R$ be an arithmetic surface, and let $C/K$ be the
generic fiber of $\mathcal{C}$.\\ (a) If $\mathcal{C}$ is proper
over $R$, then $C(K) = \mathcal{C}(R)$.\\ (b) Suppose that the
scheme $C$ is regular, and let $\mathcal{C}^0 \subset \mathcal{C}$
be the largest subscheme of $\mathcal{C}$ such that the map
$\mathcal{C}^0 \to \spec R$ is a smooth morphism. Then
$\mathcal{C}(R) = \mathcal{C}^0(R)$.
\end{prop}
\begin{thm}
Let $R$ be a Dedekind domain with fraction field $K$, and let
$C/K$ be a non-singular projective curve of genus $g$.\\ (a)
(Resolution of Singularities for Arithmetic Surfaces) There exists
a regular arithmetic surface $\mathcal{C}/R$, proper over $R$,
whose generic fiber is isomorphic to $C/K$. We call
$\mathcal{C}/R$ a proper regular model for $C/K$.\\ (b) (Minimal
Models Theorem) Assume that $g \geq 1$. Then there exists a proper
regular model $\mathcal{C}^{\min}/R$ for $C/K$ with the following
minimality property: Let $\mathcal{C}/R$ be any other proper
regular model for $C/K$. Fix an isomorphism from the generic fiber
of $\mathcal{C}$ to the generic fiber of $\mathcal{C}^{\min}$.
Then the induced birational map $\mathcal{C} \to
\mathcal{C}^{\min}$ is an $R$-isomorphism. We call
$\mathcal{C}^{\min}$ the minimal regular model for $C/K$. It is
unique up to unique $R$-isomorphism.
\end{thm}
We can now move on to defining N\'{e}ron models and giving their
basic properties. Suppose we have a discrete valuation ring $R$
with fraction field $K$. Intuitively, the N\'{e}ron model of an
elliptic curve $E/K$ is an arithmetic surface $\mathcal{E}/R$
whose generic fiber is the given elliptic curve.
\begin{definition}
Let $R$ be a Dedekind domain with fraction field $K$, and let
$E/K$ be an elliptic curve. A N\'{e}ron model for $E/K$ is a
smooth group scheme $\mathcal{E}/R$ whose generic fiber is $E/K$
and which satisfies the following universal property (N\'{e}ron
Mapping Property): \\ Let $\mathcal{X}/R$ be a smooth $R$-scheme
with generic fiber $X/K$, and let $\phi_K: X_{/K} \to E_{/K}$ be a
rational map defined over $K$. Then there exists a unique
$R$-morphism $\phi_R: \mathcal{X}_{/R} \to \mathcal{E}_{/R}$
extending $\phi_K$.
\end{definition}
\begin{remark}
The most natural and important instance of the N\'{e}ron mapping
property is the case $\mathcal{X} = \spec(R)$ and $X = \spec(K)$.
Then the set of $K$-maps $X_{/K} \to E_{/K}$ is the group of
$K$-rational points $E(K)$, and the set of $R$-morphisms
$\mathcal{X}_{/R} \to \mathcal{E}_{/R}$ is the group of sections
$\mathcal{E}(R)$. In this situation the N\'{e}ron mapping property
asserts that the natural inclusion $\mathcal{E}(R) \hookrightarrow
E(K)$ is a bijection.
\end{remark}
We shall now present some of the more important properties of
N\'{e}ron models and we start with uniqueness.
\begin{prop}
Let $R$ be a Dedekind domain with fraction field $K$, and let
$E/K$ be an elliptic curve. Suppose that $\mathcal{E}_1 / R$ and
$\mathcal{E}_2 / R$ are N\'{e}ron models for $E/K$. Then there
exists a unique $R$-isomorphism $\psi: \mathcal{E}_1 / R \to
\mathcal{E}_2 / R$ whose restriction to the generic fiber is the
identity map on $E/K$. In other words, the N\'{e}ron model of
$E/K$ is unique up to unique isomorphism.
\end{prop}
\begin{proof}
The identity map $E/K \to E/K$ is a rational map from the generic
fiber of $\mathcal{E}_1$ to the generic fiber of $\mathcal{E}_2$,
and $\mathcal{E}_1$ is smooth over $R$, so the N\'{e}ron mapping
property for $\mathcal{E}_2$ asserts that the identity map extends
uniquely to an $R$-morphism $\varphi: \mathcal{E}_1/R \to
\mathcal{E}_2/R$. Similarly, exchanging the places of
$\mathcal{E}_1$ and $\mathcal{E}_2$ we obtain a unique
$R$-morphism $\phi: \mathcal{E}_2/R \to \mathcal{E}_1/R$ which is
the identity on the generic fiber. But then both $\phi \circ
\varphi: \mathcal{E}_1 \to \mathcal{E}_1$ and the identity map
$\mathcal{E}_1 \to \mathcal{E}_1$ are $R$-morphisms that are the
same (identity) map on the generic fiber, so the uniqueness part
of the N\'{e}ron mapping property asserts that $\phi \circ
\varphi$ equals the identity map. Thus, $\phi$ and $\varphi$ are
isomorphisms.
\end{proof}
N\'{e}ron models also behave well under unramified base extension.
\begin{prop}
Let $R$ be a Dedekind domain with fraction field $K$, and let
$E/K$ be an elliptic curve. Let $K'/K$ be a finite unramified
extension, and let $R'$ be the integral closure of $R$ in $K'$.
Let $\mathcal{E}/R$ be a N\'{e}ron model for $E/K$. Then
$\mathcal{E} \times_R R'$ is a N\'{e}ron model for $E/K'$.
\end{prop}
\begin{proof}
Let $\mathcal{X}'/R'$ be a smooth $R'$-scheme with generic fiber
$X'/K'$ and let $\phi_{K'}: X_{K'}' \to E_{/K'}$ be a rational
map. Let us consider the composition \beqa \mathcal{X}' \to \spec
R' \to \spec R.\eeqa It makes $\mathcal{X}'$ into an $R$-scheme.
Moreover, the assumptions on $K'$ imply that the map $\spec R' \to
\spec R$ is a smooth morphism and hence the composition is a
smooth morphism, so $\mathcal{X}'$ is a smooth $R$-scheme.
Now the N\'{e}ron mapping property for $\mathcal{E}/R$ tells us
that there is an $R$-morphism $\phi_R : \mathcal{X}' \to
\mathcal{E}$, whose restriction to the generic fiber is the
composition
\[
\begin{CD}
X' @> \phi_{K'}>> E \times_K K'@> p_1 >> E.
\end{CD}
\]
To establish the existence part of the N\'{e}ron mapping property
note that the two $R$-morphisms $\phi_R : \mathcal{X}' \to
\mathcal{E}$ and $\mathcal{X}' \to \spec R'$ determine an
$R$-morphism (and thus an $R'$-morphism) to the fiber product
$\phi_{R'} : \mathcal{X}' \to \mathcal{E} \times_R R'$, which is
unique, so $\mathcal{E} \times_R R'$ is a M\'{e}ron model for
$E/K'$.
\end{proof}
\section{Existence of N\'{e}ron Models}
The goal of this section is to prove the existence of a N\'{e}ron
model for elliptic curves.
\begin{thm}
Let $R$ be a Dedekind domain with fraction field $K$, let $E/K$ be
an elliptic curve, let $\mathcal{C}/R$ be a minimal proper regular
model for $E/K$ as in Theorem 2, and let $\mathcal{E}/R$ be the
largest subscheme of $\mathcal{C}/R$ which is smooth over $R$.
Then $\mathcal{E}/R$ is a N\'{e}ron model for $E/K$.
\end{thm}
The proof involves numerous steps. We start with the following
local statement that over a discrete valuation ring asserts the
existence of a group scheme with generic fiber a given elliptic
curve.
\begin{prop}
Let $R$ be a discrete valuation ring with fraction field $K$, let
$E/K$ be an elliptic curve, and choose a Weierstrass equation for
$E/K$ with coefficients in $R$, \beqa E: y^2 + a_1 xy + a_3 y =
x^3 + a_2 x^2 + a_4 x + a_6.\eeqa This Weierstrass equation
defines a scheme $\mathcal{W} \subset \PP^2_R$. Let $\mathcal{W}^0
\subset \mathcal{W}$ be the largest subscheme of $W$ which is
smooth over $R$.\\ (a) Both $\mathcal{W}/R$ and $\mathcal{W}^0/R$
have generic fiber $E/K$.\\ (b) The natural map $\mathcal{W}(R)
\to E(K)$ is a bijection. If $\mathcal{W}$ is regular, then the
natural map $\mathcal{W}^0(R) \to \mathcal{W}(R)$ is also
bijection, so in this case there is natural identification $\mathcal{W}^0(R) = E(K)$.\\
(c) The addition and negation maps on $E$ extend to $R$-morphisms
$\mathcal{W}^0 \times_R \mathcal{W}^0 \to \mathcal{W}^0$ and
$\mathcal{W}^0 \to \mathcal{W}^0$, which make $\mathcal{W}^0$ into
a group scheme over $R$.
\end{prop}
\begin{remark}
If $E/K$ has good reduction and if we take a minimal Weierstrass
equation for $E/K$, then $\mathcal{W}$ itself is smooth over $R$.
Thus in this case the above theorem asserts that $\mathcal{W} =
\mathcal{W}^0$ is a group scheme over $R$.
\end{remark}
\begin{remark}
If $E/K$ has bad reduction, then there is exactly one singular
point on the reduction $\tilde{E}$ (mod $\wp$). In other words,
the special fiber $\tilde{\mathcal{W}}$ of $\mathcal{W}$ contains
exactly one singular point, say $\gamma \in \tilde{\mathcal{W}}
\subset \mathcal{W}$. Then $\mathcal{W}^0 = \mathcal{W} - \{
\gamma \}$. In particular, $\mathcal{W}^0$ and $\mathcal{W}$ have
the same generic fiber.
\end{remark}
\begin{proof}
(a) $\mathcal{W}$ is the closed subscheme of $\PP_R^2 = \proj
R[X,Y,Z]$ defined by the single homogeneous equation \beqa
\mathcal{W}: Y^2Z + a_1 XYZ + a_3 YZ^2 = X^3 + a_2 X^2 Z + a_4
XZ^2 + a_6 Z^3.\eeqa Its generic fiber is the variety in $\PP_K^2$
defined by the same equation. Thus, the generic fiber of
$\mathcal{W}$ is exactly $E/K$. $\mathcal{W}^0$ has the same
generic fiber, so this completes part (a).
(b) Since $\mathcal{W}$ is a closed subscheme of $\PP_R^2$ then it
is certainly proper over $R$. This allows us to apply Proposition
3 to conclude that $E(K) = \mathcal{W}(R)$, which is the first
part of (b). If in addition $\mathcal{W}$ is regular, then again
by Proposition 3 we get that $\mathcal{W}(R) = \mathcal{W}^0 (R)$,
which implies $E(K) = \mathcal{W}^0 (R)$.
(c) The proof of this part is quite explicit and rather long
computation, so we refer the reader to [5], Theorem 5.3.(c) for
all details.
\end{proof}
The next ingredient in the proof is the following generalization
of a theorem of Weil. Weil's theorem asserts that a rational map
from a smooth variety to a complete group variety is automatically
a morphism, and Artin has extended it to a scheme-theoretic
setting.
\begin{prop}
Let $R$ be a Dedekind domain, let $G/R$ be a group scheme over
$R$, let $\mathcal{X}/R$ be a smooth $R$-scheme, and let $\phi:
\mathcal{X} \to G$ be a rational map over $R$. Write $\dom(\phi)$
for the domain of $\phi$, and suppose that $\dom(\phi)$ is dense
in every fiber of $\mathcal{X}/R$.\\ (a) The complement
$\mathcal{X} - \dom(\phi)$ is a subscheme of $\mathcal{X}$ of pure
codimension one.\\ (b) If $G$ is proper over $R$, then $\dom(\phi)
= \mathcal{X}$. In other words, $\phi$ is a morphism.
\end{prop}
\begin{proof}
Following [5] we shall phrase our exposition in terms of points,
but to be completely rigorous, our "points" should be $T$-valued
points for arbitrary $R$-schemes $T$.
Let us consider the rational map \beqa F: \mathcal{X} \times_R
\mathcal{X} \to G, & F(x,y) = \phi(x) \phi(y)^{-1}. \eeqa We claim
that there is a natural identification \beqa \dom(\phi)
\longleftrightarrow \Delta \cap \dom(F), & x \longleftrightarrow
(x,x),\eeqa where $\Delta$ is the diagonal in $\mathcal{X}
\times_R \mathcal{X}$. Indeed, if $x \in \dom(\phi)$ then $F(x,x)
= \phi(x) \phi(x)^{-1}$ is defined, so $(x,x) \in \dom(F)$.
Conversely, if $(x,x) \in \dom(F)$ we need to show that $x \in
\dom(\phi)$. Firstly, since $\dom(F)$ is open, there is a
non-empty open set $U \subset \mathcal{X}$ such that $x \times_R U
\subset \dom(F)$. Secondly, since $\dom(\phi)$ is open, there is a
point $y \in U \cap \dom(\phi)$, which means that $\phi(x) =
F(x,y) \phi(y)$, so $x \in \dom(\phi)$ and the claim is proved.
Let $K(\mathcal{X} \times \mathcal{X})$ be the function field of
the scheme $\mathcal{X} \times_R \mathcal{X}$, and let
$\mathcal{O}_{G,0}$ be the local ring of $G$ along with the
identity section, i.e. $\mathcal{O}_{G,0}$ is the ring of rational
functions on $G$ which are well-defined at some point of the image
of the map $\sigma_0: \spec R \to G$, where $\sigma_0$ is the
identity element of the group scheme $G$.
Then we clearly get a ring homomorphism \beqa F^* :
\mathcal{O}_{G,0} \to K(\mathcal{X} \times \mathcal{X}), & f
\mapsto f \circ F.\eeqa Our next goal is to show that \beqa x \in
\dom (\phi) \Leftrightarrow (x,x) \in \dom (F^* f) \textrm{ for
all } f \in \mathcal{O}_{G,0}.\eeqa Indeed, let $f \in
\mathcal{O}_{G,0}$ and suppose $x \in \dom (\phi)$. Then $(x,x)
\in \dom(F)$ by the claim proven above, and since $F(x,x) =
\phi(x) \phi(x)^{-1}$ is the identity element of $G$, then
$F^*(f)$ is defined at $(x,x)$. Conversely, if $F^*(f) = f \circ
F$ is defined at $(x,x)$ for all functions $f \in
\mathcal{O}_{G,0}$, then $F$ must be defined at $(x,x)$. Thus we
get that \beqa x \in \dom (\phi) \Leftrightarrow (x,x) \in \dom
(F^* f) \textrm{ for all } f \in \mathcal{O}_{G,0} \Leftrightarrow
F^*(\mathcal{O}_{G,0}) \subset \mathcal{O}_{\mathcal{X} \times
\mathcal{X}, (x,x)},\eeqa where $\mathcal{O}_{\mathcal{X} \times
\mathcal{X}, (x,x)} \subset K(\mathcal{X} \times \mathcal{X})$ is
the local ring of $\mathcal{X} \times_R \mathcal{X}$ at $(x,x)$.
The scheme $\mathcal{X} \times_R \mathcal{X}$ is smooth over $R$,
so in particular it is normal. This implies that a function $f \in
K(\mathcal{X} \times \mathcal{X})$ will be defined at $(x,x)$
unless its polar divisor $\divi_\infty(f)$ goes through $(x,x)$.
In other words, \beqa \mathcal{O}_{\mathcal{X} \times \mathcal{X},
(x,x)} = \{ g \in K(\mathcal{X} \times \mathcal{X})^* : (x,x)
\notin \divi_\infty (g)\} \cup \{0\}.\eeqa This together with the
result above describing the domain of $\phi$ gives us \beqa
\mathcal{X} - \dom (\phi) = \{x\in \mathcal{X}:
F^*(\mathcal{O}_{G,0}) \not \subset \mathcal{O}_{\mathcal{X}
\times \mathcal{X}, (x,x)}\}\eeqa \beqa = \{x \in \mathcal{X}:
(x,x) \in \divi_\infty (F^* f) \textrm{ for some } f \in
\mathcal{O}_{G,0}\}\eeqa \beqa \cong \Delta \cap \bigcup_{f \in
\mathcal{O}_{G,0}} \divi_\infty (F^* f)\eeqa \beqa = \bigcup_{f
\in \mathcal{O}_{G,0}} (\Delta \cap \divi_\infty (F^* f)).\eeqa
The diagonal $\Delta$ is a complete intersection in $\mathcal{X}
\times_R \mathcal{X}$, and each divisor $\divi_\infty (F^*f)$ has
pure codimension one in $\mathcal{X} \times_R \mathcal{X}$, so
each of the intersections $\Delta \cap \divi_\infty (F^* f)$ has
pure codimension one in $\Delta$. It follows that the union over
$f \in \mathcal{O}_{G,0}$ also has pure codimension one since we
know a priori that it is a proper closed subset of $\Delta$. Thus,
the proof of (a) is completed.
(b) We shall use the following lemma asserting that a rational map
from a smooth scheme to a proper scheme is defined off of a subset
of codimension at least two. Then this lemma together with part
(a) directly give us (b).
\begin{lemma}
Let $R$ be a Dedekind domain, let $\mathcal{X}/R$ be a smooth
$R$-scheme, let $\mathcal{Y}/R$ be a proper $R$-scheme, and let
$\phi: \mathcal{X} \to \mathcal{Y}$ be a dominant rational map
defined over $R$. Then every component of $\mathcal{X} -
\dom(\phi)$ has codimension at least two in $\mathcal{X}$.
\end{lemma}
\begin{proof}
Let $\mathcal{Z} \subset \mathcal{X}$ be an irreducible subscheme
of codimension one in $\mathcal{X}$. We need to show that $\phi$
is defined at the generic point of $\mathcal{Z}$, i.e. $\phi$ is
defined on a non-empty open subset of $\mathcal{Z}$. Let us
consider the local ring $\mathcal{O}_{\mathcal{X}, \mathcal{Z}}$
of $\mathcal{X}$ at $\mathcal{Z}$. It is a discrete valuation ring
since it is local ring of dimension one, which is regular since
$\mathcal{X}/R$ is smooth.
Now, the dominant map induces a morphism $\spec K(\mathcal{X}) \to
\spec K(\mathcal{Y})$ from the generic point of $\mathcal{X}$ to
the generic point of $\mathcal{Y}$. Thus we obtain the following
commutative diagram
\[
\begin{CD}
\mathcal{X} @> \phi >>\mathcal{Y}\\
@AA A @AA A \\
\spec \mathcal{O}_{\mathcal{X}, \mathcal{Z}}\\
@AA A @AA A \\
\spec K(\mathcal{X}) @> >> \spec K(\mathcal{Y}),
\end{CD}
\]
The discrete valuation ring $\mathcal{O}_{\mathcal{X},
\mathcal{Z}}$ has fraction field $K(\mathcal{X})$ and since
$\mathcal{Y}$ is proper over $R$, by the valuative criterion of
properness ([2], Theorem 4.7.) we get that the rational map \beqa
\spec \mathcal{O}_{\mathcal{X}, \mathcal{Z}} \to \mathcal{X} \to
\mathcal{Y}\eeqa extends to a morphism $\spec
\mathcal{O}_{\mathcal{X}, \mathcal{Z}} \to \mathcal{Y}$. This
means that $\phi$ is defined at the generic point of
$\mathcal{Z}$, which is what we needed to complete the proof of
the lemma and the theorem.
\end{proof}
\end{proof}
Combining the results of the last two Propositions 6 and 7 we can
establish the existence of N\'{e}ron models for elliptic curves
with good reduction.
\begin{corollary}
Let $R$ be a Dedekind domain with fraction field $K$, let $E/K$ be
an elliptic curve given by a Weierstrass equation \beqa E: y^2 +
a_1 xy + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6\eeqa having
coefficients in $R$, and let $\mathcal{W} \subset \PP^2_R$ be the
closed subscheme of $\PP^2_R$ defined by this Weierstrass
equation. Suppose that $\mathcal{W}$ is smooth over $R$ or,
equivalently, that the Weierstrass equation has good reduction at
every prime of $R$. Then $\mathcal{W}/R$ is a N\'{e}ron model for
$E/K$.
\end{corollary}
\begin{proof}
By Proposition 6 the addition law on $E/K$ extends to make
$\mathcal{W}$ into a group scheme over the localization (which is
a discrete valuation ring) of $R$ at each of its prime ideals.
Moreover, since these group laws are given by the same equations,
they together make $\mathcal{W}$ into a group scheme over $R$.
We only need to also verify that $\mathcal{W}$ has the N\'{e}ron
mapping property. Let $\mathcal{X}/K$ be a smooth $R$-scheme with
generic fiber $X/K$, take any rational map $\phi_K: X_{/K} \to
E_{/K}$ defined over $K$, and let $\phi: \mathcal{X} \to
\mathcal{W}$ be the associated rational map over $R$. Now
$\mathcal{W}$ is a closed subscheme of $\PP^2_R$ and thus is
proper over $R$, so by Proposition 7 the map $\phi$ extends to a
morphism, which establishes the N\'{e}ron mapping property. So,
$\mathcal{W}/R$ is indeed a N\'{e}ron model for $E/K$.
\end{proof}
Unfortunately, we cannot hope to have good reduction at every
prime. The strategy to go around that is the folowing. Firstly,
prove Theorem 3 for strictly Henselian discrete valuation rings.
Secondly, descend from the strict Henselization of a discrete
valuation ring down to the ring itself and thirdly, glue the
N\'{e}ron models over discrete valuation rings to get a N\'{e}ron
model over a Dedekind domain.
For completeness let us say the basics of Henselian rings.
\begin{definition}
A discrete valuation ring $R$ is called Henselian if it satisfies
Hensel's lemma, i.e. $R$ is Henselian if for any monic polynomial
$f(x)\in R[x]$ and any element $a \in R$ satisfying \beqa f(a)
\equiv 0(\modu \wp), & f'(a) \neq 0 (\modu \wp),\eeqa there exists
a unique element $\alpha \in R$ satisfying \beqa \alpha \equiv
a(\modu \wp), & f(\alpha) = 0.\eeqa The ring $R$ is called
strictly Henselian if it is Henselian and if its residue field $k
= R/\wp$ is algebraically closed (recall that we work only with
perfect residue fields).
\end{definition}
It is a well-known fact that every discrete valuation ring $R$ can
be embedded in a minimal Henselian and minimal strictly Henselian
ring in the following sense
\begin{prop}
Let $R$ be a discrete valuation ring.\\ (a) A local homomorphism
$R \to R^{h}$ with $R^{h}$ Henselian is called a Henselization of
$R$ if any other local homomorphism $R \to R'$ with $R'$ Henselian
factors uniquely into $R \to R^{h} \to R'$. The Henselization
exists.
(b) A local homomorphism $R \to R^{sh}$ with $R^{sh}$ strictly
Henselian with residue field $k^{sh}$ is called a strict
Henselization of $R$ if any other local homomorphism from $R$ into
a strictly Henselian ring $R'$ with residue field $k'$ extends to
$R^{sh}$, and, moreover, the extension is uniquely determined once
the map $k^{sh} \to k'$ on residue fields has been specified. The
strict Henselization exists.
\end{prop}
\begin{remark}
Proposition 6.5. in [5] gives an explicit description of $R^{h}$
and $R^{sh}$, which we will not need for our purposes.
\end{remark}
We are now going to prove that the scheme $\mathcal{E}$ from
Theorem 3 is a N\'{e}ron model for $E$ provided that we work over
a strictly Henselian discrete valuation ring and the group law of
$E/K$ extends to make $\mathcal{E}$ into a group scheme over $R$.
\begin{prop}
Let $R$ be a strictly Henselian discrete valuation ring with
fraction field $K$, let $E/K$ be an elliptic curve, let
$\mathcal{C}/R$ be a minimal proper regular model for $E/K$, and
let $\mathcal{E}/R$ be the largest subscheme of $\mathcal{C}/R$
which is smooth over $R$. If the group law on $E/K$ extends to
make $\mathcal{E}$ into a group scheme over $R$, then
$\mathcal{E}/R$ is a N\'{e}ron model for $E/K$.
\end{prop}
\begin{proof}
We need to verify the N\'{e}ron mapping property, so let
$\mathcal{X}/R$ be a smooth $R$-scheme with generic fiber $X/K$,
and let $\phi_K : X \to E$ be a rational map. We need to show that
$\phi_K$ extends to a morphism $\mathcal{X} \to \mathcal{E}$.
We can apply Proposition 7 with $R = K$, since $E$ is clearly a
proper group scheme over $K$ and $X$ is smooth over $K$, to get
directly that $\phi_K$ is a morphism $X \to E$. So, the rational
map $\phi : \mathcal{X} \to \mathcal{E}$ induced by $\phi_K$ is a
morphism on the generic fiber.
To prove that $\phi$ is a morphism, let us assume the contrary and
get a contradiction. As we assumed in the statement of the theorem
$\mathcal{E}$ is a group scheme, so we can apply Proposition 7 to
get that the set of points where the rational map $\phi$ is not
defined is a set of pure codimension one in $\mathcal{X}$. Hence
these is an irreducible closed subscheme $\mathcal{Z} \subset
\mathcal{X}$ such that $\phi$ is not defined on the generic point
$\eta_\mathcal{Z}$ of $\mathcal{Z}$. From algebraic geometry,
since $\mathcal{X}$ is regular and $\mathcal{Z}$ is of codimension
one, the local ring $\mathcal{O}_{\mathcal{X}, \mathcal{Z}}$ is a
discrete valuation ring and $\eta_{\mathcal{Z}} = \spec
\mathcal{O}_{\mathcal{X}, \mathcal{Z}}$. Thus, we obtain the
following diagram
\[
\begin{CD}
\mathcal{X} @> \phi >>\mathcal{E} @> \subset >> \mathcal{C}\\
@AA A @AA A @AA A\\
\eta_{\mathcal{Z}} = \spec \mathcal{O}_{\mathcal{X}, \mathcal{Z}}\\
@AA A @AA A @AA A\\
\spec K(\mathcal{X}) @> \phi_K >> \spec K(\mathcal{E}) @> = >>
\spec K(\mathcal{C}).
\end{CD}
\]
Now the scheme $\mathcal{C}$ is proper over $R$ and
$\mathcal{O}_{\mathcal{X}, \mathcal{Z}}$ is a discrete valuation
ring, so by the valuative criterion of properness ([2], Theorem
4.7.) we have that $\phi$ extends to a morphism $\phi :
\eta_{\mathcal{Z}} \to \mathcal{C}$. In other words, if we are
mapping to the larger scheme $\mathcal{C}$, then $\phi$ is defined
generically on $\mathcal{Z}$.
However, we area assuming that $\phi : \mathcal{X} \to
\mathcal{E}$ does not extend generically to $\mathcal{Z}$, or
equivalently that $\phi(\eta_{\mathcal{Z}}) \in \mathcal{C}$ is
not contained in $\mathcal{E}$. In particular, if $k$ is the
residue field of $R$ and $x_0 \in \mathcal{Z}(k)$ is such that
$\phi: \mathcal{X} \to \mathcal{C}$ is defined at $x_0$, then
$\phi(x_0) \notin \mathcal{E}$.
Since $R$ is strictly Henselian, by [5], Proposition 6.4., the set
of $R$-valued points $\mathcal{X}(R)$ maps to a dense set of
points in the special fiber of $\mathcal{X}$. In particular, we
can find a point $x \in \mathcal{X}(R)$ which intersects
$\mathcal{Z}$ at a point, call it $x_0 \in \mathcal{Z}(k)$, at
which the map $\phi: \mathcal{X} \to \mathcal{C}$ is defined.
Composing $x$ with $\phi$ we obtain a rational map $\spec R \to
\mathcal{C}$, which by the valuative criterion of properness
extends to a morphism $\spec R \to \mathcal{C}$. On the other hand
$\phi \circ x \in \mathcal{C}(R)$ and by our construction we have
$\phi \circ x \notin \mathcal{E}(R)$. However, by Proposition 3 we
have that $\mathcal{C}(R) = \mathcal{E}(R)$, so we have a
contradiction. Thus $\phi$ extends to a morphism $\mathcal{X} \to
\mathcal{E}$, and so $\mathcal{E}$ has the N\'{e}ron mapping
property.
\end{proof}
The next result completes the proof that $\mathcal{E}/R$ is a
N\'{e}ron model for $E/K$ at least over strictly Henselian
discrete valuation ring. For a complete proof we refer to [5],
Proposition 6.10.
\begin{prop}
Let $R$ be a strictly Henselian discrete valuation ring with
fraction field $K$, let $E/K$ be an elliptic curve, let
$\mathcal{C}/R$ be a minimal proper regular model for $E/K$, and
let $\mathcal{E}/R$ be the largest subscheme of $\mathcal{C}/R$
which is smooth over $R$. Then the group law on $E/K$ extends to
make $\mathcal{E}$ into a group scheme over $R$.
\end{prop}
We are now ready to prove the existence of N\'{e}ron models for
elliptic curves over Dedekind domains, i.e. to prove Theorem 3.
\begin{proof}
[Proof of Theorem 3] If $R$ is strictly Henselian $\mathcal{E}/R$
is a N\'{e}ron model for $E/K$ by Proposition 9 and Proposition
10.
Otherwise, let us suppose first that $R$ is a discrete valuation
ring, and let $R^{sh}$ be the strict Henselization of $R$. Then we
claim that $\mathcal{C}^{sh} = \mathcal{C} \times_R R^{sh}$ is a
minimal proper regular model for $E/R^{sh}$. Indeed,
$\mathcal{C}^{sh}$ is proper over $R^{sh}$ since proper morphism
are stable under base extension. Next, $\mathcal{C}^{sh}$ is
regular since $\mathcal{C}$ is regular and $R^{sh}$ is flat and
unramified over $R$. Finally, the minimality of $\mathcal{C}^{sh}$
is a consequence of the construction of the minimal proper regular
model in terms of a regular model with all exceptional curves
blown down.
Now let $\mathcal{E}^{sh} = \mathcal{E} \times_R R^{sh}$. Then
$\mathcal{E}^{sh}$ is the largest open subscheme of
$\mathcal{C}^{sh}$ which is smooth over $R^{sh}$. Thus,
$\mathcal{E}^{sh}/R^{sh}$ is a N\'{e}ron model for $E/K^{sh}$ by
the previous case. We shall use this to prove that $\mathcal{E}/R$
is a N\'{e}ron model for $E/K$.
We need to verify that $\mathcal{E}/R$ has the N\'{e}ron mapping
property. Let $\mathcal{X}/R$ be a smooth $R$-scheme with generic
fiber $X/K$ and let $\phi_K: X_{/K} \to E_{/K}$ be a rational map
defined over $K$. We need to show that $\phi_K$ extends to a
unique $R$-morphism $\mathcal{X} \to \mathcal{E}$. Let us consider
the extension of $\mathcal{X}$ and $\phi_K$ to $R^{sh}$, let
$\mathcal{X}^{sh} = \mathcal{X} \times_R R^{sh}$ and $\phi_K^{sh}:
X_{/K^{sh}}^{sh} \to E_{/K^{sh}}$. Then the scheme
$\mathcal{X}^{sh}$ is smooth over $R^{sh}$ since smoothness is
closed under base change, so by the N\'{e}ron mapping property for
$\mathcal{E}^{sh}$ we have that $\phi_{K}^{sh}$ extends to a
unique morphism $\phi_R^{sh}: \mathcal{X}_{/R^{sh}}^{sh} \to
\mathcal{E}_{/R^{sh}}^{sh}$. Thus we obtain the following
commutative diagram
\[
\begin{CD}
\mathcal{X}^{sh} @> \phi_{R}^{sh}>>\mathcal{E}^{sh}\\
@VV V @VV V \\
\mathcal{X} @> \phi_K>> \mathcal{E},
\end{CD}
\]
where the top row is obtained from the bottom row using the base
extension $\spec R^{sh} \to \spec R$. The strict Henselization
$R^{sh}$ is faithfully flat over $R$ (see [1], 2.4, corollary 9),
so we are exactly in a situation to apply faithfully flat descent
(see [1], Chapter 6) to conclude that the rational map on the
bottom row is, in fact, a morphism. Therefore $\mathcal{E}/R$ has
the N\'{e}ron mapping property, so $\mathcal{E}/R$ is a N\'{e}ron
model for $E/K$ in the case when $R$ is a discrete valuation ring.
Let us now consider the general case when $R$ is a Dedekind
domain. By our previous considerations for each prime $\wp \in
\spec R$, the localization $\mathcal{E} \times_R R_{\wp}$ is a
N\'{e}ron model for $E$ over $K_{\wp}$. By Corollary 1 if we fix a
Weierstrass equation $\mathcal{W}/R$ for $E/K$ and if we let $S
\subset \spec R$ be the set of primes for which $\mathcal{W}$ has
bad reduction, then the part of $\mathcal{W}$ lying over $R_{S}$
$\mathcal{W} \times_R R_S$ is a N\'{e}ron model for $E$ over
$R_S$. This gives a N\'{e}ron model over a dense open subset of
$\spec R$ and gluing it to the localized models over the finitely
many bad fibers we get a N\'{e}ron model over all of $\spec R$.
Thus the proof of Theorem 3 is complete.
\end{proof}
\section{Mazur's Theorem}
Let $H^r(X,\ )$ denote the cohomology with compact support taken
over the $f p \ q f$ site. The goal of this section is to relate
the Shafarevich-Tate group with the groups $H^1(X, \mathcal{A})$
and $H^1(X, \mathcal{A}^0)$, which we shall define shortly. We
shall work with arbitrary abelian variety since the argument that
follows works in this generality, something which is certainly not
the case in the preceding sections.
Let $A$ be an abelian variety over a number field $K$. Let $K_v$
be the completion of $K$ with respect to the valuation $v$. We can
consider the standard maps that serve to define the
Shafarevich-Tate group \beqa w_v: H^1(\spec K, A_{/K}) \to
H^1(\spec K_v, A_{/K_v}).\eeqa We can now form the following two
subgroups of $H^1(\spec K, A_{/K})$ \beqa \Sigma = \cap \ker (w_v)
& \textrm{nonarchimedean } v,\eeqa \beqa \Sha = \cap \ker (w_v) &
\textrm{all } v.\eeqa As we can see from the definitions $\Sha$ is
the Shafarevich-Tate group and $\Sigma$ is slightly bigger as we
exclude the archimedean valuations. More specifically, from the
definitions above we see that there is an exact sequence \beqa 0
\to \Sha \to \Sigma \to \bigoplus_v H^1(\bar{K}_v/K_v,
A(\bar{K}_v)),\eeqa where the sum on the right is taken over all
real archimedean valuations $v$.
When the topological group $A(\bar{K}_v)$ is connected
$H^1(\bar{K}_v/K_v, A(\bar{K}_v)) = 0$, so if $A(\bar{K}_v)$ is
connected for all real valuations $v$ of $K$, then $\Sha =
\Sigma$. In any case the exact sequence above implies that the
quotient of $\Sigma$ by $\Sha$ is a group of exponent 2, the order
of which depends on the structure of $A_{K_v}$ for all real
valuations $v$ of $K$, but certainly bounded. We are going to give
a cohomological description of $\Sigma$.
Let $\mathcal{O}_K$ be the ring of integers of $K$, let $X = \spec
\mathcal{O}_K$, and let $\mathcal{A}/\mathcal{O}_K$ be the
N\'{e}ron model of $A_{/K}$.
For each closed point $x \in X$, the fiber $\mathcal{A}_x$ is a
smooth commutative group scheme over $k(x)$. Let us denote by
$\mathcal{A}_x^0 \subset \mathcal{A}_x$ its connected component
and by $\mathcal{Z}_x \subset \mathcal{A}_x$ its complement. Then
$\mathcal{Z}_x$ is non-empty for only finitely many points $x$ and
thus $\mathcal{Z} = \cup \mathcal{Z}_x$ is a closed subscheme of
$\mathcal{A}$. Let us denote by $\mathcal{A}^0 \subset
\mathcal{A}$ its open complement. Then $\mathcal{A}^0$ is an open
subgroup scheme of $\mathcal{A}$. Let $F$ be the quotient of
$\mathcal{A}$ by $\mathcal{A}^0$ regarded as shaves for the $f p \
q f$ topology. We get \beqa 0 \to \mathcal{A}^0 \to \mathcal{A}
\to F \to 0.\eeqa Since $\mathcal{A}$ and $\mathcal{A}^0$ are
smooth group schemes the cohomology of the above sequence remains
the same if computed for the $f p \ q f$, smooth, or \'{e}tale
topologies. By its definition, $F$ is a skyscraper sheaf - it is
zero outside the finite set of $x \in X$ such that $\mathcal{A}_x$
is disconnected.
Now since $X$ is normal, from the N\'{e}ron mapping property it
follows that there is an inclusion \beqa i: H^1(X,A)
\hookrightarrow H^1(\spec K, A_{/K}).\eeqa
Here is the main result about $\Sigma$ that we are going to prove.
\begin{thm}
The inclusion $i$ sends the image of $H^1(X, \mathcal{A}^0) \to
H^1(X, \mathcal{A})$ isomorphically to $\Sigma$.
\end{thm}
\begin{proof}
Let $I$ denote the image of $H^1(X, \mathcal{A}^0)$ in $H^1(X,
\mathcal{A})$. Our goal is to show that $I = \Sigma$ and to do
that we shall show both inclusions. Let us first show that $I
\subset \Sigma$. Fix a nonarchimedean valuation $v$ and let us
consider the following commutative diagram
\[
\begin{CD}
H^1(X_v, F) @< \cong << H^1(X_v, \mathcal{A}) @> >> H^1(K_v, A)\\
@AA A @AA A @AA A \\
H^1(X, F) @< << H^1(X, \mathcal{A}) @> >> H^1(K, A).
\end{CD}
\]
The left-hand square comes from the long exact sequence on
cohomology associated to the sequence $0 \to \mathcal{A}^0 \to
\mathcal{A} \to F \to 0$. The right-hand square comes from the
inclusion map $i$ and its analogous when we consider the
completion with respect to $v$.
Now, given the definition of $\Sigma$, to show that $I \subset
\Sigma$ we need to show that $I$ goes to zero under the
composition $H^1(X, \mathcal{A}) \to H^1(K,A) \to H^1(K_v, A)$.
But from Lang's Theorem ([3]) it follows that $H^1(X_v, F)
\leftarrow H^1(X_v, \mathcal{A})$ is an isomorphism. Also, $I$
goes to zero in $H^1(X,F)$ because of the long exact sequence on
cohomology asspciated to $0 \to \mathcal{A}^0 \to \mathcal{A} \to
F \to 0$. Thus, $I$ goes to zero in $H^1(X_v, \mathcal{A})$, and
consequently in $H^1(K_v, A)$, which completes the proof of $I
\subset \Sigma$.
Let us now also show that $\Sigma \subset I$. Take an element $y
\in \Sigma$. According to Mazur there is a finite set of primes $S
\subset X$ containing all primes of bad reduction for
$\mathcal{A}$ such that $y \in H^1(X-S, \mathcal{A})$. Then Mazur
asserts that we have the following diagram
\[
\begin{CD}
0 @> >> H^1(X, \mathcal{A}) @> >> H^1(X - S, \mathcal{A}) @> >> \bigoplus_{p \in S} H^2_.(X_p, \mathcal{A})\\
@V VV @V a VV @V b VV @V \cong VV \\
0 @> >> \bigoplus_{p \in S} H^1(\hat{X}_p, \mathcal{A}) @>
>> \bigoplus_{p \in S} H^1(K_p, A) @> >> \bigoplus_{p \in S} H^2_.(\hat{X}_p, \mathcal{A})
\end{CD}
\]
The horizontal lines come from relative cohomology exact sequences
and the zeros on the left and the right-hand vertical isomorphism
follow from properties of relative cohomology. Moreover, Lang's
Theorem ([3]) and the fact that $F$ has trivial support on $X - S$
give an isomorphism $H^1(X,F) \to \bigoplus_{p \in S}
H^1(\hat{X}_p, \mathcal{A})$.
We need to show that $y \in I$. But $b(y) = 0$ since $y \in
\Sigma$, so using the right-hand vertical isomorphism we see that
$y$ goes to zero in $\bigoplus_{p \in S} H^2_.(X_p, \mathcal{A})$.
Consequently, it comes from an element $z \in H^1(X,
\mathcal{A})$. Then $a(z)$ must be zero by the exactness of the
second row, so $z$ goes to zero in $H^1(X, F)$ using the
isomorphism we discussed at the end of the previous paragraph. But
this is exactly the condition for $y \in I$, so $\Sigma \subset I$
and the proof of the theorem is completed.
\end{proof}
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{\bf References:} \\
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$[1]$ Bosch, S., Lutkebohmert, W., Raynaud, M., \textit{N\'{e}ron Models}, Springer, Berlin, 1990. \\
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$[2]$ Harthshorne, R., \textit{Algebraic Geometry}, Springer-Verlag, New York, 1977. \\
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$[3]$ Lang, S., \textit{Algebraic groups over finite fields}, Amer. J. Math 78, no. 3, 530 - 561, 1959.\\
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$[4]$ Mazur, B., \textit{Rational Points of Abelian Varieties with
Values in Towers of Number Fields }, Inventiones math. 18, 183 -
266, 1972.\\
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$[5]$ Silverman, J., \textit{Advanced Topics in the Arithmetic of Elliptic Curves}, Springer, New York, 1994. \\
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$[6]$ Silverman, J., \textit{The Arithmetic of Elliptic Curves}, Springer, New York, 1986. \\
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