Schinzel Hypothesis H and its quantitative version Bateman-Horn’s conjecture give a general setting for a family of polynomials with integral coefficients to have simultaneous prime values in . This post will discuss the genus zero function field analog, which behaves in an interesting way, as the naive analog fails. Then we will focus on one result of Pollack and myself.

### Prime values of polynomials

We are motivated by the questions: Do there exist infinitely many primes such that

- ?
- ?
- ?
- ?
- ? (Twin prime conjecture)
- ?

Euclid answered the first question, while Dirichlet’s theorem gives a positive answer for the second question provided . Questions 3 and 5 are unknown. Now the answer of the remaining two is *obviously no* since there are local obstruction ( divides all the values of while divides ).

Schinzel gave a general conjecture that provides answers to these type of questions.

**Schinzel’s Hypothesis H:**

Let be polynomials with positive leading coefficients. Then there exists infinitely many such that all are prime *provided* are irreducible, and there is no local obstruction, i.e., there is no dividing all the values of .

The prime number theorem gives a quantitative answer to the first question above. Namely the number of primes up to $x$ is approximately .

The Bateman-Horn generalizes this in the general setting of the Schnizel hypothesis H:

**Bateman-Horn’s conjecture:**

Let be irreducible polynomials with positive leading coefficients and let be the number of such that all are prime. Then

where if , then

Note that

- If and , then . Therefore each factor in the product is $1$, and hence . Therefore, in this case, we retain the prime number theorem.
- If some prime divides all the values of , then , and hence the corresponding factor in is zero, so .
- Otherwise, one can show that converges to a positive constant.

### Analog over finite fields

Replacing with , for some prime power , and a finite field , we get a naive analog (in genus zero).

**Naive analog of Schinzel hypothesis H:**

irreducible, having no local obstructions, many simultaneous prime values in .

Denote by

the number of degree monic polynomials such that all are irreducible. Then the naive Bateman-Horn conjectures says that

,

where if for an irreducible polynomial with a root $\alpha$ we let , then

Surprisingly the naive analogs of the above conjectures over finite fields fails:

**Swan’s example:**

The polynomial is irreducible, has no local obstruction, but either has as a factor, or an even number of factors. In particular all of its values are composite.

Similar examples can be constructed over any finite field. Conrad, Conrad, and Gross study when the conjecture fail. Their findings lead them to revised the naive conjecture. Very briefly if , then the naive conjecture remains the same. If , , or , , then additional factor $\Lambda (f_i,n)$ is entered. In the case , no satisfactory conjecture is given.

### A result over large finite fields

From now on we consider the special case of constant coefficients, that is, we take . This special case includes, for example, the *twin prime conjecture* when taking and $f_2(X) = X+1$, or more generally, , for some .

Since finite fields are perfect, the irreducibility of implies that , so . Therefore we are in the scope of the naive conjectures. We will consider large finite fields what means that we assume to be large w.r.t. and . The conjecture constant becomes simpler under this assumption:

**Lemma (Pollack):** Let be a fixed positive integer. Then for any irreducible non-associate with we have

Therefore we have the following conjecture for large finite fields:

**Conjecture:**

Let be fixed. Then for every and for every irreducible non-associate with we have

Pollack proves this conjecture when , i.e., in odd characteristic and under the assumption .

**Theorem (Pollack):** If , the conjecture holds. More precisely

Remarks:

- The error term dependence in is of bounded by an expression of order of magnitude
- The theorem extends to the case odd and and to the case and odd. This will be discussed below.

Let’s give a very brief idea of the proof. In the first we calculate a Galois group:

The number of such that for we have

- The map has only simple ramification (i.e.\ is Morse). This means that the ramification type of every ramified point is .
- The ramification loci of are distinct. Here $\alpha$ runs over the roots of .
is . (That is almost all polynomials $g_0$ satisfy these two conditions.)

For this calculation one uses . Fix satisfying these conditions. Since , the ramification of at infinity is tame. Therefore the first property implies that is a transitive group generated by transpositions, hence . Here is the algebraic closure of $\mathbb{F}_q$. From the second property we get that the splitting fields of are linearly disjoint over (since the intersection field of such splitting fields is ramified only at infinity, and this ramification is tame). Here runs over the roots of . Noting that one can show that

We emphasize that the Galois group is the **biggest possible**.

It remains to count, for this , how many there are such that are all irreducible. This happens if and only if the Frobenius element at in the splitting field of acts transitively on the roots of for all . Thus, by Chebotarev’s theorem for finite fields we get that there are

many such . This finishes the proof since .

### Pseudo algebraically closed fields

A field $latex $K$ is called PAC if every non-void -variety that is geometrically irreducible has a -rational points.

Algebraically closed and separably closed fields are the trivial examples of PAC fields, while finite fields and number fields are not PAC. (E.g., has no rational points, so is not PAC.)

A consequence of Weil’s bound for the number of points on curves over finite fields is that a ultraproduct of finite fields is PAC (Ax).

A non-trivial consequence of Hilbert’s irreducibility theorem is that for a fixed integer , if are automorphism of the field of algebraic numbers , then the probability that the fixed field

is PAC is (Jarden). Here the probability measure is induced from the Haar measure of the compact group .

It is now appropriate to give an explicit example. Let be the field of all totally real numbers. (Totally real is a number all the conjugates of which are real.) Then is *PRC*, hence is PAC (Pop). (It is interesting to note that the fact that is *PRC* is used to prove potential modularity, a key step in the proof of Serre’s modularity conjecture.)

Hypothesis H over PAC fields cannot hold as is. Indeed, let , and let $f=f_1 = X$. Then for every , there is no irreducible . Let us introduce an obvious necessary condition.

Assume that is irreducible for some $f,g\in K[t]$. Let be a root of $f(g(t))$. Then is a root of , since . Then and

So we conclude . So has an extension of degree .

We will consider separable polynomials, in this case, all the above extensions are separable. Over PAC fields the above necessary condition suffices for Hypothesis H result, provided some condition on the characteristic, which is less restrictive than Pollack’s. Namely, if we assume that is odd:

**Theorem (Bary-Soroker):** Let

- be a PAC field of characteristic ,
- odd if ,
- separable irreducible with respective roots .

Assume has a separable extension of degree , for all . Then there exists a Zariski dense set of such that for all are separable irreducible.

The proof of this theorem starts with computing a Galois group, similarly to Pollack’s proof. This time we compute the Galois group of over , where is an -tuple of variables, and is the corresponding generic polynomial. We show that if or if and is odd, then is the biggest possible: If denotes the set of roots of , then

.

We will not get into the details of the proof. The details appear in ArXiv:1005.4528, and probably in a near future post. We will just mention, that since the Galois group of over is , for every , to show the is the biggest possible reduces to discriminant calculations (in characteristic , one should consider generalized discriminants). We use the formula for the discriminant given by the resultant when divides . In characteristic , not only the formulas become more complicated, but also there are obstructions for to be maximal: If , is the biggest possible if and only if every *even* sum of elements in does not vanishes. We do not know what happens when .

The second step in the proof is to find a Zariski dense set of simultaneous irreducible specializations for . Namely, a Zariski dense set of -tuple such that are all irreducible, where . We study this in much more generality, and give an exact criterion for a tuple of irreducible to have a Zariski dense set of simultaneous irreducible specializations over a PAC field, with no restriction on the characteristic. More details in *loc. cit.*, and again, in a future post.

### Back to finite fields

We will explain how, using the elementary theory of finite fields, Hypothesis H over PAC fields implies a result over large finite fields. In particular, this extends Pollack’s theorem to the case and and to the case and is odd. To explain how this is achieved, we have to discuss pseudo finite fields.

A field is called pseudo finite if

- is PAC,
- is perfect, and
- has a unique extension of degree , for every .

Note that the latter condition is equivalent to .

Ax showed that a consequence of Riemann Hypothesis for curves over finite field is that ultraproducts of finite fields are pseudo finite. Jarden showed that a consequence of Hilbet’s irreducibility theorem is that the fixed field of a Galois automorphism in is pseudo finite with probability one.

The significance of pseudo finite fields is that they are the infinite models of finite fields:

**Theorem (Ax):** Let be an elementary statement in the theory of rings. Then all pseudo finite fields satisfy if and only if all but finitely many finite fields satisfy .

We apply the theorem for PAC fields to a pseudo finite field . Since is perfect, we drop the separability conditions. Since has a unique separable extension of degree , for any , the necessary condition is automatically satisfied. Hence we get:

**Theorem (Bary-Soroker):** Let

- be a pseudo finite field of characteristic ,
- odd if ,
- irreducible polynomials.

Then there exists a Zariski dense set of such that for all are separable irreducible.

Note that, for a fixed , the statement: *For every irreducible polynomials satisfying if , then there exists such that all are irreducible.* Is elementary. Hence, using Ax theorem, we get an existence theorem over finite fields.

However, to get the stronger quantitive theorem for finite fields, we need to use a stronger statement that is proven. Namely over pseudo finite fields we have a variety rational points of which parametrize the coefficients of ‘s for which all are irreducible.

Therefore we reprove Pollack’s theorem and extends it to the case and and to the case and odd:

**Theorem:** For positive integers , a prime power, and non-associate irreducible polynomials such that we have

provided is odd if .