We had , , and
Theorem. [Theorem B] Assume Schinzel's hypothesis, and the finiteness of . Let , . Let . Assume: , and .
Then is infinite, and is Zariski dense.
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:
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 , .
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
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
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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