\chapter{Visibility Theory}

The main goal of this chapter is to introduce and study a subgroup of the Shafarevich-Tate
group, which is called the \emph{visible subgroup} and was first introduced by B. Mazur 
\cite{cremona-mazur}. 
One of the reasons why this subgroup is interesting to study is that it is always a finite 
subgroup of the Shafarevich-Tate group, so in
particular, it can be used to produce elements of finite order. Another strong motivation to
look at the visible subgroup is the fact that every element of the Shafarevich-Tate group can
be visualized somewhere (the visualization theorem). We provide a method 
(the visibility theorem) for producing visible elements of the Shafarevich-Tate group. 

\section{Visible Subgroups of $H^1(K, A)$ and $\Sha(A / K)$}
Suppose that $A$ and $J$ are abelian varieties, defined over a number field $K$ and let
$i : A \ra J$ be a morphism of abelian varieties over $K$.\footnote{Note that $J$ is not
necessarily a Jacobian of a curve. The only reason we use the letter $J$ is that in most of
the computational examples that we will provide later on, we will be using Jacobians of modular
curves.}

\begin{defn}
The \emph{visible subgroup} of $H^1(K, A)$ with respect to $i$ and
$J$ as
$$
\textrm{Vis}_J^{(i)}(H^1(K, A)) := \textrm{ker }\{H^1(K, A) \ra H^1(K, J)\},
$$
where $H^1(K, A) \ra H^1(K, J)$ is the map on cohomology, induced from $i$.\footnote{In some
papers, no superscript $(i)$ is used. The only reason we use it here is because we
do not necessarily require that the morphism $i$ be an embedding.}
\end{defn}

The notion is useful, because it relates to the geometric interpretation of the elements of
$H^1(K, A)$ as elements of the Weil-Ch\^atelet group in the case when $A$ is a subvariety of
$J$, i.e. when the morphism $i : A \ra J$ is a closed immersion. To see the relation,
consider the short exact sequence of abelian varieties
$$
0 \ra A \ra J \ra C \ra 0,
$$
where $C$ is the quotient $J / A$. Write the long exact sequence on Galois cohomology
$$
0 \ra A(K) \ra J(K) \ra C(K) \ra H^1(K, A) \ra H^1(K, J) \ra \dots.
$$
Using the definition of the visible subgroup, we can extract the following short exact
sequence
$$
0 \ra J(K) / A(K) \ra C(K) \ra \textrm{Vis}_J^{(i)}(H^1(K, A)) \ra 0.
$$
Let $c$ be a visible cohomology class. Then there exists a $K$-rational point $x \in C(K)$,
which maps to $c$. The fiber over $x$ for the map $J \ra C$ is a subvariety of $J$, which
when equipped with the natural action of $A$ becomes a principal homogeneous space, and
therefore represents an element of $WC(A/K)$. This element corresponds to the cohomology
class $c$. Thus, the element $x \in C(K)$ \emph{visualizes} the cohomology class $c$. The
following statement follows almost directly from the above remarks.

\begin{cor}
For any embedding $i : A \ra J$, the visible subgroup Vis$_J^{(i)}(H^1(K, A))$ is finite.
\end{cor}

\begin{proof}
The group $H^1(K, A)$ is a torsion group, and $C(K)$ is finitely generated, so the surjectivity
of the homomorphism $C(K) \ra \textrm{Vis}_J^{(i)}(H^1(K, A))$ implies that
$\textrm{Vis}_J^{(i)}(H^1(K, A))$ must be finite.
\end{proof}

Next, we define the visible subgroup of $\Sha(A/K)$ with respect to the morphism
$i : A \ra J$.

\begin{defn}
The \emph{visible subgroup} of $\Sha(A/K)$ with respect to the map $i$ is defined as
$$
\textrm{Vis}_J^{(i)}(A) := \textrm{Vis}_J^{(i)}(H^1(K, A)) \cap \Sha(A/K).
$$
\end{defn}

The above corollary implies that the visible subgroup of $\Sha(A/K)$ is always finite.

\section{The First Property of Visibility}
The first interesting property, related to visibility is the fact that each element
of $H^1(K, A)$ becomes visible for some abelian variety $J$ and a closed immersion
$i : A \hookrightarrow J$.

\begin{thm}[Visualization Theorem]
Let $c \in H^1(K, A)$ be any cohomology class. Then there exists an abelian variety $J$ and
a closed immersion $i : A \hookrightarrow J$, such that $c \in \textrm{Vis}_J^{(i)}(H^1(K, A))$.
\end{thm}

An essential ingrediant of the proof will be the existence of Weil restriction of scalars.
We will sketch the important points of the construction. A thorough discussion of this
subject is presented in \cite[\S7.6]{neronmodels}.


\begin{prop}
Suppose that $L / K$ is a field extension of number fields.\footnote{The same argument
applies for the general case, but since we will be considering abelian varieties over number
fields, we will not worry too much about the general case.}
Let $X'$ be a scheme of finite type over $L$. There exists a scheme $X$ of finite type over
$K$, such that the scheme $X$ represents the following contravariant functor
$$
\textrm{Res}_{L / K}(X') : (\textrm{Sch} / K) \ra \textrm{\{Sets\}},\
S \mapsto \textrm{Hom}_L(S \times_K L, X).
$$
Moreover, smoothness is carried over from the scheme $X'/L$ to $X / K$, i.e. if $X'$ is smooth,
then so is $X$. We denote the scheme $X$ by Res$_{L/K}(X')$.
\end{prop}

Before proceeding further, we will explain the intuition behind the construction of
Weil Restriction of scalars. For simplicity, we will work with varieties over fields, and even
the simpler case of elliptic curves.
Consider the elliptic curve
$$
E : Y^2Z = X^3 + XZ^2 + Z^3
$$
over the number field $K = \Q(\sqrt{5})$.
The goal is to construct a variety over $\Q$, whose $\Q$-rational are in one-to-one
correspondence to the $K$-rational points of $E$.
 
First, it suffices to look at the affine patch, corresponding to $Z \ne 0$ and to construct the 
restriction of scalars only for that patch. So we can assume $Z = 1$, $X = x$ and $Y = y$. 
One can use the following idea: choose any $\Q$-basis for $K$, e.g.
$\{1, \sqrt{5}\}$ and write $\ds x = x_1 + \sqrt{5}x_2$, $\ds y = y_1 + \sqrt{5}y_2$
for some indeterminates $x_1, x_2, y_1, y_2$. After plugging $x$ and $y$ into the
Weierstrass equation and equating the coefficients in front of the elements of the basis,
we get the variety
$$
X = \langle x_1^3 + 5x_1x_2^2 - y_1^2 - 5y_2^2 + x_1 + 1,\ 3x_1^2x_2 + 5x_2^3-2y_1y_2-x_2\rangle
$$
defined over $\Q$ whose $\Q$-rational points are in one-to-one correspondence with the
$K$-rational points on the affine patch of the elliptic curve $E$ fixed above. 
Indeed, we have the correspondence $(x, y) \leftrightsquigarrow (x_1, x_2, y_1, y_2)$.

This example strongly suggests how one can generalize the construction for arbitrary
schemes of finite type. We will only sketch the construction. For more formal and rigorous
treatment of the existence of Weil restriction of scalars, we refer the reader to 
\cite[\S7.6]{neronmodels}.

\begin{proof}[Sketch of Construction:]
Let $[L : K] = n$. We can reduce the question to the case when $X'$ is affine (for the general
case, we glue local data), i.e.
$$
X' = \Spc L[y_1, y_2, \dots , y_k]/I.
$$
Let $I = \langle f_1, f_2, \dots, f_s\rangle$ for polynomials $f_i \in L[y_1, \dots, y_k]$.
Choose a basis $\{e_1, \dots, e_n\}$ of $L / K$ and write $\ds y_i = \sum_{j = 1}^n x_{ij}e_j$,
where $x_{ij}$'s are indeterminates. We express the coefficients of the $f_i$'s as linear
combinations of the basis elements. Next, we plug the $y_j$'s into each of the
polynomials $f_i$ and after using the multiplication table for the basis
$\{e_1, \dots, e_n\}$ and taking the coefficients in front of each of the basis elements,
we obtain $n$ polynomials $g_{i1}, g_{i2}, \dots, g_{in}$ from $f_i$, such that
$g_{ij} \in K[x_{11}, \dots, x_{1n}, \dots, x_{kn}]$.
Consider the ideal $J = {\langle g_{ij}\rangle_{i=1}^k}_{j=1}^n$ and look at the scheme
$$
X = \Spc K[x_{11}, \dots, x_{kn}] / J.
$$
We claim that $X$ represeents the functor Res$_{L/K}(X')$. All we have to check is that
for any $K$-algebra $A$, there is a bijection
$$
\textrm{Hom}_L(\Spc A \otimes_K L, X') \ra \textrm{Hom}_K(\Spc A, X),
$$
which is functorial in $A$.\footnote{Hom$_K$ denotes morphisms of schemes over $K$} But
this is easy to check from the definition of $X$, because
it is equivalent to construct a bijection
$$
\textrm{Hom}_L(L[y_1, y_2, \dots, y_k]/I, A \times_K L) \ra
\textrm{Hom}_K(K[y_{11}, y_{12}, \dots, y_{1n}, \dots y_{kn}]/J, A),
$$
which can be constructed explicitely.\footnote{In this case Hom$_K$ denotes homomorphisms
of $K$-algebras}
\end{proof}

We are now ready to give a proof of the visualization theorem. This proof was discovered
recently by William Stein and the author and makes an explicit use of the simple transitive
action of an abelian variety on its principal homogeneous spaces.

\begin{proof}[Proof of Theorem 4.2.1.]
Recall that a cohomology class $c \in H^1(K, A)$ is trivial if and only if
the corresponding principal homogeneous space $C$ to $c$ has a $K$-rational point. Intuitively,
to trivialize $c$, it is enough to consider an extension $L / K$, so that $C$ has an
$L$-rational point. After choosing such an extension $L/K$, we obtain res$_{L/K}(c) = 0$,
where res$_{L/K} : H^1(K,A) \ra H^1(L,A)$ is the restriction map on
Galois cohomology.

Let $C/K$ be a principal homogeneous space (a torsor) for $A/K$, such that the class
$[C] \in WC(A/K)$ corresponds to the element $c \in H^1(K, A)$. Let $A \times C \ra C$
be the simple, transitive action of $A$ on $C$.\footnote{In order to avoid confusion with the group 
law on $A$, we denote the action of $A$ on $C$ by $\oplus$.} Let $P \in C(L)$ be the $L$-rational point.
We can define a morphism $\varphi : A_L \ra C_L$ be the morphism, defined by
$\varphi(a) = a \oplus P$.\footnote{If $X$ is a scheme over $K$ and $L$ is an extension
of $K$, then $X_L$ will denote the scheme $X \times_K L$} Since $A$ acts simply transitively
on $C$, then $\varphi$ is an isomorphism. Let $\psi = \varphi^{-1}$.

The first important step of the proof is to recover the group law on $A_L$ in terms of
the morphisms $\varphi$ and $\psi$. The main idea for proving this is to use
rigidity theorem for abelian varieties. Define a morphism
$$
\phi : A_L \times A_L \ra A_L
$$
by $\phi(a, a') = \psi(a \oplus \varphi(a'))$. We compute $\phi(a, 0) = \psi(a \oplus P) = a$. Also,
$\phi(0, a) = \psi(0 \oplus \varphi(a)) = \psi(a \oplus P) = a$. Therefore, if $\mu_L : A_L \times A_L
\ra A_L$ is the multiplication map, then $\phi - \mu_L$ satisfies the hypothesis for the
rigidity theorem, therefore $\phi = \mu_L$.

Let $J = \textrm{Res}_{L/K}(A_L)$. The isomorphism $\psi : C_L \ra A_L$ induces an inclusion
$C(K) \hookrightarrow C(L) \cong A_L(L) \cong J(K)$ and the identity morphism
$\textrm{id} : A_L \ra A_L$ induces an inclusion $A(K) \hookrightarrow A_L(L) \cong J(K)$.
These inclusions correspond via Yoneda's lemma to injective morphisms $A \ra J$ and $C \ra J$.
Since these morphisms are proper, then \cite[\S8.11.5]{ega4_3} implies that they are closed immersions.

Since $A$ is defined over $K$, we may view $J_L$ as a product of
$n$ copies of $A_L$, i.e.
$$
J_L \cong \prod_{i=1}^n A_L.
$$
The closed immersion $C \hookrightarrow J$ base extended to $L$
gives a morphism $C_L \ra J_L$. This morphism is the map, sending
$$
x \mapsto (\psi_1(x), \psi_2(x), \dots, \psi_n(x)),
$$
where $\psi_i : C_L \ra A_L$ are the conjugates of $\psi : C_L \ra
A_L$, obtained by applying the $n$ different embedding $L
\hookrightarrow \overline{K}$ to $\psi$, which fix the field $K$.
Note that each of the $\psi_i$'s is a morphism $C_L \ra A_L$,
since both $C$ and $A$ are defined over $K$.

We claim that the image of $C_L$ inside $J_L$ is a translate of
$A_L$. The morphism $\ds A_L \hookrightarrow J_L \cong
\prod_{i=1}^n A_L$ is precisely the diagonal embedding. To
determine the image of $C_L$, we consider the morphism
$$
A_L \xrightarrow{\phi} C_L \ra \prod_{i=1}^n A_L,
$$
which maps $a \mapsto (\psi_1(\varphi(a)), \psi_2(\varphi(a)),
\dots \psi_n(\varphi(a)))$. The image of $a \in A_L(\overline{K})$
is the unique $b$, such that $b \oplus \sigma_i(P) = a \oplus P$. But
the action is transitive, so we get $(- b + a) \oplus P = \sigma_i(P)$,  
which means that $b = a - \psi(\sigma_i(P))$. This shows
that the image of $C_L$ in $J_L$ is a translate of $A_L$ by
$(- \psi(\sigma_1(P)), -\psi(\sigma_2(P)), \dots,
- \psi(\sigma_n(P)))$, so we are done.
\end{proof}

\section{Producing Upper Bound on the Visibility Dimension}

The theorem from the previous section gives rise to an interesting question,
relating the order of a cohomology class $c$ and the minimal dimension of an abelian
variety $J$, for which $c$ is visible, under a closed immersion $i : A \hookrightarrow J$.

\begin{defn}
Let $c \in H^1(K, A)$. The \emph{visibility dimension} of $c$ is the minimal dimension of
an abelian variety $J$, such that $c \in \textrm{Vis}_J^{(i)}(H^1(K, A))$.
\end{defn}

First of all, we produce an upper bound for the visibility dimension of a cohomology class
$c \in H^1(K, A)$, in terms of the order $n$ of that element and the dimension of $A$.

\begin{lem}
Suppose that $G$ is a group and $M$ is a finite $G$-module. Let $c \in H^1(G, M)$ be any
cohomology class. Then there is a subgroup $H \subseteq G$, such that the restriction of
$c$ to $H^1(H, M)$ is trivial and $[G : H] \leq |M|$.
\end{lem}

\begin{proof}
Consider $H = \textrm{ker}(f)$, where $f : G \ra M$ is any representative of
the cohomology class $c$. The map $f$ satisfies the cocycle condition
$$
f(\sigma \tau) = \sigma f(\tau) + f(\sigma).
$$
Clearly, the restriction of $c$ to $H^1(H, M)$ is trivial. To bound
the dimension, we construct an injection $G / H \hookrightarrow M$, by sending $\tau H \mapsto
f(\tau)$. By the definition of $H$, this map is well defined. Suppose $f(\sigma) = f(\tau)$.
Then, the cocycle condition
$$
f(\tau) = f(\sigma (\sigma^{-1}\tau)) = \sigma f(\sigma^{-1}\tau) + f(\sigma).
$$
Thus, $\sigma f(\sigma^{-1}\tau) = 0$, i.e. $f(\sigma^{-1}\tau) = 0$, which means that
$\sigma^{-1}\tau \in H$. This proves injectivity of the map $G / H \hookrightarrow M$, and so
the bound follows.
\end{proof}

\begin{prop}
The visibility dimension of any $c \in H^1(K, A)$ is at most $d \cdot n^{2d}$, where $n$ is the
order of $c$ in $H^1(K, A)$ and $d$ is the dimension of $A$.
\end{prop}

\begin{proof}
It follows from the proof of the Visualization Theorem (Theorem 4.2.1) that the dimension 
of $J$, which was constructed using Weil restriction of scalars is at
most $[L : K] \cdot \textrm{dim }A$. Thus, we need an upper bound
for the degree of the extension $[L : K]$. To get this, consider
the surjective map $H^1(K, A[n]) \ra H^1(K, A)[n]$, which is
induced from the long exact sequence on Galois cohomology. Since
$c \in H^1(K, A)[n]$, it suffices to trivialize one of its
preimages. By the above lemma, there exists a finite index
subgroup of Gal$(\overline{K} / K)$ (which by Galois theory
corresponds to some field extension $L / K$), such that $c$ is
trivialized in $H^1(L, A[n])$ and the index of the subgroup $[L :
K]$ is at most $|A[n]| = n^{2d}$. Thus, one can choose $L$, so
that $[L : K] \cdot \textrm{dim }A \leq d \cdot n^{2d}$, so we get
an upper bound for the dimension.
\end{proof}

\section{Smooth and Surjective Morphisms}

\subsection{Flat, Smooth and \'Etale Morphisms}
In order to make sense of the notion of continuous family of
schemes, we need to introduce the notion of flatness.

Recall from commutative algebra that a morphism $f : A \ra B$ of rings is \emph{flat} if
the functor $M \mapsto M \otimes_A B$ from $A$-modules to $B$-modules is exact. Since
\emph{flatness} is a local property, in order to check that $f$ is flat, it suffices to
check that the homomorphism of local rings $A_{f^{-1}(\mathfrak m)} \ra B_{\mathfrak m}$ is
flat for every maximal ideal $\mathfrak m \subset B$.

\begin{defn}
A morphism
$\varphi : Y \ra X$ of schemes is \emph{flat}, if the homomorphisms of
local rings $\shf_{X, \varphi(y)} \ra \shf_{Y, y}$ is flat for every $y \in Y$.
\end{defn}

\begin{example}
Any finite, surjective morphism $f : X \ra Y$ of nonsingular varieties over algebraically closed 
fields is flat. 
\end{example}

The notion of flatness corresponds to continuous family of manifolds in differential topology.
Indeed, $\varphi : Y \ra X$ being flat implies that all points $x \in X$, such that
$\varphi^{-1}(x)$ is nonempty behave like regular values, i.e. if $Y_x := \varphi^{-1}(x)$, then
$$
\textrm{dim } Y_x = \textrm{dim } Y - \textrm{dim } X.
$$

Next, we use the notion of flatness to define the relative notion of nonsingular varieties.

\begin{defn}
A morphism $\varphi : Y \ra X$ of schemes of finite type over a field $k$ is
\emph{smooth of relative dimension $n$},
if:\\
(1) $\varphi$ is flat; \\
(2) if $Y' \subseteq Y$ and $X' \subseteq X$ are irreducible components, such
that $\varphi(Y') \subseteq X'$, then
$$
\textrm{dim }Y' = \textrm{dim }X' + n;
$$
(3) for each point $y \in Y$, one has
$$
\textrm{dim}_{k(y)}(\Omega_{Y / X} \otimes k(y)) = n.
$$
\end{defn}

\begin{example}
If $Y = \Spc k$ and $k$ is algebraically closed, then $X$ is smooth 
over $k$ of relative dimension $n$ if and only if $X$ is regular of dimension 
$n$. In particular, if $X$ is irreducible and separated, then $X$ is smooth if and only 
if $X$ is a variety.  
\end{example}

Finally, we can define the notion of an \'etale morphism.
\begin{defn}
A morphism $\varphi : Y \ra X$ is called \emph{\'etale} if it is
smooth and of relative dimension zero.
\end{defn}

\begin{example}
Open immersions are smooth morphisms of relative dimension zero, so they are \'etale. 
\end{example}


\subsection{Henselian Rings and Strictly Henselian Rings}
We start by introducing a special class of rings $R$, called \emph{henselian} rings.
Roughly speaking, those are rings, for which the Hensel lemma is true.

\begin{defn}
Let $R$ be a local ring with residue field $k$. The ring $R$ is called \emph{henselian}
if, for every monic polynomial $P \in R[T]$, all $k$-rational zeros of the residue class
$\overline{P} \in k[T]$ lift to $R$-rational zeros of $P$. If, in addition, the residue
field $k$ is separably closed, the ring $R$ is called \emph{strictly henselian}.
\end{defn}

The following statement deals with some properties of strictly henselian rings.
\begin{prop}
Let $R$ be a local ring. The following are equivalent: \\
(1) $R$ is a strictly henselian ring. \\
(2) For all \'etale morphisms $f : X \ra \Spc R$ and for all points $x \in X$, such that
$f(x) = s$ is a closed point of $\Spc R$, there exists a section $u : \Spc R \ra X$
(i.e. $S$-morphism), such that $u(s) = x$.
\end{prop}

\begin{proof}
The proposition follows from \cite[\S18.8.1]{ega4_4}.
\end{proof}

\subsection{Surjectivity of $[n] : G(R) \ra G(R)$}

The main goal of this whole section is to prove the following theorem:

\begin{prop}
Let $A$ be an abelian variety over a field $K$, which is the residue field of a strictly
henselian discrete valuation ring $R$. Let $x \in A(K)$, such that the reduction
of $x$ lies in the identity component of the closed fiber of the N\'eron model
$\mathcal A$ of $A$. Then for any integer $n$, prime to the residue characteristic of $R$,
one has $x \in nA(K)$.
\end{prop}


We start by several technical lemmas which will be used in the proof of the proposition.
\begin{lem}
Let $G$ be a smooth, commutative group scheme of finite type over an arbitrary base scheme $S$.
Let $n$ be an integer which is not divisible by the residue characteristic of
the local ring at every point $s \in S$. Then the multiplication by $n$ morphism
$n_G : G \ra G$ is \'etale.
\end{lem}

\begin{proof}
The lemma follows from Lemma 2(b) of \cite[\S7.3]{neronmodels}.
\end{proof}


\begin{lem}
Suppose that $U \subseteq G$ is an open dense group subscheme of $G$. Then $U = G$.
\end{lem}

\begin{proof} Since the underlying topological spaces for $G$ and $G \times_R K$ are the same,
it suffices to prove the statement for a commutative group scheme $G$ over a field $K$. Let
$G$ be a commutative group scheme over a field $K$.  It suffices to
prove that
$G \times_K \overline{K} = (U \times_K \overline{K}) \cdot (V \times_K \overline{K})$.
Therefore, one can assume that $K$ is algebraically closed. In this case, it suffices to prove
that $U$ contains all closed points of $G$. Indeed, $U$ contains all generic points; otherwise
$U^c$ will be the closure of a generic point, which is impossible.

Suppose that $x \in G$ is a closed point. Since $K$ is algebraically closed, then $x$ is
rational, so $U$ and $U \cdot x$ are both open. Thus, they have at least one common
closed point $v$, so there is $u \in U$, such that $ux = v$, i.e.
$x = u^{-1}v \in U$.
\end{proof}


\begin{lem}
Suppose that $G$ is a finite-type commutative group scheme over a strictly henselian
local ring $R$ and the fibers of $G$ over $R$ are geometrically connected.\footnote{We say
that a scheme $X$ over a field $K$ is geometrically connected if $X \times_K \overline{K}$
is connected} The multiplication map by $n$ map
$$
[n] : G(R) \ra G(R)
$$
is surjective when $n \in R^{\times}$.
\end{lem}

\begin{proof} Choose a point $x \in G(R)$. Then $x$ corresponds to a morphism
$x : \textrm{Spec }R \ra G$. Form the following pullback diagram

\begin{equation*}
\xymatrix{
Y_x \ar[r]^{\psi}\ar[d] & \textrm{Spec }R \ar[d]^{x} \\
G \ar[r]^{n_G} & G \\
}
\end{equation*}

The surjectivity of $[n] : G(R) \ra G(R)$ will follow if we prove that there is a
section $\Spc R \ra Y_g$. Indeed, note that $\Spc R \ra Y_g \ra G$ corresponds to a
$R$-rational point on $G$, which is mapped to $x \in G(R)$ under $n_G$.

Since $R$ is strictly henselian and by Proposition 4.4.8, it suffices to prove that $Y_g \ra \Spc R$
is \'etale and the closed fiber of $Y_g$ is nonempty. The last two statements would
evidently follow if we prove that $n_G : G \ra G$ is \'etale and surjective. \'Etaleness follows 
from Lemma 4.4.10. The surjectivity of
$n_G$ follows from the fact that the image of $n_G$ must be an open, dense subgroup 
scheme, so by Lemma 4.4.11 the morphism $n_G$ must be surjective. 
\end{proof}

\begin{lem}
Let $X$ be a connected scheme over an arbitrary field $K$. Suppose that
there exists at least one $K$-rational point of $X$. Then the scheme $X$ is
geometrically connected (i.e. the scheme $X \times_K \overline{K}$ is connected).
\end{lem}

\begin{proof}
This is proved in \cite[\S4.5.13]{ega4_2}. 
\end{proof}

\begin{proof}[Proof of Proposition 4.4.9.]
According to the basic properties of the N\'eron model, $A(K) \cong \mathcal A(R)$.
The image of $x \in A(K)$ under this isomorphism is a point of $\mathcal A^0(R)$.
Since $\mathcal A^0(R)$ is connected and has a $R$-rational point, the fibers of $A^0$
over $\Spc (R)$ are geometrically connected by Lemma 4.4.13. Therefore, we can apply
Lemma 4.4.12 to obtain that the multiplication by $n$ map $[n] : G(R) \ra G(R)$ is
surjective. This gives us a point $z \in A(K)$, such that $nz = x$ and we are done.
\end{proof}

\subsection{Surjectivity of the Induced Map on Generic Fibers}
Here we discuss the last bit of algebraic geometry that will be needed for the main result in
this chapter. Suppose that $\mathcal A$ and $\mathcal B$ are commutative, smooth, group
schemes over strictly Henselian local ring $R$, which are the N\'eron models of abelian
varieties $A$ and $B$ (both defined over the fraction field $K$ of $R$) and
$\phi : \mathcal A \ra \mathcal B$ is a
morphism. We discuss a condition for $\phi$, under which the induced map on the generic
fibers is always surjective.

\begin{prop}
Suppose that $\phi : \mathcal A \ra \mathcal B$ is smooth and surjective. Then the
induced morphism $\phi_K : A(K) \ra B(K)$ is surjective.
\end{prop}

\begin{proof}
The idea is very similar to the one that we used in the proof of Lemma 4.4.12.
It suffices to show that the induced map $\phi_R : \mathcal A(R) \ra \mathcal B(R)$ is
surjective. Choose a point $x \in \mathcal B(R)$ and consider the corresponding morphism
$\Spc R \ra \mathcal B$. As in the previous proof, form the pullback diagram
\begin{equation*}
\xymatrix{
Y_x \ar[r]^{\psi}\ar[d] & \textrm{Spec }R \ar[d]^{x} \\
\mathcal A \ar[r]^{\phi} & \mathcal B \\
}
\end{equation*}
It will suffice to check that the morphism $\psi : Y_x \ra \Spc R$ has a section. To do this, 
we only need to check that the closed fiber of $\psi$ has a section. But the closed fiber is smooth and 
nonempty (since $\phi$ is surjective); also, its base field is separably closed, since $R$ is strictly henselian. 
Hence, the closed fiber has an $R$-rational point by \cite[\S2.2.13]{neronmodels}.  
\end{proof}

\section{Producing Visible Elements of the Shafarevich-Tate Group}
Suppose that $A$ is an abelian variety over a number field $K$.
We describe a technique which produces visible elements of $\Sha(A / K)$. The basic
idea is that under certain conditions it will be possible to inject a weak Mordell-Weil
group of some abelian variety into $\Sha(A / K)$, so we will produce element of finite order
of $\Sha(A/K)$. The precise statement is the following

\begin{thm}[Visibility Theorem]
Let $A/K$ and $B/K$ be abelian subvarieties of $J/K$ which have finite intersection.\footnote{As before, $J$ is not
necessarily a Jacobian of a curve; the notation is used only because we will often apply
the theorem for $J$ being a Jacobian of a modular curve.} Let $N$ be the product of the
residue characteristics of the non-archimedian places of bad reduction for $B$. Suppose that
$p$ is a prime number, which satisfies the following conditions: \\
(i) $p \nmid N \cdot |(J/B)(K)_{tors}| \cdot |B(K)_{tors}|
\cdot \prod_{\nu}c_{A, \nu} \cdot c_{B, \nu}$, where $c_{A, \nu}$ and $c_{B, \nu}$ are
the Tamagawa numbers (or the orders of the component groups of the fibers 
of the N\'eron models at $\nu$);  \\
(ii) $B[p] \subset A$; \\
(iii) If $e_{\mathfrak p}$ is the ramification index of the prime ideal $\mathfrak p$, then
$e_{\mathfrak p} < p-1$ for any prime ideal $\mathfrak p$ lying above $p$. \\
Under these hypothesis, there is a natural map
$$
\varphi : B(K) / pB(K) \ra \Sha(A/K),
$$
such that the order of the kernel of $\varphi$ is at most $p^r$, where $r$ is
the Mordell-Weil rank of $A$. In particular, the map is injective if the Mordell-Weil
rank of $A$ is 0.
\end{thm}

\begin{proof}  There are two major steps for the proof of the theorem. First, we construct
a map from the weak Mordell-Weil group $B(K) / pB(K)$ to the
visible part of $H^1(K, A)$, using the hypothesis that $B[p] \subset
A$. The second step is proving that the image of $B(K) / pB(K)$
in $H^1(K, A)$ consists of locally trivial cohomology classes, which immediately implies
that this image is contained in Vis$_J^{(i)}(\Sha(A/K))$. \\

\noindent \textit{\bf Step I:} \textit{Constructing a map $B(K) / pB(K) \ra H^1(K, A)$.}

The argument we will use is purely algebraic and is based on diagram chasing. Start with the
short exact sequence
$$
0 \ra A \ra J \ra C \ra 0,
$$
where $C$ is simply the quotient $J / A$, considered over $K$. The associated long exact
sequence on Galois cohomology is

\begin{equation}\label{eqn:les}
0\ra A(K) \ra J(K) \ra C(K) \xrightarrow{\,\delta\,}
          H^1(K,A) \ra \cdots.
\end{equation}

One can construct a map $\psi : B \ra C$ by composing the inclusion map $B \hookrightarrow J$
and the map $J \ra C$. Since $B[p] \subset A$ and $A$ is the kernel of the map $J \ra C$, then
the map $\psi : B \ra C$ factors through the multiplication by $p$ map
$B \xrightarrow{\, \cdot p\,} B$. This gives us the following commutative diagram:
\begin{equation*}
\xymatrix{
& & B \ar[d] \ar[r]^{.p} \ar[dr]^{\psi} & B\ar[d] \\
0 \ar[r] & A \ar[r] & J \ar[r] & C
}
\end{equation*}

We still have not used the fact that $B(K)[p]$ is empty. We take
$K$-rational points and use this fact to get the following diagram, with exact rows
and columns:

\begin{equation*}
\xymatrix{
         & K_0\ar[d] & K_1\ar[d]& K_2\ar[d]\\
0 \ar[r] & B(K) \ar[r]^{.p}\ar[d] & B(K)\ar[dr]^{\pi} \ar[r]\ar[d]
         & B(K)/pB(K)\ar[r]\ar[d]^{\varphi}  & 0\\
0 \ar[r] & J(K)/A(K)\ar[r]\ar[d] & C(K) \ar[r] & \delta(C(K)) \ar[r] & 0\\
         & K_3
}
\end{equation*}

By definition, the visible part of $H^1(K, A)$ is the kernel of the
map $H^1(K, A) \ra H^1(K, J)$, which by the long exact sequence on Galois cohomology is
exactly the image of the map $H^0(K, C) \xrightarrow{\,\delta\,} H^1(K, A)$, which is
$\delta(C(K))$. Now, we apply the snake lemma to get an exact sequence
$$
K_0 \ra K_1 \ra K_2 \ra K_3.
$$
We can analyze further the sequence, by observing that $K_1$ is finite. This is because
the kernel of $\psi : B \ra C$ is $A \cap B$, which is finite.
Therefore, $K_1 \subset B(K)_{\textrm{tors}}$. But $B(K)$ does not contain
$p$-torsion elements. Since $K_2 \subset B(K) / pB(K)$ is a $p$-group,
then $K_1 \ra K_2$ must necessarily be the zero map. Therefore, $K_2$ injects into
$K_3 \cong J(K)/(A(K)+B(K)) \cong (J(K)/B(K))/A(K)$. Since torsion of $J(K)/B(K)$ is
contained in $(J/B)(K)_{\textrm{tors}}$, then $J(K)/B(K)$ has no $p$-torsion. Therefore,
if $A(K)$ is a torsion group, then $(J(K)/B(K))/A(K)$ has no $p$-torsion and so $K_2$ (which
is a $p$-group) is trivial, i.e. $\varphi : B(K)/pB(K) \ra H^1(K, A)$ must be injective.

More generally, suppose that the Mordell-Weil rank of $A(K)$ is $r$. The order of the
kernel of $\varphi$ is bounded from above by the order of $(J(K)/(A(K)+B(K)))[p]$.
The last group is precisely the $p$-torsion part of the cokernel of the map
$\phi : A(K) \ra J(K)/B(K)$, so the bound on that kernel follows from 

\begin{lem}
Suppose that $G$ and $H$ are finitely generated abelian groups, $G$ is
of rank $r$, and $H$ has no $p$-torsion elements. Suppose that $f : G
\ra H$ is a group homomorphism. Then $\textrm{coker}(f)[p] \leq p^r$.
\end{lem}

\begin{proof}
We may consider that $H$ is a torsion-free, since no torsion of order prime to $p$ contributes
to the order of $\textrm{coker}(f)[p]$. Thus, all the torsion of $G$ is mapped to 0 via $f$ and
so we might as well assume that $G$ is torsion-free. For free groups, we can see this by considering 
the Smith normal form of the integer matrix, corresponding ot $f$. Indeed, the new matrix we obtain 
consists of at most $r$ diagonal entries $[d_1, d_2,\dots ]$, such that $d_1 \mid d_2 \dots$, which immediately 
implies that the order of the $p$-torsion of the cokernel is at most $p^r$. 
\end{proof}

\noindent {\it {\bf Step II:} The Local Analysis.}

In the previous step we constructed a map $\varphi : B(K)/pB(K) \ra H^1(K, A)$ and
proved that the kernel has order at most $p^r$, where $r$ is the Mordell-Weil rank of $A$ (the
last statement follows from Lemma 4.5.2). We should also prove that the image of $\varphi$
consists of locally trivial cohomology classes in order to conclude that this image
lies in $\Sha(A/K)$. Consider the composition $\pi : B(K) \ra H^1(K, A)$ of the quotient
map $B(K) \ra B(K)/pB(K)$ and the map $\varphi$. Let $x \in B(K)$ be a $K$-rational point.
We want to show that for each place $\nu$, the restriction res$_\nu(\pi(x)) = 0$.

We prove this by considering the different possibilities for the place $\nu$.

\noindent {\it {\bf Case 1:} $\nu$ is archimedian.}

There is nothing to prove in the case when $\nu$ is complex
archimedian, because the local cohomology group is trivial.

If $\nu$ is real archimedian, we have
$H^1(\textrm{Gal}(\overline{K}_\nu)/K_\nu, A(K_\nu)) =
H^1(\textrm{Gal}(\C / \R), A(\R))$. But Gal$(\C/\R) \cong \Z/2\Z$
and since the order of the group kills any element of the first
cohomology, then res$_\nu(\pi(x))$ is 2-torsion. But $\pi(x)$ is
also $p$-torsion and $p$ is odd. Therefore, res$_\nu(\pi(x))$ is
both $p$-torsion and 2-torsion, which means that res$_\nu(\pi(x)) = 0$. \\

\noindent {\it {\bf Case 2:} $p \ne char(\nu)$.}

Let $m$ be the order of the component group $\Phi_{B, \nu}(k_\nu)$ of the closed
fiber $\mathcal B_{k_\nu}$ at $\nu$ of the Neron model $\mathcal B$
(i.e. the Tamagawa number $c_{B, \nu}$). Then $mx$ is in the identity component
$\mathcal B_{k_\nu}^0$. Hence, we can apply Proposition 4.4.9 for the point $mx$ and the
local field $K_\nu^{\textrm{ur}}$ (whose valuation ring is strictly Henselian) to get
that there exists $z \in B(K_\nu^{\textrm{ur}})$,
such that $pz = mx$. Now, look at res$_\nu(\pi(mx)) \in H^1(K_\nu, A(K_\nu))$.
By the discussion of the Kummer pairing in Chapter 1.1, this cohomology class is represented
by the 1-cocycle
$$
f : \textrm{Gal}(\overline{K}_\nu/K_\nu) \ra A(\overline{K}_\nu),\ \sigma \mapsto \sigma(z) - z.
$$
Since $z \in B(K_\nu^{\textrm{ur}})$, it follows that $f$ is unramified cocycle, i.e.
res$_\nu(\pi(mx))$ is an unramified cohomology class. Thus, res$_\nu(\pi(mx)) \in
H^1(K_\nu^{\textrm{ur}} / K_\nu, A(K_\nu^{\textrm{ur}}))$.

Next, we use the following relationship between the unramified cohomology and
the the cohomology of the component group.

\begin{lem}
Let $A$ be an abelian variety over a local field $K$, which is the fraction field of a
discrete valuation ring $R$ with residue field $k$.
Let $\mathcal A$ be the N\'eron model of $A$ and $\Phi(\overline{k})$ be the component
group of the closed fiber $\mathcal A_k$ of $\mathcal{A}$, i.e.
$\Phi(\overline{k}) := \mathcal{A}_{k}(\overline{k}) / \mathcal{A}_{k}^0(\overline{k})$.
Then
$$
H^1(K^{\textrm{ur}}/K, A(K^{\textrm{ur}})) = H^1(K^{\textrm{ur}}/K, \Phi(\overline{k})),
$$
where $K^{\textrm{ur}}$ denotes the maximal unramified extension of $K$.
\end{lem}

The lemma is proved in \cite[Prop.3.8]{milne:duality}. It implies that
$$
H^1(K_\nu^{\textrm{ur}}/K_\nu, A(K_\nu^{\textrm{ur}})) = H^1(K_\nu^{\textrm{ur}}/K_\nu,
\Phi_{A, \nu}(\overline{k_\nu})).
$$

The cohomology of the component group is easier to work with, because the component
group is a finite Gal$(K_\nu^{ur}/K_\nu)$-module and Gal$(K_\nu^{ur}/K_\nu)$ is a cyclic,
so we can apply the following

\begin{lem}
Suppose that $G$ is a cyclic group and $A$ is a finite $G$-module. Let
$$
h(A) := |H^0(G, A)| / |H^1(G, A)|
$$
be the Herbrand quotient of the $G$-module $A$. Then $h(A) = 1$.
\end{lem}

\begin{proof}
Let $g$ be a generator for $G$ and $A^G$ denote the fixed submodule of $A$. Let
$I_G$ denote the kernel of the homomorphism $\Z[G] \ra G$, which maps $g \ra 1$.
There is an exact sequence
$$
0 \ra A^G \ra A \xrightarrow{\cdot(g-1)}  A \ra A / I_GA \ra 0.
$$
Since $A$ is a finite module, it follows that $|A/I_GA| = |A^G|$. Next, let
$N : A \ra A$ be the homomorphism, obtained by multiplication by
$\ds N = \sum_{h \in G}h$. This homomorphism induces a homomorphism
$N^* : A / I_GA \ra A^G$. Moreover, we have an exact sequence
$$
0 \ra H^1(G, A) \ra A/I_GA \xrightarrow{N^*} A^G \ra H^0(G, A) \ra 0,
$$
which immediately implies that $|H^1(G, A)| = |H^0(G, A)|$, so $h(A) = 1$.
\end{proof}

The lemma implies that the order of $H^1(K_\nu^{\textrm{ur}}/K_\nu,
\Phi_{A, \nu}(\overline{k}_\nu))$ is equal to the order of the
component group $\Phi_{A, \nu}(\overline{k}_\nu)$. But $p \nmid
c_{A, \nu} = |\Phi_{A, \nu}(\overline{k}_\nu)|$ by assumption.
Since the order of res$_\nu(\pi(mx))$ divides $p$, it follows that
res$_\nu(\pi(mx))$ is trivial. Therefore, $m$res$_{\nu}(\pi(x)) =
0$. But $p\pi(x) = 0$ and $p \nmid m = c_{B, \nu}$, then it
follows that res$_{\nu}(\pi(x)) = 0$, i.e. $\pi(x)$ is a
locally trivial element. \\

\noindent {\it {\bf Case 3:} $char(\nu) = p$.}

Consider the maximal unramified extension $K_\nu^{\textrm{ur}}$ of $K_\nu$.
Let $\mathcal A$, $\mathcal J$ and $\mathcal C$ be the
Neron models of $A, J$ and $C$ respectively.

The first observation is that the induced sequence on the N\'eron models
$$
0 \ra \mathcal A \ra \mathcal J \xrightarrow{\psi} \mathcal C \ra 0
$$
is exact, which is a consequence of the following lemma, proved in \cite[\S7.5, Thm 4.]{neronmodels}

\begin{lem}
Suppose that $0 \ra A' \ra A \ra A'' \ra 0$ is an exact sequence of abelian
varieties over a field $K$, which is the fraction field of a discrete valuation ring $R$.
Assume that the ramification index $e = \nu(p)$ satisfies $e < p-1$, where $p$ is the residue
characteristic and $\nu$ is the normalized valuation on $R$. Let $\mathcal A'$, $\mathcal A$
and $\mathcal A''$ be the N\'eron models of $A'$, $A$ and $A''$ respectively. If $\mathcal A$
has abelian reduction, then the induced sequence
$$
0 \ra \mathcal A' \ra \mathcal A \ra \mathcal A'' \ra 0
$$
is exact and consists of abelian $R$-schemes.
\end{lem}

Hence, $\psi : \mathcal J \ra \mathcal C$ is a flat morphism, which is surjective and which has
a smooth kernel $\mathcal A$. This is enough to claim that $\psi$ is smooth
\cite[\S 2.4, Prop.8]{neronmodels}. Next, using Lemma 4.4.14, it follows that $\mathcal J(R) \ra
\mathcal C(R)$ is surjective, and therefore
$\mathcal J(K^{\textrm{ur}}_\nu) \ra \mathcal C(K^{\textrm{ur}}_\nu)$ is surjective. Therefore,
res$_\nu(\pi(x))$ is a unramified cohomology class. Using Lemma 4.5.3, we have 
$$
H^1(K^{ur}_\nu / K_\nu, A) \cong H^1(K^{ur}_\nu / K_\nu, \Phi_{A, \nu}(\overline{k}_\nu))
$$

But $A$ has good reduction at $\nu$, since $p \nmid N$, so $\Phi_{A, \nu}(\overline{k}_\nu)$ is trivial. 
Therefore $H^1(K^{\textrm{ur}}_\nu / K_\nu, \Phi_{A, \nu}(\overline{k}_\nu))$
is trivial, so res$_\nu(\pi(x)) = 0$, which completes the proof of the visibility theorem.
\end{proof}