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posicDate: 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_3and 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.
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
(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.
