**Письмо Богомолову о моей расширенной версии его гипотезы**

Date: Tue, 16 Sep 2003 00:55:49 +0200

From: Leonid Positselski

To: Bogomolov

cc: Voevodsky

Subject: a generalization of your conjecture

Dear Fedya,

I want to ask your opinion about a generalization of your

conjecture about freeness of commutator subgroups of Galois

groups of fields which I invented. Here is the formulation

of this conjecture and some supporting evidence.

Let F be an arbitrary field and l a prime number. Let K be

the field obtained by adjoining to F all roots of l-power

degree of all elements of F. That is, for each a\in F and

each N>0 we adjoin all roots of the equation x^{l^N}-a=0.

If F contains all the l-power roots of unity, then K will

be just the maximal abelian pro-l-extension of F.

Otherwise it is something different -- the maximal l-power

radical extension. Its Galois group Gal(K/F) will be an

extension of a subgroup of Z_l^* with an abelian kernel.

I conjecture that the Sylow pro-l-subgroup of K is a free

pro-l-group. This conjecture implies your conjecture

about the commutator subgroups of maximal pro-l-quotient

groups, but is stronger in two ways:

(1) it applies to arbitrary fields and not just fields

containing roots of unity or an algebraically closed

subfield; and

(2) it claims freeness of the l-Sylow subgroup, which is

a stronger property than freeness of the maximal quotient

pro-l-group.

Here is the supporting evidence that I have:

1. The conjecture is true in the equal characteristic

case, l = char F. Indeed, in this case K will be just

the perfect closure of F, so Galois groups of K and F

are the same. Of course, the Galois group of any field

of char=l has l-cohomological dimension 1.

(In addition, a perfect field of char=l has no l-torsion

in its Brauer group.)

2. The conjecture is true for number fields. Indeed, it

sufficed to adjoin to Q all the l-power roots of unity

(one doesn't even need to take roots of other numbers)

in order to obtain a field with free Sylow l-subgroup.

This follows from the computation of Brauer groups in

class field theory.

3. The conjecture is true for a Henselian discrete

valuation field F if and only if it is true for its

residue field f. In particular, if char F = 0 and

char f = l, the conjecture holds for the field F.

This follows from the computation of Brauer groups of

discrete valuation fields in Serre's book "Local

Fields". That's because the field K has a perfect

residue field and an l-divisible valuation group.

(There are some technical complications and my

understanding of the valuation theory is very limited,

which is why I was unable to generalize this to

non-discrete valuations with residue char = l. Any

valuations with residue characteristic not l are OK.)

4. Let K/F be the above-defined extension. Then any

pro-l-subgroup of the Galois group G_K with no more

than 3 generators is a free pro-l-group. In particular,

G_K contains no abelian pro-l-subgroups of rank more

than 1. Moreover, let L be any algebraic extension

of K; then the maximal pro-l-quotient group of G_L has

the same property: any its subgroup with no more than

3 generators is free.

This follows from some results of Jochen Koenigsmann and

Ido Efrat about pro-l Galois groups with small numbers

of generators. They classified pro-l Galois groups with

no more than 4 generators; and for 2 or 3 generators

they prove that either such group is free, or the field

admits a valuation with non-l-divisible value group

(assuming the square root of -1 in the field if l=2).

But it is clear that any valuation on any algebraic

extension of our field K has an l-divisible value group,

and K contains the l-power roots of unity.

(For 4 generators, there is a Demuskin group defined by

the relation (x,y)(u,v)=1 in their list, about which

it is not known whether it can be the pro-l Galois

group of a field at all, nor whether such a field

would have a nontrivial valuation, etc.)

5. Haran-Jarden-Koenigsmann proved that the free product

of absolute Galois groups is an absolute Galois group:

for any fields F_1 and F_2 there is a field F such that

the Galois group G_F is freely generated by two subgroups

isomorphic to G_{F_1} and G_{F_2}. Unlike your original

conjecture, my generalized conjecture's statement seems to

depend not just on the Galois group G_F, but on the field

F itself -- however, it can be reformulated in terms of

the group G_F and its cyclotomic character. Using

a pro-finite version of Kurosh's theorem about subgroups

of free products, one can see that the conjecture holds

for the field F of the Haran-Jarden-Koenigsmann

construction if it holds for F_1 and F_2.

The free product is the only kind of binary operation

on pro-finite groups that which preserves the class of

absolute Galois groups (as far as I know).

6. Finally, there is your argument with abelian groups

of the form Z/l+...+Z/l, which can be extended to

the new situation as follows. The generalized conjecture

says that for any algebraic extension E of F the image

of the group H^2(G_E, Z/l) in H^2(G_KE, Z/l) vanishes

(where KE is the composite of K and E over F). Your

argument proves this when E/F is a Galois extension

whose Galois group is the direct sum of several copies

of Z/2, or the sum of several copies of Z/3, or, more

generally, an abelian group killed by multiplication

with a number q such that the field K has no nontrivial

finite extensions of degree less or equal to q/2. Here

it does not matter whether q is divisible by l or not.

The point is that it suffices to have the Bass-Tate

property for Milnor K-groups with rational

coefficients for E/F: if K_2(E)\otimes Q is generated

by products of elements of K_1(F) and K_1(E), then

K_2(E)\otimes Q/Z_(l) dies in K_2(KE)\otimes Q/Z_(l).

Now K_2(E)\otimes Q is a representation of the finite

group Gal(E/F) over a field of characteristic 0, so

it is the direct sum of irreducibles. Moreover, any

element invariant under the action of a subgroup of

Gal(E/F) comes from K_2\otimes Q of the corresponding

subfield. If Gal(E/F) is abelian, the irreducibles

are actually representations of cyclic quotient groups

of Gal(E/F), so their elements come from cyclic

extensions of F of degree no more than q. Now the

Bass-Tate lemma holds (even with integral coefficients)

for any extension of degree q of any field F having

no extensions of degree less or equal to q/2.

This argument can be extended to very small nonabelian

Galois groups like S_3 ~~and A_4~~, but not much farther

(without imposing conditions on the field F).

This is about all the supporting evidence I have.

So what do you think about all of this?

Best wishes,

Lenya.