Field arithmetician's blog

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.

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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 $1\leq i\leq m(n)$

$H_i$ is normal in $G$ ,
if $i$ is even, then $H_{i-1}/H_i$ is abelian,
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

finitely generated,
abelian,
pro-nilpotent but not pro- $p$ for every prime $p$ , or
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

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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

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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:

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 $f_2 = X+1$ .
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.
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…

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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

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

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

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.
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 $\mathfrak{S}=0$ .
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:

The error term dependence in $n,B$ is of bounded by an expression of order of magnitude $(n!)^B$
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

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 $(2,1,\ldots, 1)$ .

The ramification loci of $t\mapsto t-\alpha$ are distinct. Here $\alpha$ runs over the roots of $f:=f_1\cdots f_r$ .

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 $p=2$ .

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Galois Representations and The Inverse Galois Problem

August 18, 2010

This post tries to summarize Gabor Wiese’s talk given in the GTEM day held on the 11-12 of August 2010 at the IEM in Essen, organized by Gabor and me. All details are available on arXiv:0905.1288.

Introduction

Consider $\rho\colon {\rm Gal}(\mathbb{Q})\to {\rm GL_n}(\bar{\mathbb{F}}_l)$ continuous group homomorphism, where ${\rm Gal}(\mathbb{Q})$ is the absolute Galois group of the rational numbers endowed with the profinite Krull topology, and ${\rm GL_n}(\bar{\mathbb{F}}_l)$ is endowed with the discrete topology. Then $\ker \rho = {\rm Gal}(K)$ , where $K/\mathbb{Q}$ is a number field Galois over $\mathbb{Q}$ and ${\rm Gal}(K/\mathbb{Q}) \subseteq GL_n(\bar{\mathbb{F}}_l)$ .

The idea is to

Vary $\rho$ in order to realize “many” groups $G$ as Galois group/ $\mathbb{Q}$ .
Modular forms give rise to Galois representations.
Vary modular forms in order to realize “many” groups $G$ as Galois group/ $\mathbb{Q}$ .

From now on we focus on $GL_2$ .

Vertical Theorem (Wiese). Fix a prime $l$ . There are infinitely many $n$ such that $PSL_2(\mathbb{F}_{l^n})$ occurs as a Galois group over $\mathbb{Q}$ such that the corresponding number field ramifies at most at $l$ and one more prime.

Horizontal Theorem (Dienlefait-Wiese). Fix $n$ . The set of primes $l$ such that $PSL_2(\mathbb{F}_{l^n})$ occurs as a Galois group over $\mathbb{Q}$ with $l$ and at most three other primes ramified has a positive density.

Schematic description for the proved cases of $PSL_2(\mathbb{F}_{l^n})$ occuring as Galois group over $\mathbb{Q}$

Remark. The vertical theorem was generalized a lot by Khare, Larsen and Savin to many other groups, e.g., $PSp_{2m}$ .

Modular Forms and Galois Representations

A (cuspidal) modular form of weight $k$ and level $N$ is a holomorphic function $f\colon \mathbb{H} \to\mathbb{C}$ , such that

$f (\frac{az+b}{cz+d}) = (cz+d)^kf(z)$ , for any matrix $\begin{pmatrix} a& b \\ c& d\end{pmatrix} \in SL_2(\mathbb{Z})$ such that $\begin{pmatrix} a& b \\ c& d\end{pmatrix}\equiv \begin{pmatrix}1& x \\ 0& 1\end{pmatrix} \mod N$ .
“ $f$ is holomorphic at cusps”
(“ $f$ vanishes at the cusps”)

Here $\mathbb{H} = \{ x+iy\in \mathbb{C} \mid y>0\}$ is the upper half plain.

When we take $\begin{pmatrix} 1& 1\\ 0& 1\end{pmatrix}$ , we have $f(z+1) = f(z)$ . So we have an expansion to a Fourier series

$\displaystyle f(z) = \sum_{n\geq 0} a_n(f) q^n, \qquad q=e^{2\pi z}.$

Let $S_k(N)$ be the $\mathbb{C}$ vector space of cusp forms of level $N$ and weight $k$ . It is finite dimensional.

There is a family of commutating operators, the Hecke operators, $T_n\in {\rm End}(S_k(N))$ . Those are given by explicit formulas in terms of the Fourier coefficients.

A cusp form $f$ which is an eigenfunction for all $T_n$ is called a Hecke eigenform. A Hecke eigenform is normalized if $a_1(f) = 1$ . (We have $a_0=0$ since it’s a cusp form.) Theorems we will use are:

Theorem (Shimura). Let $f$ be a normalized Hecke eigenform. Then $\mathbb{Q}_f = \mathbb{Q}(a_n(f)\mid n>0)$ is a number field.

Theorem (Shimura-Deligne). Let $\Lambda$ be any prime ideal in $O_f=$ integers of $\mathbb{Q}_f$ . There is a Galois representation

$\displaystyle\rho_{f,\Lambda} \colon {\rm Gal}(\mathbb{Q}) \to GL_2(O_f/\Lambda)$

such that

$\rho_{f,\Lambda}(I_p)=0$ for all $p\nmid Nl$ (“unramified outside $Nl$ “)

${\rm char.poly}(\rho_{f,\Lambda}(Frob_p))\equiv X^2 - a_p(f) X + \epsilon(p) p ^{k-1}$ modulo $\Lambda$ , for all $p\nmid Nl$ . Here $\epsilon$ is a fixed Dirichlet’s character $\epsilon \colon (\mathbb{Z}/N\mathbb{Z})^\times \to \bar{\mathbb{Z}}^\times$ .

$det(\rho_{f,\Lambda}(complex conj.)) =-1$ “odd”.

Theorem (Ribet). For almost all $\Lambda$ , $\rho^{proj}_{f,\Lambda}({\rm Gal}(\mathbb{Q}))$ is $PSL_2(O_f/\Lambda)$ or $PGL_2(O_f/\Lambda)$ , provided that $f$ does not have CM nor inner twists, where

$\displaystyle \rho^{proj}_{f,\Lambda} \colon {\rm Gal}(\mathbb{Q}) \to GL_2(O_f/\Lambda) \to PGL_2(O_f/\Lambda).$

Theorem (Carayol). At $p\mid N$ , $p\neq l$ , $\rho_{f,\Lambda}|_{D_p}$ is also determined by $f$ .

We are left with the following tasks:

“make $O_f/\Lambda$ big”
Avoid the exceptional cases of Ribet’s theorem.

The method has the chance to succeed (if the group is realizable) because of

Serre’s modularity conj. (Thm by Khare, Witenberg, Kisin). Let $\rho \colon {\rm Gal}(\mathbb{Q}) \to GL_2(\bar{\mathbb{F}_l}$ be irreducible and odd. Then $\exists f, \Lambda$ such that $\rho \cong \rho_{f,\Lambda}$ .

As a consequence one gets the following

Corollary. Let $K/\mathbb{Q}$ be a totally imaginary Galois number field, ${\rm Gal}(K/\mathbb{Q})$ is $PSL_2$ of $PGL_2$ . Then $K$ is cut out by some $\rho_{f,\Lambda}$ , i.e. ${\rm Gal}(K) = \ker(\rho_{f,\Lambda})$ .

Proof. This extension induces an irreducible representation, and totally imaginary implies odd. (Indeed, the complex conjugation maps to a non-scalar involution, hence is odd.) QED

Idea of proof

Main point: Choose $N,k, f$ in such a way that $\rho_{f,\Lambda}|_{I_p}$ has a “desired” shape at some $p\mid N$ . (These are the bad places, we cannot control the “good” places, i.e., the unrmified primes.)

Recall: Let $G\subseteq PGL_2(\bar{\mathbb{F}}_l)$ be a finite subgroup. Then $G$ is conjugated to one of the following

$PSL_2(\mathbb{F}_{l^n})$ , $PGL_2(\mathbb{F}_{l^n})$
dihedral groups
$A_4, S_5, A_5$

(these are the irreducible representations)

Borel $\begin{pmatrix}* & *\\ 0& *\end{pmatrix}$
cyclic groups

(these are the reducible representations)

Definition. Given $\rho$ or $\rho_{f,\Lambda}$ or just $f$ , a prime $q$ is called tamely dihedral of order $r$ if $\rho|_{D_q}\sim {\rm Ind}_{\mathbb{Q}_{q^2}}^{\mathbb{Q}_q}(\chi)$ with some $\chi \colon {\rm Gal}(\mathbb{Q}_{q^2}) \to \bar{\mathbb{Z}}^\times$ of order $r$ .

We shall use

Level raising. Let $f\in S_k(N)$ be a normalized eigenform + conditions that we won’t get into them. Let $\Lambda$ . Then there is a prime $q$ and $g\in S_k(Nq^2)$ such that $\rho_{f,\Lambda} \cong \rho_{g,\tilde\Lambda}$ for some $\tilde\Lambda\subseteq O_g$ and $g$ is tamely dihedral of order a power of $l$ for $g$ .

By level raising we can force nice structure at bad primes.

Idea of the vertical result.

Fix $l$ . Want $PSL_2(\mathbb{F}_{l^n})$ for big $n$ .

Start with any $f$ , Hecke eigenform (e.g. $f\in S_2(11)$ ).
Use level raising modulo $p$ , get $g\in S_2(11 q^2)$ , for some $q$ , s.t. $q$ is tamely dihedral of order $p$ .
Choose any $\Lambda \subseteq O_g$ , $\Lambda\mid l$ .

Then

$\rho_{g,\Lambda}$ is irreducible, because $\rho_{g,\Lambda}|_{D_q} \cong {\rm Ind}\chi$ irreducible.
$\rho_{g,\Lambda}|_{I_l} \sim \begin{pmatrix}1& *\\ 0 & 1\end{pmatrix}$ , where $*$ denote a nonzero element. Hence the image $\rho^{proj}_{g,\Lambda}({\rm Gal}(\mathbb{Q}))$ is $PGL_2$ or $PSL_2$ . (That matrix kills the possibility of dihedral and we can get rid also of the $A_4,S_4,A_5$ .)
$\rho_{g,\Lambda}(I_q)$ contains $\begin{pmatrix}\zeta& 0 \\ 0 & \bar\zeta\end{pmatrix}$ , where $\zeta$ is a primitive $p$ -th root of unity. This follows from that it comes from 2 points above. This implies $\zeta + \bar\zeta$ sits in $O_g/\Lambda$ , hence our field is big. So we choose $p$ such that this element generates a big field. If $p>5$ we exclude $A_4,S_4, A_5$ , since this matrix is of order $p$ .

1 Comment | Galois theory, Representation theory | Permalink
Posted by barylior

Inverse Galois problem in modern Galois theory

May 19, 2010

The inverse Galois problem

Galois theory attaches to a polynomial $f(X)$ with integral coefficients its Galois group, denoted by

$\displaystyle Gal(f,\mathbb{Q}) = Gal(\Omega_f/\mathbb{Q}):= Aut_{\mathbb{Q}}(\Omega_f).$

Here $\Omega_f$ is the splitting field of $f$ . This finite group comes with a natural permutation representation on the roots of $f$ , (assuming $f$ has no multiple root). Then properties of $f$ reflects in the Galois group, most famously: there exists a radical root formula if and only if $Gal(f,\mathbb{Q})$ is solvable. After one is introduced to the notion of Galois groups, maybe the first question that comes to one’s mind is:

Inverse Galois Problem (over $\mathbb{Q}$ ) Can every finite group can be realized as a Galois group over $\mathbb{Q}$ ?

The IGP is widely open and it is the motivating problem of Galois theory. Some results:

Shafarevich’s theorem: Every solvable group can be realized as Galois group. For this
some deep number theoretical results were obtained.
Many simple groups can be realized as Galois groups via Hilbert’s irreducibility theorem. This theorem transfer the problem to Geometry. We will elaborate on this below.
A quite recent approach due to Gabor and Larsen et.al. is via Modular representations.

We note that the IGP does not make sense over any field (e.g., the complex numbers, the real numbers, finite fields, local fields, etc.). It does make sense over number fields, function fields, and more generally over Hilbertian fields.

Hilbert’s irreducibility theorem asserts that

For every irreducible $f(t,X)\in \mathbb{Q}[t,X]$ there exist infinitely many irreducible specializations: $a\in \mathbb{Q}$ such that $f(a,X)$ is irreducible.

A field having irreducible specializations (for every separable in $X$ irreducible polynomial) is called Hilbertian. Over a Hilbertian field $K$ , a stronger specialization property, the group-preserving specialization, follows, namely there are $a\in \mathbb{Q}$ for which

$\displaystyle Gal(f(t,X),\mathbb{Q}(t)) \cong Gal(f(a,X),\mathbb{Q})$ .

In particular, affirmative solution of IGP $/\mathbb{Q}$ follows from affirmative solution of IGP $/\mathbb{Q}(t)$ .

Essentially any field extension of $\mathbb{Q}(t)$ is composed out of two types of extensions: regular and arithmetic. Arithmetic extensions are those that comes from extensions of $\mathbb{Q}$ , e.g., $\mathbb{Q}(t)(\sqrt{2})$ . Regular extensions are those that contain no new algebraic elements over $\mathbb{Q}$ , e.g., $\mathbb{Q}(t,\sqrt{t^3+1})$ , or more generally $\mathbb{Q}(t,x)$ , where $f(t,x)=0$ , for some polynomial $f(t,X)\in \mathbb{Q}[t,X]$ that is absolutely irreducible, i.e. irreducible in the ring $\overline{\mathbb{Q}}[t,X]$ , where $\overline{\mathbb{Q}}$ is the algebraic closure of $\mathbb{Q}$ . In more geometric terms, regular extensions correspond to covers $C\to \mathbb{P}^1$ where $C$ is geometrically irreducible.

This leads to the regular inverse Galois problem that makes sense over any field $K$ .

Regular Inverse Galois Problem (over $K$ )
Does every finite group $G$ occurs as the Galois group of a regular extension of $K(t)$ ?

Let me give one example for a positive result: Harbater show that the RIGP has an affirmative solution over $\mathbb{Q}_p$ , (although the IGP does NOT hold true over $\mathbb{Q}_p$ ).

No field is known to have a negative answer to the RIGP.

Free absolute Galois groups

Another central motivating problem in Galois theory is the following conjecture of Shafarevich about the absolute Galois group of $\mathbb{Q}_{ab}$ . The maximal abelian extension $\mathbb{Q}_{ab}$ of $\mathbb{Q}$ is the compositum of all extensions of $\mathbb{Q}$ with abelian Galois group. By Kronecker-Webber this field is generated by all roots of unity. The absolute Galois group of a field $K$ is $Gal(K) = Aut_K(\bar{K})$ . It is a compact topological group, moreover a profinite group.

Shafarevich’s Conjecture:
$Gal({\mathbb{Q}_{ab}})$ is a free profinite group.

A more vague question is to what interesting fields have free absolute Galois group.

Some results:

The absolute Galois group of $\mathbb{C}(t)$ is free (Douady). This essentially follows from the fact that the fundamental group of the Riemann sphere minus $n+1$ points is free on $n$ elements.
If $K$ is a separably closed field, then $K(t)$ is free (Harbater, Pop, also proved by Haran-Jarden). For this different method of patching were developed (resp. formal, rigid, and algebraic).

By a theorem of Kuyk, Hilbertianity is inherited by abelian extensions, hence $\mathbb{Q}_{ab}$ is Hilbertian. This will allow us to formulate a geometric conjecture, similar to the RIGP, over an arbitrary field, a positive solution of which over $\mathbb{Q}_{ab}$ would imply Shafarevich’s conjecture.

Embedding problems.

Let us start by an example. Assume you want to realize $\mathbb{Z}/4\mathbb{Z}$ over some field $K$ . Since we have a map $\alpha\colon \mathbb{Z}/4\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$ and since quadratic extensions are fairly well understood, a good strategy will be to first realize $\mathbb{Z}/2\mathbb{Z}$ over $K$ , say by $L/K$ . This is easy, we just take a square root of a non-square element of $K$ (in characteristic $2$ we take an Artin-Schreier extension). Then what left is to embed $L$ into a quadratic extension $N$ such that $N/K$ is Galois with $Gal(N/K)\cong \mathbb{Z}/4\mathbb{Z}$ AND such that the following diagram commutes: $\xymatrix{ Gal(N/K) \ar[r]^{res} \ar[d]\ar@{}[dr]|*+[o][F-]{c} &Gal(L/K)\ar[d]\\ \mathbb{Z}/4\mathbb{Z}\ar[r]^{\alpha} &\mathbb{Z}/2\mathbb{Z}}$ An interesting exercise (for you the reader ^^) is to show that if $K=\mathbb{Q}$ , and if $L=\mathbb{Q}(\sqrt{d})$ , then the above embedding problem is solvable for exactly one $d\in \{3,5\}$ .

So $(L/K, \alpha)$ is an embedding problem for $K$ and $N$ together with the identification $Gal(N/K)\cong \mathbb{Z}/4\mathbb{Z}$ is a solution. A general finite embedding problem $(L/K, \alpha)$ is composed of an epimorphism of finite groups $\alpha\colon H\to G$ and a Galois extension $L/K$ with $Gal(L/K)\cong G$ . A solution is a Galois extension $N/K$ with $Gal(N/K)\cong H$ and a $K$ -embedding $L\to N$ under which the diagram $\xymatrix{ Gal(N/K) \ar[r]^{res} \ar[d]\ar@{}[dr]|*+[o][F-]{c} &Gal(L/K)\ar[d]\\ H\ar[r]^{\alpha} &G }$ commutes.

The embedding problem is trivial if $\ker \alpha =1$ (then $N=L$ is the unique and trivial solution). If $\alpha$ splits, i.e. has a group theoretical section, then the embedding problem is called split.

Note that if one considers an embedding problem with $L=K$ (or equivalently $G=1$ ), then to solve it is the same as to find a realization of $H$ as Galois group.

Going back to Shafarevich’s conjecture. Iwasawa proved that a profinite group that is countably generated (in the profinite sense) is free if and only if every finite embedding problem is solvable. Moreover $Gal({\mathbb{Q}_{ab}})$ is projective, so in order to show it is free it suffices to solve only finite split embedding problems. Note that an embedding problem $(L/K,\alpha\colon H\to G)$ for $K$ induces an embedding problem for $K(t)$ , namely $(L(t)/K(t),\alpha)$ , because $Gal(L/K)\cong Gal(L(t)/K(t))$ via the restriction map. Thus we have two conjectures (coined by Dèbes and Dechamps):

Conjectures. (RDD/ $K$ )

DD/ $K$

Every finite split embedding problem $(L/K,\alpha\colon H\to G)$ for $K$ has a solution. (Does not make sense for all fields, but does make sense for Hilbertian $K$ ).

RDD/ $K$

Every finite split embedding problem $(L/K,\alpha\colon H\to G)$ for $K$ has a regular solution, that is, a solution of $(L(t)/K(t),\alpha\colon H\to G)$ with solution field that is regular over $L$ .

The first conjecture implies the IGP, and the second conjecture implies the RIGP. There is no single field for which a negative answer to RDD is known.

Results:

If $K$ is PAC of characteristic $0$ , then RDD has an affirmative solution. (Fried-Völklein) By PAC (=pseudo algebraically closed) field we mean a field over which every absolutely irreducible variety has a rational point.
If $K$ is ample (of any characteristic), then RDD has an affirmative solution. (Pop) Ample means that if an absolutely irreducible variety has a smooth rational point it has infinitely many smooth rational points. This family includes both PAC fields, Henselian fields (in particular $\mathbb{Q}_p$ ), real closed fields, and many more. Shafarevich’s conjecture will follow from a positive answer to the following open problem:

Is $\mathbb{Q}_{ab}$ ample?
Harbater and Stevenson took this conjecture one step further: Let $K$ be an arbitrary field, let $E=K((x,y))$ be the fraction field of the ring of formal Taylor series $R=K[[x,y]]$ in $2$ variables. One may think of $E$ as a dimension $2$ local field. They showed that every non-trivial finite split embedding problem for $E$ has as many solutions as possible:

Theorem. (Harbater-Stevenson)
Every non-trivial finite split embedding problem for $E$ has $card(E)$ distinct solutions.

This result is interesting because in a sense it is free without projectivity. More precisely, a profinite group with this property (that every non-trivial finite split embedding problem has $m$ distinct solutions, for some infinite cardinal $m$ ) is free if and only if it is projective. Thus they called such profinite groups quasi-free (of rank $m$ ). So this notion enables one to break the proof of freeness into to steps quasi-freeness and projectivity. Each one of these properties is approached by totally different machinery. A small remark is that if the condition is satisfied for $m=1$ , then it is satisfied for $m=\aleph_0$ , so this definition does not make sense for finite $m$ . I won’t get into details but the only reasonable definition for quasi-free group of finite rank is that of a free (profinite) group.

Using quasi-free groups Harbater was able to prove the following

Theorem.
Let $E = K((x,y))$ , where $K$ is separably closed. Then $Gal(E_{ab})$ is free.

Projectivity is achieved by generalizing the work of Colliot-Thélène, Ojanguren, and Parimala from characteristic $0$ to an arbitrary characteristic.
Harbater proved that quasi-freeness is inherited by the commutator subgroup. Since $Gal(E)$ is quasi-free by Harbater-Stevenson, and $Gal(E_{ab})$ is the commutator of $Gal(E)$ , we get that it is quasi-free, and hence free.

One is now led to ask what other permanence criteria for quasi-free groups are there. I would remark that unlike abstract free groups (every subgroup of which is free) in the profinite category the problem is much more subtle. For example, $\mathbb{Z}_p$ is a subgroup of the free profinite group of rank $1$ . Another example, the intersection of open normal subgroups $N$ of a free profinite group $F$ of rank at least $2$ with solvable quotient $F/N$ is not free.

Since projectivity is inherited by subgroups, any permanence criterion for quasi-free groups will imply a criterion for free profinite groups. Therefore, as a first step, it is natural to ask whether the permanence criteria that are known for free profinite groups also hold true for quasi-free groups. In a joint work with Harbater and Haran, we show that some permanence criteria do not carry over to quasi-free groups. We construct a quasi-free profinite group of rank $m>\aleph_0$ and a closed subgroup $M$ such that the weight of $F/M$ is $\aleph_0$ but $M$ is not quasi-free. I will not describe here the construction in favor of the big picture.

Semi-free profinite groups

The bad behavior of quasi-free profinite groups led us (Haran, Harbater, and me) to modify the definition of quasi-free groups. We define the so-called semi-free profinite groups which are more restrictive than quasi-free. In order that this definition will be interesting we obtain the following desirable properties:

free implies semi-free implies quasi-free but not vice versa. In particular semi-free + projective = free.
The absolute Galois group of $E= K((x,y))$ is semi-free.
permanence criteria for free profinite groups can be extended to semi-free groups, including the so-called Diamond theorem. In fact, the proofs of the free case can be extended to the semi-free case.

The definition of a semi-free profinite group is the following. For quasi-free group we demanded that every finite split embedding problem has $m$ distinct solutions. Now we demand that the solutions are independent meaning as distinct as possible (in terms of fields, if the embedding problem is some $(L/E,\alpha)$ then two solutions are independent if solution fields are linearly disjoint over $L$ ).

After setting this new definition the following vast generalizations of Harbater-Stevenson’s theorem has been proved.

Theorem. (Pop, independently Paran-me)
Let $R$ be a complete local domain of dimension at least $2$ . Then the absolute Galois group of the fraction field of $R$ is semi-free of rank $|R|$ .

Two remarks:

Jarden has kindly informed me that from Pop’s proof it follows that the assumption that $R$ is local can be dropped. We are looking for this in the upcoming book of Jarden on algebraic patching.
In our proof, we develop the theory of a stronger notion of Hilbertianity, the so-called fully Hilbertian fields. This connects semi-free with Hilbertianity, thus giving new insights to the Jarden-Lubotzky twinning principle. On the one hand this connection gives the above theorem. On the other hand, we manage, to the best of my knowledge, for the first time, to derive permanence criteria for Hilbertian fields using purely group theoretic theorems.

Yet another result:

Theorem. (Haran-Harbater-me, Jarden)
Let $K$ be ample. Then the absolute Galois group of $K(t)$ is semi-free.

The proof of HHm uses ideas of Harbater-Stevenson (who proved this theorem with quasi-free replacing semi-free) while Jarden, in an independent work, gives a new proof.

Applications

The last two theorems lead to some interesting constructions of fields with free absolute Galois groups. But it seems that this post is too long already, hence I will write about it in another post.

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Hilbertian fields and rational embedding problems

February 11, 2010

In this post I introduce the so-called rational embedding problem notion which is a central tool in my research. I’ll also describe an application to the theory of Hilbertian fields.

The embedding problems for a given field is a group theoretic notion that is used to the study of the absolute Galois group of the field. The rational embedding problems are embedding problems coming from geometric/arithmetic objects. These embedding problems enable one to study geometric/arithmetic properties of a field using group theoretic tools. To demonstrate the efficiency of this, we will give a criterion for Hilbertiainity, solving a problem of Debes and Haran.

Embedding problems for a field

Let $K$ be a field. A (finite) embedding problem for $K$ consists on (1) a Galois extension $L/K$ with Galois group $G=G(L/K)$ (2) an epimorphism $\alpha\colon H\to G$ of finite groups. The problem is whether it is possible to embed $L$ in an $H$ -Galois extension $N$ such that $\alpha$ coincides with the restriction map. More precisely a proper solution is an $H$ -Galois extension $N/K$ , a $K$ -embedding $L\to N$ under which $\alpha (\bar \theta(\sigma)) = \sigma|_{L}$ , for all $\sigma\in G(N/K)$ .

In terms of the absolute Galois group $G(K)$ of $K$ , an embedding problem for $K$ consists on two finite groups $H,G$ and two (continuous) epimorphisms $\nu\colon G(K) \to G$ and $\alpha\colon H\to G$ . The problem, then, translates to a lifting problem, i.e., can $\alpha$ be lifted to an epimorphism $\theta \colon G(K) \to H$ (be lifted means that $\alpha\circ \theta = \nu$ ). We call such surjective $\theta$ proper solution. Then $N$ is the fixed field of $\ker \theta$ , we have $G(K)/\ker(\theta) \cong G(N/K)$ , so the above $\bar\theta$ is the epimorhism induced by $\theta$ .

If $\theta$ satisfies $\alpha\circ \theta = \nu$ , but is not surjective, we call it solution. I should remark on the notation: usually non-surjective $\theta$ is called weak solution, and the term solution can be either weak solution or proper solution, as is more convenient in context. I decided to use the term “solution” for weak solutions in this post.

Rational embedding problems

Now I want to define embedding problems in which $H$ is the Galois group of some geometric object (or if one prefer arithmetic object). We shall use the language of function fields and places, but everything can be expressed in the language of varieties and rational points.

Let $E = K(T_1, \ldots, T_n)$ be a rational function field over $K$ and let $F/E$ be a finite Galois extension. If $K_s$ denotes the separable closure of $K$ , then $L = F\cap K_s$ is a Galois extension of $K$ ( $L$ is the constant field of $F$ ). So we have the restriction of automorphisms map $\alpha \colon G(F/E) \to G(L/K)$ . The Galois correspondence says that the image of $\alpha$ is $G(L/F\cap L) = G(L/K)$ , so $\alpha$ is surjective. Thus $F/E$ induces an embedding problem, $\mathcal{E} (F/E)$ (here $\alpha$ and $\nu$ are the corresponding restriction maps); we call it rational embedding problem.

In this post we always assume that a place of a function field is trivial on the constant field (i.e., acts as the identity on the constant field.) Let $\varphi$ be a $K$ -rational place of $E$ that is unramified in $F$ . Extend $\varphi$ to a place $\Phi$ . Note that $\Phi(x)$ is finite iff $\Phi(\tau x)$ is finite, $\tau \in G(F/E)$ . Since $\Phi/\varphi$ is unramified the following formula defines a homomorphism $\Phi^* \colon G(K) \to G(F/E)$ :

$\displaystyle \Phi(\Phi^*(\sigma) x) = \sigma\Phi(x), \qquad \forall x\in F,\ \Phi(x)\neq \infty. \qquad (1)$

Since $\Phi$ fixes $L$ we have $\Phi^*(\sigma) x = \sigma x$ , for $x\in L$ and $\sigma\in G(K)$ . In other words $\alpha \circ \Phi^* = \nu$ . Therefore $\Phi^*$ is a solution of the rational embedding problem $\mathcal{E}(F/E)$ . We call a solution $\theta$ geometric if there exists an unramified $\Phi/\varphi$ as above with $\Phi^* = \theta$ . In general, not every solution is geometric, and in fact, there is a characterization of the family of fields for which any solution of a rational embedding problem is geometric: they are the PAC fields.

Splitness of rational embedding problems

We say that an embedding problem

$\mathcal{E} = ( \nu \colon G(K) \to G, \alpha\colon H\to G)$

splits if $\alpha$ has a group theoretical section, that is, $\beta\colon G\to H$ such that $\alpha\circ \beta= {\rm id}_G$ . We shall see that every rational embedding problem is dominated by a split rational embedding problem:

Lemma. Let $\mathcal{E} (F/E)$ be a rational embedding problem, then there exists a Galois extension $N/K$ such that $\mathcal{E}(FN/K)$ is a split raional embedding problem.

The proof of this lemma goes as follows. Choose some $K$ -rational place $\varphi$ of $E$ that is unramified in $F$ and extend it to a place $\Phi$ of $F$ . Let $N$ be the residue field of $\Phi$ , or in other words $N$ is the fixed field of the kernel of $\Phi^*$ . The Galois group $G(FN/E)$ is the fiber product of $G(F/E)$ and $G(N/K)$ over $G(L/K)$ (recall that $L=F\cap K_s$ ). Here $(FN) \cap K_s = N$ , so the “ $\alpha$ ” of the embedding problem $\mathcal{E}(FN/K)$ is the restriction map $G(FN/E) \to G(N/K)$ which coincides with the corresponding projection of the fiber product. A section of this map is given by $\sigma \mapsto (\Phi^*(\sigma), \sigma)$ (which is an element of the fiber product, thus induces an element of $G(FN/E)$ ). $\square$

If $\Psi^*$ is a geometric solution of $\mathcal{E}(FN/K)$ , then $(\Psi|_{F})^*(\sigma) = \Psi^*(\sigma)|_{F}$ , and it is a geometric solution of $\mathcal{E} (F/E)$ . Since the map $G(FN/E) \to G(F/E)$ is surjective, we have that $(\Psi_{F})^*$ is proper whenever $\Psi^*$ is proper. Thus we have

Corollary. Every rational embedding problem has a proper geometric solution iff every split rational embedding problem has a proper geometric solution.

Rational embedding problems induced by a polynomial

Assume that $F$ is the splitting field of a polynomial $f(X) \in E[X]$ . (Here $f$ is a polynomial in $X$ whose coefficients are rational functions in $T_1, \ldots, T_n$ .) This gives us a permutation representation of the group $G(F/E)$ . We denote the embedding problem with the permutation structure by $\mathcal{E}(f,K)$ .

Assume $a_i := \varphi(T_i)\in K$ and the polynomial, call it $\bar f$ , that we get by substituting $a_i$ for $T_i$ in the coefficients of $f$ is well defined, separable, and of the same degree as ${\rm deg} f$ ; in this case we abbreviate and say that the specialization is nice. Then $\Phi$ maps the roots of $f$ injectively, and hence bijectively, onto the roots of $\bar f$ . Eq. (1) implies that

the Galois action of $G(K)$ on the roots of $\bar f$ is isomorphic to the Galois action of $\Phi^*(G(K))$ on the roots of $f$ .

(Note that $\Phi^*(G(K))$ is the decomposition group of $\Phi/\varphi$ .) In particular, we get that $\bar f$ is irreducible if and only if $G(K)$ acts transitively if and only if the image of $\Phi^*$ is transitive. In particular, if $f$ is irreducible over $E$ , then the group $G(F/E)$ acts transitively on the roots of $f$ . so we have the following

Lemma. Let $f(T_1, \ldots, T_n, X)\in K(T_1,\ldots, T_r)[X]$ be irreducible and separable. Let $(T_1, \ldots, T_n) \mapsto (a_1, \ldots, a_n)$ be a nice specialization, and let $\Phi$ be a place of the splitting field of $f$ extending $T_i\mapsto a_i$ . Then if $\Phi^*$ is a proper solution, the $\bar f$ is irreducible.

Applying the corollary above we get a criterion for a field to be Hilbertian (that is, for all polynomials as above to have irreducible specialization).

Criterion. A field $K$ is Hilbertian if and only if every rational split embedding problem is properly solvable. In fact, it suffices to consider the case where $(T_1, \ldots, T_n) = T$ .

This criterion, more or less, can be found in the work of many. Some, in alphabetical order, are Bary-Soroker (that’s me), Debes, Dechamps, Fried, Haran, Jarden, Pop, Roquette, and Voelklein. It is possible that I missed someone, you can comment on that below.

An application – a criterion for Hilbertianity

Now I want to give one application of the above criterion. Let me start with a brief discussion to motivate the application.

Fried and Voelklein introduced the RG-Hilbertian fields (R=regular and G=Galois). That is $K$ is called RG-Hilbertian if for every Galois extension $F/K(T)$ that is regular over $K$ , there exist infinitely many place extensions $\Phi/\varphi$ whose residue field extensions $N/K$ are of the same degree as the degree of $F/K(T)$ . In terms of polynomials, a field is RG-Hilbretian if and only if for every separable in $X$ polynomial $f(T,X)$ that is Galois over $K(T)$ (i.e., a root of whose generate the splitting field) and absolutely irreducible (that is, remains irreducible when replacing $K$ with its algebraic closure) there exist infinitely many irreducible specializations. This notion is interesting in light of the inverse Galois problem, because many groups are realized as RG extensions. Fried and Voelklein showed that RG-Hilbertianinty is strictly weaker than Hilbertianity by group theoretically classifying these two notions when $K$ is PAC.

Theorem (Fried-Voelklein). Let $K$ be a PAC field. Then

$K$ is Hilbertian if and only if every embedding problem is properly solvable.

$K$ is RG-Hilbertian if and only if every finite group is realizable as a Galois extension over $K$ .

Debes and Haran further studied the RG-Hilbertian fields, finding more examples of them, which are, in a sense, nicer. They also introduced other variants. Notably, they introduced the R-Hilbertian fields, fields having irreducible specialization for absolutely irreducible polynomials, and showed that if $K$ is PAC, then $K$ is Hilbertian if and only if $K$ is R-Hilbertian. They asked whether it is true in general. Here is an affirmative answer I gave to this problem (published in IMRN).

Theorem. A field is Hilbertian if and only if it is R-Hilbertian.

That is, to have irreducible specializations for any irreducible $f(T,X)\in K(T)[X]$ that is separable in $X$ it suffices to merely have irreducible specialization when $f$ is absolutely irreducible.

Let me describe the proof. Clearly Hilbertianity is stronger than R-Hilbertianity, hence we assume $K$ is R-Hilbertian and show $K$ is Hilbertian. By the above criterion it suffices to properly and geometrically solve a rational split embedding problem. That is, we have an embedding problem

$\displaystyle \mathcal{E} = (\nu\colon G(K) \to G(L/K), \alpha\colon G(F/K(T)) \to G(L/K))$

with $L\subseteq F$ , $F$ regular over $L$ , and $\nu,\alpha$ the corresponding restriction maps. We also assume $\alpha$ splits, i.e., there exists $\beta \colon G(L/K) \to G(F/K(T))$ such that $\alpha\circ \beta = {\rm id}_{G}$ . Let $G$ be the image of $\beta$ , and let $E$ be the fixed field of $G$ . It follows from the Galois correspondence that $E\cap L = K$ , so $E$ is regular and that $EL = F$ .

Because $E$ is regular over $K$ , by assumption, there exists a specialization $T\mapsto a\in K$ under which the residue field $E(a)$ of $E$ over $K$ has the same degree as $E/K(T)$ . Since $F=EL$ , the residue field $F(a)$ of $F$ equals $E(a) L$ (for almost all $a$ ‘s) and one may expect that

$\displaystyle [F(a):K] = [E(a):K][L:K]=[F:K(T)].$

In other words, the corresponding geometric solution is proper. The problem is that it may happen that $E(a)$ and $L$ are not linearly disjoint, which would imply that $[F(a):K] < [E(a):K][L:K]$ .

To solve this difficulty, we take many linearly disjoint copies of $E$ , say $E_1, \ldots, E_r$ (we shall say how large $r$ should be in a second) such that $\hat{E} = E_1 \cdots E_r$ is still regular over $K$ . (To do this, e.g., we apply the isomorphism $T\mapsto a_iT+b_i$ to $K(T)$ for good choices of $a_i,b_i\in K$ , extend it to $E$ , and let $E_i$ be the image.) We specialize $T\mapsto a$ such that $[\hat{E}(a):K]=[\hat{E}:K(T)]$ . Then, it is easy to see that $[E_i(a):K]=[E:K(T)]$ and that the $E_i(a)$ are linearly disjoint.

If you take $r$ to be bigger than the number of subfields of $L/K$ , by the pigeonhole principle, $E_i(a)\cap L = E_j(a) \cap L$ , for some $i\neq j$ . But $E_i(a) \cap E_j(a) = K$ , so $E_i(a) \cap L = K$ , and $E_i(a)$ is linearly disjoint of $L$ over $K$ . This finishes the proof because $E_i(a) = E(aa_i + b_i)$ . $\square$

Further applications

Here I chose one application that might be done without the language of rational embedding problems.

In the future, I plan to post more applications of rational embedding problems. Some buzz words: Fully Hilbertian fields, fields having free absolute Galois groups, Jarden-Lubotzky twinning principle, PAC extensions, and more…

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Power series rings and irreducible specializations

January 5, 2010

This post describes a new work of Elad Paran on Hilbertianity of the fraction fields of power series rings, as he reported in Amitsur algebra seminar of Jerusalem on December 24, 2009.

Introduction and motivation

The inverse Galois problem and Hilbertian fields

The Inverse Galois Problem asks for every finite group $G$ to find a $G$ -Galois extension $L/\mathbb{Q}$ . One fruitful approach to attack the IGP is Hilbert’s irreducibility theorem. This theorem says that $\mathbb{Q}$ is Hilbertian: every irreducible polynomial $f(T_1, \ldots, T_n,X) \in \mathbb{Q}[T_1, \ldots, T_n,X]$ admits $a_1, \ldots, a_n\in \mathbb{Q}$ s.t. $f(a_1, \ldots, a_n,X)$ is irreducible in $\mathbb{Q}[X].$

The property of being Hilbertian reduces the IGP over $K$ to the IGP over $K(T_1, \ldots, T_r)$ , which is easier since there are more degrees of freedom. Now, as a generalization, one can study the IGP over $K(T_1, \ldots, T_r)$ for arbitrary $K$ . For example, the IGP holds over $\mathbb{C}(T)$ . (In fact $\mathbb{C}(T)$ has a much stronger Galois theoretic property, namely the absolute Galois group of $\mathbb{C}(T)$ is the free profinite group based on $\mathbb{C}$ . This was proved by Douady, but is essentially due to Riemann.)

Conjecturally a stronger property than IGP holds over Hilbertian $K$ :

Conjecture (Debes-Deschamps). If $K$ is Hilbertian, then any finite split embedding problem for $K$ is solvable.

This conjecture was proved for a large family of fields, the so-called ample fields. (Alternatively, …. for an ample family of fields, the so-called large fields.)

Theorem (Pop). If $K$ is ample, then any finite split embedding problem over $K(T)$ is solvable. In particular, if $K$ is also Hilbertian, then the conjeture holds.

Let’s give the definition of these fields:

Definition. A field $K$ is called ample if every smooth curve over $K$ with a $K$ -rational point, has infinitely many $K$ -rational points.

(I would like to remark that Pop called these fields large and later some other names were suggested, including “ample”. I feel that the adjective “large” is too general, and should not be used for a permanent definition in mathematics, for example can you imagine that somebody will define “simple group” and the problems it will cause Therefore I decided to used the adjective “ample” instead of “large”.)

This definition captures the common property of seemingly very different types of fields (notably complete and pseudo algebraically closed) and at the same time is extremely interesting, as Pop’s theorem demonstrates. (A very interesting open problem is to determine whether the maximal cyclotomic extension of $\mathbb{Q}$ is ample. A positive answer will imply a conjecture of Shafarevich that says the absolute Galois group of this field is free profinite.)

Studying Galois theory of power series (or local complete fields)

We want to study the fraction field of a ring of power series $R[[X]] = \{\sum_{i=0}^\infty r_i X_i\}$ , where $R$ is a ring. This ring is complete w.r.t. to the $X$ -adic topology. However if $R$ is not a field, its fraction field, denoted by $E$ , is not complete! In 2005, Harabater and Stevenson studied the Galois theory of such fields.

Theorem (Harabater-Stevenson). If $R$ is a complete dvr then the conjecture holds over $E$ . (dvr=discrete valuation ring)

Notably, if we apply the theorem for $R = K[[X]]$ over an arbitrary field $K$ , we get the conjecture for the fraction field of the ring of power series in two variables $K[[X,Y]] := K[[X]][[Y]]$ over $K$ . In 2006, Paran generalized the theorem and proved the following

Theorem (Paran). If $R$ is a Krull domain, then the conjecture holds over $E$ .

We postpone the definition of Krull domains, and mention in the meanwhile that normal+Noetherian, in particular dvr, implies Krull. Since the Krull property of a ring $R$ is preserved by $R[[X]]$ , Paran’s theorem implies the conjecture to the fraction field of $K[[X_1, \ldots, X_n]]$ , for any $n\geq 2$ . (It clearly false for $n=1$ .)

Later, in 2007, Pop showed that $E$ is ample in a greater generality.

Theorem (Pop). If $R$ is an arbitrary complete domain, then $E$ is ample.

By a theorem of Weissauer (see below) $E$ is Hilbertian in the cases discussed by Harabater-Stenvenson and Paran. Hence these results falls into Pop’s theorem mentioned at the beginning. All we discussed so far motivates the following

Question. When is the fraction field of $R[[X]]$ Hilbertian?

Generalized Krull Domains and Hilbertianity

If $R$ is a Krull domain, then $R[[X]]$ is also Krull. This together with the following result of Weissauer (1982), gives ab affirmative answer when $R$ is a Krull domain. Let us start with definitions.

Definition (Ribenboim 1956). We say that $R$ is a generalized Krull domain (GKD) if there exists a non-empty family, $\mathcal{F}$ , of rank- $1$ valuations on its fraction field $K$ satisfying

$\mathcal{F}$ is of finite type (i.e., every element has only finitely many valuation on which it is not trivial),

$R$ is defined by $\mathcal{F}$ , i.e., $R$ is the intersection of all the valuation rings,

each $v\in \mathcal{F}$ is essential on $R$ , that is to say, the valuation ring $R_v$ equals the localization of $R$ by the center of $v$ , i.e., $\mathfrak{p}(v) = \{ x\in R \mid v(x) >0\}$ .

This family of valuations $\mathcal{F}$ is determined by the ring $R$ : it is the family of all valuations of $K$ that are essential on $R$ . We call this family the essential family.

Definition. If all the valuations in the essential family are discrete, we say that $R$ is a Krull domain.

Noetherian normal domains and UFD’s are Krull. The simplest example of a GKD which is not Krull is a rank- $1$ real valuation ring which is not discrete.

One needs to think about GKD as rings with arithmetic; the essential family plays the role of primes in these rings. Weissauer proved:

Theorem (Weissauer). The fraction field of a GKD of dimension $\geq 2$ is Hilbertian.

This answer the posed question when $R$ is Krull because if $R$ is Krull, then $R[[X]]$ is also Krull and its dimension is $1 + \dim R\geq 2.$

Corollary. If $R$ is Krull, then the fraction field of $R[[X]]$ is Hilbertian.

This led to two open problems (appearing in Fried-Jarden’s Field Arithmetic) for a GKD $R$ :

Is $R[[X]]$ GKD?
Is the fraction field of $R[[X]]$ Hilbertian?

Note that 1 implies 2 by Weissauer. The first problem was answered by Paran and Temkin: very NO. More precisely,

Theorem (Paran-Temkin). Let $R$ be a GKD. Then $R[[X]]$ is a GKD if and only if $R$ is Krull (and then $R[[X]]$ is Krull).

Paran solved the second problem affirmatively, and we shall describe the proof below.

Theorem (Paran). If a domain $R$ is contained in a rank- $1$ valuation ring of its fraction field $K$ , then the fraction field of $R[[X]]$ is Hilbertian.

(It is obvious that a GKD satisfies the condition of the theorem.) It is interesting to note that although there are no GKD in the formulation of the theorem, the proof uses this notion of GKD and Weissauer theorem.

Nice idea: Instead of trying to somehow generalize the proof of Weissauer, Paran finds a GKD between $R[[X]]$ and its fraction field.

Ugly part: Is the intersection of two GKDs with the same fraction field is GKD? There is a candidate for the essential family of the intersection, namely the union of the essential families of each of the GKDs. And indeed, conditions 1+2 of the definition will hold, but sometimes 3 fails, so the answer is NO. (This does work for Krull domains.)

Therefore we replace 3 with

3′. Each $v\in\mathcal{F}$ is either discrete or $v(R) = v(R_v)$ . (The latter is called well-centered.)

If we have (1)+(2)+(3′), then $R$ is GKD with essential family inside the given family. We won’t prove this criterion.

Proof of Paran’s Theorem

An element in $f\in R[[X]]$ can be thought of as a function on the open unit disc on the projective $K$ -line (e.g., if $R$ is complete). It might have an infinite number of roots (finite in each disc with a smaller radius). The ring of convergent power series behave differently, because of Weirstrass preparation theorem.

Let $R$ be a domain and $v$ valuation on the fraction field of $R$ . Then we define

$R\{ X\} = \{\sum_{i=0}^\infty a_i X^i \mid v(a_i) \to \infty\}$ .

The valuation $v$ extends to a valuation $v^*$ on the fraction field of $R\{X\}$ defined by $v^*(\sum a_i X^i ) = \min v(a_i)$ .

Lemma. If $R_v$ is a valuation ring, then $R_v\{X\}$ is GKD with essential family consisting on discrete valuations and $v^*$ .

The proof of this lemma is again based on Weirstrass preparation theorem, and we skip it. Let’s prove the theorem.

Choose $a\in R$ such that $v(a)>0$ . We have an embedding $\varphi\colon R[[X]]\to R\{X\}$ given by $X\mapsto aX$ . Let $E = {\rm Frac}(R[[X]])$ and $F = {\rm Frac}(R_v\{X\})$ . Then $\varphi$ extends to $\varphi\colon E\to F.$ Let $D = \varphi(E) \cap R_v\{X\}$ and $F'$ the fraction field of $D$ . Since $D$ contains $\varphi (R[[X]])$ , we get that $\varphi(E)= F'$ , so it suffices to prove that $F'$ is Hilbertian. By Weissauer’s theorem we need to prove that $D$ is GKD of dimension $\geq 2$ .

The ring $D$ is an intersection of a GKD and a field, so we have a candidate for the essential family: the restriction of the essential family of $R_{v}\{X\}$ . In this family there are discrete valuations that come from minimal primes of $R_v\{X\}$ and $v^*$ . Conditions 1+2 of the definition of GKD are immediate. We have

$v(R_v) =v^*(R_v) \subseteq v^*(D) \subseteq v^*(R_v\{X\}) \subseteq v(R_v).$

Therefore $v^*$ is well-centered on $D$ , hence $D$ is GKD.

We are left to show that the dimension of $D$ is at least $2$ . We have a map $D\to R_v\{X\} \to R_v \to 0$ (the latter comes from the substitution $X\mapsto 0$ ) which is identity on $R_v\subseteq D$ . So $D\to R_v$ is surjective, but not injective. Moreover both the image and the kernel are not fields. So $\dim S = \dim \ker + \dim R_v \geq 2$ . $\square$

Concluding remark

One can think that there should be a notion of what is “good” arithmetic on a ring, this notion should be inherited by the power series ring and should imply Hilbertianity. This will give a direct proof of the theorem, and not via the trick of replacing our ring with some “better” ring.

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Diamond theorems for profinite groups and Hilbertian fields

December 30, 2009

I decided to write this post for several reasons: The first two papers starting this subject are the first research papers I’ve read. There is some interesting activity in the last few years evolving from the diamond theorem (I’ll elaborate below). But in fact I’m about to give a talk about this in Amitsur algebra seminar in the Hebrew University, so it may well serve me as a kind of preparation

Proper disclosure: Haran was my Master and PhD supervisor.

I. Free profinite groups

In the abstract world free subgroups well behave under taking subgroups:

Nielsen-Schreier Theorem. Every subgroup of a free group is free.

If $M$ is a finite index subgroup of $F$ , then we have a formula of the rank of $M$ :

N-S Formula. ${\rm rank}( M )= ({\rm rank} (F) - 1)\cdot (F:M) + 1$ .

In the pro- $p$ category the same result holds. However in the profinite category things are not so well-behaved. It should be stressed that while working in the profinite category we assume all subgroups to be closed and all homomorphisms to be continuous, unless explicitly stated.

$\hat{\mathbb{Z}}$ is the free profinite group of rank 1 and it contains $\mathbb{Z}_p$ which is not free.
More generally, because we have $p$ -sylow theorems in the profinite category, if $F$ is free profinite, then its $p$ -sylow subgroups are not free.
The $p$ -sylow subgroups are not normal. Here is an example with a normal subgroup: Consider the kernel, say $M(p)$ , of the natural epimorhism from a free nonabelian profinite group, say of countable rank, to the free pro- $p$ group. Then $M(p)$ has no $\mathbb{Z}/p$ quotients, hence is not free.

These examples shows that the N-S theorem fails in the profinite category. However the N-S formula does hold:

N-S formula for profinite groups: Let $F$ be a free profinite group and $M$ an open subgroup. Then $M$ is free and the N-S formula holds.

Another surprising result of v.d.Dries and Lubotzky asserts that although a normal subgroup is not necessarily free, any proper open subgroup of a normal subgroup is free! Melnikov gave a full classification of the normal subgroups of free profinite groups, by their so-called $S$ -rank ( $S$ being finite simple group): For a profinite group $G$ and a finite simple group $S$ we let $r_G(S)$ be the maximal cardinal such that $S^m$ is a quotient of $G$ . Then the function $r_G$ is called the $S$ -rank of $G$ .

Theorem (Melnikov). The normal subgroups of a free profinite group are determined by their $S$ -rank. Namely, if $M,N$ are normal subgroup of a free profinite group, then $M\cong N$ if and only if $r_M = r_N$ .

Melnikov proves this theorem in the more generalized category of pro- $\Sigma$ groups, where $\Sigma$ is some nice family of finite groups, but I don’t want to get into it classes of groups in this post. I’ll just mention that all results we shall discuss work in a pro- $\Sigma$ categories, for suitable classes of finite groups $\Sigma$ . Moreover, Melnikov, determined what are the exact possible $S$ -rank for normal subgroups of free profinite groups, again I won’t elaborate on it now.

II. Haran’s diamond theorem for free profinite groups

Now we come to Haran’s diamond theorem. The first distinguish of this theorem is that it deals with arbitrary (closed) subgroups. Moreover, much more important than the result itself, is the method of proof, that I call The Haran-Shapiro Induction, that is so flexible it applies in many different setting: Hilbertian fields, the stronger recently introduced fully Hilbertian fields, semi-free profinite groups, surface groups. The theorem itself has some very recent new applications, to the construction of fields with free absolute Galois groups and to fields coming from torsion points of abelian varieties. I intend one day to discuss them in a new post….

We say that a subgroup $M$ of a profinite group $F$ is contained in an $F$ -diamond if there exist normal subgroups $N_1, N_2$ of $F$ such that $N_1\cap N_2\leq M$ but $M$ contains neither of the $N_i$ ‘s. $\xymatrix{ &F/N\ar[d]^{\nu}\\ \bar{A}\wr_{G_1} G\ar[r]^{\bar\alpha} &G } \] \[ \xymatrix{ &F\\ N_1\ar@{-}[ur] &M\ar@{-}[u] &N_2\ar@{-}[ul]\\ &N_1\cap N_2 \ar@{-}[ur]\ar@{-}[ul]\ar@{-}[u] }$

The diamond theorem for free profinite groups (Haran, B-S). Let $F$ be a non-abelian free profinite group and let $M$ a subgroup that is contained in an $F$ -diamond. Then $M$ is free. Moreover if the index of $M$ in $F$ is infinite, then the rank of $M$ is $\max\{\mbox{rank} F, \aleph_0\}$ .

Haran proved this theorem for free groups of infinite rank, and I extended Haran’s proof (which will be discussed below) to the finite rank case using some combinatorial arguments. Before going to the method I want to give several special instances of this theorem, so you might have a feeling of the contained in an $F$ -diamond condition.

v.d.Dries-Lubotzky result follows from this theorem. Here is a more general version: Let $N\leq M\leq F$ with $N$ normal in $F$ . Then any proper open subgroup $U$ of $M$ not contained in $N$ is free (of infinite rank equals to $\max\{\mbox{rank} F, \aleph_0\}$ ). Indeed, let $N'$ be an open normal subgroup with $N'\cap M\leq U$ . Then the $F$ -diamond is composed from the normal subgroups $N$ and $N'$ . This criterion implies an analog of Karras-Solitar theorem for profinite groups (first proved by Jarden): A finitely generated subgroup of infinite index of a non-abelian free profinite group contains no normal nontrivial subgroup.
The commutator subgroup $F'$ of a free profinite group is free: Indeed, any abelian profinite group is either pro-cyclic or a direct product $A\times B$ . Since $F/F'$ is not cyclic (it has $(\mathbb{Z}/2\mathbb{Z})^2$ as a quotient) we get that $F'$ is the intersection of two normal subgroups each not contained in the other. These subgroups form the $F$ -diamond. Note: We actually proved that any subgroup with abelian non-pro-cyclic quotient is free. The pro-cyclic case is in fact easier and follows directly from the Haran-Shapiro induction.
The following special case of the diamond (applies to semi-free groups) was used (with $N_2= F'$ ) in a joint work with Haran and Harbater to prove that certain fields have free absolute Galois groups: Let $N_1,N_2$ be a normal subgroups of $F$ . Then any subgroup lying between $N_2$ and $N_1\cap N_2$ is free.
Assume that $F/N$ is a product of finite groups of bounded order. Then any subgroup between $F$ and $N$ is contained in an $F$ -diamond. This was used by Jarden to prove that, for an abelian variety $A$ , any subfield of $\mathbb{Q}(A_{tor})$ is Hilbertian, where $A_{tor}$ are the torsion points of $A$ .

III. Haran’s diamond theorem for Hilbertian fields

Recall that a Hilbertian field is a field $K$ having the irreducible specialization property: for every irreducible $f(T,X)\in K[T,X]$ that is separable in $X$ , there exists $a\in K$ for which $f(a,X)$ is irreducible in the ring $K[X]$ . I plan to write a long post on Hilbertian fields in the (near?) future. In any rate, Hilbert proved his irreducibility theorem that says $\mathbb{Q}$ is Hilbertian in order to realize finite groups over $\mathbb{Q}$ (that is to attack the inverse Galois problem). The theorem has many other applications, but they will be posted later.

The analogous problem here is when an extension of a Hilbertian field is Hilbertian. A finitely generated transcendental extension of an arbitrary field is Hilbertian, thus the problem reduces to algebraic extensions. A purely transcendental extension of a Hilbertian field is again Hlibertian, so we reduced to separable algebraic extensions. There were several results that preceded the diamond theorem. Hilbertianity is preserved, e.g., under finite extensions (Hilbert) and abelian extensions (Kuyk).

It is interesting to note that a similar phenomenon as in profinite groups occurs for fields: If $K$ is Hilbertian and if $N$ is a Galois extension of $K$ , then $N$ is not Hilbertian in general, but any proper finite extension is. This theorem is of Weissauer who proved it using non-standard methods. The similarity is no coincidence, but a reflection of Jarden-Lubotzky twining principle, that is not fully understood yet, but recently was approached by Paran and myself. Later Fried gave a field theoretic proof of the theorem. And Haran gave a fairly group theoretic proof, inventing the Haran-Shapiro induction, that led to the diamond theorem (both the field theoretic and the group theoretic).

It is important to note that in the field theoretic setting we do not have an analog of Melnikov’s theorem or the combinatorial method of v.d.Dries-Lubotzky. (I believe that Melnikov’s theorem will not hold for Hilbertian fields.) So the diamond theorem is the most general permanence criterion.

To have simple notation, we say that a separable extension $L$ of a field $K$ is contained in a $K$ -diamond if there exist Galois extensions $N_1, N_2$ such that $L\subseteq N_1 N_2$ but $L$ is not contained in neither of the $N_i$ ‘s. So $L$ is contained in a $K$ -diamond if and only if its absolute Galois group $G(L)$ is contained in a $G(K)$ -diamond.

The diamond theorem for Hilbertian fields (Haran). Let $K$ be a Hilbertian field and $L/K$ a separable extension that is contained in a $K$ -diamond. Then $L$ is Hilbertian.

IV. The Haran-Shapiro induction

We shall try to explain the method in the case of free groups and for the sake of simplicity we stick to the case when $F$ is the free profinite group of countable rank. (This means that $F$ has a countable free basis that converges to $1$ .) We start with some general properties of profinite groups.

IV.1. A freeness criterion of Iwasawa

An embedding problem for a profinite group $F$ consists of two profinite groups $G,H$ and two epimorphisms $\mu \colon F\to G$ and $\alpha\colon H\to G$ . A solution is a lifting of $\alpha$ to an epimorphism $\psi \colon F \to H$ . If we have a lifting which is not surjective we shall call it weak solution.

(The reason we call it embedding problem and not a lifting problem is historic, namely it came from embedding problems of fields. )

$\xymatrix{ &F\ar@{->>}[d]^{\mu}\ar@{.>}[dl]_{\psi}\\ H\ar@{->>}[r]^{\alpha} &G }$

An embedding problem is called finite if $G,H$ are finite, it is called split if $\alpha$ has a group theoretical section. In this case $H$ is isomorphic to $A \rtimes G$ , $A =\ker \alpha$ and under this isomorphism $\alpha$ is the projection on $G$ map.

Iwasawa criterion. A countably generated profinite group is free of countable rank if and only if every finite embedding problem is solvable.

A general idea (a post on it will come?) is that one can `factor’ an embedding problem to a Frattini followed by split. This imply the following criterion

Theorem. A countably generated profinite group is free if and only if it is projective and every finite split embedding problem is solvable.

I believe that this formulation of the theorem firstly appeared in the work of Harbater-Stevenson. Since a free group is projective, projectiveness is inherited by subgroups, and the countably generated is also inherited we get the following

Corollary. A subgroup $M$ of the free profinite group of countable rank is free of countable rank if and only if every finite split embedding problem for $M$ is solvable.

IV.2. Brief description of the method

We have a subgroup $M$ of the free group $F$ and a finite split embedding problem

$\mathcal{E}_0: \quad (\nu_0 \colon M\to G_0, \alpha_0 \colon A\rtimes G_0 \to G_0)$

for $M$ . In order to solve the embedding problem we shall first induce a new embedding problem, say $\mathcal{E}$ , for $F$ with the property that a weak solution $\psi$ of $\mathcal{E}$ induces a weak solution $\psi_0$ of $\mathcal{E}_0$ . This is the Shapiro part of the method. This part is very much related to the non-abelian Shapiro lemma.

However there is an essential problem: We need that a solution induces a solution, but in general the surjectivity of $\psi$ does not imply the surjectivity of $\psi_0$ . Here comes Haran’s part. He gives a sufficient condition to preserve surjectivness under the induction of $\psi_0$ from $\psi$ .

IV.3. The induced embedding problem

We start by choosing an epimorphism $\nu\colon F \to G$ onto a finite group $G$ such that $\ker \nu\cap M\leq \ker \nu_0$ . It shall become crucial later on that we can take $G$ as large as we want!

Let $G_1=\nu_1(M)$ and $\nu_1\colon M \to G_1$ be the restriction of $\nu$ to $M$ . Then $\nu_0$ factors through $\nu_1$ , so we have a map $\mu\colon G_1\to G_0$ such that $\nu_0 = \mu\nu_1$ . Then $\mu$ defines an action of $G_1$ on $A$ and we have

$\xymatrix{ {\rm Ind}_{G_1}^G(A) \rtimes G_1\ar[r]\ar[d]^{s}&G_1\ar@{=}[d]\\ A\rtimes G_1\ar[r]^{\alpha_1}\ar[d]^{\pi} &G_1\ar[d]^{\mu}\\ A\rtimes G_0 \ar[r]^{\alpha_0} &G_0 }$

commutative diagram

Her $\pi(a,g) = (a,\mu(g))$ . Let $\mathcal{E}_1$ be the embedding problem for $M$ coming from the upper row, namely $(\nu_1,\alpha_1)$ . It suffices to solve $\mathcal{E}_1$ , because then composing with $\pi$ will give a solution of $\mathcal{E}_0$ .

So far we extended $\nu_0$ to $F$ . Next we need to lift the embedding problem. Although $G$ does not act on $A$ it does act on the `induced’ group

$\displaystyle {\rm Ind}_{G_1}^G(A) = \{ f\colon G\to A \mid f(\sigma \tau) = f(\sigma)^{\tau},\ \forall\sigma\in G,\ \tau\in G_1\}.$

(We assume $G_1$ acts on $A$ from the right.) So $\sigma\in G$ acts on $f$ by translating from the left, i.e.

$\displaystyle f^{\sigma}(\rho) = f(\sigma\rho), \qquad \forall \sigma,\rho\in G.$

Then we have the twisted wreath product: $A\wr_{G_1} G := {\rm Ind}_{G_1}^G(A) \rtimes G$ together with the projection map $\alpha\colon A\wr_{G_1} G\to G$ . So the induced embedding problem is

$\mathcal{E}:\quad (\nu\colon F\to G, \alpha\colon A\wr_{G_1} G\to G).$

The twisted wreath product comes with the Shapiro map, which we shall denote by $s\colon{\rm Ind}_{G_1}^G(A)\rtimes G_1 \to A\rtimes G_0$ which is defined by

$s(f,\sigma) = (f(1),\sigma).$

It’s a straightforward to check that $s$ is an epimorphism.

$\xymatrix{ A\rtimes G_0 \ar[d]^{\alpha_0} &A\rtimes G_1\ar[d]^{\alpha_1}\ar[l]_{\pi} &{\rm Ind}_{G_1}^G(A) \rtimes G_1\ar[d]\ar@{->>}[l]_-{s}\ar@{^(->}[r] &A\wr_{G_0}G\ar[d]^{\alpha}\\ G_0 &G_1\ar[l]^{\mu} &G_1\ar@{=}[l]\ar@{^(->}[r] &G }$

Now a weak solution $\psi \colon F \to A\wr_{G_1} G$ induces the weak solution $\psi_0 = \pi\circ s\circ \psi|_{M}$ because $\psi(M)$ is contained in the inverse image of $\nu(M) = G_1$ . It will be easier to consider $\psi_1 = s\circ \psi|_{M}$ , as we shall do from now on.

IV.4. A necessary condition for surjectiveness

Haran’s condition comes after we chose $\nu\colon F\to G$ with $\ker \nu\cap M \leq \ker \nu_0$ (in applications one needs to take $G$ to be sufficiently large in a smart way). Assume there exists a normal subgroup $N$ of $F$ such that $N\leq M\cap \ker \nu$ , so $\nu$ factors via $F/N$ . Assume further that a wreath product embedding problem

$\xymatrix{ &F/N\ar[d]^{\nu}\\ \bar{A}\wr_{G_1} G\ar[r]^{\bar\alpha} &G }$ wreath product embedding problem for F/N

has no solution whenever $\bar A\neq 1$ is a quotient of $A$ on which $G_1$ acts. Then any solution of $\mathcal{E}$ induces a solution of $\mathcal{E}_1$ (and hence of $\latex \mathcal{E}_0$). What do we mean by $G_1$ acts on $\bar A$ ? We mean that $\bar A = A/A_1$ , for some $G_1$ invariant normal subgroup $A_1$ of $A$ . Equivalently, $A_1$ is normal in $A\rtimes G_1$ Why this condition gives that $\psi_1$ surjective? Let $A_1 = s \circ \psi(N) \leq A\rtimes G_1$ and let $\bar A = A/A_1$ . The technical condition $G_1$ acts on $\bar A$ is satisfied because $N$ is normal in $F$ . We have

$\psi(N) \leq \{ f\colon G\to A \mid f(1) \in A_1\},$

and conjugating this equation by all $\sigma \in G$ and taking intersection we get

$\psi(N) \leq \{ f\colon G\to A \mid f(\sigma) \in A_1\}.$

Thus $\psi(N)$ is contained in the kernel of the map $A\wr_{G_1} G \to \bar A\wr_{G_1} G$ , so $\psi$ induces a solution of the wreath product embedding problem with $\bar A$ that is trivial on $N$ . By the condition above this implies $\bar A = 1$ , so $\psi_1(N) =A$ , so $\psi_1(M) = A\rtimes G_1$ , as needed.

V. Applications of the Haran-Shapiro induction

Let us first prove that an open subgroup $M$ of a free profinite group $F$ of countable rank is free. (The countable condition is for simplicty.) Take $G$ so large such that $\ker (\nu\colon F\to G)$ contains $M$ . Let $N = \ker \nu$ , so $F/N\to G$ is an isomorphism. Thus the wreath product embedding problem for $F/N$ has no solution.

$\square$

The proof of the diamond theorem is more involved and contains some more ideas. Assume $M$ is contained in an $F$ -diamond, say composed of normal subgroups $N_1,N_2$ . The ket point is that, if $N$ is a normal subgroup of the form $N = N_1\cap N_2 \cap L$ , for an open normal subgroup $L$ of $F$ , then the quotient $F/N$ has some `abeian’ features w.r.t. the map $F/N\to G:=F/L$ . On the other hand, the group $A \wr_{G_1} G\to G$ is far from being abelian (if, e.g., $(G:G_1)$ is large$. Thus there could not be a solution of wreath product embedding problem for $F/N$ .

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Preparations:

Proof of Thornhill’s theorem.

Embedding probelms

Geometric embedding problems

Examples

Application to pseudo algebraically closed fields

Idea of the proof

A special case of Chebotarev’s theorem

Connection to irreducibility

Reducing to calculation of Galois groups

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

Prime values of polynomials

Analog over finite fields

A result over large finite fields

Pseudo algebraically closed fields

Back to finite fields

Introduction

Modular Forms and Galois Representations

Idea of proof

Idea of the vertical result.

The inverse Galois problem

Free absolute Galois groups

Embedding problems.

Semi-free profinite groups

Applications

Embedding problems for a field

Rational embedding problems

Splitness of rational embedding problems

Rational embedding problems induced by a polynomial

An application – a criterion for Hilbertianity

Further applications

Introduction and motivation

The inverse Galois problem and Hilbertian fields

Studying Galois theory of power series (or local complete fields)

Generalized Krull Domains and Hilbertianity

Proof of Paran’s Theorem

Concluding remark

I. Free profinite groups

II. Haran’s diamond theorem for free profinite groups

III. Haran’s diamond theorem for Hilbertian fields

IV. The Haran-Shapiro induction

IV.1. A freeness criterion of Iwasawa

IV.2. Brief description of the method

IV.3. The induced embedding problem

IV.4. A necessary condition for surjectiveness

V. Applications of the Haran-Shapiro induction

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