Zero-cycles of degree one on Skorobogatov's bielliptic surface

TL;DR

This paper proves that Skorobogatov's bielliptic surface, which counters the Hasse principle without Brauer-Manin obstruction, indeed has a zero-cycle of degree one, confirming a conjecture by Colliot-Thélène.

Contribution

It demonstrates the existence of a zero-cycle of degree one on Skorobogatov's bielliptic surface, supporting the conjecture that such cycles exist even when the Hasse principle fails.

Findings

Confirmed the existence of a zero-cycle of degree one on the surface.

Supported Colliot-Thélène's conjecture regarding zero-cycles.

Provided insights into the arithmetic of bielliptic surfaces.

Abstract

Skorobogatov constructed a bielliptic surface which is a counterexample to the Hasse principle not explained by the Brauer-Manin obstruction. We show that this surface has a $0$-cycle of degree 1, as predicted by a conjecture of Colliot-Th\'el\`ene.

Equations

$$\mathcal{S}:(x^2+1)y^2=(x^2+2)z^2=3(t^4-54t^2-117t-243).$$

$$x_0$$

$$y_0$$

$$z_0$$

$$t_0$$

$$\mathcal{C}:U^2=g(T)=3(T^4-54T^2-117T-243),$$

$$\mathcal{D}:\{\begin{array}{ll}{Y}^2=p(X)=X^2+1& \\ Z^2=q(X)=X^2+2.& \end{array}$$

$$(T,U)\mapsto (T,-U)\text{and}(X,Y,Z)\mapsto (X,-Y,-Z).$$

$$\mathcal{S}:\{\begin{array}{ll}{y}^2=g(t)p(x)& \\ z^2=g(t)q(x)& \end{array}$$

$${\mathcal{C}}^a:a{U}^2=g(T)\text{and}{\mathcal{D}}^a:\{\begin{array}{ll}a{Y}^2=p(X)& \\ a{Z}^2=q(X)& \end{array}$$

$$1\to {\mu }_2\to \mathcal{D}(K)\stackrel{\varphi }{\to }{\mathcal{D}}^{\prime }(K)\stackrel{{\delta }_{\varphi }}{\to }{H}^1(K,{\mu }_2)\simeq K^{\times }/K^{\times 2}.$$

Full text

Consider the smooth projective surface ${\mathcal{S}}/{\mathbb{Q}}$ given by the affine equations,

[TABLE]

Skorobogatov [Skorobogatov99] showed that the set ${\mathcal{S}}({\mathbb{Q}})$ of rational points on ${\mathcal{S}}$ is empty, despite there being adelic points on ${\mathcal{S}}$ that are orthogonal to all elements in the Brauer group of ${\mathcal{S}}$. A fortiori the Brauer group does not obstruct the existence of ${\mathbb{Q}}$-rational [math]-cycles of degree $1$ on ${\mathcal{S}}$. In this short note we show that, as predicted by a conjecture of Colliot-Thélène [Colliot-Thelene99], the surface ${\mathcal{S}}$ does in fact possess a ${\mathbb{Q}}$-rational [math]-cycle of degree $1$.

To wit,

[TABLE]

where $\theta$ satisfies $\theta^{3}+\theta+1=0$ are the coordinates of a closed point of ${\mathcal{S}}$ whose residue field is the cubic number field $L={\mathbb{Q}}[\theta]$. As there are obviously [math]-cycles of degree $4$ on ${\mathcal{S}}$, this shows that there is a [math]-cycle of degree $1$ on ${\mathcal{S}}$.

We note, however, that the far more important question of whether the Brauer-Manin obstruction to the existence of [math]-cycles of degree $1$ is the only one for bielliptic surfaces remains open.

Let us briefly explain how this point was discovered. Consider the genus one curves

[TABLE]

and

[TABLE]

There are actions of $\mu_{2}=\{\pm 1\}$ on ${\mathcal{C}}$ and ${\mathcal{D}}$ given by

[TABLE]

The quotient of ${\mathcal{C}}\times{\mathcal{D}}$ by the diagonal action of $\mu_{2}$ is the $\mu_{2}$-torsor $\rho:{\mathcal{C}}\times{\mathcal{D}}\to{\mathcal{S}}$ given by $s^{2}=g(t)$ over ${\mathcal{S}}$. After a change of variables, ${\mathcal{S}}$ may be written

[TABLE]

and $\rho$ is defined by $x=X,t=T,y=UY,z=UZ$.

Lemma

There is a [math]-cycle of degree $1$ on ${\mathcal{S}}$ if and only if there is an odd degree number field $K$ and $a\in K^{\times}$ such that genus one curves over $K$ given by

[TABLE]

both possess a $K$-rational point.

Proof.

It is clear that there is a [math]-cycle of degree $1$ on ${\mathcal{S}}$ if and only if ${\mathcal{S}}(K)\neq\emptyset$ for some odd degree number field $K$. Let $\rho_{K}:({\mathcal{C}}\times{\mathcal{D}})_{K}\to{\mathcal{S}}_{K}$ denote the base change to $K$. By descent theory every element of ${\mathcal{S}}(K)={\mathcal{S}}_{K}(K)$ lifts to a $K$-rational point on some twist of the $\mu_{2}$-torsor $\rho_{K}:({\mathcal{C}}\times{\mathcal{D}})_{K}\to{\mathcal{S}}_{K}$ by a cocycle in $H^{1}(K,\mu_{2})=K^{\times}/K^{\times 2}$. A straightforward computation shows that the twist of $({\mathcal{C}}\times{\mathcal{D}})_{K}$ by $a\in K^{\times}/K^{\times 2}$ is the product of the curves in the statement of the proposition. ∎

The quotient of ${\mathcal{D}}$ by $\mu_{2}$ is the $\mu_{2}$-torsor $\phi:{\mathcal{D}}\to{\mathcal{D}}^{\prime}$ given by $U^{2}=p(X)$ over the genus one curve ${\mathcal{D}}^{\prime}$ given by $W^{2}=p(X)q(X)$. This curve has two rational points at infinity. Fixing either as the identity endows ${\mathcal{D}}^{\prime}$ with the structure of an elliptic curve such that the other point is $2$-torsion. There are rational points on ${\mathcal{D}}$ above both of these, and so we may view $\phi:{\mathcal{D}}\to{\mathcal{D}}^{\prime}$ as an isogeny of elliptic curves. The twists ${\mathcal{D}}^{a}$ which have a $K$-rational point correspond to the image of ${\mathcal{D}}^{\prime}(K)$ under the connecting homomorphism $\delta_{\phi}$ in the exact sequence

[TABLE]

In particular, for a given $K$, one can compute by means of an explicit $2$-isogeny descent a finite set $A\subset K^{\times}/K^{\times 2}$ such that every twist with a $K$-point is ${\mathcal{D}}^{a}$ for some $a\in A$. Thus the determination of ${\mathcal{S}}(K)$ is reduced to the determination of the set of $K$-rational points on an explicit finite set of genus $1$ curves over $K$, namely the curves ${\mathcal{C}}^{a}$ and ${\mathcal{D}}^{a}$ with $a\in A$. Conjecturally, this is a finite computation. For $K$ of small degree and discriminant it is often possible in practice.

We carried out these computations for various number fields using the Magma Computational Algebra System [Magma]. There are no points on ${\mathcal{S}}$ defined over the cubic field $L_{1}$ of smallest absolute discriminant. The map $\phi:{\mathcal{D}}(L_{1})\to{\mathcal{D}}^{\prime}(L_{1})$ is surjective and so the only twist with $L_{1}$-points is ${\mathcal{D}}={\mathcal{D}}^{1}$. But as ${\mathcal{C}}={\mathcal{C}}^{1}$ has no points over any odd degree number field we see that ${\mathcal{S}}(L_{1})=\emptyset$. The point $(x_{0},y_{0},z_{0},t_{0})\in{\mathcal{S}}(L)$ is defined over the cubic field of second smallest absolute discriminant. The group ${\mathcal{D}}^{\prime}(L)$ has rank $1$ and is generated by the points with $X$-coordinate equal to $x_{0}=\theta^{2}+1$. This corresponds to the twist by $a=6\theta^{2}-4\theta+9\in L^{\times}$ and one finds that ${\mathcal{C}}^{a}$ has a point with $T$-coordinate equal to $t_{0}=-42\theta^{2}+24\theta-54$. In fact, ${\mathcal{C}}^{a}(L)$ and ${\mathcal{D}}^{a}(L)$ are both infinite, showing that ${\mathcal{S}}(L)$ is Zariski dense in ${\mathcal{S}}$.

Acknowledgements

The author would like to thank J.-L. Colliot-Thélène for helpful comments and suggestions.