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%% On the Hasse principle for bielliptic surfaces
%% Carmen Laura Basile and Alexei Skorobogatov
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 \markboth{\qquad On the Hasse principle for bielliptic surfaces
\hfill}{\hfill Carmen Laura Basile and Alexei Skorobogatov \qquad}

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\begin{document}

\title{On the Hasse principle for bielliptic surfaces}
\author{Carmen Laura Basile and Alexei Skorobogatov}
\date{\em To Sir Peter Swinnerton-Dyer}

%\thanks{Notes from Rational and Integral Points on Higher Dimensional
%Varieties, American Institute of Mathematics, December 11--20, 2002, by John
%Voight, \texttt{jvoight@math.berkeley.edu}}

\maketitle
\addcontentsline{toc}{chapter}{Carmen Laura Basile and Alexei
Skorobogatov, On the Hasse principle for bielliptic surfaces}
% \tableofcontents

{From} the geometer's point of view,
bielliptic surfaces can be described as 
quotients of abelian surfaces by freely acting finite
groups, that are not abelian surfaces themselves. 
Together with abelian, K3 and 
Enriques surfaces they exhaust the class of smooth
and projective minimal surfaces of Kodaira dimension $0$.
Because of their close relation to abelian surfaces,
bielliptic surfaces are particularly amenable
to computation. At the same time they display phenomena
not encountered for rational, abelian or K3 surfaces, for example, 
torsion in the N\'eron--Severi group, finite geometric Brauer group,
non-abelian fundamental group. This curious geometry is reflected
in amusing arithmetical properties of these surfaces 
over number fields. 

The behaviour of rational points on bielliptic surfaces was first studied
by Colliot-Th\'el\`ene, Swinnerton-Dyer and the second author [CSS] in
relation with Mazur's conjectures on the connected components of the real
closure of $\bfQ$-points. The second author then constructed a bielliptic
surface over $\bfQ$ that has points everywhere locally but not globally;
moreover, this counterexample to the Hasse principle cannot be explained
by the Manin obstruction [S1] (see also [S2], Ch.~8).  D. Harari [H]
showed that bielliptic surfaces give examples of varieties with a Zariski
dense set of rational points that do not satisfy weak approximation;
moreover this failure cannot be explained by the Brauer--Manin
obstruction. 

A discrete invariant of a bielliptic surface is 
the order $n$ of the canonical class in the Picard group. The
possible values of $n$ are 2, 3, 4 and 6.
The surface contructed in [S1] has $n=2$. Until now this was 
the only known 
counterexample to the Hasse principle that cannot be
explained by the Manin obstruction.
In this note we construct a similar example in 
the case $n=3$. The difference is that 
we now need to consider elliptic curves with complex multiplication.
The actual construction turns out to be somewhat
simpler than in [S1].
In contrast, for the bielliptic surfaces with $n=6$ 
we prove that the Manin obstruction to the Hasse principle
is the only one (under the assumption that the Tate--Shafarevich
group of its Albanese variety is finite).


\section{Bielliptic surfaces}

Let $k$ be a field of $\opchar k=0$, and $\kbar$ be an 
algebraic closure of $k$. For a $k$-variety $X$ we write
$\ov{X}=X\times_k\ov k$.

\begin{defn}\rm
A \emph{bielliptic surface} $X$ over $k$ is a smooth projective
surface such that $\ov X$ is a minimal surface of 
Kodaira dimension $0$, and
is not a K3, abelian or Enriques surface.
\end{defn}

Bielliptic surfaces over $\ov k$ were classified by Bagnera and
de Franchis (see [B], VI.20). Their theorem says that
any bielliptic surface over $\ov k$ can be obtained as the quotient
of the product of two elliptic curves
$E \times F$ by a freely acting finite abelian group. 
%where $\Gamma$ acts on $E$ by translations.
The geometric genus of any bielliptic surface is 0.
For a bielliptic surface $X$ let $n$ be the order
of $K_{\ov X}$ in $\Pic \ov X$. It follows from the 
Bagnera--de Franchis classification
that $n$ can be $2,3,4$ or $6$ ({\it loc.cit.}).

\begin{prop} \label{1}
Let $X$ be a bielliptic surface over $k$.
There exists an abelian surface $A$, 
a principal homogeneous space $Y$ of
$A$, and a finite \'etale morphism $f\colon Y \to X$
of degree $n$, that is a torsor
under the group scheme $\mu_n$.
\end{prop}

\begin{proof}
The natural map $\Pic X\to\Pic\ov X$ is injective, hence
$nK_X$ is a principal divisor.
We write $nK_X=(\phi)$, where $\phi\in k(X)^*$. 
Let $Y$ be the normalization of 
the covering of $X$ given by $t^n=\phi$. 
Then the natural
map $f\colon Y\to X$ is unramified, and is a torsor
under $\mu_n$ (cf.\ [CS], 2.3.1, 2.4.1). 
This implies that
$K_Y=f^*K_X=0$. By the classification of surfaces,
$\ov{Y}$ is an abelian surface. (It is not K3 as the only
unramified quotients of K3 surfaces are Enriques surfaces.)
Let $A$ be the Albanese variety of $Y$, defined over $k$
(see [L], II.3). Then $\ov A$ is the Albanese variety of $\ov Y$.
The choice of a base point makes $\ov Y$ an abelian variety
isomorphic to $\ov A$, so that $\ov Y$ is naturally 
a principal homogeneous space of $\ov A$.
Choose $\ov y_0\in Y(\ov k)$, then we have
an isomorphism $\ov Y\to \ov A$ that sends
$\ov y$ to $\ov y-\ov y_0$. 
Then $\rho(g)={}^g\ov y_0-\ov y_0$
is a continuous 1-cocycle
of $\Gal(\kbar/k)$ with coefficients in $A(\ov k)$.
Let $A^\rho$ be the principal homogeneous space of $A$
defined by $\rho$; it corresponds to the twisted Galois action
$(g,\ov a)\mapsto {}^g\ov a+\rho(g)$, where $\ov a\in A(\ov k)$
(see [S], III.1, or [S2], 2.1).
Then the above $\ov k$-isomorphism $\ov Y\to \ov A$ descends to a
$k$-isomorphism $Y\to A^\rho$. 
\end{proof}

Note that the analogue of the proposition
fails in higher dimension because there are many more
possibilities for $Y$.


\section{Group action on principal homogeneous\\
 spaces of abelian varieties}

Let $A$ be an abelian variety over $k$, and $Z$
a principal homogeneous space of $A$. Suppose that
a $k$-group scheme $\Gamma$ acts on $Z$.
This gives rise to
a Galois-equivariant action of the group $\Gamma(\kbar)$
on the set $Z(\ov k)$.
The action of $\Gamma$ on $Z$ naturally defines
an action of $\Gamma$ on $A$, the Albanese variety of $Z$.
Then the action of the group $A(\kbar)$ on the set $Z(\kbar)$
is both Galois and $\Gamma$-equivariant. Let $A^\Gamma$ be
the $\Gamma$-invariant group subscheme of $A$. Similarly, let
$Z^\Gamma\subset Z$ be the closed subscheme consisting 
of points fixed by $\Gamma$.

\begin{prop}
Suppose that a $k$-group scheme $\Gamma$ acts 
on $Z$ in such a way that $Z^\Gamma$ is a nonempty scheme 
(i.e., some $\ov k$-point of $Z$ is fixed
by $\Gamma(\kbar)$). 
Then $[Z] \in {\rm Im}[H^1(k,A^{\Gamma}) \to H^1(k,A)]$.
\end{prop}

\begin{proof}
Take $\ov{x} \in Z(\kbar)$, fixed by $\Gamma(\kbar)$. Then
a 1-cocycle of $\Gal(\kbar/k)$ sending $g \in \Gal(\kbar/k)$ to 
${}^g\ov{x}-\ov{x} \in A(\kbar)$ represents the class
$[Z] \in H^1(k,A)$. 
For any $\gamma\in \Gamma(\kbar)$ we have 
$$\gamma({}^g\ov{x}-\ov{x})=\gamma\cdot{}^g\ov x-\gamma\cdot \ov x=
{}^g({}^{g^{-1}}\gamma\cdot \ov x)-\ov x={}^g\ov x-\ov x.$$
Therefore, ${}^g\ov{x}-\ov{x} \in A^\Gamma(\kbar)$.
\end{proof}

It is easy to see that $Z^\Gamma$ is a principal homogeneous 
space of $A^\Gamma$.
The $A$-torsor $Z$ is the push-forward of the $A^\Gamma$-torsor
$Z^\Gamma$ with respect to the natural injection of group schemes
$A^\Gamma\to A$. This gives an alternative proof of the proposition.

\begin{cor} \label{cc}
Let $A_1=A/A^\Gamma$, and $\alpha\colon A \to A_1$ the 
natural surjection.  Then $[Z]\in H^1(k,A)[\alpha_*]$,
where 
$$H^1(k,A)[\alpha_*]=\ker[\alpha_*\colon H^1(k,A)\to H^1(k,A_1)].$$
\end{cor}

We now consider the case when $Z=C$ is a curve of genus 1 equipped
with a faithful action of $\Gamma$ that has a fixed point. 
Then $A=E$ is
the Jacobian of $C$. We shall write ${\rm Aut}_0(\ov E)$
for the automorphism group of $\ov E$ as an elliptic curve.
Now $\Gamma(\ov k)\subset {\rm Aut}_0(\ov E)$,
hence $\Gamma(\ov k)$ is a cyclic group of
order $n$, where $n$ can be 1, 2, 3, 4 or 6. A straightforward
calculation shows that, excluding
the trivial case $n=1$, we have one of the following possibilities:
\[
 \renewcommand{\arraystretch}{1.3}
 \begin{array}{c|c}
n & \# E^{\Gamma}(\kbar) \\
\hline
6 & 1 \\
4 & 2 \\
3 & 3 \\
2 & 4
\end{array} \]
By Corollary~\ref{cc}
the first line of this table shows that if a cyclic
group scheme of order 6 acts on a curve of genus 1,
then this curve has a $k$-point. As a consequence of this fact
we obtain in the next section a simple description of bielliptic surfaces
with $n=6$.

\section{A case when the Manin obstruction to the Hasse 
principle is the only one}

\begin{prop}
Let $X$ be a bielliptic surface over $k$ such that
the order of $K_{\ov X}$ in $\Pic \ov X$ is $6$.
There exist an elliptic curve $E$ and a
curve $D$ of genus $1$ such that the group scheme $\mu_6$
acts on $E$ by automorphisms of an elliptic curve (in particular,
preserving the origin), and acts on $D$ by translations, 
in such a way that $X=(E\times D)/\mu_6$.
\end{prop}

\begin{proof}
The Bagnera--de Franchis classification ([B], VI.20)
says that for any bielliptic surface $\ov X$ with $K_{\ov X}$
of order 6 in $\Pic\ov X$ there exist elliptic curves 
$C_1$ and $C_2$ over $\kbar$ such that: 
 \begin{enumerate}
 \renewcommand{\labelenumi}{(\arabic{enumi})}
 \item $\mu_6$ acts on $C_1$ by automorphisms of an elliptic curve
(in particular, preserving the origin); 

 \item the group scheme $\mu_6$ is a subgroup of $C_2$; 

 \item $X=(C_1\times C_2)/\mu_6$.
 \end{enumerate}
The free action of $\mu_6$ on $C_1\times C_2$
makes the finite \'etale map $C_1\times C_2\to \ov X$
a torsor under $\mu_6$.
Let us compare it with the torsor $\ov Y\to\ov X$
constructed in Proposition \ref{1}. 

Recall that the type of a $Z$-torsor under a 
group of multiplicative type $S$ is a certain functorial map 
$\widehat S\to\Pic\ov Z$, where $\widehat S$ is the module of characters of $S$
(see [S2], Definition~2.3.2).
A torsor under a group of 
multiplicative type over an integral projective
$\kbar$-variety is uniquely determined up to isomorphism 
by its type (this follows from the fundamental 
exact sequence of Colliot-Th\'el\`ene and Sansuc, see [CS], [S2], (2.22)).
Therefore it is enough to compare the respective types.
There is an exact sequence 
$$0\to\Hom(\mu_6,\ov k^*)=\bfZ/6\to\Pic\ov X\to\Pic\ov Y,$$
where the second arrow is the type of 
the torsor $\ov Y\to\ov X$,
and a similar sequence for $C_1\times C_2\to \ov X$ 
([S2], (2.4) and Lemma 2.3.1). Since the canonical class
of an abelian surface is trivial, 
$K_{\ov X}$ is in
the image of $\bfZ/6$ in $\Pic\ov X$, and hence it is
a generator of that image.
Thus the types of both torsors
are the same (up to sign). 
Hence the pair ($\ov Y$, the action of $\mu_6$)
can be identified with the pair ($C_1\times C_2$, the action of $\mu_6$).

Let $A$ be the Albanese variety of $Y$. This is an
abelian surface defined over $k$. 
Let $s$ be the $k$-endomorphism of $A$
given by $s=\sum_{\sigma\in\mu_6}\sigma$.
Let $A_1$ (respectively $A_2$) be the connected component of $0$
in $\ker(s)$ (respectively in $A^{\mu_6}$). 
Note that $s$ acts as $0$ on $J_1={\rm Jac}(\ov C_1)\subset \ov A$,
and as multiplication by $6$ on $J_2={\rm Jac}(\ov C_2)\subset \ov A$.
Therefore, $\ov A_1=J_1$, $\ov A_2=J_2$. Now the map 
$$A_1\times A_2\to A, \quad (x,y)\mapsto x+y,$$
is an isomorphism, since over $\kbar$ it is the
natural isomorphism $J_1\times J_2\to \ov A$. This proves that
$A$ is a product of two elliptic curves over $k$. Hence
$Y$, which is a principal homogeneous space of $A$, 
is a product of two curves of genus 1 over $k$: $Y=E\times D$,
where $C_1\simeq\ov E$, $C_2\simeq\ov D$.

By the Bagnera--de Franchis theorem the group scheme $\mu_6$ 
acts on $E$ with a fixed point.
By the remark preceding the statement of the proposition,
this point is unique, and hence is $k$-rational.
Hence $E$ is an elliptic curve (isomorphic to $A_1$). 
\end{proof}

See the beginning of the next section (or, in more generality,
[S2], 5.2) for the definition of the Manin obstruction.

\begin{cor} \label{c}
Let $k$ be a number field.
The Manin obstruction is the only obstruction
to the Hasse principle on
the bielliptic surfaces $X$ over $k$ such that
the order of $K_{\ov X}$ in $\Pic \ov X$ is $6$, and the Tate--Shafarevich
group of the Albanese variety of $X$ is finite.
\end{cor}

\begin{proof}
By the previous proposition we have 
$X=(E\times D)/\mu_6$. Consider the curve $D'=D/\mu_6$ of genus 1,
and let $p\colon X\to D'$ be the natural surjective map.
Let $J'$ be the Jacobian of $D'$. It is known ([B], VI)
that the Albanese variety of any bielliptic surface
has dimension 1.
Using the universal property of the Albanese variety (see [L], II.3)
and the connectedness of the fibres of $p$
one easily checks that $J'$ is the Albanese variety of $X$.

Let $\{Q_v\}$ be a collection of local points on $X$, for all places
$v$ of $k$, that satisfies the Brauer--Manin conditions.
Then $\{p(Q_v)\}$ satisfies the Brauer--Manin conditions on $D'$.
If $\Sha(J')$ is finite, then 
$D'$ has a $k$-point by a theorem of Manin (see [S2], Theorem~6.2.3). 
Call this point $Q$. The inverse image of $Q$
in $D$ defines a class $\rho\in H^1(k,\mu_6)=k^*/k^{*6}$.
Consider the twisted torsor $E^\rho\times D^\rho\to X$.
Now $D^\rho$ has a $k$-point over $Q$. But the action of $\mu_6$ on
$E$ preserves the origin, hence the twisted curve
$E^\rho$ has a $k$-point. Therefore, we obtain a $k$-point
on $E^\rho\times D^\rho$, and hence on $X$. 
\end{proof}

Note that for the bielliptic surfaces of Corollary~\ref{c}
the quotient of $\Br X$ by the image of $\Br k$
is infinite, but in the proof 
we only used the Brauer--Manin conditions given by the elements
of the conjecturally finite group $\Sha(J')$.

Corollary~\ref{c} is a particular case of a more general
situation. Let $\Gamma$ be an algebraic group acting on varieties
$V$ and $W$ such that the action on $W$ is free. Suppose that
$V$ has a $k$-point fixed by $\Gamma$. If the Manin obstruction
to the Hasse principle is the only one on $W/\Gamma$, then
the same is true for $(V\times W)/\Gamma$.


\section{Main construction and example}

 
Now assume $k=\bfQ$, and let $\bdA_\bfQ$ be the ring of  ad\`eles of
$\bfQ$. For a projective variety $X$ we have $X(\bdA_\bfQ)=\prod_{v}
X(\bfQ_v)$, where $v$ ranges over all places of $\bfQ$ including the real
place.
Let $X(\bdA_\bfQ)^{\Br}$ be the subset of $X(\bdA_\bfQ)$
consisting of the families of local points $\{P_v\}$ satisfying all
the Brauer--Manin conditions. 
These conditions, one for each $A\in \Br X$, are
$$\sum_{\mathrm{all\ } v}{\rm inv}_v \, A(P_v)=0,$$
where ${\rm inv}_v$ is the local invariant at the place $v$,
which is a canonical map $\Br\bfQ_v \to\bfQ/\bfZ$ provided
by local class field theory. The Brauer--Manin conditions
are satisfied for any $\bfQ$-point of $X$ by the Hasse
reciprocity law, so that we have
$X(\bfQ)\subset X(\bdA_\bfQ)^{\Br}$. If the last
set is empty, this is an obstruction to the existence of a $\bfQ$-point
on $X$; it is called the Manin obstruction.

We now give a construction of bielliptic
surfaces $X$ for which $X(\bdA_\bfQ)^{\Br} \neq \emptyset$, 
but $X(\bfQ)=\emptyset$. Then $X$
is a counterexample to the Hasse principle that is not
explained by the Manin obstruction. 

\begin{thm}
Let $E$ be an elliptic curve over $\bfQ$
with a nontrivial action of the group scheme $\mu_3$. 
Let $\alpha\colon E\to E_1$ be the degree $3$ isogeny with kernel
$E^{\mu_3}$. Let $D$ be an elliptic curve with a group
subscheme isomorphic to $\mu_3$.
Assume that:
\begin{enumroman}
\item $\Gal(\bfQbar/\bfQ)$ acts nontrivially on $E^{\mu_3}$; 
%which is isomorphic to $\bfZ/3\bfZ$ as an abstract group;
\item $\#\Sha(E)[\alpha_*]=3$;
\item $C$ is a principal homogeneous space of $E$
representing a nontrivial element of $\Sha(E)[\alpha_*]$;
\item $\Sel(D,\mu_3)=0$, that is, 
for any principal homogeneous space of $D$
obtained from a nontrivial class in $H^1(\bfQ,\mu_3)=
\bfQ^*/\bfQ^{*3}$, 
there exists a place $v$ where it has no $\bfQ_v$-point.
\end{enumroman}
Then $X=(C \times D)/\mu_3$ is a counterexample to the Hasse principal not
explained by the Manin obstruction.
\end{thm}

Let us give an example
of curves $C$ and $D$ satisfying the conditions of the
theorem. Let $\zeta$ be a primitive cubic root of unity.

Let $C$ be the plane cubic curve $x^3+11y^3+43z^3=0$, where the root of
unity $\zeta$ acts by $(x:y:z) \mapsto (x:y:\zeta z)$. 
The Jacobian $E$ of $C$ is the plane curve $x^3+y^3+473z^3=0$,
with the \pagebreak action of $\mu_3$ given by the same formula.
One easily checks that Condition (i) is satisfied.
Condition (ii) is verified in Example 4.3 of [F2]. The
elements of $H^1(\bfQ, E)[\alpha_*]$ are given by the curves
$mx^3+m^2y^3+473z^3=0$ with $m$ a cube-free integer.
The curve $C$ corresponds to $m=11$. 
It has been known for some time [Se] that $C$ has points everywhere
locally but not globally. This gives Condition (iii). (See also [Ba], VI.)

Let $D$ be the elliptic curve
$u^3+v^3+w^3=0$, with $(1:-1:0)$ as the origin. 
The group subscheme of $D$ generated by $(1:-\zeta:0)$
is isomorphic to $\mu_3$. The translation
by this element is $(u,v,w) \mapsto (u:\zeta v:\zeta^2 w)$.
The elements of the Selmer group $\Sel(D,\mu_3)$
are represented by the principal
homogeneous spaces $D_a$ defined by $u^3+av^3+a^2u^3=0$,
where $a$ is a cube free integer. Let $p$ be a
prime factor of $a$. Then $D_a$ has no $\bfQ_p$-point.
Therefore, the only curve $D_a$ with points everywhere locally
is $D$ itself, so that $\Sel(D,\mu_3)=0$, which is
our Condition (iv).

\begin{rmk} \rm
On changing some of the conditions of the theorem one obtains
bielliptic surfaces for which the Manin
obstruction to the Hasse principle is the only one. We
replace Condition (ii) by the condition $\Sha(E)[\alpha_*]=0$,
and instead of Condition (iii) we require that $C$ is any 
principal homogeneous space of $E$
whose class is in $H^1(\bfQ,E)[\alpha_*]$.
We drop Condition (i) and keep Condition (iv). Then
the Manin obstruction is the only obstruction to
the Hasse principle for the surfaces $(C \times D)/\mu_3$.
For the proof, consider the torsor $C \times D\to (C \times D)/\mu_3$
under $\mu_3$. Under our assumptions the class of 
twists $C^\rho\times D^\rho$, $\rho\in \bfQ^*/\bfQ^{*3}$,
satisfies the Hasse principle. By descent theory ([S2],
Corollary~6.1.3 (2)) this implies our statement.
\end{rmk}

\section{Proof of the theorem}


Consider the alternating Cassels pairing 
$\Sha(E)\times \Sha(E) \to \bfQ/\bfZ$.
Its restriction to $\Sha(E)[\alpha_*]$ gives
an alternating pairing
$$\Sha(E)[\alpha_*] \times \Sha(E)[\alpha_*] \to \bfQ/\bfZ.\eqno{(1)}$$ 
The kernel of the last pairing is the image of
$\alpha_*^t\colon \Sha(E_1)\to\Sha(E)$, where $\alpha^t\colon E_1 \to E$ is
the dual isogeny. (This seems to be part of the folklore;
see [F1] for a proof.) Since $\Sha(E)[\alpha_*] \cong \bfZ/3\bfZ$
by Condition (ii),
the pairing (1) must be zero. Therefore, there exists
a principal homogeneous space $C_1$ of $E_1$ with points
everywhere locally, that lifts $C$. This means that the map
$\alpha_*^t\colon H^1(\bfQ,E_1)\to H^1(\bfQ,E)$ sends $[C_1]$ to $[C]$.
There is a finite \'etale morphism
$C_1\to C$ that represents $C$ as the quotient of $C_1$
by the action of $\ker(\alpha^t)$. Let $Y=C\times D$,
$Y_1=C_1\times D$. This gives rise \pagebreak to a finite
\'etale morphism $Y_1\to Y$ which is the identity on $D$.
Let $f_1$ be the composition of the finite \'etale maps
$Y_1 \to Y \to X$, and let $\pi\colon Y_1\to D$ be the projection
to the second factor. In this notation we have the following
key property analogous to ([S1], Theorem~1):
$$f_1^*(\Br X) \subset \pi^*(\Br D).\eqno{(2)}$$
To prove this we note that
for any smooth and projective surface $X$ 
with $p_g=0$, in particular, for a bielliptic surface,
we have an isomorphism of Galois modules
$\Br \ov{X}=\Hom(\NS(\ov{X})_{\tors},\bfQ/\bfZ)$
(see [G], II, Corollary~3.4, III, (8.12)). 
As in the proof of Corollary~\ref{c} one shows that
the Albanese variety of $X$ is $D/\mu_3$.
The same argument as in 
([S1], pp.~403--404) works in our situation, and we obtain
$\NS(\overline{X})_{\tors}=E^{\mu_3}$.  Then (i) implies
that $(\Br \overline{X})^{\Gal(\bfQbar/\bfQ)}=0$.
Therefore, $\Br X=\ker[\Br X\to \Br\ov X]$.
A well known Leray spectral sequence shows that
the quotient of this group by the image of $\Br\bfQ$
is naturally isomorphic to $H^1(\bfQ,\Pic\ov X)$ ([S2], (2.23);
here we use the fact that $H^3(\bfQ,\ov\bfQ^*)=0$).
The analysis of the morphism of Galois modules
$f_1^*\colon \Pic\ov X\to \Pic\ov Y_1$ is carried out
in the same way as in the proof of Lemma 2 of [S1],
where the multiplication by 2 on $E$ has now to be
replaced by the isogeny $\alpha\colon E\to E_1$. The result 
is that the image $f_1^*(H^1(\bfQ,\Pic\ov X))$ in
$H^1(\bfQ,\Pic\ov Y_1)$ is contained in 
$\pi^*(H^1(\bfQ,\Pic\ov D))$. Formula (2) now
follows from the functoriality of the Leray spectral
sequence.

Let us construct an adelic point on $X$ satisfying all
the Brauer--Manin conditions. Take a rational point 
$R \in D(\bfQ)$, and a
collection $\{P_v\} \in C_1(\bdA_\bfQ)$.
Then $f_1(\{(P_v,R)\})\in X(\bdA_\bfQ)^{\Br}$, as follows
from (2) and the Hasse reciprocity law.

It remains to show that there are no $\bfQ$-points on $X$.
Indeed, rational points on $X$ come from twists of $Y$
given by $a\in H^1(\bfQ,\mu_3)=\bfQ^*/\bfQ^{*3}$. 
Any such twist of $Y$ is the product $C_a\times D_a$,
where $C_a$ and $D_a$ are curves of genus 1.
Moreover, $D_a$ is a principal homogeneous space of $D$
of the kind described in Condition (iv) of the theorem.
By that condition, if $D_a$ has points everywhere locally, then
$a$ is trivial, so that $D_a=D$. Thus there are no
$\bfQ$-points on the nontrivial twists of $Y$. On the other hand,
$Y$ has no $\bfQ$-points since by Condition (iii)
there are no $\bfQ$-points on $C$.
Therefore, $X(\bfQ)=\emptyset$. This completes the proof.

More details can be found in the thesis of the first author [Ba]. 
The preparation of this paper
was speeded up by John Voight's notes of the conference 
``Rational and Integral Points on Higher Dimensional
Varieties" at the American Institute of Mathematics
in December, 2002. We thank him for the notes, and
the organizers for stimulating atmosphere. We are very grateful
to Tom Fisher for telling us about the curve $x^3+11y^3+43z^3=0$.
We thank Ekaterina Amerik for useful discussions.

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\clearpage
\noindent
Carmen Laura Basile, \\
Department of Mathematics, \\
Imperial College London, \\
South Kensington Campus, \\
London SW7 2AZ, England \\
e-mail: laura.basile@ic.ac.uk

\medskip
\noindent
Alexei Skorobogatov, \\
Department of Mathematics, \\
Imperial College London, \\
South Kensington Campus, \\
London SW7 2AZ, England \\
e-mail: a.skorobogatov@ic.ac.uk

\end{document}

The Leray
spectral sequence 
$$H^p(\mu_6,H^q(\ov Y,{\bf G}_m))
\Rightarrow H^{p+q}(\ov X,{\bf G}_m)$$ 
gives rise to the exact sequence
$$0\to\Hom(\mu_6,\ov k^*)=\bfZ/6\to\Pic\ov X\to\Pic\ov Y.$$
The second arrow in this sequence is the type of 
the torsor $\ov Y\to\ov X$ (see [S2]).
