Counting torsion points on subvarieties of the algebraic torus
Abstract:
We estimate the growth rate of the function which counts the number of torsion points of order at most on an algebraic subvariety of the algebraic torus over some algebraically closed field. We prove a general upper bound which is sharp, and characterize the subvarieties for which the growth rate is maximal. For all other subvarieties there is a better bound which is power saving compared to the general one. Our result includes asymptotic formulas in characteristic zero where we use Laurent’s Theorem, the Manin-Mumford Conjecture. However, we also obtain new upper bounds for the algebraic closure of a finite field.
1 Introduction
Main results.
Let be an integer and be an algebraically closed field. We denote by the -dimensional algebraic torus with base field . We will identify with , the group of its -points. We denote by the subgroup of points of finite order therein, which we call torsion points. A linear torus is an irreducible algebraic subgroup and if is a torsion point and is a linear torus, we call a torsion coset. We often identify a subvariety of with the set of its -points.
Lukas Fink considered the following question in his master thesis [Fin08]. Let and . What can we say about the asymptotic growth of as ? Considering the first coordinate yields . He shows in Proposition 2.7 that . Fink’s result is therefore better than the trivial bound. Based on computations he conjectures that . Our first result, Theorem 1.1 below, includes a power saving over the trivial bound .
We consider the average rather than the number of torsion points of order exactly since the latter behaves erratic. In Fink’s example the set of such that there is no torsion point of order has natural density one. The number of points on Fink’s curve whose order divides is also bounded by a constant times if (4) in [Moh20] holds for the set of points such that . On the other hand, this number is for all .
An algebraic subset is the zero set of finitely many Laurent polynomials in . The natural generalisation of the question of Fink is as follows. Take an algebraic set , and denote by the set of torsion points in of order bounded by . How fast does grow?
In characteristic zero, the Manin-Mumford Conjecture for , a theorem of Laurent [Lau84] reduces this problem to the case where is a torsion coset as we will see below in Corollary 1.5. Altough many of our results are true in any characteristic, we highlight the case where is an algebraic closure of a finite field. In this case all points of have finite order. To illustrate the kind of results we get and compare them to the work of Fink, we look at the curve defined by :
Theorem 1.1.
Let be a prime and . Then we have for all
To state the main result, we have to introduce more terminology. An algebraic subset is called admissible if it is finite or does not contain a top dimensional torsion coset. The stabilizer of an algebraic subset is the set of such that . It is well known that the stabilizer is an algebraic subgroup of . The main result is
Theorem 1.2.
Let be an algebraically closed field and be irreducible, admissible and of dimension . Let be the dimension of the stabilizer . Then we have for all .
As in the example there is an analogue of the trivial bound given by
Proposition 1.3.
Let be an algebraically closed field and an algebraic subset of dimension . Then for all .
So again we have a power saving with respect to the general bound.
If is a torsion coset, we can give an asymptotic formula whose main term depends only on the characteristic of the field, the dimension and the order . In the theorem below denotes the Riemann zeta function.
Theorem 1.4.
Let be an algebraically closed field of characteristic and let be a torsion coset of dimension and order . Then if we have
and if we have
Theorem 1.4 implies that the exponent in Proposition 1.3 is optimal. We also see that torsion cosets are those subvarieties which in some sense contain the most torsion points if we compare among varieties of the same dimension.
Recall that in characteristic zero the Manin-Mumford Conjecture for reduces the general case to torsion cosets and again we are able to give an asymptotic formula. Let be an algebraic set. There exist torsion cosets such that is the decomposition into irreducible components. Set
Then we have
Corollary 1.5.
Let be an algebraic field of characteristic zero, let be algebraic, and . Then we have if and
What about lower bounds for ? Fink proved in Proposition 2.7 in [Fin08] that for all . Using the famous theorem of Lang-Weil, we can prove
Theorem 1.6.
Let be a prime and . Let be algebraic of dimension . Then there exist and such that for all .
Hence if is admissible, the correct exponent, if it exists, is in , where . In the example of Fink, numerical analysis suggests that the smallest exponent is possible.
Overview of the proofs.
Let us give two examples. First we look at the curve and prove Theorem 1.1. Let and assume that is of order . By Minkowski’s first theorem there exists such that and . Since we have . After multiplying by suitable powers of and we see that is a root of a nonzero polynomial of degree at most . Hence the number of such that is at most . Thus we have the bound . For a fixed there are vectors such that . Thus we find
Now let us consider . Note that Stab is one dimensional. Hence the exponent in Theorem 1.2 equals . Then we observe that any is contained in the torsion coset The order of such a torsion coset is given by . Thus any is contained in a torsion coset for some . Therefore we can bound . Denote by the order of . We claim that . If is of order , the order of is the least common multiple of and . Thus if and only if . The number of elements of order in is given by Euler’s totient except if the characteristic divides . Since in positive characteristic the order is always coprime to , there are no elements of order , whenever divides . Thus in any case there are at most elements of order in . We therefore have
If and is as in the inner sum, we can write for some . And since we see that for all . It is also well known, that for all . Therefore we get
Let denote the number of elements of order in . Then we can bound and we know that for every . To get sums of this kind, we write for all . Thus we find
using that and hence .
Now let us consider a hypersurface . We divide into two parts: the union of all torsion cosets contained in of positive dimension and its complement . To bound we generalise the proof for : Instead of Minkowski’s first theorem we can use the second theorem to find short linearly independent vectors such that . The such that for all are given by torsion cosets for some and finitely many ; we refer to Section 2 for our notation. Let be such that is the vanishing locus of in . Then is in if and only if . If is not the zero polynomial, the number of such is bounded in terms of the degree of and . This is always the case if . Here we use the fact that is a hypersurface. Summing over all possible -tuples gives us a bound .
For the union of positive dimensional torsion cosets we show that if we consider the maximal cosets, then the number of corresponding algebraic groups is finite and every torsion coset in is contained in one of them. However in positive characteristic it is possible that for a fixed linear torus there are infinitely many torsion cosets which are contained in . So let us fix . To control the number of cosets which are contained in , we map to for some matrix such that for all . Note that . We can find equations for the image and use induction on . Thus we have a bound for the number of cosets of contained in . Together with a bound for cosets we get the claimed result for hypersurfaces.
The generalisation to arbitrary algebraic sets follows by projecting to some coordinates and doing algebraic geometry.
For Theorem 1.4 we can assume that for some of order , say. Since the order of is , we have to count the whose order is bounded and belongs to a fixed congruence class modulo . With the Möbius inversion formula one can see that the number of elements of order is given by Jordan’s totient , where is the Möbius function and . Once we have the asymptotics of over arithmetic progressions, Theorem 1.4 is not hard to deduce. In positive characteristic the orders are always coprime to . If is coprime to , we still have elements of order . Thus the case follows from the case .
Organisation of the article
The algebraic subgroups of play an important role in the results as well as in the proofs. Therefore we need a characterisation of them. This is already known in characteristic zero (Theorem 3.2.19 in [BG06]). In positive characteristic it is possibly known. But we could not pin down a suitable reference and have therefore added an appendix.
The next goal is to estimate . The most technical part is to bound the degree of . Some known preparations are done in the second appendix and used to prove the bound in section 4.
Acknowledgements
I want to thank my advisor Philipp Habegger for good discussions and advice, and Pierre le Boudec for his help concerning the sum of Jordan’s totient in arithmetic progressions. This will be part of my PhD thesis. I have received funding from the Swiss National Science Foundation grant number 200020_184623.
2 Notation
We write and denotes always an algebraically closed field of any characteristic. Let and the ring of Laurent polynomials in variables. Let be the units of . Then for and we define . This definition is compatible with matrix multiplication in the sense that for all and . The letter denotes the element and hence we can write a general Laurent polynomial as . The support of such a polynomial is the finite set of such that and we denote it by .
For we set which together with the coordinate-wise multiplication is a group. We write for the elements of finite order in and call them torsion points. If it is clear what is, we usually omit in the notation. Elements are called roots of unity. For we write or for the roots of unity in such that .
We denote by the identity matrix in the ring of -matrices . We write and for matrices whose entries are all zero or one respectively. We write for the -th vector of the standard basis of . The will be clear from the context.
We denote by the affine space endowed with the Zariski topology.
For we denote by the maximum of the absolute vaues of its coordinates, where on the right hand side is the usual absolute value. We use the Vinogradov notation and for some set of variables.
3 Jordan’s totient function and torsion cosets
In characteristic zero the -th Jordan’s totient function counts the number of points of order in . Thus it is a natural generalization of Euler’s totient function to higher dimensions. In this section we prove an asymptotic formula for the sum of Jordan’s totient function in arithmetic progressions and deduce an asymptotic formula for , where is a torsion coset.
Definition 3.1.
Let be the set of arithmetic functions. We say that is completely multiplicative if for all . The Dirichlet convolution of is defined as We can associate to the formal Dirichlet series , where is an indeterminate.
Definition 3.2.
Define by and for all and by for all . The Möbius function is nonzero only on squarefree numbers and if is a product of distinct primes we have .
For we define as and Jordan’s totient function given by , a generalisation of Euler’s totient function .
Any character gives rise to defined by if and zero otherwise. We will sometimes abuse notation and write for .
The Riemann zeta function is the series corresponding to and the Dirichlet -functions are defined by .
Remark 3.3.
Together with the pointwise addition the Dirichlet convolution endows with the structure of a commutative unitary ring with multiplicative identity . If is completely multiplicative we have for all . Here denotes the pointwise multiplication. For all we have
as formal series. In particular if all three series converge absolutely for some the equality above holds in .
Remark 3.4.
Möbius inversion is equivalent to . The functions and are completely multiplicative for all and characters respectively. The explicit formula for Jordan’s totient function reads where the product is taken over all primes dividing and as usual the empty product equals one. The Riemann -function as well as the Dirichlet -functions converge absolutely if the real part of is strictly larger than .
Lemma 3.5.
Let and . Then
Proof.
Let such that . Then every which is congruent to modulo is of the form for some . Thus we have
By exercise 6.9 in [McC86] we know that
So if then and hence
Lemma 3.6.
Let . Then
Proof.
Let be the principal character modulo , and recall that . Since is completely multiplicative we have . Therefore the infinite sum satisfies . We have and hence . For the error term we find
Theorem 3.7.
Let , and . Then
Proof.
We claim that for natural numbers the following statements are equivalent: and and and , where and . So assume that there are such that and . Since we find and is equivalent to . Since divides both and , it also has to divide and hence also . Dividing through we find . Since there is such that . If we multiply the congruence by we get the congruence of . For the other direction we multiply the congruence by and find for some integer . Multiplying this equation by we find that .
Recall that . Hence we find
where and are defined as in the claim. By Lemma 3.5 the inner sum equals . Since is a divisor of , we can find an absolute constant which holds for all and therefore the sum of the error terms is where we need the log term only if since the sum is if and if . Therefore we find
where if and if .
Observe that for all we have . This allows us to write
since if . By Lemma 3.6 the sum is Hence
Finally we plug this into the equation above and find
since . The error term is if and if . ∎
Lemma 3.8.
Let . Then .
Proof.
Fix a divisor of . If satisfies , then for some coprime to and hence the number of such equals . Thus we can write the sum in question as , which is nothing else than . Thus we want to prove . Since is completely multiplicative we have
Lemma 3.9.
Let be an algebraically closed field of characteristic and . Then the number of such that is if or and is if and .
Proof.
Note that for all we have if and only if . Let be the number of elements in of order . The number of solutions of equals . If we have and if and we have and thus if and only if . Since and its formal derivative have no common roots, the equation has distinct roots. Therefore . Let . Then counting solutions in two ways yields So by Möbius inversion we have .
If we have and therefore and we are left to prove the lemma in positive characteristic.
So let us assume and write where . Since and are coprime we can factor any divisor of uniquely in a product of coprime divisors of and respectively. Thus we find
using that . If then and hence since . And if we have and hence .∎
Definition 3.10.
Let and an algebraic subgroup. Then we define
Proof of Theorem 1.4.
Since a monoidal transform preserves orders, we can assume that the coset is of the form for some of order by Lemma A.17. Then the order of equals for all . Fix and assume that and is of order bounded by . Thus and hence .
We first assume that . Then by Lemma 3.9 the number of elements of order in is given by . We use Theorem 3.7 and Lemma 3.8 to find
where if and if .
For the case let us start with some remarks. Since is the order of it is coprime to . Thus there exists an integer such that . Then for natural numbers we have and if and only if . Let us also compute since . By Lemma 3.9 there are also torsion points of order if . Otherwise there are none. Thus we have to subtract
from the term in characteristic above. Here is as above and we used again Theorem 3.7 for the second and Lemma 3.8 for the last equation. Since the claimed result follows.∎
Remark 3.11.
Let . Then we have
In particular if we compare the contribution in congruence classes mod , the proportions are and hence the proportion of all the coprime congruence classes is , which coincides with the factor found in Theorem 1.4.
4 Outside torsion cosets
We count the torsion points which are not contained in a positive dimensional torsion coset. We use Minkowski’s second theorem to construct a one dimensional subgroup which is defined by equations with small coefficients. In the first part we estimate the number of torsion points lying in such a subgroup, and then sum over all such groups. As usual always denotes an algebraically closed field.
Definition 4.1.
Let and . Then we call a factorisation Smith normal form of if and for some such that for all .
Remark 4.2.
That every matrix has a Smith normal form is a classical result established in [Smi61].
Definition 4.3.
Let be a Laurent polynomial. There exists a unique polynomial which is coprime to and such that . We define the Laurent degree as .
Remark 4.4.
Since the zero locus of and above in are equal, a Laurent polynomial of degree has at most roots.
Lemma 4.5.
Let , and . Suppose that . Then the Laurent degree of is bounded by .
Proof.
Let , and . Then we have . By definition there exists and of degree such that . Let and suppose that . Then there exists such that and hence such that . We can bound . Thus there exist bounded by such that . Then we find .∎
Definition 4.6.
We call a set algebraic over if there exists polynomials such that
Let be an algebraic subset of dimension . We call it admissible if it is finite or and it does not contain a torsion coset of dimension . We define
For we let be the zero locus of in
Lemma 4.7.
Let and be linearly independent and and . Then
Proof.
A suitable version of Bézouts Theorem would suffice for our purposes. However we follow a more elementary and self-contained approach. Let . Let be a Smith normal form of with and note that . Let such that . We claim that
Note that is in the intersection on the left if and only if and since has an inverse in this is equivalent to . Let . Then if and only if for . There is no condition on . Therefore any such that is of the form where and Note that is primitive since it is a row of . We also find
Since the kernel of is one dimensional this shows that primitivity and being in the kernel of determines up to sign.
Let and Since , the vector is primitive. Then for any we have
and hence by the Leibniz formula. Lemmas B.5 and B.7 imply that .
Now fix . If , then for all the element is contained in a torsion coset of positive dimension and hence .
In the other case an element is of the form for some root of the Laurent polynomial . By Lemma 4.5 we have . Since we have
Summing over all - which are at most distinct elements - we get the claimed bound.∎
Lemma 4.8.
Let and . Then for all we have
Proof.
Comparing the sum with the integral yields that as for all real . The proof is by induction on . If we have
So assume that and the lemma holds for all . Let . Then
by induction since .
We claim that the exponent . This is equivalent to and since also to which is true since the are in . Thus we can bound
So the exponent in the bound equals
Definition 4.9.
Let and a subgroup of rank . Then the successive minima of with respect to are defined as
where the norm of a vector is given by .
Lemma 4.10.
Let and be a subgroup of rank . Then .
Proof.
This follows from Minkowski’s second theorem (Theorem V on page 218 in [Cas97]) and the fact that the volume of equals . Furter the determinant of is given by the absolute value of the determinant of a matrix whose columns form a basis of . Since the determinant is given by .∎
Definition 4.11.
For we define .
Lemma 4.12.
Let be of order . Then is a subgroup of of rank and .
Proof.
Since contains , it has full rank. Consider the group homomorphism , given by . Then and . By the isomorphism theorem we find The last equality holds clearly in characteristic zero. It is also true in positive characteristic since the order of an element in is always coprime to .∎
Definition 4.13.
For a set and we define
Definition 4.14.
A hypersurface is the zero set of in .
Remark 4.15.
Let Then the dimension of equals .
Lemma 4.16.
Let be a hypersurface in . Then
Proof.
Let be of order . Since there are only finitely many points in of finite norm, the infimum in the definition of is a minimum. Thus there exists a system of linear independent vectors such that for all . Then by Lemmas 4.10 and 4.12 we have . Since we have for all which yields . Therefore
The number of vectors satisfying is bounded by a constant depending on times . By Lemma 4.7 have
in the first two sums the satisfy , in the last step we sum over such that . The last sum can be bounded by Lemma 4.8 with and . We have and thus the exponent equals .∎
5 Common roots and admissibility
This section contains technical statements on algebraic subgroups of . As we work in arbitrary characteristic and as the literature (cf. [BG06]) concerns mainly characteristic zero we give full proofs.
Lemma 5.1.
Let , of rank and be a torsion point. Then the set of such that is either empty or a finite union of torsion cosets of dimension . Moreover the union is not empty if .
Proof.
Suppose that there exists satisfies . Let . Then if and only if and we can conclude with Lemma A.4. For the second statement let be a Smith normal form of , where . Let and . Let . Then . Since is algebraically closed we can find such that . Then satisfies ∎
Definition 5.2.
We call primitive if its entries are coprime.
Lemma 5.3.
Let and . Then is irreducible if and only if is primitive.
Proof.
Note that for all the map is an automorphism of . By the theorem of elementary divisors (Theorem III.7.8 in [Lan02]) there exists and a basis of such that . Let . Then is irreducible if and only if is irreducible too. It is not hard to show that this is the case if and only if .∎
Lemma 5.4.
Let each be a product of factors of the form , where is a root of unity and is primitive. Assume that and are coprime. Then the set of their common roots is a finite union of torsion cosets of dimension at most .
Proof.
If there are empty products there are no common roots, therefore we can assume that there are no which equal . Suppose that is a common root. Then for each there exist a primitive and a root of unity such that and . Let and . Then . Let be a smith normal form of . The rank of is the largest index such that . Since the are primitive we have and therefore there exists such a . Assume that . Then we find
Since the are primitive, we have . Thus . So for all there is some root of unity such that and , since . Since the are coprime, not all are the same and hence there is no common root. So we can assume that . Then the set of such that is a finite union of torsion cosets of dimension by Lemma 5.1. Since is from a finite set, this shows the claim.∎
Lemma 5.5.
Let and . Assume that . Then there exist and such that
Proof.
Let be the Laurent polynomial such that . Let and be the coefficients of and . Then we have for all . Fix any and for all let . Let and . Then and therefore . Similarly let and . Then and therefore . Then . Since the scalar is a natural number.∎
Lemma 5.6.
Let be a subgroup of codimension , a basis of and . For the map is a character on . For a character we put . For those such that we fix . For all we have . Thus there is a unique such that . Finally we define .
Then we have and if is such that , then for all characters such that .
Proof.
The argument has similarities with the proof of Proposition 3.2.14 in [BG06]. First we show that the are well defined. Let . Since we have for all and hence .
Let us abbreviate . Then we have
Suppose that is such that . Let . Then we have
By Artin’s Lemma (Theorem 4.1 in chapter VI in [Lan02]) we must have for all .
We compute
∎
Lemma 5.7.
Let , and . Then the following are equivalent
-
1.
is admissible.
-
2.
There are no primitive and root of unity such that .
-
3.
There are no and root of unity such that .
Proof.
If , is not admissible and every divides . And if we have which is admissible. Since is never a unit if and , the second and the third statement are also true. Thus we can assume that and hence the dimension of is The implication follows directly from Lemma 5.1 by contraposition.
We prove the contraposition and assume that there exists such that . Write for a primitive and a . Since and we can write for some we find and hence there exists a primitive and a root of unity such that
We show the contraposition and assume that there is an -dimensional algebraic subgroup and a torsion point such that . By Theorem A.12 the rank of equals one. Thus there exists which is a basis of . We can apply Lemma 5.6 and find that is a root of and hence for all characters such that . So let be such that . Then again by Lemma 5.6 we find
Therefore , where and is a torsion point.∎
6 Bound for cosets
Although we already esablished an asymptotic formula for if is a torsion coset, we prove an upper bound which is nice to work with, since it is very simple and there is no error term.
Definition 6.1.
For an algebraic subgroup we denote by the linear torus equal to the connected component of which contains the identity.
Lemma 6.2.
Let and let be an algebraic subgroup of dimension . Then we have .
Proof.
Let us first assume that is a linear torus. In this case we can assume that for some of order by Lemma A.17. Thus the order of is for any . Hence if and only if ord. By Lemma 3.9 the number of elements of order in is if or and zero otherwise. In particular is always an upper bound. Hence we find
With the formula one can see that for all . And since , we have for all by the Möbius inversion formula. If and is as in the inner sum, then we can write for some . Thus we find
For the general case we decompose into a disjoint union of torsion cosets and observe that since . Therefore
7 Hypersurfaces
In this section we prove the main theorem for hypersurfaces. It remains to understand the points which are contained in a positive dimensional torsion coset. Since any such torsion coset is contained in a maximal one, we have to study them. Let be a maximal coset. There are only finitely many possibilities for . But it may happen that there are infinitely many torsion cosets which are contained in the hypersurface. We construct a map from the torsion cosets of to torsion points of a variety of lower dimension and get a bound by induction.
Lemma 7.1.
Let and an admissible hypersurface. Let be a linear torus of dimension . Let and such that . Then there exist an admissible hypersurface and finitely many torsion cosets of dimension at most with the property that for every the map is an injection from the torsion cosets of order bounded by to .
Proof.
Without loss of generality we can assume that there exist torsion cosets of which are contained in . Thus by admissibility we have and hence . Let such that . Let be defined as in Lemma 5.6 for all such that . For any such let be the product of all factors of of the form for some primitive and of finite order, which are irreducible by Lemma 5.3, counted with multiplicity. Let such that
Define and let . Since we can apply Lemma 5.7 and see that is admissible by construction of .
We claim that the are coprime. Assume by contradiction that they have a common factor , where and is a root of unity and write . Then we find by Lemma 5.6 that
Thus divides . We now show that . Let be such that . Then . Since and there exist and a natural number such that by Lemma 5.5. We have since . Therefore and which contradicts the fact that is admissible by Lemma 5.7. So we showed that the are coprime. We can therefore apply Lemma 5.4 and see that the common zeros in is a finite union of torsion cosets of dimension at most .
Let and be such that and We claim that is well defined and injective. This is a well defined map, because for we have since the columns of are in . To prove injectivity assume that . Let and . Then there exists such that . Therefore which shows that by Theorem A.12 and hence . Furthermore the order of the image is bounded by the order of the coset.
By Lemma 5.6 we have . If there is such that the element is a root of of order bounded by and hence in . If not, it is a common root of the and hence contained in the finite union of the torsion cosets of dimension at most .∎
Lemma 7.2.
Let be a sequence, a real number and suppose that there is such that for all . Then there is such that for all .
Proof.
Note that for all we have . Comparing the sum with the integral and using we can write
Lemma 7.3.
Let an admissible hypersurface in and . Assume that we know that for all admissible hypersurfaces in , where . Let be a linear torus of dimension . Then
Proof.
Let us denote by the quantity on the left. Note that implies that . Thus we have only to consider cosets of order bounded by . Let be the number of torsion cosets such that and Then we can use Lemma 6.2 to find the bound . Let and and be the admissible hypersurface and the torsion cosets from Lemma 7.1. By hypothesis there exists a constant such that and by Lemma 6.2 we have for all . So by Lemma 7.1 for all we have
Since we have and therefore by Lemma 7.2. So we get ∎
Lemma 7.4.
Let and a hypersurface in . Then If we even have .
Proof.
The proof is by induction on . If the set is finite and hence the lemma is true in this case. So assume that and the lemma is true in dimension smaller than . Write . Let be a nonzero Laurent polynomial which exists since . To estimate we divide into two sets: and . First we bound . Fix and assume that . Then there are at most solutions satisfying . We can bound by Lemma 6.2. If , then we have and the claimed bound holds.
Otherwise we also have to bound and is a hypersurface. Let . Then
The order of a coset equals . Let be a constant such that for all which exists by induction. Let be the number of points of order in . Thus by Lemma 6.2 we have to bound the sum
By induction we know that if and if . So if we find . If we have , so we can apply Lemma 7.2 and find . Thus we have and hence .∎
Theorem 7.5.
Let , be an admissible hypersurface in . Then
Proof.
Since any torsion coset in is contained in a maximal one, we find
Let which is finite by Corollary A.8. Then we have
and since is finite it is enough to show the bound for each summand separately. The proof is by induction on . If we have and the theorem follows from Lemma 4.16. Let and be of dimension . We have by admissibility. Thus the hypothesis of Lemma 7.3 is satisfied by induction. We can apply it and find . Thus we have .∎
8 Generalisation
We deduce the general theorem from the case where the subvariety is a hypersurface using algebraic geometry.
Lemma 8.1.
Let , be an algebraic set of dimension and the projection to the first coordinates. Let . Then there exists such that for all we have or . The set is an algebraic set of dimension at most . The Zariski closure of the image is of dimension at most .
Proof.
Let . The number of irreducible components in the fibres is uniformly bounded and hence there exists such that all finite fibres, contain at most points. Note that the fibres of positive dimension are of the form and therefore for some . By Theorem 14.112 in [GW20] we have that and hence also is closed. Thus is closed. Since we have . That the dimension of the closure of a morphism is at most the dimension of the domain is well known.∎
Proof of Proposition 1.3.
We prove . The first inequality follows immediately. It suffices to show the second one.
Let be the irreducible components of . Then and for all . Therefore we can assume that is irreducible.
If then is finite and the claim is immediate. If and , then and the sum is bounded by . So we may assume and
We continue by induction on , the case is already done.
Let denote the projection to the first coordinates. The set
is Zariski closed and of dimension at most by Lemma 8.1. Let and observe that the sum in question is at most , where
Then
By induction applied to we have .
If there are at most possibilities for such that by Lemma 8.1. We apply the current lemma by induction to the Zariski closure which has dimension at most . So .
∎
Definition 8.2.
Let be a topological space. Then we denote by the set of irreducible components of .
Remark 8.3.
If an algebraic subset is irreducible and of dimension , it is a hypersurface. This can be deduced from the same fact for subvarieties of , which is stated in Theorem 1.21 in [Sha13].
Theorem 8.4.
Let be an algebraically closed field and irreducible and admissible of dimension . Then we have for all .
Proof.
As the conclusion of the theorem is invariant under permuting coordinates
we may freely do so. As itself is not admissible we may assume . If , the set is finite and hence we have . So we can assume that .
Next we fix a good projection. To this end we consider the coordinate functions
restricted to . They are denoted and members of the
function field of . By dimension theory, the function field has
transcendence degree over . After permuting coordinates we may assume that
are algebraically independent.
We claim that there exists such that are multiplicatively independent. Let us assume the contrary and deduce a contradiction. For each such there exist not all zero such that
(1) |
identically on . By algebraic independence we must have . So the vectors for are linearly independent. They generate a subgroup of rank . The algebraic subgroup is of dimension . By (1) we have . But since is irreducible and of the same dimension as , it is an irreducible component of and hence a torsion coset. This contradicts the admissibility of .
After permuting coordinates we may assume . Let denote the projection onto the first coordinates. Then the Zariski closure of in is irreducible and . For all we have . Thus must be algebraically inedependent elements of the function field of and hence we have . Next we show that is admissible. Assume that it is not. Then there exists a torsion coset of dimension . Since both and are irreducible we have . Then there exists such that for all . Let be the order of . Then since for all we find identically on , a contradiction. Hence is admissible. By Remark 8.3 we have that is a hypersurface.
Let
It is an algebraic subset of by Theorem 14.112 in [GW20]. The set is algebraic and so is . By Corollary 14.116 in [BG06] there is a nonempty open of such that for all . Since is dense in , its intersection with is nonempty and therefore is a proper subset of . In particular we find .
Let . Then where
Applying Proposition 1.3 we find for all . Let be a uniform bound for the number of irreducible components in a fibre. For any we have and is an irreducible component of . Thus we find
Since is an admissible hypersurface in we can apply Theorem 7.5 to bound for all . The theorem follows from .∎
Lemma 8.5.
Let be a surjective morphism of algebraic groups. Let be a subgroup and . Then .
Proof.
Let and which maps to . Since is surjective, is surjective too. We have and since the lemma follows. ∎
Proof of Theorem 1.2.
If , and are both finite. The inequality is satisfied, since the exponent equals . Thus we can assume that . If let and be the connected component of the identity of . The coset is irreducible and of dimension . Thus we find . Since is admissible is not a torsion point and hence . Thus we can assume that .
Let be a basis of . Let given by . Note that is surjective since . Let be the Zariski closure of . Then is irreducible and hence either admissible or a torsion coset. Since the fibres of are cosets of we have . Assume by contradiction that is a torsion coset. Then there exists a torsion point of order and a linear torus such that . Then is an algebraic subgroup of of dimension . By Lemma 8.5 we have . Since and we see that is an irreducible component of which are torsion cosets by Lemma A.4. Since this contradicts admissibility. Thus is admissible.
Let be the number of cosets of of order contained in . We have
using Lemma 6.2. If is a coset of order , it is the preimage of which is of order at most . Therefore induces an injection from the cosets of which are contained in and of order bounded by into . Thus we find for all by Theorem 8.4. Since the exponent is strictly larger than and using Lemma 7.2 we find
Corollary 8.6.
Let be an algebraically closed field and admissible of dimension . Let . Then .
Proof.
Note that and apply Theorem 1.2 to each summand.∎
Remark 8.7.
The number in Corollary 8.6 is also equal to
9 Characteristic 0
In characteristic zero we can say much more than in general: The Manin Mumford Conjecture allows us to reduce the general case to the case of torsion cosets. Thus we can give an asymptotic formula for any subvariety. In this section denotes a field of characteristic .
The following theorem is sometimes called the Manin-Mumford- conjecture. In the current setting it was proved by Laurent.
Theorem 9.1.
Let be an algebraic set. Then there are finitely many torsion cosets such that .
Proof.
This is Théorème 2 in [Lau84].∎
Proof of Corollary 1.5.
By Theorem 9.1 there exist finitely many torsion cosets in such that . Intersecting both sides with we get . If , there are at most finitely many torsion points in and thus we have in this case. If the corollary follows directly from Theorem 1.4 since the intersection of the torsion cosets is a finite union of torsion cosets of strictly lower dimension.∎
10 Lower bounds in
We use the Lang-Weil estimates to establish lower bounds if .
Theorem 10.1.
Let be a geometrically irreducible variety defined over a finite field of positive characteristic with elements. Assume that is of dimension . Then there exists such that .
Proof.
This is Theorem 1 in [LW54].∎
Proof of Theorem 1.6.
Considering an irreducible component of of dimension we can assume that is irreducible. Let be the Zariski closure of in . Then is irreducible and of dimension . Since is irreducible, the proper and closed subset is of dimension at most and we have .
Let be such that is defined over . Let be a multiple of . Then is a subfield of and hence and are defined over too. By Theorem 10.1 there exist constants independent of such that and . Let be large enough and a multiple of such that . Note that for all we have since for all nonzero we find . Therefore we get
Thus if and hence is large enough, we have .∎
Appendix A Algebraic subgroups and subgroups of
We verify that many results of chapter 3 in [BG06] remain valid over arbitrary algebraically closed fields. These facts may be well-known, but we are not aware of a reference. In this section denotes an algebraically closed field.
Definition A.1.
For an algebraic subgroup we define
and for a subgroup we set
Definition A.2.
Let be a subgroup. The saturation of is defined as
We call primitive if .
Definition A.3.
Let be a prime or and a subgroup. We say that is -full if or if is prime and for all we have that implies that .
Lemma A.4.
Let and be a -full subgroup of of rank . Then is an algebraic subgroup of of dimension , which is the union of torsion cosets of dimension . We also have .
Proof.
By the theorem of elementary divisors (Theorem III.7.8 in [Lan02]) there exists a basis of and such that is a basis of . Let . Then the monoidal transform gives an isomorphism between and where since if and only if and thus if and only if satisfies . Then we see that is a union of linear tori of dimension .
We claim that . Since the inclusion holds. For the other let . There exists such that Hence we can write and the comparison of coefficients yields . Hence we have and the claim is true.
Consider the surjective group homomorphism given by and compose it with the reduction . Let . Then the kernel of the composition is . Thus we find . Since is -full the are coprime to , if and therefore we have in all cases. Thus we find .
Let . In a first step we reduce the last statement to . We have seen above that is an isomorphism between and . Therefore we have
So if we assume that we find .
In the second step we prove and let . Then we have for all and hence .
For the other inclusion let . Let . Since is coprime to if there always exists a root of unity of order . We have where is the -th coordinate. Hence which shows that . Now let . Then for all we have where is the -th coordinate and hence . This implies that , and thus and the other inclusion holds too. ∎
Definition A.5.
A torus coset is a coset of a linear torus.
Lemma A.6.
Let , be an algebraic set defined by Laurent polynomials and let be the set of exponents appearing in the monomials in . Let be a maximal coset in . Then , where is a subgroup generated by vectors of type with for .
Similarly suppose that a linear torus has a torus coset which is maximal among torus cosets contained in . Then , where is a subgroup generated by vectors of type with for .
Proof.
Suppose that is maximal. Let us define . By Lemma 5.6 we have
for every and such that .
Let . By definition of we find . On the other hand note that for all and all and the term is always the same, say , for all by definition of . Therefore we can write
and find . This holds in both cases. Since is a maximal coset in we have in the first case. In the second case we note that is an irreducible subset of which contains the identity. Thus the connected component of the identity of contains . Thus we have . Since is a linear torus and by the maximality of among torus cosets in we find . Since is also a linear torus inside with finite index, we have .∎
Corollary A.7.
Every algebraic subgroup of is of type for some subgroup of .
Proof.
Apply Lemma A.6 to and its maximal coset .∎
Corollary A.8.
Let be an algebraic subset. Then there are only finitely many linear tori with the property that there exists such that is a maximal torus coset in .
Proof.
Assume that is a maximal torus coset in . By Lemma A.6 there exists generated by vectors of type with in a finite set such that . Since the generators come from a finite set, there are also only finitely many and hence is from a finite set.∎
Definition A.9.
Let be a subgroup and a prime or . The -saturation is defined by
Remark A.10.
The -saturation of a subgroup is -full.
Lemma A.11.
Let and char We have
Proof.
If we have and the equality obviously holds. Thus we may assume that . Since it suffices to show . Let and . There exists a natural number such that . Therefore . Since the characteristic of the field is we have and hence we find which implies that and thus .∎
Theorem A.12.
The map is a bijection between -full subgroups of and the algebraic subgroups of . The inverse is given by . Moreover we have and is a linear torus if and only if is primitive.
Proof.
Definition A.13.
For a prime let be the localisation of at the powers of .
Lemma A.14.
Let and be a prime. Then there is a bijection between the submodules of and the -full subgroups of . The maps are given by and .
Proof.
We abbreviate . First we show that both maps are well defined. Let be a submodule of . Clearly is a subgroup of . Assume that for some . There exists such that and therefore since and hence . Thus the first map is well defined. That is a submodule of is straight forward.
Now we look at the compositions of the maps and show first that . The inclusion is immediate since . For the other direction let . Then there exists and such that . We have , so and hence .
Secondly we want to show that . Here the inclusion is direct. For the other direction let where and . If , then . Else we write for an integer and . Then we find that . Since is -full we find also in this case that .
∎
Corollary A.15.
If char, the map is a bijection between the submodules of and the algebraic subgroups of .
Definition A.16.
A monoidal transform is an automorphism given by for some
Lemma A.17.
Let be a torsion coset of dimension . Then there exists a monoidal transformation and such that
Proof.
Let be the linear torus in and such that . Let . Let such that . Let and . Then using one can show that doing straight forward computations. ∎
Appendix B The gcd of the determinats of the minors of a matrix
The aim of this apppendix is to give two lemmas which imply that the product of the first entries of the diagonal matrix in the Smith Normal Form of is equal to the gcd of the determinats of the -minors of . This is used to bound the number of isolated torsion points.
Definition B.1.
For , a matrix , and we define as the matrix with entries where and range over and , respectively, and in increasing order. Similarly denotes the matrix with entries where and range over and , respectively, and in increasing order. Finally we see that and denote this matrix by .
Remark B.2.
For natural numbers matrices and we have and .
Definition B.3.
For and let us define for
the gcd of the determinants of the -minors of .
Theorem B.4.
Let and . Then we have
Proof.
This is the Cauchy-Binet formula. A proof can be found in [Gan86], page 27-28.∎
Lemma B.5.
Let , and . Then for all we have .
Proof.
Since are invertible it is enough to show that . So let such that . By Theorem B.4 we have
Therefore and hence divides their greatest common divisor . ∎
Definition B.6.
For and we denote by the matrix where if and for all .
Lemma B.7.
Let such that for all . Then for all .
Proof.
Let and such that . If there is a row of zeros in since the entries are where and and hence the determinant vanishes. If the determinant of is given by . Therefore we find that , where we use the divisibility properties of the .∎
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