Colliot-Thelene 1: Rational points on surfaces with a pencil of curves of genus one

Symmetrizing the Computation of the Selmer Group

Let $ k$ be a field, $ {\mathrm{char}}k = 0$, $ E:y^2=(x-e_1)(x-e_2)(x-e_3)$ an elliptic curve with $ e_i \in k$ so that all $ 2$-torsion of $ E$ is rational. We have the exact sequence

$\displaystyle 0 \to E[2] \to E \xrightarrow{2} E \to 0 $

where $ E[2] \cong (\mathbb{Z}/2\mathbb{Z})^2$. The long exact sequence in Galois cohomology gives

$\displaystyle 0 \to E(k)/2E(k) \to H^1(k,E[2]) \to H^1(k,E)[2] \to 0 $

where $ H^1(k,E[2]) \cong (k^*/k^{*2})^2$ classifies $ 2$-coverings: that is, given $ (\alpha_1,\alpha_2) \in (k^*/k^{*2})^2$, we have the $ 2$-cover defined by the equations: $ x-e_1=\alpha_1u_1^2$, $ x-e_2=\alpha_2u_2^2$, $ x-e_3=(\alpha_1\alpha_2)^{-1}u_3^2$. The group $ H^1(k,E)$ classifies principal homogeneous spaces.

Now let $ k$ be a number field, $ \Omega$ the set of places of $ k$. We have the diagram

$\displaystyle \xymatrix @C=12pt{
& & {\mathrm{Sel}}(E,2) \ar[r] \ar[d] & \text{...
...v) \ar[r] & \prod_v H^1(k_v,E[2]) \ar[r] &
\prod_v H^1(k_v,E)[2] \ar[r] & 0
} $

so $ {\mathrm{Sel}}(E,2)$ gives $ 2$-covers with points everywhere locally, and {\tencyr\cyracc Sh}$ ^2(k,E)$ measures the difference.

Over a local field, we have a pairing

$\displaystyle H^1(k_v,E[2]) \times H^1(k_v,E[2]) \to H^2(k_v,\mu_2) $

induced by the Weil pairing, which is nondegenerate and alternating, so that $ (x,x)=0$.

Fact. [Tate] $ E(k_v)/2E(k_v) \hookrightarrow H^1(k_v,E[2])$ is maximal isotropic for the above pairing.

If $ E/k$ has good reduction at $ k_v$, then $ E(k_v)/2E(k_v) = E(\mathcal{O}_v)/2E(\mathcal{O}_v) \simeq H^1(\mathcal{O}_v,E[2])$, which is $ (\mathcal{O}_v^*/\mathcal{O}_v^{*2})^2 \subset (k_v^*/k_v^{*2})^2$, the maximal isotropic subgroup.

Suppose $ S \subset \Omega$ is a finite set of places, and suppose $ S$ contains the primes above $ \infty,2$, and the primes of bad reduction. Then

$\displaystyle \xymatrix @C=0pt{
\bigoplus_{v \in S}E(k_v)/2E(k_v) \ar[d] & \tim...
...r[rr]^>>>>>>>>>>{\sum e_v} & \ \ \ \ \ \ \ \ \ \ \ \ & \mathbb{Z}/2\mathbb{Z}} $

Then $ {\mathrm{Sel}}(E,2)$ is the right kernel of $ e$.

Fact. If $ {\mathrm{Cl}}(\mathcal{O}_S)=0$, then $ i$ is injective.

This follows from class field theory. So we choose $ S_0$ such that $ {\mathrm{Cl}}(\mathcal{O}_{S_0})=0$, and take $ S \supset S_0$. If $ {\mathrm{Cl}}(\mathcal{O}_S)=0$, $ i$ is an injection, and the image of $ i$ is a maximal isotropic subgroup of $ V_S=\bigoplus_{v \in S}V_v$, $ V_v=H^1(k_v,E[2])$.

What we have achieved: the Selmer group $ {\mathrm{Sel}}(E,2)$ is now a kernel of `a square matrix', since $ \bigoplus_{v \in S}E(k_v)/2E(k_v)$ and $ H^1(\mathcal{O}_S,E[2])$ have the same dimension over $ \mathbb{F}_2$. Letting $ \mathfrak{h}(A)=(A^*/A^{*2})^2$, we have

$\displaystyle W_S \times \mathfrak{h}(\mathcal{O}_S) \to \mathbb{Z}/2\mathbb{Z}$


$\displaystyle W_S=\bigoplus_{v \in S} W_v,\ W_v=E(k_v)/2E(k_v). $

Proposition. Assume $ S \supset S_0$ (containing primes above $ \infty,2$ and those of bad reduction and such that $ {\mathrm{Cl}}(\mathcal{O}_{S_0})=0$) is a finite set of places. Suppose that $ \mathfrak{h}(\mathcal{O}_S) \subset V_S$ is a maximal isotropic subgroup. Then there exist $ K_v \subset V_v$ for $ v \in S$ maximal isotropic such that $ K_v=\mathfrak{h}(\mathcal{O}_v)$ for $ v \in S \setminus S_0$, and

$\displaystyle \mathfrak{h}(\mathcal{O}_S) \oplus \bigoplus_{v \in S} K_v=V_S. $

This is purely a result in linear algebra.

Recall we have $ W_S \times \mathfrak{h}(\mathcal{O}_S) \to \mathbb{Z}/2\mathbb{Z}$ from $ W_S \subset V_S$ and $ \mathfrak{h}(\mathcal{O}_S) \subset V_S$.

Definition. We let $ {\mathrm{Sel}}(E,2) \subset \mathcal{I}^S \subset \mathfrak{h}(\mathcal{O}_S)$, where

$\displaystyle \mathcal{I}^S=\{\xi \in \mathfrak{h}(\mathcal{O}_S):$for all $ v \in S$, $ \xi \in K_v+W_v$$\displaystyle \}. $

We let

$\displaystyle \mathcal{W}_S=\bigoplus_{v \in S} W_v/(W_v \cap K_v). $

We then have a pairing

$\displaystyle \mathcal{W}_S \times \mathcal{I}^S \xrightarrow{e_S} \mathbb{Z}/2\mathbb{Z}. $

Proposition. The Selmer group is the kernel of $ e_S$. The map $ \tau:\mathcal{I}^S \simeq \mathcal{W}_S$ is an isomorphism. For $ \mu \in \mathcal{I}^S$, we define a map

$\displaystyle \mu=\textstyle{\sum}_{v \in S} \alpha_v+\beta_v \mapsto \textstyle{\sum}_{v \in S} \alpha_v; $

via $ \tau$, the new pairing $ \mathcal{W}_S \times \mathcal{I}^S \to \mathbb{Z}/2\mathbb{Z}$ is symmetric.

Example. For $ E:y^2=(x-e_1)(x-e_2)(x-e_3)$, $ e_i \in \mathcal{O}_v$, $ v(\prod(e_i-e_j))=1$ (reduction is of type $ I_2$). Then

$\displaystyle \xymatrix{
E(k_v)/2E(k_v) \ar@{->}[r] \ar[dr] & (k_v^*/k_v^{*2})^...
...] \ar[u] \ar@{->>}[r] & \mathbb Z/2\mathbb Z\subset (\mathbb Z/2\mathbb Z)^2

Here $ W_v/(W_v \cap K_v)=\mathbb{Z}/2\mathbb{Z}$, $ W_v \cong (\mathbb{Z}/2\mathbb{Z})^2$.

Algebraico-Geometric version of Selmer group

Let $ k$ be a field, $ {\mathrm{char}}k = 0$, $ E:y^2=(x-c_1(t))(x-c_2(t))(x-c_3(t))$, so $ E$ is defined over $ F=k(t)$. To simplify, we assume that all $ p_i=c_i-c_j$ are of the same even degree. We also assume that $ r(t)=\prod(c_i-c_j)$ is separable, $ r(t)=\rho \prod_{M \in \mathcal{M}} r_M(t)$ where $ \rho \in k^*$, $ r_M$ monic irreducible, so $ E$ has reduction type $ I_2$.

Now assume $ k$ is a totally imaginary number field. The Neron model $ \mathcal{E}/\mathbb{P}^1$ has

$\displaystyle 0 \to \mathcal{E}^* \to \mathcal{E}\to \textstyle{\bigoplus}_{M \in \mathcal{M}} i_{M*} \mathbb{Z}/2\mathbb{Z}\to 0. $

Let $ V \subset \mathbb{A}^1={\mathrm{Spec}}k(t)$ be $ V=\mathbb{A}^1 \setminus \{r=0\}$.

Let $ \mathcal{S}\subset \mathfrak{h}(k(V))$ consist of triples $ (m_1,m_2,m_3)$, $ m_i \in k(t)$ squarefree, $ (\prod_{i=1}^{3} m_i)$ is a square which divides $ r(t)^2$, and $ \gcd(m_i,p_i)=1$.

Put another way, we have

$\displaystyle \xymatrix{
0 \ar[r] & \mathcal{E}(F)/2\mathcal{E}(F) \ar[r] & \ma...
...{E}) \ar[r] \ar[d] & 0 \\
& & (F^*/F^{*2})^2 \ar[r] & H^1(F,E)[2] \ar[r] & 0

Corresponding to $ m=(m_1,m_2,m_3)$, we have the surface $ X_m$ defined by the equations $ x-e_i(t)=m_i(t)u_i^2$, $ i=1,2,3$. Then $ X_m/\mathbb{P}^1$ has a minimal model if and only if $ X_m/\mathbb{P}^1$ is locally isomorphic for the étale topology with $ \mathcal{E}/\mathbb{P}^1$.

We have a map

$\displaystyle \mathcal{S}\xrightarrow{\delta} \bigoplus_{M \in \mathcal{M}} (k_M^*/k_M^{*2}) $

where $ k_M=k[t]/r_M(t)$. We see that $ E(F)/2E(F) \subset \ker \delta$.

Theorem. [Theorem A] Suppose that $ \ker \delta$ is the image of $ E(F)[2]$. (In particular, the generic rank is zero.) Assume Schinzel's hypothesis. Then there exist infinitely many $ x \in \mathbb{P}^1(k)$ such that $ {\mathrm{rk}}E_x(k)=0$.

Theorem. [Theorem B] Let $ m \in \mathcal{S}$, and assume that

$\displaystyle \ker\left(\mathcal{S}\to \bigoplus_M k_M^*/(k_M^{*2},\delta_M(m))\right)=E(F)[2] \oplus \mathbb{Z}/2\mathbb{Z}. $

Assume Schinzel's hypothesis. Assume $ X_m(\mathbf A_k)^{{\mathrm{Br}}_1(X)} \neq \emptyset$. Then there exist infinitely many $ x \in \mathbb{P}^1(k)$ such that $ E_x(k)$ is of rank one and $ X_{m,x}(k) \neq \emptyset$.

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