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

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

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 $F^{*2}$. $\mathcal{X}_m/\mathbb{P}^1$, $x-c_i(t)=m_i(t)u_i^2$, $i=1,2,3$.

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

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

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

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 . 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.

Here,

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

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:

• $X_x(\mathbf A_k) \neq \emptyset$;
• $\dim_{\mathbb{F}_2}({\mathrm{Sel}}(E_x,2))=3$, spanned by $E_x[2]$ and $m(x)$.

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

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

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

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

Here

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

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

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

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

Proposition.

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

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 and Cassels-Tate pairing, $\dim($$(E_x)[2]) \leq 1$ implies $(E_x)[2]=0$, so the rank is $1$.

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