Colliot-Thelene 2: Rational points on surfaces with a pencil of curves of genus one

We had $ y^2=(x-c_1(t))(x-c_2(t))(x-c_3(t))$, $ E/F=k(t)$, and

$\displaystyle r(t)=\prod_{i<j}(c_i-c_j)=\rho \prod_{M \in \mathcal{M}} r_M(t) $

separable, all $ r_M$ irreducible, even degree, $ \mathcal{S}\subset \mathfrak{h}(F)=(F^*/F^{*2})^2$, $ m=(m_1,m_2,m_3)$, $ m_3=m_1m_2$ up to $ F^{*2}$. $ \mathcal{X}_m/\mathbb{P}^1$, $ x-c_i(t)=m_i(t)u_i^2$, $ i=1,2,3$.

$\displaystyle \delta:\mathcal{S}\to \bigoplus_{M \in \mathcal{M}}k_M^*/k_M^{*2} $

and $ \delta'$ is the composition with the map

$\displaystyle \bigoplus_{M \in \mathcal{M}}k_M^*/k_M^{*2} \to \bigoplus_{M \in \mathcal{M}}k_M^*/(k_M^{*2},\delta_M(m)). $

where $ k_M=k[t]/r_M$.

Fact. $ E(F)/2E(F) \subset \ker\delta$.

Theorem. [Theorem B] Assume Schinzel's hypothesis, and the finiteness of {\tencyr\cyracc Sh}. Let $ m \in \mathcal{S}$, $ m \not \in E(F)[2]$. Let $ X=\mathcal{X}_m/\mathbb{P}^1$. Assume: $ \ker\delta'=E(F)[2] \oplus \mathbb{Z}/2\mathbb{Z}(m)$, and $ X(\mathbf A_k)^{{\mathrm{Br}}_{\text{vert}}(X)} \neq \emptyset$.

Then $ R=\{x \in \mathbb{P}^1(k):\char93 X_x(k)=\infty\}$ is infinite, and $ X(k)$ is Zariski dense.


$\displaystyle {\mathrm{Br}}_{\text{vert}}(X)=\{\xi \in {\mathrm{Br}}(X):\text{$...
...\vert _{X_\eta}$\ comes from ${\mathrm{Br}}(F)$}\} \subset {\mathrm{Br}}_1(X). $

Schinzel's hypothesis: Let $ P_i(x) \in \mathbb{Z}[x]$ for $ i=1,\dots,r$ be distinct irreducible polynomials with leading coefficient positive (plus technical condition, to exclude polynomials like $ X^2+X+2$); then there exist infinitely many values $ n \in \mathbb{N}$ such that each $ P_i(x)$ is a prime.

Remark. The assumption that $ \ker\delta'=E(F)[2] \oplus \mathbb{Z}/2\mathbb{Z}(m)$ is satisfied for general $ c_i$. For the assumption that $ X(\mathbf A_k)^{{\mathrm{Br}}_{\text{vert}}(X)} \neq \emptyset$, in general we have $ {\mathrm{Br}}_{\text{vert}}(X)={\mathrm{Br}}k$ so this reduces to $ X(\mathbf A_k) \neq \emptyset$.

The proof of this theorem will take up the rest of these notes.

We shall define a finite set $ S \subset \Omega$ of `bad places'. For each $ v \in S$, we have some $ x_v \in k_v$, and we look for $ x \in \mathcal{O}_S$, with $ x$ very close to $ x_v$ for $ v \in S$, and find one $ x$ such that:

We have $ S \supset S_0$, containing $ 2$-adic places and those over $ \infty$ (note we are being sloppy for real places), and places of bad reduction of $ \mathcal{E}/k$, of $ \mathcal{X}_m=X/k$, and such that $ {\mathrm{Cl}}(\mathcal{O}_{S_0})=0$. If $ v \not\in S$, then $ r_M(t) \in \mathcal{O}_S[t]$ is separable (finite étale cover).

We look at $ r_M(x)$ and look at its prime decomposition; it will have some part in $ S$ and another part $ T_M^x$ of primes of multiplicity $ 1$, and one prime $ v_M$, the `Schinzel prime'. We realize $ (T_M \cup v_M) \cap (T_N \cup v_N)=\emptyset$ for $ M \neq N$. Then $ \mathcal{E}_x/k$ has bad reduction in $ S \cup (\bigcup_M T_M) \cup (\bigcup v_M)$.

First we find $ x$ such that $ X_x(\mathbf A_k) \neq \emptyset$. For $ M \in \mathcal{M}$, we introduce the algebra $ A_M(t)={\mathrm{cores}}_{k_M/k}(K_M/k_M,t-\theta_M)$, where $ K_M/k_M$ is the quadratic extension connected to $ m$. Then $ r_M(t)=N_{k_M/k}(t-\theta_M)$. A priori, $ A_M(t) \in {\mathrm{Br}}k(t) \to {\mathrm{Br}}X_\eta$. Since $ r_M$ is even, $ A_M \in {\mathrm{Br}}(X)$.

Let $ (P_v) \in X(\mathbf A_k)^{{\mathrm{Br}}_{\text{vert}}(X)}$, with projection $ (x_v) \in \mathbb{A}^1(k_v)$. We know that $ \sum_{v \in \Omega} A_M(x_v)=0$. For almost all places of $ k$, $ X(k_v) \xrightarrow{A_M} {\mathrm{Br}}(k_v)$ is trivial.

Fix $ S$ as before plus places where some $ A_M$ is not trivial on $ X(k_v)$. Now $ \sum_{v \in S} A_M(x_v)=0$. Fix $ x_v \in k_v$ for $ v \in S$ with this property. Note $ X_{x_v}(k_v) \neq \emptyset$. Suppose $ x \in \mathcal{O}_S$ is very close to $ x_v$ for $ v \in S$ and the decomposition of $ r_M(x)$ has all primes in $ T_M$ split in $ K_M/k_M$.

Claim. For such $ x$, $ X_x(\mathbf A_k) \neq \emptyset$.

Proof. The only places where it will fail to have a point are those of bad reduction. For $ v \in S$, $ X_x(k_v) \neq \emptyset$, as $ x$ is close to $ x_v$. For $ v \in T_M$, $ X_x(k_v) \neq \emptyset$ since the two rational curves are defined. For $ v=v_M$, write

$\displaystyle 0=\sum_{v \in \Omega} A_M(x)_v=\sum_{v \in S} A_M(x)_v+\sum_{v \in T_M} A_M(x)_v + A_M(x)_{v_M}=0+0+A_M(x)\vert _{v_M} $

the second because $ v$ splits in $ K_M$; therefore the prime that is left over forces the prime $ v_M$ to split in $ K_M/k_M$, we again have points locally. $ \qedsymbol$

For the second part, we now need to control the Selmer groups uniformly in the family $ \mathcal{E}_x$, for $ x$ satisfying $ (*)$, namely, $ x$ is very close to $ x_v$ for $ v \in S$, $ r_M(x)$ decomposes into primes in $ S$, $ T_M$ primes splitting in $ K_M/k_M$, and the Schinzel prime $ v_M$.

Let $ T=T_x=S \cup \bigcup_{M \in \mathcal{M}}T_m$, and $ \widetilde{T}=\widetilde{T_x}=S \cup \bigcup T_M \cup \bigcup {v_M}$. We have

$\displaystyle \xymatrix{
0 \ar[r] & \mathfrak{h}(\mathcal{O}_T) \ar[r] \ar[d] &...
...{ord}}_M} &
\bigoplus_{M \in \mathcal{M}}(\mathbb Z/2\mathbb Z)^2 \ar[r] & 0


$\displaystyle {\mathrm{Sel}}(E_x) \subset \mathcal{I}^{\widetilde{T}} \subset \mathfrak{h}(\mathscr{O}_{\widetilde{T}}), $

and $ N_x$ the Cartesian square from $ \mathfrak{h}(\mathcal{O}_T[u]) \xrightarrow{{\mathrm{ev}}_x} \mathfrak{h}(\mathcal{O}_{\widetilde{T}})$. Then $ {\mathrm{Sel}}(E_x)$ is the kernel of a symmetric pairing on $ N_x$.

We find two `constant' subgroups $ N_x$. First, we find $ N_0=\mathfrak{h}(\mathcal{O}_S) \cap \mathcal{I}^{\widetilde{T_x}}$, fixed because $ x$ very close to $ x_v$ for $ v \in S$. For the second pair:

Proposition. For each $ M \in \mathcal{M}$, there exists a unique $ (a_M,b_M) \in \mathfrak{h}(\mathcal{O}_S)$ such that for any $ x$ with $ (*)$, $ (a_Mr_M(x)^{f_M},b_Mr_M(x)^{g_M}) \in \mathcal{I}^{\widetilde{T_x}}$, and for each $ v \in S$, its component in $ V_v$ belongs to $ K_v$.

We define the subgroup

$\displaystyle A=\bigoplus_{M \in \mathcal{M}} \mathbb{Z}/2(a_M r_M(t)^{f_M},b_Mr_M(t)^{g_M}). $

Note $ (f_M,g_M) \in \{(0,1),(1,0),(1,1)\}$.


$\displaystyle N_x=N_0 \oplus A \oplus \phi_x^{-1}\left(\bigoplus_{v \in T \setminus S}(W_v/W_v \cap K_v)\right). $

Proposition. On $ N_0 \oplus A \subset N_x$, the restriction of the pairing $ e_x$ is independent of $ x$.

To prove this, use various reciprocity laws.

To conclude, write $ N_0 \oplus A=B_0 \oplus B_1 \oplus B_2 \subset \mathfrak{h}(k(t))$, where $ B_0=E(F)[2] \oplus \mathbb{Z}/2\mathbb{Z}(m)$, $ B_0 \oplus B_1 = \ker e_x\vert _{N_0 \oplus A}$, and $ B_2$ is the supplement. Use the assumption that $ \ker \delta'=(\mathbb{Z}/2\mathbb{Z})^3$ to get rid of $ B_1$...

Now use finiteness of {\tencyr\cyracc Sh} and Cassels-Tate pairing, $ \dim(${\tencyr\cyracc Sh}$ (E_x)[2]) \leq 1$ implies {\tencyr\cyracc Sh}$ (E_x)[2]=0$, so the rank is $ 1$.

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