## Irreducible speciailizations give group preserving specializations

June 12, 2013

In this post I discuss the classical connection between irreducible and group preserving specializations in Hilbert’s irreducibility theorem.

A field $K$ is Hilbertian if every $f\in K[t,X]$ that is separable and of positive degree in $X$ and irreducible in $K[t,X]$ admits infinitely many irreducible specializations, that means, there exist infinitely many $a\in K$ such that $f(a,X)$ is irreducible and separable as a polynomial in $K[X]$.

A group preserving specialization is a stronger type of specialization. For $f(t,X)$ separable in $X$, let $F$ be its splitting field over $K(t)$, and let $G= Gal(f,K(t)) = Gal(F/K(t))$ be the Galois group of $f$ over $K(t)$. For any $a\in K$ with $f(a,X)$ separable, let $F_a$ be the splitting field of $f(a,X)$ and $G_a = Gal(f_a,K)=Gal(F_a/K)$ be the Galois group. One always has that $G_a$ is a subquotient of $G$ (that is $G_a$ is isomorphic to the decomposition group modulo the inertia group of a prime of $F$ that lies over the prime $(t-a)$ of $K(t)$).

A specialization $t\mapsto a$ is called group preserving specialization if $G\cong G_a$, or equivalently if $|G|=|G_a|$.

If $t\mapsto a$ is an irreducible specialization for $f$ it is NOT necessarily group preserving. For example, the Galois group of $X^4+X^3+X^2+ X+t$ is $S_4$, the specialization $t\mapsto 1$ gives $X^4+X^3+X^2+X+1$ – which is irreducible but its Galois group is $G_1 = (\mathbb{Z}/5\mathbb{Z})^\times \cong \mathbb{Z}/4\mathbb{Z}$ (since its the minimal polynomial of the root of unity $\zeta= e^{2 \pi i/5}$).

Nevertheless, if $K$ is Hilbertian we can always find group preserving specializations:

Theorem (Hilbert): If $K$ is Hilbertian, then  for every separable-in-$X$ polynomial $f(t,X)$ there is a group preserving specialization.

A proof with all details can be found, e.g., in Galois theory of Serre or in Field Arithmetic of Fried-Jarden. Here’s a

Sketch of Proof:
Let $F$ be the splitting field of $f(t,X)$ over $K(t)$ and let $=Gal(F/K(t))$. Since $F/K(t)$ is separable, by the primitive element theorem, there exists $\xi \in F$ such that $F=K(t)(\xi)$. Let $g(t,X)$ be the minimal polynomial of $\xi$ over $K(t)$. Now apply the Hilbertianity property for $g$: There exist infinitely many $a\in K$ such that $g(a,X)$ is irreducible. Choose such an $a\in K$ with the extra properties that (1) $\deg_X f(t,X) = \deg f(a,X)$ and (2) $\deg_X g(t,X) = \deg g(a,X)$. Now since $F$ is generated by the roots of $f(t,X)$ on the one side and by $\xi$ on the other, one can show that the splitting field $F_a$ of $f(a,X)$ equals the field generated by a root of $g(a,X)$. So

$\begin{array}{ll} |G| &= [F:K(t)]=\deg_X g(t,X)\\ & = \deg g(a,X) =[F_a:K] =|G_a|.\end{array}$

So the groups are of the same size, as needed.

QED.

## Hardy-Littlewood tuple conjecture over large finite fields

November 27, 2012

In this post I will describe the solution to the Hardy-Littlewood tuple conjecture for polynomials over large finite fields.
Let’s first recall the

## Classical Setting

The prime number theorem says that

$\displaystyle \pi(x) := \sum_{0.

(In fact, the inverse logarithmic function, $\textnormal{Li}(x) = \int_{2}^x \frac{1}{\log t}dt$, better approximates $\pi(x)$, but we will not get into these subtleties  in this post.)

Let $k>0$. We will say that a $k$-tuple $(n_1, \ldots, n_k)$ of integers is a prime tuple if $n_i$ is a prime for every $i$.

For a $k$-tuple of integers, say $a=(a_1,\ldots, a_k)$, let $\pi_a(x)$ be the number of $0 such that $(n+a_1, \ldots, n+a_k)$ is a prime tuple.

The Hardy-Littlewood tuple conjecture asserts that

$\displaystyle \pi_a(x) \sim \mathfrak{S}_a\frac{x}{\log^k x}, \quad x\to\infty$,

where, if $\nu(p)$ denotes the number of mod $p$ residue classes  in $\{a_1, \ldots, a_k\}$, then the singular series, $\mathfrak{S}_a$, is given by

$\displaystyle \mathfrak{S}_a = \prod_{p} \left(1-\frac{\nu(p)}{p}\right)$.

First we note that if $\nu(p)=p$, for some $p$, then $p$ divides one of the entries of $(n+a_1, \ldots, n+a_k)$, for every $n$, so $\pi_a(x) = O(1)$. This is consistent with the Hardy-Littlewood conjecture, since in this case, $\mathfrak{S}_a=0$.

If $\nu(p) for all primes, then one can show that the product converges, so $\mathfrak{S}_a>0$. Therefore, in this case, the events that $n+a_1$ is prime, $\ldots$, $n+a_k$ is prime, are, in a sense, independent up to the factor of the singular series.

A notable case of The Hardy-Littlewood tuple conjecture is when $a=(0,2)$, in which $\pi_a(x)$ counts twin primes.

The only case of Hardy-Littlewood tuple conjecture that is settled is $k=1$, which is the prime number theorem. All other cases are hopelessly open.

## Polynomials over Finite Fields

In the ring $\mathbb{F}_q[t]$ of polynomials over a finite field with $q$ elements, the monic irreducible polynomials play the role of (positive) prime numbers. To get the analog of the prime number theorem, we let $\displaystyle \Pi(q,n)$ be the number of irreducible monic polynomials $f\in \mathbb{F}_q[t]$ of degree $n$. Then the analogous prime number theorem tells us that

$\displaystyle \Pi(q,n)=\frac{q^n}{n}+O(\frac{q^{n/2}}{n}) \sim \frac{q^n}{n}, \quad q^n \to \infty$.

Here $q^n$ plays the role of $x$ in the classical setting, since there are $q^n$ monic polynomials of degree $n$, and $n=\log_q (q^n)$ plays the role of $\log x$.

We note that here, in contrast to the classical case, there are several ways $q^n$ can tend to $\infty$, notably $q\to \infty$ and $n\to \infty$. Since we have a good error term in the prime number theorem in polynomial rings, the theorem holds in all limits.

However, it happens that some questions in number theory in polynomial rings can be solved only in certain limits. In what follows I’m going to explain a solution of the Hardy-Littlewood tuple conjecture in polynomial rings in the limit $q\to \infty$, $q$ odd.

For positive integers $n,k$, a prime power $q$, and a $k$-tuple $a=(a_1, \ldots, a_k)$ of polynomials in $\mathbb{F}_q[t]$ each of degree $, we let $\Pi_a(q,n)$ be the number of monic polynomials $f \in \mathbb{F}_q[t]$ of degree $n$ such that $(f+a_1, \ldots, f+a_k)$ is a $k$-tuple of irreducible polynomials.

For an analogous Hardy-Littlewood tuple conjecture we shall have to introduce a singular series $\mathfrak{S}_a$, which is defined analogously to the classical case. That is

$\displaystyle \mathfrak{S}_a =\prod_{f}\left(1-\frac{\nu(f)}{q^{\deg f}}\right)$,

where $f$ runs over all monic irreducibles and $\nu(f)$ is the number of residues of $\{a_1, \ldots, a_k\}$ mod $f$. We note that $\nu(p)\leq k$, so if $q\to \infty$, then each factor will tend to $1$, so $\mathfrak{S}_a\to 1$ (I wasn’t very rigorous here…).

Therefore, in the limit $q\to \infty$, the Hardy-Littlewood tuple conjecture should say that $\Pi_a(q,n) \sim \frac{q^n}{n^k}$, as $q\to \infty$.

At least if $q$ is odd this is true:

Theorem (Bary-Soroker 2012): Let $n,k$ be positive integers, let $q$ be an odd prime power, and let $(a_1,\ldots, a_k)$ be a $k$-tuple of distinct polynomials of degree $. Then

$\displaystyle \Pi_a(q,n) = \frac{q^n}{n^k} + O_{n,k}(q^{n-1/2})$

Here the $O_{n,k}$ notation means that there exists a constant $C = C(n,k)$, depending only on $n$ and $k$ – and NOT on $q$ or on $a$ – such that $|\Pi_a(q,n)-\frac{q^n}{n^k}|\leq C q^{n-1/2}$.

Before going to the points in the proof several remarks are in place.

1. The interesting case $k=2$ is due to Bender and Pollack, see here.
2. Another interesting special case is when the polynomials in $a$ are constants. In this case there are stronger results by Pollack and by myself.

Let’s explain the main steps of the proof

## Basic Strategy

Let $F$ be the generic polynomial of degree $n$ over $\mathbb{F}_q$. That is to say, $F_U(t)=t^n + U_1 t^{n-1} + \cdots +U_n$, where $U = (U_1, \ldots, U_n)$ is an $n$-tuple of variables. Then, we can think of $\Pi_a(q,n)$ as the number of specializations $U\mapsto u\in \mathbb{F}_q^n$ such that $(F_u+a_1, \ldots, F_u+a_k)$ is a $k$-tuple of irreducibles. (Here $F_u = t^n + u_1 t^{n-1} + \cdots + u_n$.)

This point of view allows us to break the problem into two parts. The first part is an analog of Hilbert’s irreducibility theorem: The study of irreducible specializations for polynomials. We will see that unlike the classical setting, over finite fields the existence, and the number of irreducible specializations is governed by the Galois group. We will elaborate more below.

For now we shall say that the above consideration leads to the study of the Galois groups of the polynomials $F_U + a_1, \ldots, F_U+a_k$. Since the coefficients of $F_U$, hence of $F_U + a_i$, are variables we get that $F_U+a_i$ is irreducible (in the ring $\mathbb{F}_q[U,t]$, and even ${\rm Gal}(F_U+a_i, \mathbb{F}_q(U)) = S_n$). This, however, does not suffice for our needs. To apply the above mentioned Hilbert’s irreducibility theorem, we have to know the Galois group of $\hat{F}_U = \prod_{i=1}^k (F_U + a_i)$.

## Calculating a Galois Group

Let $\mathbb{F}$ be an algebraic closure of $\mathbb{F}_q$.  Then the restriction of automorphisms map gives an embedding

$\displaystyle {\rm Gal}(\hat{F}_U,\mathbb{F}(U))\to {\rm Gal}(\hat{F}_U,\mathbb{F}_q(U))$.

Since $\hat{F}_U = \prod_{i=1}^k (F_U + a_i)$ we have

$\displaystyle {\rm Gal}(\hat{F}_U,\mathbb{F}_q(U)) \leq \prod_{i=1}^k {\rm Gal}(F_U+a_i,\mathbb{F}_q(U))\cong S_n^k$.

Thus if we show that $G = {\rm Gal}(\hat{F}_U,\mathbb{F}(U))\cong S_n^k$, we will get that also  ${\rm Gal}(\hat{F}_U,\mathbb{F}_q(U))\cong S_n^k$.

For each $k$ we have an epimorphism $\varphi_i\colon G \to {\rm Gal}(F_U+a_i, \mathbb{F}(U))\cong S_n$. Composing each $\varphi_i$ with the signature map $S_n\to S_2$, we get $\nu_i\colon G\to S_2$.

Lemma. The map $\prod_{i=1}^k \varphi_i \colon G\to S_n^k$ is surjective if and only if the map $\prod_{i=1}^k \nu_i \colon G\to S_2^k$ is surjective.

By this lemma it suffices to study the maps $\nu_i$.

Since $q$ is odd (here is the only place we are using the oddness of $q$), by Kummer theory, each $\nu_i$ corresponds to a square class of an element in the field $\mathbb{F}(U)$. By its definition, $\nu_i$ corresponds to the class of the discriminant of $F_U+a_i$. Note that $\prod_{i} \nu_i$ is surjective if and only if

• the discriminant of $F_U+a_i$ is  non-square (this is obvious since the corresponding Galois group is $S_n$) for each $i$.
• The product of any subset of discriminants is non-square.

The latter condition can be achieved by applying a very nice argument of Carmon and Rudnick, taken from here. The details will be given in a later post.

## Irreducible Specializations

Let me state a type Hilbert’s irreducibility theorem for finite fields in the setting we have here. I should say that this is not the most general theorem.

Theorem. Let $U=(U_1, \ldots, U_n)$ be a set of variable, with $n\geq 1$, let $F_1,\cdots, F_k$ be polynomials in $\mathbb{F}_q[U,t]$ of respective degrees $d_i$ and $t$-degrees $n_i$. Assume that the Galois group of the product $F=F_1\cdots F_k$ over $\mathbb{F}(U)$ is isomorphic to $\prod_{i=1}^k S_{n_i}$. Then the number of simultaneous irreducible specializations, i.e. of $u\in \mathbb{F}_q^n$ for which $F_i(u,t)$ is irreducible for $i=1, \ldots, k$, is

$\displaystyle \frac{q^n}{n^k} + O_{D}(q^{n-1/2})$,

where $D = \sum_{i=1}^k d_i$.

A few remarks:

1. If the Galois groups of $F$  over $\mathbb{F}(U)$ and over $\mathbb{F}_q(U)$ are isomorphic; denote by $G\leq \prod_{i=1}^k S_{n_i}$ , i.e., if the splitting field of $F$ over $\mathbb{F}_q(U)$ is regular, then the same result as in the above theorem holds with a factor $c=c(G)=C(G)/|G|$ before $\frac{q^n}{n}$. Explicitly, $C(G)$ is the number on elements $g=(g_1, \ldots, g_k)$ in $G\leq \prod_{i=1}^k S_{n_i}$ such that each $g_i$ is an $n_i$-cycle.
2. The theorem also has a version when the splitting field is not regular.
3. The theorem straightforwardly follows from an explicit Chebotarev’s theorem for function fields. The case $n=1$ was proved by Geyer and Jarden, and perhaps a future post will discuss it.

## Conclusion

As explained before $\Pi_a(q,n)$ is the number of specializations $U\to u\in \mathbb{F}_q^n$ such that $F_{u}+a_i$ is irreducible for every $i=1, \ldots, k$. The calculation of the Galois group described above tells us that we can apply the theorem on irreducible specializations for $F_i = F_U+a_i$. Since the degree of $F_U+a_i \leq n$, we get that the sum $D$ of the degrees is bounded by $nk$, so

$\displaystyle \Pi_a(q,n) = \frac{q^n}{n^k}+O_{n,k}(q^{n-1/2})$,

as needed.

## square-independent numbers

March 21, 2012

We know that the number of non-square integers in $I(x) = \{1, \ldots, x\}$ is approximately $\sqrt{x}$. Thus almost all numbers are not squares. Similar result holds when one looks for sets whose elements and product of two elements are all non-square.

Let’s say that a set of integers $A$ is square independent if for every distinct elements $a_1, \ldots, a_n$ of $A$ we have that $a_1 \cdots a_n$ is not a square. An interesting question is

What is the maximal size ${\rm si}(x)$ of square-independent subset $A$ of $I(x)$?

Clearly, the prime numbers in $I(x)$ are square-independent, thus ${\rm si}(x)\geq \pi(x)$.

At a first glance, one might be led to the conclusion (as I did) that ${\rm si}(x)$ might be much bigger, however this is not the case. Namely

${\rm si}(x) = \pi(x)$.

After discussing this matter with Lior Rosenzweig, he came up with the following proof (we didn’t check if it appears in the literature, I’ll be glad to have a reference).

Let $V$ be the set of all integers whose prime factors are $\leq x$ modulo the square relation, i.e. $n\sim m$ iff $nm = \square$. Then $V$ is a vector space over $\mathbb{F}_2$. Since the set of primes $\leq x$ is a basis of $V$, we get that $\dim V = \pi(x)$. Since a square-independent set $A$ is an independent set in $V$, we get that $|A| \leq \pi(x)$. This finishes the proof.

## A simplified proof of a theorem of Thornhill

January 29, 2012

In this post I wish to present a simplified proof of a theorem of Christopher Thornhill that settles a part of a conjecture of Moshe Jarden.

Conjecture (Jarden): Let $K$ be a Hilbertian field, let $n$ be an integer, and for each prime number $\ell$ let $\rho_\ell \colon G_K \to GL_n(\mathbb{Z}_\ell)$ be a Galois representation. (Here $G_K$ is the absolute Galois group of $K$.) Let $K_\ell$ be the fixed field of $\ker \rho_\ell$ in the algebraic closure of $K$ and let $N = \prod_\ell K_\ell$ be the compositum of $K_\ell$. Then every extension  $M/K$ that is contained in $N$ is Hilbertian.

Jarden proved the conjecture for $K$ a number field and abelian variety $A/K$, that is, $\rho_\ell$ is the family of Galois representations induced from the action of the Tate module $T_\ell = \varprojlim A[\ell^n]$. The original conjecture of Jarden is about abelian varieties over arbitrary Hilbertian fields.

Thornhill proved the above conjecture under the extra assumption that $M/K$ is Galois.

Theorem (Thornhill): Under the notation of the conjecture of Jarden, if $M/K$ is Galois, then $M$ is Hilbertian.

Thornhill’s proof contains many innovations, in particular his definition of Hilbertian pairs of profinite groups. In this post I wish to give a new and simplified proof of Thornhill’s Theorem.

## Preparations:

The first result we need is a consequence of the Larsen-Pink Theorem that we formulate as a fact:

Fact 1(Larsen-Pink’s Theorem):
There exists a constant $J(n)$, depending only on $n$, such that for every $G\leq GL_n(\mathbb{F}_\ell)$ there exist normal subgroups $G_3\leq G_2\leq G_1\leq G$ of $G$ such that $[G:G_1]\leq J(n)$, $G_1/G_2$ is a product of finite simple groups of Lie type in characteristic $\ell$, $G_2/G_3$ is abelian, and $G_3$ is an $\ell$ group.

Lemma 1: There exists a constant $m(n)$ such that for every $G\leq GL_n(\mathbb{Z}_\ell)$ there exists a series of open subgroups

$H_{m(n)} \leq \cdots \leq H_0 = G$

such that  $H_m(n)$ is pro-$\ell$, and for every $1\leq i\leq m(n)$

1. $H_i$ is normal in $G$,
2. if $i$ is even, then $H_{i-1}/H_i$ is abelian,
3. if $i$ is odd, then $H_{i-1}/H_i$ is a product of finite simple groups.

Proof. The kernel $N$ of $GL_n(\mathbb{Z}_\ell)\to GL_n(\mathbb{F}_\ell)$ is pro-$\ell$ (See Page 51 of Analytic pro-p groups). Since $G/G\cap N\leq GL_n(\mathbb{F}_\ell)$, Fact 1 gives normal subgroups $N\cap G \leq G_3\leq G_2\leq G_1\leq G$ of $G$ such that $[G:G_1]\leq J(n)$, $G_1/G_2$ is a product of finite simple groups, $G_2/G_3$ is abelian, and $G_3/G\cap N$ is an $\ell$ group. Let $G_1=K_r\leq \cdots \leq K_0=G$ be normal subgroups of $G$ such that for each $i$ the group $K_{i-1}/K_i$ is a minimal normal subgroup of $G/K_{i}$. Then for each $1\leq i\leq r$ we have $K_{i-1}/K_i$ is a product of finite simple groups. Note that $r\leq \log_2 J(n)$.

For $i=0, \ldots, r$ set $H_i = K_i$ and set $H_{r+1}=G_2$, $H_{r+2} = G_3$. Then $H_{i-1}/H_{i}$ is either isomorphic to an abelian group of isomorphic to a product of finite simple groups, and $H_{r+3}$ is pro-$\ell$. Extend the sequence by adding the same groups, if necessary, to assume that $H_{i-1}/H_i$ is abelian for even $i$ and a product of finite simple groups for odd $i$. Then the length of the series can at most be doubled, hence this length $u$ is bounded by $2(r+2) \leq 2(\log_2 J(n)+2) =:m(n)$. We extend the series even further by setting $H_{k} = H_u$, for $u< k\leq m(n)$, to conclude that the length is exactly $m(n)$. QED

The second result we need is a Hilbertianity criteria that are well known to experts:

Lemma 2: Let $K$ be Hilbertian and $L/K$ a Galois extension such that $Gal(L/K)$ is either

1. finitely generated,
2. abelian,
3. pro-nilpotent but not pro-$p$ for every prime $p$, or
4. a product of finite simple groups.

Then $L$ is Hilbertian.

Proof. The first three cases appear in Jarden-Lubotzky.

If $Gal(L/K)$ is a product of one finite simple group, then it is finitely generated and thus we are done by (1). If $Gal(L/K)$ is a product of finite simple groups that has at least two factors, then we have  $Gal(L/K)= H\times G$, where $H,G\neq 1$. If we let $N,M$ be the respective fixed fields of $H,G$ in $L$, then by the Galois correspondence, $L=NM$ and $N\cap M = L$, so the assertion follows by Corollary 13.8.4 of Field Arithmetic. QED

Here is a very well known result about products of simple groups.

Lemma 3: Let $G = \prod_{i\in I} S_i$, where $S_i$ is a non-abelian finite simple groups for every $i\in I$. Then for every normal subgroup $N$ of $G$ there exists $I_0\subseteq I$ such that $N = \prod_{i\in I_0} S_i \times \prod_{i\in I\smallsetminus I_0} 1$.

Lemma 4: Let $\{L_{i}/K\mid i\in I\}$ be a family of Galois extensions and let $L=\prod_{i\in I} L_i$. If $L_i/K$ is abelian for each $i\in I$, then $L/K$ is abelian. If $Gal(L_i/K)$ is isomorphic to a product of non-abelian finite simple groups for each $i\in I$, then $Gal(L/K)$ is isomorphic to a product of non-abelian  finite simple groups.

Proof. The abelian case is trivial. Assume thus that $G_i=Gal(L_i/K) = \prod_{j\in J_i} S_{ij}$, where $S_{ij}$ is a non-abelian finite simple groups for each $i\in I$ and $j\in J_i$. Take $i\neq i'$. Then $Gal(L_iL_{i'}/K) = G_i \times_A G_{i'}$, where $A$ is a common quotient of $G_i$ and $G_{i'}$. Let $N_i$ be the  kernel of $G_i\to A$. Then there is a partition $J_i = J_{0i} \cup J_{1i}$ such that $N_{i}=\prod_{j\in J_{0i}} S_{ij} \times \prod_{j\in J_{1i}}1$ (Lemma~\ref{lem:norsimp}). Thus $A\cong \prod_{j\in J_{1i}} S_{ij}$ and $G_{i} \cong N_{i} \times A$. Similarly $G_{i'} \cong N_{i'}\times A$, where $N_{i'}=\prod_{j\in J_{0i'}} S_{i'j} \times \prod_{j\in J_{1i'}}1$.  So $Gal(L_{i}L_{i'}/K)=G_i\times_{A} G_{i'} \cong N_{i}\times N_{i'}$ is isomorphic to a product of finite simple groups. This proves the assertion for $|I|=2$. In general one applies transfinite induction. QED

Fact 2: A closed subgroup of $GL_n(\mathbb{Z}_\ell)$ is finitely generated.

## Proof of Thornhill’s theorem.

By Lemma 1, for each $\ell$, we have a series of normal subgroups

$H_{m(n)} \leq \cdots \leq H_0 = {\rm im}(\rho_{\ell}) = Gal(K_{\ell}/K)$

satisfying the properties of the lemma. Let $K_{\ell,i}$ be the fixed field of $H_{i}$ in $K_\ell$. Then we have a tower of Galois extensions of $K$,

$K= K_{\ell,0} \subseteq \cdots \subseteq K_{\ell,m(n)} \subseteq K_{\ell}$

such that $Gal(K_{\ell,i}/K_{\ell,i-1})$ is abelian for even $i$ and isomorphic to a product of non-abelian finite simple groups for odd $i$ and such that $Gal(K_\ell/K_{\ell, m(n)})$ is pro-$\ell$.

Let $K_i = \prod_{\ell} K_{\ell,i}$, $i=0, \ldots, m(n)$; in particular $K_0=K$. Let $N=\prod_{\ell} K_\ell$.
First note that

$\begin{array}{lcl} \Gamma_\ell &=& Gal(K_\ell K_{m(n)}/K_{m(n)}) \cong Gal(K_\ell/ K_{m(n)} \cap K_\ell)\\ &\leq& Gal(K_\ell/K_{\ell,m(n)}) \leq GL_n(\mathbb{Z}_\ell).\end{array}$

Thus $\Gamma_\ell$ is a pro-$\ell$ group that is finitely generated (the latter follows from Fact 2). We also get that $Gal(N/K_{m(n)}) \cong \prod_{\ell} \Gamma_\ell$ is pro-nilpotent.

Next, we show, by induction on $i$, that $Gal(K_i/K_{i-1})$ is either abelian or a product of non-abelian finite simple groups. Indeed, since each $K_{\ell,1}/K$ has Galois group a product of non-abelian finite simple groups, by Lemma 4, $Gal(K_1/K_0)$ is isomorphic to a product of finite simple groups, hence the induction basis. The Galois group $Gal(K_{\ell, i+1}/K_{i})$ is isomorphic to a normal subgroup of $Gal(K_{\ell, i+1}/K_{\ell,i})$. Hence either $Gal(K_{\ell, i+1}K_{i}/K_{i})$ is abelian for every $\ell$ or  $Gal(K_{\ell, i+1}K_{i}/K_{i})$ is a product of non-abelian finite simple groups for every $\ell$ (Lemma 3). By Lemma 4 we get that $Gal(K_{i+1}/K_{i})$ is either abelian or a product of non-abelian finite simple groups, as needed.

Now let $M$ a Galois extension of $K$ that is contained in $N=\prod_{\ell} K_\ell$. Let $M_i = M\cap K_{i}$. Then $Gal(M/M_{m(n)})$ is a quotient of $Gal(N/K_{m(n)})$, hence pro-nilpotent.  Moreover, if $Gal(M/M_{m(n)})$ is a pro-$p$ group for some prime $p$, then it is a quotient of the $p$-sylow subgroup $\Gamma_p$ of $Gal(N/K_{m(n)})$. Therefore, since $\Gamma_p$ is finitely generated, so is $Gal(M/M_{m(n)})$.

For each $1\leq i\leq m(n)$$Gal(M_{i}/M_{i-1})$ is a quotient of $Gal(K_i/K_{i-1})$. Thus it is either abelian or a product of non-abelian finite simple groups (Lemma 3). Applying Lemma 2 repeatedly yields that $M_1, M_2, \ldots, M_{m(n)}$ are all Hilbertian.

Since we have showed that $M/M_{m(n)}$ satisfies either Condition 1 or Condition 3 of Lemma 2, we get that $M$ is Hilbertian.
QED

## Geometric embedding problems

December 20, 2011

### Embedding probelms

Let $K$ be a field. A finite embedding problem for a field $K$ is composed of two finite groups $G,H$ and continuous epimorphisms $\alpha\colon G_K\to H$ and $\beta \colon G\to H$. Here $G_K$ denotes the absolute Galois group of $K$.

The goal of this post is to explain a construction of embedding problem for a field $K$ that come from geometric objects. I’ll give the definitions and discuss some applications. Probably in later posts I’ll discuss other applications.

A weak solution of the embedding problem

$\mathcal{E} = (\alpha\colon G_K \to H, \beta\colon G\to H)$

is a continuous homomorphism $\gamma \colon G_K \to G$ such that $\alpha =\beta\circ \gamma$. A weak solution is called proper solution if it is surjective.

Note that $\alpha$ factors through the restriction map $G_K\to Gal(L/K)$, where $L$ is the fixed field of $\ker \alpha$. Thus $\alpha$ induces an isomorphism (via the first isomorphism theorem)  $\bar\alpha\colon Gal(L/K) \to H$. Therefore we may assume that $H = Gal(L/K)$ and $\alpha$ is the restriction map, when it is convenient.

If $\gamma$ is a weak solution of $\mathcal{E}$, the the fixed field $F$ of $\ker \gamma$ is called the solution field, and $L$ embeds inside $F$ in such a way that the restriction map $Gal(F/E) \to G(L/K)$ coincides with the map induced by $\beta$.

### Geometric embedding problems

We now construct a special type of embedding problems coming from covers of algebraic varieties. The construction is generic in a sense, hence the embedding problems, actually, are defined on the level of function fields, and we shall use the terminology of function fields.

Let $K$ be a field. Assume that $E$ is a finitely generated regular extension of $K$ and that $F/E$ is a finite Galois extension with Galois group $G$. Let $L$ be the algebraic closure of $K$ in $F$. Then $L/K$ is a Galois extension, and if we denote $H=Gal(L/K)$ we have a natural restriction map $\beta \colon G\to H$. This map $\beta$ is surjective because $E$ is regular over $K$. Thus

$\mathcal{E}(F/E) = (\alpha \colon G_K \to H, \beta\colon G\to H),$

in which $\alpha$ is the restriction map, is a finite embedding problem for $K$. We call such an embedding problem geometric.

Assume that $R\subseteq S$ is an integral ring extension whose respective quotient field extension is $E\subseteq F$. Let us assume that $R$ is finitely generated over $K$, i.e. $R=K[x_1, \ldots, x_n]$, for some $x_1, \ldots, x_n \in E$. Further assume that $L\subseteq S$. Choose generators $y_1, \ldots, y_k$ of $S$ over $R$, i.e. $S = R[y_1, \ldots, y_k]$. Write $d_i$ for the discriminant of $y_i$ over $E$. Then $d_i \in R$.

Consider a homomorphism $\phi \colon R\to K$ that is identity on $K$ with the property that $\phi(d_i)\neq 0$. Extend $\phi$ to a homomorphism $\Phi \colon S \to K_s$ that is trivial on $L$. Here $K_s$ is the separable closure of  $K$.

For every $\sigma \in G_K$ the equation

$\sigma\Phi( x) = \Phi(\Phi^{*}(\sigma x), \quad \forall x\in S$

uniquely defines a element $\Phi^*(\sigma)\in G$. Moreover the map $\Phi^*\colon G_K \to G$ is a continuous homomorphism and the fixed field of the kernel is $\Phi(S)$. Since $\Phi$ is the identity on $L$ we have $\alpha = \beta\circ \Phi^*$. Thus every such $\Phi$ defines a weak solution of $\mathcal{E}(F/E)$. We call solutions obtained from such homomorphisms geometric solutions. (In general any $\phi$ whose kernel is etale in $S$ defines a geometric solution, this is a special case.)

### Examples

Example 1. Assume $E = \mathbb{Q}(x)$ and $F = \mathbb{Q}(\sqrt x)$. Then $L=\mathbb{Q}$ and

$\mathcal{E}(F/E) = (G_\mathbb{Q} \to 1, \mathbb{Z}/2\mathbb{Z} \to 1)$.

Take $R = \mathbb{Q}[x]$ and $S = R[\sqrt x]$. The discriminant of $\sqrt x$ is $4x$. Consider the homomorphism $\phi\colon R\to \mathbb{Q}$ defined by $\phi(x) = d$, where $d\neq 0$. Then $\Phi^* \colon G_K \to \mathbb{Z}/2\mathbb{Z}$ is defined by $\Phi^*(\sigma\sqrt{x}) = \pm \sqrt{x}$  iff $\sigma \sqrt{d} = \pm \sqrt{d}$.

Example 2. Assume $E = \mathbb{Q}(x)$ and $F = \mathbb{Q}(x,\sqrt 2)$. Here $L = \mathbb{Q}(\sqrt{2})$ and

$\mathcal{E}(F/E) = (G_\mathbb{Q} \to \mathbb{Z}/2\mathbb{Z}, \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z})$.

In particular any homomorphism will yield the same solution.

Example 3. Assume $E = \mathbb{Q}(x)$ and $F = \mathbb{Q}(\sqrt x + \sqrt 2)$. Then

$\mathcal{E}(F/E) = (G_\mathbb{Q} \to \mathbb{Z}/2\mathbb{Z}, \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z})$.

What are the geometric solutions here?

### Application to pseudo algebraically closed fields

We call a field $K$ pseudo algebraically closed, or in short PAC, if every non-void absolutely irreducible $K$-variety has a $K$ rational point. Note that $K$ is PAC if and only if for every finitely generated regular extension $E$ over $K$ and for every finitely generated ring $R$ over $K$ whose quotient field is $E$ there exists a homomorphism $\phi \colon R\to K$.

Lemma: A field $K$ is PAC if and only if every geometric embedding problem for $K$ has a geometric solution.

Proof. Assume $K$ is PAC and let $\mathcal{E}(F/E)$ be a geometric embedding problem for $K$. Let $x_1, \ldots, x_n$ be generators of $E$ over $K$, let $R_0 = K[x_1,\ldots, x_n]$. Let $y\in F$ be a generator of $F/E$ that is integral over $R$  and let $d\in E$ be the discriminant of $y$ over $E$.

We set $R = R_0[d^{-1}]$ and $S = R[y]$. As $K$ is PAC, we have a homomorphism $\phi \colon R\to K$ (recall that $E$, the quotient field of $R$, is regular over $K$, by the definition of geometric embedding problems). Extend $\phi$ to a homomorphism $\Phi \colon S\to K_s$, to get that $\Phi^*$ is a geometric solution of $\mathcal{E}(F/E)$.

Next assume that every geometric embedding problem has a geometric solution. Let $E$ be a regular extension of $K$. The geometric embedding problem $\mathcal{E}(E/E)$ has a geometric solution that corresponds to a homomorphism $\phi \colon R \to K$, for some finitely generated $K$-algebra $R$. Hence $K$ is PAC. QED

Corollary: The absolute Galois group of a PAC field is projective, i.e. every finite embedding problem for $K$ has a weak solution.

Proof: For an epimorphism $\beta\colon G\to H$ we define a function field extension $F/E$. Let $L$ be the fixed field of the kernel of $\alpha$. Let $\{x_\sigma\mid \sigma\in G\}$ be $|G|$ variables and let $F = L(x_\sigma\mid \sigma\in G)$. Define the “diagonal” action of $G$ on $F$ by

$\tau x_\sigma = x_{\sigma \tau}, \quad \sigma, \tau \in G$

and

$\tau l = \beta (\tau) l, \quad \tau\in G$ and $l\in L$.

Let $E$ be the fixed field of $G$ in $F$ under the above action. The action is free, hence $F/E$ is Galois with Galois group $G$. Moreover it is a rather easy exercise in Galois theory to check that $E$ is regular over  $K$. Clearly $L$ is the algebraic closure of $K$ in $F$. Summing up $\mathcal{E}(F/E) = (\alpha,\beta)$. Thus it has a weak geometric solution, in particular a weak solution. QED

Theorem: Let $K$ be a PAC field, let $\mathcal{E}(F/E)$ be a geometric embedding problem for $K$, and let $\theta$ be a weak solution. Then $\theta$ is geometric. Moreover, for every $0\neq e\in E$ there exist a $K$-algebra $S$ with quotient field $F$ and a homomorphism $\Phi\colon S\to K_s$ such that  $\Phi$ induces a geometric solution, $\Phi^* = \theta$ and $\Phi (e) \neq 0$.

I will skip the proof in the meanwhile, I’ll just say that it involves the field crossing argument which deserves a post on itself.

Corollary: Let $K$ be a PAC field of characteristic $0$. Then every separable extension of degree $n$ is generated by a root of a polynomial $X^n + X^{n-1} + a$.

Proof: The polynomial $f(T,X) = X^n + X^{n-1} + T$ has Galois group $S_n$ over $K_s(T)$ (this is classical result in Galois theory that appears, e.g., in Serre’s Topic in Galois Theory). Let $F$ be the splitting field of $f(T,X)$ over $K(T)$. Let $R = K[T]$ and $S = R[x_1, \ldots, x_n]$, where $x_1, \ldots, x_n$ are the roots of $f$ in $F$. Then $\mathcal{E}(F/K(T)) = (G_K\to 1, S_n \to 1)$.

Let $L/K$ be a separable extension of degree $n$ of $K$. Then the action of $G_K$ on the cosets of $G_L$ yields a homomorphism $\theta\colon G_K \to S_n$. By the theorem we have a homomorphism $\Phi\colon S\to K_s$ such that $\Phi(R) = K$ and $\Phi^*=\theta$.

The formula $\sigma \Phi(x) = \Phi(\Phi^*(\sigma)x)$ for $\sigma \in G_K$ and $x\in S$ implies that $\sigma\in G_L$ if and only if $\Phi^{*} (\sigma)$ stabilizes $1$ if and only if $\Phi^*(\sigma) \in Gal(F/K(T,x_1))$ if and only if $\sigma \in G_{\Phi(K[T,x_1]}$. Thus $L = \Phi(K[T,x_1]) = K(\Phi(x_1))$. This finishes the proof since $\Phi(x_1)$ is a root of $X^n + X^{n-1} + \Phi(T)$. QED

## Twin polynomails theorem

December 14, 2011

The twin prime conjectures asserts that there exist infinitely many primes $p$ such that $p+2$ is also a prime. The Hardy-Littlewood conjecture predicts how many twin prime are there up to $x$: Let $\pi_2(x)$ denotes the number of primes $p\leq x$ such that $p+2$ is also a prime. Then

Conjecture: $\pi_2(x) \sim 2 \prod_{p\geq 3} \frac{p(p-2)}{(p-1)^2} \frac{x}{\log^2 x} \approx 1.32 \frac{x}{\log^2 x}$, as $x\to\infty$.

We can use the usual scheme of going from $\mathbb{Z}$ to $\mathbb{F}_q[t]$, $q$ a prime power, and to state a similar conjecture. Here it is possible that $f(t), f(t)+1$ are both prime, i.e. irreducible.  Hence we shall denote by $\Pi_1(q,N)$ the number of irreducible monic polynomials $f(t) \in \mathbb{F}_q[t]$ of degree $N$ such that both $f$ and $f_1$ are irreducible.

Conjecture: $\Pi_1(q,N) \sim C\frac{q^N}{N^2}$, as $N\to \infty$. Here $C = \log q \prod_{p}\frac{q^{\deg p} (1-2q^{\deg p})}{(q^{\deg p} - 1)^2}$, where $p$ runs over all irreducible polynomials in $\mathbb{F}_q[t]$ with order induced by the degree.

In this post I wish to discuss the asymptotic of $\Pi_1(q,N)$ when $N$ is fixed and $q\to \infty$.

Main Theorem: $\Pi_1(q,N) = \frac{q^N}{N^2} + O(q^{N-\frac12})$, when $N$ is fixed and $q\to \infty$, provided that $q$ is odd or $q$ is even and $N$ is odd.

Remarks:

1. This theorem is in fact a special case of a theorem of Pollack and of an extension of it by myself. The general theorem considers the simultaneous irreducible values in $\mathbb{F}_q[t]$ of irreducible $f_1,\cdots, f_r\in \mathbb{F}_q[X]$. This theorem is in the case $f_1 = X$ and $f_2 = X+1$.
2. Pollack proved the theorem under the assumption that $\gcd(q,2N)=1$. Later I gave a different proof, that is close to Pollack’s, but differ in several crucial points. With this second proof the extra cases – when $q$ is odd $\gcd(q,N)>1$ and when $q$ is a power of $2$ and $N$ is odd – were proved.
3. In this post for simplicity I restrict the discussion to the theorem above about twin polynomials.

### Idea of the proof

Let $A=(A_1, \ldots, A_N)$ be an $N$-tuple of variables. Clearly the polynomial $g(A,t) = t^N + A_1 t^{N-1} + \cdots + A_N$ is irreducible in the ring $\mathbb{F}_q(A)[t]$.  The polynomial $g+1$ is also irreducible. Therefore to calculate $\Pi_1(q,N)$ is the same as to count how many specializations $A\mapsto a=(a_1,\ldots, a_N)\in \mathbb{F}_q^N$ such that the irreducibility is preserved.

The point is that the latter can be answered, when $q$ is large, using Chebotarev’s theorem for function fields if we know the Galois group.

### A special case of Chebotarev’s theorem

Let $X\to \mathbb{P}^N_{\mathbb{F}_q}$ be a finite separable cover of absolutely irreducible smooth varieties that is generically Galois. Basically, if $F/\mathbb{F}_q(A)$ is the corresponding function field we assume that it is a finite Galois extension and that when extending scalars to the algebraic closure $\mathbb{F}$ of $\mathbb{F}_q$, then $G=Gal(F/\mathbb{F}_q(A)) = Gal(F\mathbb{F}/ \mathbb{F}(A))$.

Out of the $q^N$ elements of $a\in \mathbb{P}^N(\mathbb{F}_q)$ only $O(q^{N-1})$ are not etale in $X$. Each etale point $a\in \mathbb{P}^N(\mathbb{F}_q)$ and a point $x\in X$ in the fiber of $a$ give-rise to the Frobenius element $Fr_x$. Namely $Fr_x\in G$ is the unique element such that

$Fr_x b \equiv b^{q_x} \mod m_x$,

where $m_x$ is the maximal ideal of the local ring $O_{X,x}$ and $q_x$ is the number of element in the residue field $K(x) = O_{X,x}/m_x$. If we vary $x$ over $a$, we get a conjugacy class $Fr_a = \{ Fr_x \mid x\in X\mapsto a\}$ of $G$.

We are ready to present a special case of Chebotarev’s theorem:

Chebotarev’s Theorem: Let $X\to \mathbb{P}^N_{\mathbb{F}_q}$ be a finite separable cover of absolutely irreducible smooth varieties that is generically Galois with Galois group $G$. Let $D>0$ and assume $\deg X \leq D$.  Then for every conjugacy class $C$ in $G$ we have

$\# \{a\in \mathbb{P}^N(\mathbb{F}_q) \mid Fr_a=C\} = \frac{|C|}{|G|}q^N + O_D(q^{N-\frac12}).$

### Connection to irreducibility

Under the assumptions and notation of the previous part, assume that $f,g$ are irreducible polynomials in $\mathbb{F}_q[A,t]$ whose roots lie in $F$. For $q^N + O(q^{N-1})$ points of $a\in \mathbb{F}_q^N$ we have that $f(a, t), g(a,t)$ is irreducible in $\mathbb{F}_q[t]$ if and only if the conjugacy class $Fr_a\subseteq G$ acts transitively on the roots of $f$ and acts transitively on the roots of $g$. So a consequence of the above Chebotarev’s theorem implies the following result.

Corollary: Let $C \subseteq G$ be the set of all elements of $G$ that acts transitively on the roots of $f$ and on the roots of $g$. Then $C$ is a union of conjugacy classes and

$\# \{a\in \mathbb{F}_q^N) \mid f(a,X) \mbox{ irreducible}\} = \frac{|C|}{|G|}q^N + O(q^{N-\frac12}).$

### Reducing to calculation of Galois groups

Recall that $g(A,t) = t^N + A_1 t^{N-1} + \cdots +A_N$ is a polynomial with variable coefficients. Let $F_0$ be the splitting field of $g$ over $\mathbb{F}_q(A)$. A classic exercise in Galois theory shows that $S_N = Gal(F_0/\mathbb{F}_q) = Gal(F_0\mathbb{F}/\mathbb{F}(A))$. Here $S_N$ is the permutation group on $N$ letters, $\mathbb{F}$ is the algebraic closure of $\mathbb{F}_q$, and the isomorphisms are given via the action on the roots of $g$.

The same is true for the splitting field $F_1$ of $g+1$ over $\mathbb{F}_q(A)$. Let $F = F_1F_2$ and choose $X\to \mathbb{P}^N_{\mathbb{F}_q}$ that corresponds to the extension $F/\mathbb{F}_q(A)$.

Reduction Lemma: If $Gal(F_0F_1\mathbb{F}/\mathbb{F}(A)) = S_N\times S_N$, then the Main Theorem holds.

Proof. By Galois theory $G_{12}=al(F_0F_1\mathbb{F}/\mathbb{F}(A))$ is isomorphic to the fiber product of $Gal(F_i\mathbb{F}/\mathbb{F}(A))$ over the Galois group of the intersection $F_0\cap F_1$. In particular $G_{12} = S_N\times S_N$ if and only if $G_{12} = Gal(F_1\mathbb{F}/\mathbb{F}(A))\times Gal(F_2\mathbb{F}/\mathbb{F})$. In particular an element $(\sigma,\tau)$  acts transitively on the roots of $g$ and on the roots of $g+1$ if and only if both $\sigma$ and $\tau$ are $N$-cycles. Note that there are $\frac{1}{N^2}$ such pairs in $S_N\times S_N$.  Thus by the corollary $\Pi_1(q,N) \sim \frac{q^N}{N^2}$, as $q\to \infty$. qed

### Idea of the calculation of the Galois group of g(g+1)

By the reduction lemma it suffices to calculate $Gal(F_1F_2\mathbb{F}/\mathbb{F}(A))$ and to show it equals $S_N\times S_N$.

We shall use the following

Lemma: If $E_1, E_2, \ldots, E_r$ are Galois extensions of a field $E$ with Galois groups $S_N$ and if $U_1, \ldots, U_r$ are respective  subfields of $E_1,\ldots, E_r$ that correspond to $A_N$, then $E_1, \ldots, E_r$ are linearly disjoint if and only if $U_1, \ldots, U_r$ are linearly disjoint.

So the lemma allows us to reduce to quadratic fields.

Galois extensions of a field $E$ of degree $\leq 2$ are in a natural correspondence with the elements of $Hom(G_E,\mathbb{Z}/2\mathbb{Z}) = H^1(G_E,\mathbb{Z}/2\mathbb{Z})$.

If the characteristic of $E$ is not $2$, then by Kummer’s theory it’s nothing else than $E^* /(E^*)^2$. In characteristic $2$, Artin-Schreier theory gives us it’s $E/\wp(E)$, where $\wp(x) = x^2-x$.

We will discuss how to continue in another post…

## Irreducible values of polynomials

December 1, 2010

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 $\mathbb{Z}$. 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 $p$ such that

1. $p=n$?
2. $p=an+b$?
3. $p=n^2+1$?
4. $p=n^2+n+2$?
5. $p_1=n, p_2 = n+2$? (Twin prime conjecture)
6. $p_1=n, p_2 = n+2, p_3=n+4$?

Euclid answered the first question, while Dirichlet’s theorem gives a positive answer for the second question provided $\gcd(a,b)=1$. Questions 3 and 5 are unknown. Now the answer of the remaining two is obviously no since there are local obstruction ($2$ divides all the values of $n^2+n+2$ while $3$ divides $n(n+2)(n+4)$).

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

Schinzel’s Hypothesis H:
Let $f_1(X), \ldots, f_r(X)\in \mathbb{Z}[X]$ be polynomials with positive leading coefficients. Then there exists infinitely many $n\in \mathbb{Z}$ such that all $f_i(n)$ are prime provided $f_i(X)$ are irreducible, and there is no local obstruction, i.e., there is no $p$ dividing all the values of $f_1(X)\cdots f_r(X)$.

The prime number theorem gives a quantitative answer to the first question above. Namely the number of primes up to $x$ is approximately $\frac{x}{\log x}$.
The Bateman-Horn generalizes this in the general setting of the Schnizel hypothesis H:

Bateman-Horn’s conjecture:
Let $f_1(X), \ldots, f_r(X)\in \mathbb{Z}[X]$ be irreducible polynomials with positive leading coefficients and let $N$ be the number of $n>0$ such that all $f_i(n)$ are prime. Then

$\displaystyle N \sim \frac{x}{\log^r x}\frac{\mathfrak{S}(f_1,\ldots, f_r)}{\prod_{i=1}^r \deg f_i},$

where if $\rho(p) = \#\{n\mod p\mid f_1(n)\cdots f_r(n)\equiv 0 \mod p\}$, then

$\displaystyle \mathfrak{S}(f_1,\ldots, f_r):= \prod_{p} \frac{1-\rho(p)/p}{\left(1-1/p \right)^{r}}$

Note that

1. If $r=1$ and $f_1(X) = X$, then $\rho(p)=1$. Therefore each factor in the product $\mathfrak{S}(X)$ is $1$, and hence $\mathfrak{S}(X)=1$. Therefore, in this case, we retain the prime number theorem.
2. If some prime $p$ divides all the values of $f_1\cdots f_r$, then $\rho(p)=p$, and hence the corresponding factor in $\mathfrak{S}$ is zero, so $\mathfrak{S}=0$.
3. Otherwise, one can show that $\mathfrak{S}$ converges to a positive constant.

### Analog over finite fields

Replacing $\mathbb{Z}$ with $\mathbb{F}_q[t]$, for some prime power $q=p^\nu$, and a finite field $\mathbb{F}_q$, we get a naive analog (in genus zero).

Naive analog of Schinzel hypothesis H:
$\forall f_1(t,X), \ldots, f_r(t,X) \in \mathbb{F}_q[t][X]$ irreducible, having no local obstructions, $\exists^\infty$ many simultaneous prime values in $\mathbb{F}_q[t]$.

Denote by

$N(n, q, f_1, \ldots, f_r)$

the number of degree $n$ monic polynomials $g(t) = t^n + \cdots \in \mathbb{F}_q[t]$ such that all $f_i(g(t))$ are irreducible. Then the naive Bateman-Horn conjectures says that

$\dispalystyle N(n,q,f_1, \ldots, f_r)\sim \frac{q^n}{n}\frac{\mathfrak{S}(f_1,\ldots, f_r)}{\prod_{i=1}^r \deg_X f_i}$,

where if for an irreducible polynomial $p(t)$ with a root $\alpha$ we let $\rho(p) = \#\{x\in \mathbb{F}_{q^{\deg p}} \mid f_1(\alpha,x)\cdots f_r(\alpha,x)=0\}$, then

$\displaystyle \mathfrak{S}(f_1,\ldots, f_r):= \prod_{p} \frac{1-\rho(p)/p}{\left(1-1/p \right)^{r}}$

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

Swan’s example:
The polynomial $X^8 + t^3 \in \mathbb{F}_2[t][X]$ is irreducible, has no local obstruction, but $g(t)^8 + t^3$ either has $t$ 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 $f_i(t,X)\not\in \mathbb{F}_q[t][X^p]$, then the naive conjecture remains the same. If $f_i(t,X)\in \mathbb{F}_q[t][X^p]$, $p>2$, or $f_i(t,X)\in \mathbb{F}_q[t][X^4]$, $p=2$, then additional factor $\Lambda (f_i,n)$ is entered. In the case $f_i(t,X)\in \mathbb{F}_q[t][X^2]\smallsetminus\mathbb{F}_q[t][X^4]$, 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 $f_i(t,X) = f_i(X)\in \mathbb{F}_q[X]$. This special case includes, for example, the twin prime conjecture when taking $f_1(X) = X$ and $f_2(X) = X+1$, or more generally, $f_i(X) = X+\alpha_i$, for some $\alpha_i\in \mathbb{F}_q$.

Since finite fields are perfect, the irreducibility of $f_i(X)$ implies that $f_i'\neq 0$, so $f_i(X) \not \in \mathbb{F}_q[X^p]$. Therefore we are in the scope of the naive conjectures. We will consider large finite fields what means that we assume $q$ to be large w.r.t. $n$ and $\sum \deg f_i$. The conjecture constant becomes simpler under this assumption:

Lemma (Pollack): Let $B$ be a fixed positive integer. Then for any irreducible non-associate $f_1(X), \ldots, f_r(X)\in \mathbb{F}_q[X]$ with $\sum \deg f_i\leq B$ we have

$\displaystyle \frac{\mathfrak{S}(f_1, \ldots, f_r)}{\prod_{i=1}^r \deg f_i} = 1 + O_B(1/q).$

Therefore we have the following conjecture for large finite fields:

Conjecture:
Let $n,B$ be fixed. Then for every $q$ and for every irreducible non-associate $f_1(X), \ldots, f_r(X)\in \mathbb{F}_q[X]$ with $\sum \deg f_i\leq B$ we have

$\displaystyle N(n,q,f_1, \ldots, f_r) \sim \frac{q^n}{n^r}, \qquad q\to \infty.$

Pollack proves this conjecture when $\gcd(q,2n)=1$, i.e., in odd characteristic and under the assumption $p\nmid n$.

Theorem (Pollack): If $\gcd(q,2n)=1$, the conjecture holds. More precisely

$\displaystyle N(n,q,f_1, \ldots, f_r) \sim \frac{q^n}{n^r} + O_{n,B}(q^{n-1/2}).$

Remarks:

1. The error term dependence in $n,B$ is of bounded by an expression of order of magnitude $(n!)^B$
2. The theorem extends to the case $p$ odd and $p\mid n$ and to the case $p=2$ and $n$ 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 $(a_1,\ldots,a_{n-1})\in \mathbb{F}_q^{n-1}$ such that for $g_0(t)=t^n+a_1t^{n-1} + \cdots a_{n-1}t$ we have

1. The map $t\mapsto g_0(t)\colon \mathbb{A}_1\to \mathbb{A}_1$ has only simple ramification (i.e.\ $g_0$ is Morse). This means that the ramification type of every ramified point is $(2,1,\ldots, 1)$.
2. The ramification loci of $t\mapsto t-\alpha$ are distinct. Here $\alpha$ runs over the roots of $f:=f_1\cdots f_r$.
3. is $q^{n-1} + O(q^{n-2})$. (That is almost all polynomials $g_0$ satisfy these two conditions.)

For this calculation one uses $p\neq 2$. Fix $g_0$ satisfying these conditions. Since $p\nmid n$, the ramification of $t\mapsto U=g_0(t)$ at infinity is tame. Therefore the first property implies that $Gal(g_0(t) - U, \mathbb{F}(U))$ is a transitive group generated by transpositions, hence $=S_n$. Here $\mathbb{F}$ is the algebraic closure of $\mathbb{F}_q$. From the second property we get that the splitting fields of $g_0(t)-U-\alpha$ are linearly disjoint over $\mathbb{F}(U)$ (since the intersection field of such splitting fields is ramified only at infinity, and this ramification is tame). Here $\alpha$ runs over the roots of $f$. Noting that $f(g_0(t)-U) = \prod_{\alpha} g_0(t)-U-\alpha)$ one can show that

$\displaystyle Gal(f(g_0(t)-U),\mathbb{F}_q) = S_n \wr \gal(f,\mathbb{F}_q).$

We emphasize that the Galois group is the biggest possible.

It remains to count, for this $g_0$, how many $a_n\in \mathbb{F}_q$ there are such that $f_i(g_0(t) + a_n)$ are all irreducible. This happens if and only if the Frobenius element at $(U+a_n)$ in the splitting field of $f(g_0(t) - U)$ acts transitively on the roots of $f_i(g_0(t)-U)$ for all $i$. Thus, by Chebotarev’s theorem for finite fields we get that there are

$\frac{q}{n^r} + O(q^{n-1/2})$

many such $a_n's$. This finishes the proof since $(q^{n-1} + O(q^{n-2}))(\frac{1}{n}q + O(q^{\frac12})) = \frac1nq^{n} + O(q^{n-\frac12})$.

### Pseudo algebraically closed fields

A field $latex$K$is called PAC if every non-void $K$-variety that is geometrically irreducible has a $K$-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., $\{X^2 + Y^2 + 1\}$ has no rational points, so $\mathbb{Q}$ 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 $e\geq 1$, if $\sigma_1, \ldots, \sigma_e$ are automorphism of the field of algebraic numbers $\widetilde{\mathbb{Q}}$, then the probability that the fixed field $\displaystyle\widetilde{\mathbb{Q}}(\sigma_1, \ldots, \sigma_e) = \{ x\in \widetilde{\mathbb{Q}} \mid \sigma_i(x) = x \ \forall i\}$ is PAC is $1$ (Jarden). Here the probability measure is induced from the Haar measure of the compact group $Aut(\widetilde{\mathbb{Q}})$. It is now appropriate to give an explicit example. Let $\mathbb{Q}_{tr}$ be the field of all totally real numbers. (Totally real is a number all the conjugates of which are real.) Then $\mathbb{Q}_{tr}$ is PRC, hence $\mathbb{Q}_{tr}(\sqrt{-1})$ is PAC (Pop). (It is interesting to note that the fact that $\mathbb{Q}_{tr}$ 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 $K=\mathbb{C}$, and let$f=f_1 = X$. Then for every $n\geq 2$, there is no irreducible $g(t)=t^n+\cdots$. Let us introduce an obvious necessary condition. Assume that $f(g(t))$ is irreducible for some$f,g\in K[t]$. Let $\beta$ be a root of$f(g(t))\$. Then $\alpha=g(\beta)$ is a root of $f$, since $f(\alpha) = f(g(\beta))=0$. Then $K(\alpha)\subseteq K(\beta)$ and

$\deg(f(g(t)) = [K(\beta):K] = [K(\beta):K(\alpha)][K(\alpha):K]\leq \deg f \deg g.$

So we conclude $[K(\beta):K(\alpha)]=\deg g$. So $K(\alpha)$ has an extension of degree $\deg g$.

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 $p=2$ we assume that $n$ is odd:

Theorem (Bary-Soroker): Let

• $K$ be a PAC field of characteristic $p\geq 0$,
• $n\geq 1$ odd if $p=2$,
• $f_1, \ldots, f_r\in K[X]$ separable irreducible with respective roots $\alpha_1, \ldots, \alpha_r$.

Assume $K(\alpha_i)$ has a separable extension of degree $n$, for all $i$. Then there exists a Zariski dense set of $(a_1, \ldots, a_n)\in K^n$ such that for $g(t) = t^n + a_1 t^{n-1} + \cdots + a_n$ all $f_i(g(t))$ 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 $G$ of $f(\mathfrak{g})$ over $K(\mathbf{A})$, where $\mathbf{A}=(A_1,\ldots, A_n)$ is an $n$-tuple of variables, and $\mathfrak{g} = t^n + A_1 t^{n-1} + \cdots + A_n$ is the corresponding generic polynomial. We show that if $p\neq 2$ or if $p=2$ and $n$ is odd, then $G$ is the biggest possible: If $R$ denotes the set of roots of $f$, then

$G \cong S_n \wr_R \gal(f,K)$.

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 $\mathfrak{g}-\alpha$ over $E(\bfA)$ is $S_n$, for every $\alpha\in R$, to show the $G$ is the biggest possible reduces to discriminant calculations (in characteristic $2$, one should consider generalized discriminants). We use the formula for the discriminant given by the resultant when $p\neq 2$ divides $n$. In characteristic $2$, not only the formulas become more complicated, but also there are obstructions for $G$ to be maximal: If $p=n=2$, $G$ is the biggest possible if and only if every even sum of elements in $R$ does not vanishes. We do not know what happens when $n= 4,6, 8, \ldots$.

The second step in the proof is to find a Zariski dense set of simultaneous irreducible specializations for $f_i(\mathfrak{g})$. Namely, a Zariski dense set of $n$-tuple $\mathbf{a}\in K^n$ such that $f_i(g(t))$ are all irreducible, where $g(t) = t^n + a_1 t^{n-1} + \cdots + a_n$. We study this in much more generality, and give an exact criterion for a tuple of irreducible $\{h_i(A_1, \ldots, A_{n}, X)\}_{i=1}^r$ 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 $p\neq 2$ and $p\mid n$ and to the case $p=2$ and $n$ is odd. To explain how this is achieved, we have to discuss pseudo finite fields.

A field $K$ is called pseudo finite if

• $K$ is PAC,
• $K$ is perfect, and
• $K$ has a unique extension of degree $n$, for every $n\geq 1$.

Note that the latter condition is equivalent to $Gal(K)\cong \widehat{\mathbb{Z}}$.

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 $\widetilde{\mathbb{Q}}(\sigma)$ of a Galois automorphism $\sigma\in Gal(\mathbb{Q})$ in $\widetilde{\mathbb{Q}}$ 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 $\theta$ be an elementary statement in the theory of rings. Then all pseudo finite fields satisfy $\theta$ if and only if all but finitely many finite fields satisfy $\theta$.

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

Theorem (Bary-Soroker): Let

• $K$ be a pseudo finite field of characteristic $p\geq 0$,
• $n\geq 1$ odd if $p=2$,
• $f_1, \ldots, f_r\in K[X]$ irreducible polynomials.

Then there exists a Zariski dense set of $(a_1, \ldots, a_n)\in K^n$ such that for $g(t) = t^n + a_1 t^{n-1} + \cdots + a_n$ all $f_i(g(t))$ are separable irreducible.

Note that, for a fixed $n,B$, the statement: For every $f_1, \ldots, f_r$ irreducible polynomials satisfying $\sum \deg f_i \leq B$ if $1+1=0$, then there exists $g(t) = t^n + a_1t^{n-1} + \cdots + a_n$ such that all $f_i(g(t))$ 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 $g(t)$‘s for which all $f_i(g(t))$ are irreducible.

Therefore we reprove Pollack’s theorem and extends it to the case $p\neq 2$ and $p\mid n$ and to the case $p=2$ and $n$ odd:

Theorem: For positive integers $n,B$, $q$ a prime power, and non-associate irreducible polynomials $f_1, \ldots, f_r\in \mathbb{F}_q[X]$ such that $\sum \deg f_i \leq B$ we have

$\displaystyle N(n,q,f_1, \ldots, f_r) \sim \frac{q^n}{n^r} + O_{n,B}(q^{n-1/2}).$

provided $n$ is odd if $p=2$.