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

separable, all irreducible, even degree, , , up to . , , .

and is the composition with the map

where .

Fact. .

Theorem. [Theorem B] Assume Schinzel's hypothesis, and the finiteness of . Let , . Let . Assume: , and .

Then is infinite, and is Zariski dense.

Here,

Schinzel's hypothesis: Let for be distinct irreducible polynomials with leading coefficient positive (plus technical condition, to exclude polynomials like ); then there exist infinitely many values such that each is a prime.

Remark. The assumption that is satisfied for general . For the assumption that , in general we have so this reduces to .

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

We shall define a finite set of bad places'. For each , we have some , and we look for , with very close to for , and find one such that:

• ;
• , spanned by and .

We have , containing -adic places and those over (note we are being sloppy for real places), and places of bad reduction of , of , and such that . If , then is separable (finite étale cover).

We look at and look at its prime decomposition; it will have some part in and another part of primes of multiplicity , and one prime , the Schinzel prime'. We realize for . Then has bad reduction in .

First we find such that . For , we introduce the algebra , where is the quadratic extension connected to . Then . A priori, . Since is even, .

Let , with projection . We know that . For almost all places of , is trivial.

Fix as before plus places where some is not trivial on . Now . Fix for with this property. Note . Suppose is very close to for and the decomposition of has all primes in split in .

Claim. For such , .

Proof. The only places where it will fail to have a point are those of bad reduction. For , , as is close to . For , since the two rational curves are defined. For , write

the second because splits in ; therefore the prime that is left over forces the prime to split in , we again have points locally.

For the second part, we now need to control the Selmer groups uniformly in the family , for satisfying , namely, is very close to for , decomposes into primes in , primes splitting in , and the Schinzel prime .

Let , and . We have

Here

and the Cartesian square from . Then is the kernel of a symmetric pairing on .

We find two `constant' subgroups . First, we find , fixed because very close to for . For the second pair:

Proposition. For each , there exists a unique such that for any with , , and for each , its component in belongs to .

We define the subgroup

Note .

Proposition.

Proposition. On , the restriction of the pairing is independent of .

To prove this, use various reciprocity laws.

To conclude, write , where , , and is the supplement. Use the assumption that to get rid of ...

Now use finiteness of and Cassels-Tate pairing, implies , so the rank is .

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