We had , , and

**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 ,
.

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

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. **

**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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