Skorobogatov: Counterexamples to the Hasse Principle...

This is joint work with Laura Basile.

We work over a field $ k$ with $ {\mathrm{char}}k=0$, $ \overline{k}$ its algebraic closure.

Definition. A bielliptic surface $ X$ is a $ \overline{k}/k$-form of a smooth projective surface of Kodaira dimension 0 that is not $ K3$, neither abelian nor Enriques.

There is a complete list of such available. We have $ K_X \neq 0$, but $ nK_X=0$, for $ n=2,3,4$ or $ 6$. Over the algebraic closure, $ \overline{X}=E \times F/\Gamma$, where $ \Gamma$ acts on $ E'$ by translations.

Proposition. There exists an abelian surface $ A/k$, a principal homogeneous space $ Y$ of $ A$, and a finite étale morphism $ f: Y \to X$, $ \deg f=n$.

Proof. Since $ nK_X=(g)$, then consider $ t^n=g$ and normalize; the map is unramified, so $ K_Y=f^*K_X=0$. By the classification of surfaces, $ \overline{Y}$ is an abelian surface. ($ f$ is a torsor under $ \mu_n$.) $ \qedsymbol$

Remark. This will not hold in higher dimension; there are just many more possibilities.

Consider $ A=E \times F$, $ Y=C \times D$, $ C$ a principal homogeneous space of $ E$, and likewise $ D$ for $ F$. Now $ \mu_n$ acts on $ Y$ so that $ \mu_n$ acts on $ D$ by translations, $ \mu_n \subset F$; the action on $ \mu_n$ on $ C$ cannot be by translations or else $ X$ itself would be a principal homogeneous space, so the action has fixed points.

Proposition. $ [C] \in {\mathrm{img}}(H^1(k,E^{\mu_n}) \to H^1(k,E))$.

Proof. Take $ \overline{x} \in C(\overline{k})$, fixed by $ \mu_n$, and write down the usual cocycle: if $ g \in {\mathrm{Gal}}(\overline{k}/k)$, $ g\overline{x}-\overline{x} \in E^{\mu_n}(\overline{k})$.

(It arises from $ [C^{\mu_n}]$.) $ \qedsymbol$

Corollary. Let $ \alpha:E \to E_1$ be the isogeny with kernel $ E^{\mu_n}$. Then $ [C] \in H^1(k,E)[\alpha_*]$.

We have one of the following possibilities:

$\displaystyle \begin{array}{c\vert c}
n & \char93  E^{\mu_n} \\
6 & \text{$1$\ (so $C(k) \neq \emptyset$)} \\
4 & 2 \\
3 & 3 \\
2 & 4
\end{array} $

Now assume $ k=\mathbb{Q}$, and $ X(\mathbf A_\mathbb{Q})=\prod_v X(\mathbb{Q}_v)$; we want an example where $ X(\mathbf A_\mathbb{Q})^{{\mathrm{Br}}} \neq \emptyset$, but $ X(\mathbb{Q})=\emptyset$. We do the case $ n=3$.

With the notation as above:

Theorem. Assume that:

  1. $ {\mathrm{Gal}}(\overline{\mathbb{Q}}/\mathbb{Q})$ acts nontrivially on $ E^{\mu_3}$, which is $ \cong \mathbb{Z}/3\mathbb{Z}$ as an abstract group;
  2. $ \char93 ${\tencyr\cyracc Sh}$ (E)[\alpha_*]=3$;
  3. $ [C] \in$   {\tencyr\cyracc Sh}$ (E)[\alpha_*]$;
  4. $ \mu_3 \subset F$;
  5. $ {\mathrm{Sel}}(F,\mu_3)=0$, that is, for any principal homogeneous space of $ F$ obtained from a class in $ H^1(\mathbb{Q},\mu_3)$, there exists a place $ v$ where it has no point.
Then $ X=(C \times F)/\mu_3$ is a counterexample to the Hasse principal not explained by the Manin obstruction.

Example. If $ C:x^3+11y^3+43z^3=0$, $ \mu_3$ acts by $ (x:y:z) \mapsto (x:y:\zeta_3 z)$, $ F=D:u^3+v^3+w^3=0$, with $ \mu_3$ acting by $ (u,v,w) \mapsto (u:\zeta_3 v:\zeta_3^2 w)$; looking at the Selmer group, you look at principal homogeneous spaces of the form $ u^3+av^3+a^2u^3=0$, so if $ p \mid a$, this has no $ \mathbb{Q}_p$ point.

We have $ {\mathrm{Br}}\overline{X}={\mathrm{Hom}}(NS(\overline{X})_{{\mathrm{tors}}},\mathbb{Q}/\mathbb{Z})$ as Galois modules (this holds more generally if $ X$ is a surface and $ h^{2,0}=0$), and $ NS(\overline{X})_{{\mathrm{tors}}}=E^{\mu_3}$. Then (i) implies that $ ({\mathrm{Br}}\overline{X})^{{\mathrm{Gal}}(\overline{\mathbb{Q}}/\mathbb{Q})}=0$.

The kernel of the restricted Cassels pairing {\tencyr\cyracc Sh}$ (E)[\alpha_*] \times$   {\tencyr\cyracc Sh}$ (E)[\alpha_*] \to \mathbb{Q}/\mathbb{Z}$ consists of elements in the image of {\tencyr\cyracc Sh}$ (E_1) \xrightarrow{\alpha_*^t}$   {\tencyr\cyracc Sh}$ (E)$, where $ \alpha^t:E_1 \to E$ is the dual isogeny. Since {\tencyr\cyracc Sh}$ (E)[\alpha_*] \cong \mathbb{Z}/3\mathbb{Z}$ must be zero because it is alternating, so lift $ C_1 \to C$; then we have étale maps

$\displaystyle C_1 \times F = Y_1 \to C \times F=Y \to X; $

to find an adelic point, find a rational point $ R \in F(\mathbb{Q})$ and a collection $ \{P_v\} \in C_1(\mathbf A_\mathbb{Q})$; then $ f_1:Y_1 \to X$ has $ f_1^*({\mathrm{Br}}_1 X) \subset \pi^*({\mathrm{Br}}F)$, where $ \pi:Y_1 \to F$ is the projection.

The last condition (v) says that there are no rational points on $ X$; rational points on $ X$ comes from twists of $ Y$, but by assumption these have no point over a place $ v$, so they arise from {\tencyr\cyracc Sh}.

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