Hi,

Let K be a field. I am especially interested in finite fields, so
assume (if you wish) that K is finite.

Let G\subset SL_n be a linear algebraic group defined over K.

Let V\subset G be a proper (but not necessarily irreducible)
subvariety of G defined over the algebraic closure of K.

I would like to conclude that there is at least one point of G(K) not
on V(K),
provided that the number of elements of K is larger than a constant
depending only on n and on the degrees and number of irreducible
components of V.

Does anybody know how to prove this? It would be quite enough to give
a lower bound for the number of points m on G(K) of the form

m |K|^d,

where d is the dimension of G as an algebraic variety. Is this always
true?

A friend suggested using the unirationality of G (as proved in Borel's
Linear Algebraic Groups) but then one needs a bound on the unirational
parametrisation (not given in Borel). Would this be easy to prove? Is
there a better way? Is the result in the literature?
Harald

Harald Helfgott wrote:


Hi,

Let K be a field. I am especially interested in finite fields, so
assume (if you wish) that K is finite.

Let G\subset SL_n be a linear algebraic group defined over K.

Let V\subset G be a proper (but not necessarily irreducible)
subvariety of G defined over the algebraic closure of K.

I would like to conclude that there is at least one point of G(K) not
on V(K),
provided that the number of elements of K is larger than a constant
depending only on n and on the degrees and number of irreducible
components of V.

Two comments:

(i) I would guess that for most definitions of "subvariety"
we have V--G a monomorphism? Hence V--G is a proper monomorphism
(as V is proper and G is affine) and hence a closed immersion;
so V is both proper and affine and that makes it finite.
So V is just a bunch of points and if V(K) contains G(K) then
V had better have at least G(K) components? On the other hand,

(ii) If G is the trivial group then the size of K can be as large
as you like and n can be anything and V can be G, which only has
one component, and G(K) will equal V(K).

So I am wondering whether the OP doesn't quite mean what he
says, or perhaps I don't quite understand what he says.

Kevin

Harald Helfgott wrote:



Hi,

Let K be a field. I am especially interested in finite fields, so
assume (if you wish) that K is finite.

Let G\subset SL_n be a linear algebraic group defined over K.

Let V\subset G be a proper (but not necessarily irreducible)
subvariety of G defined over the algebraic closure of K.

I would like to conclude that there is at least one point of G(K) not
on V(K),
provided that the number of elements of K is larger than a constant
depending only on n and on the degrees and number of irreducible
components of V.

Does anybody know how to prove this? It would be quite enough to give
a lower bound for the number of points m on G(K) of the form

m |K|^d,

where d is the dimension of G as an algebraic variety. Is this always
true?

Note that V(K) does not make sense unless V is defined over K so I interpret
what you want as saying that there is a constant C_D such that for any
extension field L of K and any subvariety V of degree =D of G defined over
L, G(L)-V(L) is non-empty provided |L| = C_D.

You say that you are OK with restricting to finite fields but let me just
add that there are linear algebraic groups over non-perfect fields with
only the identity element as rational point so in that case it is not true.

For an absolutely irreducible variety of dimension e over a finite field K
we have that the number of points is |K|^e. This can be applied to the
component of the identity of G to conclude that the number of points of G
is indeed |K|^d.

I am a little bit concerned about your statement that it would be enough to
have such a lower bound. I guess you have in mind to bound the number of
points on an e-dimensional variety as O(|K|^e) with the constant depending
only on the degree. By using a finite linear projection one does indeed get
a bound D|K|^e where D is the degree. However, one may have to go to a
larger finite field in order to get such a projection and it is not clear
that one can bound the size of that field. It is however possible to prove
what one wants but the proof that I have is a little bit complicated.

Torsten


Torsten

As a follow-up to my own posting:

If G is not connected the result may not actually be true. The mistake in my
argument is that in the non-connected case a proper subvariety may also
have O(|K|^d) as it may have the dimension of G. Hence one should assume
``of lower dimension«« rather than proper which makes things OK even in the
non-connected case (and is equivalent to the original statement in the
connected case).

An example that connectedness is needed in the original formulation is given
by considering Z/3 and using the automorphism of it of order 2 to twist it
the constant algebraic group Spec F_p x Z/3 with the algebraic extension
F_p F_{p^2}. We can then consider the proper subvariety Spec F_p x {0}.
For any finite extension F_p L, there is only the rational point Spec L x
{0} of the group precisely when L is of odd degree over F_p. As there are
arbitrarily large such extension we have a counterexample.