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Counting torsion points on subvarieties of the algebraic torus

Gerold Schefer

Abstract:

We estimate the growth rate of the function which counts the number of torsion points of order at most TT on an algebraic subvariety of the algebraic torus 𝔾mn\mathbb{G}_{m}^{n} 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 KK the algebraic closure of a finite field.

1 Introduction

Main results.

Let n1n\geq 1 be an integer and KK be an algebraically closed field. We denote by 𝔾md\mathbb{G}_{m}^{d} the nn-dimensional algebraic torus with base field KK. We will identify 𝔾mn(K)\mathbb{G}_{m}^{n}(K) with (K{0})n(K\setminus\{0\})^{n}, the group of its KK-points. We denote by (𝔾mn)tors(\mathbb{G}_{m}^{n})_{\mathrm{tors}} the subgroup of points of finite order therein, which we call torsion points. A linear torus G<𝔾mnG<\mathbb{G}_{m}^{n} is an irreducible algebraic subgroup and if 𝜻(𝔾mn)tors\boldsymbol{\zeta}\in(\mathbb{G}_{m}^{n})_{\mathrm{tors}} is a torsion point and GG is a linear torus, we call 𝜻G\boldsymbol{\zeta}G a torsion coset. We often identify a subvariety of 𝔾mn\mathbb{G}_{m}^{n} with the set of its KK-points.

Lukas Fink considered the following question in his master thesis [Fin08]. Let X={(x,y)𝔾m2(𝔽¯2):x+y=1}X=\{(x,y)\in\mathbb{G}_{m}^{2}(\overline{\mathbb{F}}_{2}):x+y=1\} and c(T)=#{𝐱X:ord(𝐱)T}c(T)=\#\{\mathbf{x}\in X:\mathrm{ord}(\mathbf{x})\leq T\}. What can we say about the asymptotic growth of c(T)c(T) as TT\to\infty? Considering the first coordinate yields c(T)#{x𝔾m:ord(x)T}T2c(T)\leq\#\{x\in\mathbb{G}_{m}:\mathrm{ord}(x)\leq T\}\ll T^{2}. He shows in Proposition 2.7 that c(T)=o(T2)c(T)=o(T^{2}). Fink’s result is therefore better than the trivial bound. Based on computations he conjectures that c(T)Tc(T)\ll T. Our first result, Theorem 1.1 below, includes a power saving over the trivial bound c(T)T2c(T)\ll T^{2}.

We consider the average rather than the number of torsion points of order exactly nn since the latter behaves erratic. In Fink’s example the set of nn\in\mathbb{N} such that there is no torsion point of order nn has natural density one. The number of points on Fink’s curve whose order divides nn is also bounded by a constant times n11/560n^{1-1/560} if (4) in [Moh20] holds for the set of points x𝔽2φ(n)x\in\mathbb{F}_{2^{\varphi(n)}} such that xn=1=(x+1)nx^{n}=1=(x+1)^{n}. On the other hand, this number is n1n-1 for all n=2k1n=2^{k}-1.

An algebraic subset X𝔾mn(K)X\subset\mathbb{G}_{m}^{n}(K) is the zero set of finitely many Laurent polynomials P1,,PrK[X1±1,,Xn±1]P_{1},\dots,P_{r}\in K[X_{1}^{\pm 1},\dots,X_{n}^{\pm 1}] in 𝔾mn(K)\mathbb{G}_{m}^{n}(K). The natural generalisation of the question of Fink is as follows. Take an algebraic set X𝔾mn(K)X\subset\mathbb{G}_{m}^{n}(K), and denote by XT={𝜻X(𝔾mn)tors:ord(𝜻)T}X_{T}=\{\boldsymbol{\zeta}\in X\cap(\mathbb{G}_{m}^{n})_{\mathrm{tors}}:\mathrm{ord}(\boldsymbol{\zeta})\leq T\} the set of torsion points in XX of order bounded by TT. How fast does #XT\#X_{T} grow?

In characteristic zero, the Manin-Mumford Conjecture for 𝔾mn\mathbb{G}_{m}^{n}, a theorem of Laurent [Lau84] reduces this problem to the case where XX 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 KK is an algebraic closure of a finite field. In this case all points of 𝔾mn\mathbb{G}_{m}^{n} 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 x+y=1x+y=1:

Theorem 1.1.

Let pp be a prime and X={(x,y)𝔾m2(𝔽¯p):x+y=1}X=\{(x,y)\in\mathbb{G}_{m}^{2}(\overline{\mathbb{F}}_{p}):x+y=1\}. Then we have #XT16T3/2\#X_{T}\leq 16T^{3/2} for all T1.T\geq 1.

To state the main result, we have to introduce more terminology. An algebraic subset X𝔾mnX\subset\mathbb{G}_{m}^{n} is called admissible if it is finite or does not contain a top dimensional torsion coset. The stabilizer Stab(X)\mathrm{Stab}(X) of an algebraic subset X𝔾mnX\subset\mathbb{G}_{m}^{n} is the set of 𝐠𝔾mn\mathbf{g}\in\mathbb{G}_{m}^{n} such that 𝐠XX\mathbf{g}X\subset X. It is well known that the stabilizer is an algebraic subgroup of 𝔾mn\mathbb{G}_{m}^{n}. The main result is

Theorem 1.2.

Let KK be an algebraically closed field and X𝔾mnX\subset\mathbb{G}_{m}^{n} be irreducible, admissible and of dimension dd. Let δ\delta be the dimension of the stabilizer Stab(X)\mathrm{Stab}(X). Then we have XTXTd+11dδ+1X_{T}\ll_{X}T^{d+1-\frac{1}{d-\delta+1}} for all T1T\geq 1.

As in the example there is an analogue of the trivial bound given by

Proposition 1.3.

Let KK be an algebraically closed field and X𝔾mnX\subset\mathbb{G}_{m}^{n} an algebraic subset of dimension dd. Then #XTXTd+1\#X_{T}\ll_{X}T^{d+1} for all T1T\geq 1.

So again we have a power saving with respect to the general bound.

If CC 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 ord(C)=min{ord(𝜼):𝜼C(𝔾mn)tors}\mathrm{ord}(C)=\min\{\mathrm{ord}(\boldsymbol{\eta}):\boldsymbol{\eta}\in C\cap(\mathbb{G}_{m}^{n})_{\mathrm{tors}}\}. In the theorem below ζ\zeta denotes the Riemann zeta function.

Theorem 1.4.

Let KK be an algebraically closed field of characteristic pp and let C𝔾mnC\subset\mathbb{G}_{m}^{n} be a torsion coset of dimension dd and order gg. Then if p=0p=0 we have

#CT=Td+1(d+1)ζ(d+1)g+{Og(Tlog(T))if d=1Od,g(Td)if d2 as T\#C_{T}=\frac{T^{d+1}}{(d+1)\zeta(d+1)g}+\begin{cases}O_{g}(T\log(T))&\text{if }d=1\\ O_{d,g}(T^{d})&\text{if }d\geq 2\end{cases}\text{ as }T\to\infty

and if p>0p>0 we have

#CT=(p1)Td+1(ppd)(d+1)ζ(d+1)g+{Og(Tlog(T))if d=1Od,g(Td)if d2 as T.\#C_{T}=\frac{(p-1)T^{d+1}}{(p-p^{-d})(d+1)\zeta(d+1)g}+\begin{cases}O_{g}(T\log(T))&\text{if }d=1\\ O_{d,g}(T^{d})&\text{if }d\geq 2\end{cases}\text{ as }T\to\infty.

Theorem 1.4 implies that the exponent d+1d+1 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 𝔾mn\mathbb{G}_{m}^{n} reduces the general case to torsion cosets and again we are able to give an asymptotic formula. Let X𝔾mnX\subset\mathbb{G}_{m}^{n} be an algebraic set. There exist torsion cosets C1,,CmC_{1},\dots,C_{m} such that X(𝔾mn)tors¯=i=1mCi\overline{X\cap(\mathbb{G}_{m}^{n})_{\mathrm{tors}}}=\bigcup_{i=1}^{m}C_{i} is the decomposition into irreducible components. Set

a(X)=max1imdim(Ci) and b(X)=1(a(X)+1)ζ(a(X)+1)i=1dim(Ci)=a(X)m1ord(Ci).a(X)=\max_{1\leq i\leq m}\dim(C_{i})\text{ and }b(X)=\frac{1}{(a(X)+1)\zeta(a(X)+1)}\sum_{\begin{subarray}{c}i=1\\ \dim(C_{i})=a(X)\end{subarray}}^{m}\frac{1}{\mathrm{ord}(C_{i})}.

Then we have

Corollary 1.5.

Let KK be an algebraic field of characteristic zero, let X𝔾mnX\subset\mathbb{G}_{m}^{n} be algebraic, a=a(X)a=a(X) and b=b(X)b=b(X). Then we have #XTX1\#X_{T}\ll_{X}1 if a=0a=0 and

#XT=bTa+1+{OX(Tlog(T))if a=1OX(Ta)if a2 as T.\#X_{T}=bT^{a+1}+\begin{cases}O_{X}(T\log(T))&\text{if }a=1\\ O_{X}(T^{a})&\text{if }a\geq 2\end{cases}\text{ as }T\to\infty.

What about lower bounds for #XT\#X_{T}? Fink proved in Proposition 2.7 in [Fin08] that c(T)T/22c(T)\geq T/2-2 for all TT\in\mathbb{N}. Using the famous theorem of Lang-Weil, we can prove

Theorem 1.6.

Let pp be a prime and nn\in\mathbb{N}. Let X𝔾mn(𝔽¯p)X\subset\mathbb{G}_{m}^{n}(\overline{\mathbb{F}}_{p}) be algebraic of dimension dd. Then there exist c=c(X)>0c=c(X)>0 and T01T_{0}\geq 1 such that cTd#XTcT^{d}\leq\#X_{T} for all TT0T\geq T_{0}.

Hence if XX is admissible, the correct exponent, if it exists, is in [d,d+11dδ+1][d,d+1-\frac{1}{d-\delta+1}], where δ=dim(Stab(X))\delta=\dim(\mathrm{Stab}(X)). In the example of Fink, numerical analysis suggests that the smallest exponent d=1d=1 is possible.

Overview of the proofs.

Let us give two examples. First we look at the curve C={(x,y)𝔾m2:x+y=1}C=\{(x,y)\in\mathbb{G}_{m}^{2}:x+y=1\} and prove Theorem 1.1. Let T1T\geq 1 and assume that 𝜻=(ζ,ξ)CT\boldsymbol{\zeta}=(\zeta,\xi)\in C_{T} is of order NN. By Minkowski’s first theorem there exists 𝐚=(a,b)2{0}\mathbf{a}=(a,b)\in\mathbb{Z}^{2}\setminus\{0\} such that 𝜻𝐚:=ζaξb=1\boldsymbol{\zeta}^{\mathbf{a}}:=\zeta^{a}\xi^{b}=1 and |𝐚|:=max{|a|,|b|}N1/2|\mathbf{a}|:=\max\{|a|,|b|\}\leq N^{1/2}. Since ξ=1ζ\xi=1-\zeta we have ζa(1ζ)b=1\zeta^{a}(1-\zeta)^{b}=1. After multiplying by suitable powers of ζ\zeta and 1ζ1-\zeta we see that ζ\zeta is a root of a nonzero polynomial of degree at most |a|+|b||a|+|b|. Hence the number of 𝜻C\boldsymbol{\zeta}\in C such that 𝜻𝐚=1\boldsymbol{\zeta}^{\mathbf{a}}=1 is at most 2|𝐚|2|\mathbf{a}|. Thus we have the bound #CT2|𝐚|T1/2|𝐚|\#C_{T}\leq 2\sum_{|\mathbf{a}|\leq T^{1/2}}|\mathbf{a}|. For a fixed k1k\geq 1 there are 4(2k+1)4=8k4(2k+1)-4=8k vectors 𝐚2\mathbf{a}\in\mathbb{Z}^{2} such that |𝐚|=k|\mathbf{a}|=k. Thus we find

#XT16k=1T1/2k216T3/2.\#X_{T}\leq 16\sum_{k=1}^{T^{1/2}}k^{2}\leq 16T^{3/2}.

Now let us consider Y={(x,y,z)𝔾m3:x+y=1}Y=\{(x,y,z)\in\mathbb{G}_{m}^{3}:x+y=1\}. Note that Stab(Y)={𝟏}×𝔾m(Y)=\{\mathbf{1}\}\times\mathbb{G}_{m} is one dimensional. Hence the exponent in Theorem 1.2 equals 5/25/2. Then we observe that any (ζ1,ζ2,ζ3)Y(𝔾m3)tors(\zeta_{1},\zeta_{2},\zeta_{3})\in Y\cap(\mathbb{G}_{m}^{3})_{\mathrm{tors}} is contained in the torsion coset {ζ1}×{ζ2}×𝔾mY.\{\zeta_{1}\}\times\{\zeta_{2}\}\times\mathbb{G}_{m}\subset Y. The order of such a torsion coset is given by ord(ζ1,ζ2)\mathrm{ord}(\zeta_{1},\zeta_{2}). Thus any 𝜻YT\boldsymbol{\zeta}\in Y_{T} is contained in a torsion coset {𝝃}×𝔾m\{\boldsymbol{\xi}\}\times\mathbb{G}_{m} for some 𝝃CT\boldsymbol{\xi}\in C_{T}. Therefore we can bound #YT𝝃CT#({𝝃}×𝔾m)T\#Y_{T}\leq\sum_{\boldsymbol{\xi}\in C_{T}}\#(\{\boldsymbol{\xi}\}\times\mathbb{G}_{m})_{T}. Denote by gg the order of 𝝃CT\boldsymbol{\xi}\in C_{T}. We claim that #({𝝃}×𝔾m)TT2/g\#(\{\boldsymbol{\xi}\}\times\mathbb{G}_{m})_{T}\leq T^{2}/g. If ζ𝔾m\zeta\in\mathbb{G}_{m} is of order nn, the order of (𝝃,ζ)(\boldsymbol{\xi},\zeta) is the least common multiple of gg and nn. Thus (𝝃,ζ)({𝝃}×𝔾m)T(\boldsymbol{\xi},\zeta)\in(\{\boldsymbol{\xi}\}\times\mathbb{G}_{m})_{T} if and only if ngcd(g,n)T/gn\leq\gcd(g,n)T/g. The number of elements of order nn in 𝔾m\mathbb{G}_{m} is given by Euler’s totient φ(n)\varphi(n) except if the characteristic p>0p>0 divides nn. Since in positive characteristic the order is always coprime to pp, there are no elements of order nn, whenever pp divides nn. Thus in any case there are at most φ(n)\varphi(n) elements of order nn in 𝔾m\mathbb{G}_{m}. We therefore have

#({𝝃}×𝔾m)T1ngcd(g,n)T/gφ(n)=d|gn:gcd(g,n)=d,ndT/gφ(n).\#(\{\boldsymbol{\xi}\}\times\mathbb{G}_{m})_{T}\leq\sum_{1\leq n\leq\gcd(g,n)T/g}\varphi(n)=\sum_{d|g}\sum_{n:\gcd(g,n)=d,n\leq dT/g}\varphi(n).

If d|gd|g and nn is as in the inner sum, we can write n=kdn=kd for some 1kT/g1\leq k\leq T/g. And since φ(n)=npn(1p1)\varphi(n)=n\prod_{p\mid n}(1-p^{-1}) we see that φ(ab)aφ(b)\varphi(ab)\leq a\varphi(b) for all a,ba,b\in\mathbb{N}. It is also well known, that d|nφ(d)=n\sum_{d|n}\varphi(d)=n for all nn\in\mathbb{N}. Therefore we get

#({𝝃}×𝔾m)Td|g1kT/gφ(kd)d|gφ(d)1kT/gkg(T/g)2=T2/g.\#(\{\boldsymbol{\xi}\}\times\mathbb{G}_{m})_{T}\leq\sum_{d|g}\sum_{1\leq k\leq T/g}\varphi(kd)\leq\sum_{d|g}\varphi(d)\sum_{1\leq k\leq T/g}k\leq g(T/g)^{2}=T^{2}/g.

Let ana_{n} denote the number of elements of order nn in CC. Then we can bound #YTT2n=1Tan/n\#Y_{T}\leq T^{2}\sum_{n=1}^{T}a_{n}/n and we know that n=1Nan16N3/2\sum_{n=1}^{N}a_{n}\leq 16N^{3/2} for every N1N\geq 1. To get sums of this kind, we write 1/n=1/T+k=nT1(1k1k+1)=1/T+k=nT11k(k+1)1/n=1/T+\sum_{k=n}^{T-1}(\frac{1}{k}-\frac{1}{k+1})=1/T+\sum_{k=n}^{T-1}\frac{1}{k(k+1)} for all n<Tn<T. Thus we find

n=1Tan/n=k=1T11k(k+1)n=1kan+1/Tn=1Tan16k=1T1k3/22+16T3/2148T1/2\sum_{n=1}^{T}a_{n}/n=\sum_{k=1}^{T-1}\frac{1}{k(k+1)}\sum_{n=1}^{k}a_{n}+1/T\sum_{n=1}^{T}a_{n}\leq 16\sum_{k=1}^{T-1}k^{3/2-2}+16T^{3/2-1}\leq 48T^{1/2}

using that x1/2𝑑x=2x1/2\int x^{-1/2}dx=2x^{1/2} and hence YT48T5/2Y_{T}\leq 48T^{5/2}.

Now let us consider a hypersurface X𝔾mnX\subset\mathbb{G}_{m}^{n}. We divide XX into two parts: the union of all torsion cosets contained in XX of positive dimension and its complement XX^{*}. To bound #XT\#X^{*}_{T} we generalise the proof for CC: Instead of Minkowski’s first theorem we can use the second theorem to find short linearly independent vectors 𝐚1,,𝐚n1n\mathbf{a}_{1},\dots,\mathbf{a}_{n-1}\in\mathbb{Z}^{n} such that 𝜻𝐚i=1\boldsymbol{\zeta}^{\mathbf{a}_{i}}=1. The 𝐱𝔾mn\mathbf{x}\in\mathbb{G}_{m}^{n} such that 𝐱𝐚i=1\mathbf{x}^{\mathbf{a}_{i}}=1 for all i=1,,n1i=1,\dots,n-1 are given by torsion cosets {𝝃t𝐮:t𝔾m}\{\boldsymbol{\xi}t^{\mathbf{u}^{\top}}:t\in\mathbb{G}_{m}\} for some 𝐮n{0}\mathbf{u}\in\mathbb{Z}^{n}\setminus\{0\} and finitely many 𝝃(𝔾mn)tors\boldsymbol{\xi}\in(\mathbb{G}_{m}^{n})_{\mathrm{tors}}; we refer to Section 2 for our notation. Let PK[X1±1,,Xn±1]P\in K[X_{1}^{\pm 1},\dots,X_{n}^{\pm 1}] be such that XX is the vanishing locus of PP in 𝔾mn\mathbb{G}_{m}^{n}. Then 𝝃t𝐮\boldsymbol{\xi}t^{\mathbf{u}^{\top}} is in XX if and only if Q(t)=P(𝝃t𝐮)=0Q(t)=P(\boldsymbol{\xi}t^{\mathbf{u}^{\top}})=0. If QQ is not the zero polynomial, the number of such tt is bounded in terms of the degree of PP and |𝐮||\mathbf{u}|. This is always the case if 𝝃t𝐮X\boldsymbol{\xi}t^{\mathbf{u}^{\top}}\in X^{*}. Here we use the fact that XX is a hypersurface. Summing over all possible (n1)(n-1)-tuples 𝐚1,,𝐚n1\mathbf{a}_{1},\dots,\mathbf{a}_{n-1} gives us a bound XTXTn1/nX^{*}_{T}\ll_{X}T^{n-1/n}.

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 XX is contained in one of them. However in positive characteristic it is possible that for a fixed linear torus GG there are infinitely many torsion cosets 𝜻G\boldsymbol{\zeta}G which are contained in XX. So let us fix GG. To control the number of cosets 𝜻G\boldsymbol{\zeta}G which are contained in XX, we map 𝜻G\boldsymbol{\zeta}G to 𝜻L\boldsymbol{\zeta}^{L} for some matrix LL such that 𝐠L=𝟏\mathbf{g}^{L}=\mathbf{1} for all 𝐠G\mathbf{g}\in G. Note that ord(𝜻L)ord(𝜻G)\mathrm{ord}(\boldsymbol{\zeta}^{L})\leq\mathrm{ord}(\boldsymbol{\zeta}G). We can find equations for the image and use induction on nn. Thus we have a bound for the number of cosets of GG contained in XX. Together with a bound for cosets we get the claimed result for hypersurfaces.

The generalisation to arbitrary algebraic sets follows by projecting XX to some coordinates and doing algebraic geometry.

For Theorem 1.4 we can assume that G={𝜻}×𝔾mdG=\{\boldsymbol{\zeta}\}\times\mathbb{G}_{m}^{d} for some 𝜻(𝔾mnd)tors\boldsymbol{\zeta}\in(\mathbb{G}_{m}^{n-d})_{\mathrm{tors}} of order gg, say. Since the order of (𝜻,𝝃)(\boldsymbol{\zeta,\xi}) is lcm(g,ord(𝝃))=g(g,ord(𝝃))ord(𝝃)\mathrm{lcm}(g,\mathrm{ord}(\boldsymbol{\xi}))=\frac{g}{(g,\mathrm{ord}(\boldsymbol{\xi}))}\mathrm{ord}(\boldsymbol{\xi}), we have to count the 𝝃(𝔾md)tors\boldsymbol{\xi}\in(\mathbb{G}_{m}^{d})_{\mathrm{tors}} whose order is bounded and belongs to a fixed congruence class modulo gg. With the Möbius inversion formula one can see that the number of elements of order nn is given by Jordan’s totient Jd(n)=μId(n)J_{d}(n)=\mu\star I_{d}(n), where μ\mu is the Möbius function and Id(n)=ndI_{d}(n)=n^{d}. Once we have the asymptotics of JdJ_{d} over arithmetic progressions, Theorem 1.4 is not hard to deduce. In positive characteristic the orders are always coprime to pp. If nn is coprime to pp, we still have Jd(n)J_{d}(n) elements of order nn. Thus the case p>0p>0 follows from the case p=0p=0.

The proofs of Corollary 1.5 and Theorem 1.6 are short and straight forward.

Organisation of the article

The algebraic subgroups of 𝔾mn\mathbb{G}_{m}^{n} 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.

In section 3 we prove Theorem 1.4.

The next goal is to estimate #XT\#X^{*}_{T}. The most technical part is to bound the degree of Q(t)Q(t). Some known preparations are done in the second appendix and used to prove the bound in section 4.

For the proof of Theorem 1.2 when XX is a hypersurface, we first deal with common roots and admissibility in section 5, then we establish an explicit bound when XX is a coset. In section 7 we give the proofs of Theorems 1.2 and 1.3 when XX is a hypersurface.

Finally we are able to prove Theorems 1.2 and 1.3 in full generality in the next section.

In the last two sections we establish Corollary 1.5 and Theorem 1.6.

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 ={1,2,3,}\mathbb{N}=\{1,2,3,\dots\} and KK denotes always an algebraically closed field of any characteristic. Let nn\in\mathbb{N} and R=K[X1±1,Xn±1]R=K[X_{1}^{\pm 1},\dots X_{n}^{\pm 1}] the ring of Laurent polynomials in nn variables. Let RR^{*} be the units of RR. Then for r,s,𝐱=(x1,,xn)(R)nr,s\in\mathbb{N},\mathbf{x}=(x_{1},\dots,x_{n})\in(R^{*})^{n} and A=(ai,j)Matn×m()A=(a_{i,j})\in\mathrm{Mat}_{n\times m}(\mathbb{Z}) we define 𝐱A=(x1a1,1xnan,1,,x1a1,mxnan,m)(R)m\mathbf{x}^{A}=(x_{1}^{a_{1,1}}\cdots x_{n}^{a_{n,1}},\dots,x_{1}^{a_{1,m}}\cdots x_{n}^{a_{n,m}})\in(R^{*})^{m}. This definition is compatible with matrix multiplication in the sense that 𝐱AB=(𝐱A)B\mathbf{x}^{AB}=(\mathbf{x}^{A})^{B} for all k,m,n,𝐱(R)k,AMatk×m()k,m,n\in\mathbb{N},\mathbf{x}\in(R^{*})^{k},A\in\mathrm{Mat}_{k\times m}(\mathbb{Z}) and BMatm×n()B\in\mathrm{Mat}_{m\times n}(\mathbb{Z}). The letter 𝐗\mathbf{X} denotes the element (X1,,Xn)(R)n(X_{1},\dots,X_{n})\in(R^{*})^{n} and hence we can write a general Laurent polynomial PRP\in R as P=𝐢np𝐢𝐗𝐢P=\sum_{\mathbf{i}\in\mathbb{Z}^{n}}p_{\mathbf{i}}\mathbf{X^{i}}. The support of such a polynomial is the finite set of 𝐢n\mathbf{i}\in\mathbb{Z}^{n} such that p𝐢0p_{\mathbf{i}}\neq 0 and we denote it by Supp(P)\mathrm{Supp}(P).
For nn\in\mathbb{N} we set 𝔾mn(K)=(K{0})n\mathbb{G}_{m}^{n}(K)=(K\setminus\{0\})^{n} which together with the coordinate-wise multiplication is a group. We write 𝔾mn(K)tors\mathbb{G}_{m}^{n}(K)_{\mathrm{tors}} for the elements of finite order in 𝔾mn(K)\mathbb{G}_{m}^{n}(K) and call them torsion points. If it is clear what KK is, we usually omit KK in the notation. Elements ζ(𝔾m)tors\zeta\in(\mathbb{G}_{m})_{\mathrm{tors}} are called roots of unity. For NN\in\mathbb{N} we write μN(K)\mu_{N}(K) or μN\mu_{N} for the roots of unity ζ\zeta in KK such that ζN=1\zeta^{N}=1.
We denote by EnE_{n} the identity matrix in the ring of n×nn\times n-matrices Matn()\mathrm{Mat}_{n}(\mathbb{Z}). We write 𝟎\mathbf{0} and 𝟏\mathbf{1} for matrices whose entries are all zero or one respectively. We write 𝐞i\mathbf{e}_{i} for the ii-th vector of the standard basis of n\mathbb{Z}^{n}. The nn will be clear from the context. We denote by 𝔸n(K)\mathbb{A}^{n}(K) the affine space KnK^{n} endowed with the Zariski topology.
For 𝐚=(a1,,an)n\mathbf{a}=(a_{1},\dots,a_{n})\in\mathbb{Z}^{n} we denote by |𝐚|=max1in|ai||\mathbf{a}|=\max_{1\leq i\leq n}|a_{i}| the maximum of the absolute vaues of its coordinates, where |||\cdot| on the right hand side is the usual absolute value. We use the Vinogradov notation \ll and I\ll_{I} for some set II of variables.

3 Jordan’s totient function and torsion cosets

In characteristic zero the dd-th Jordan’s totient function JdJ_{d} counts the number of points of order nn in 𝔾md\mathbb{G}_{m}^{d}. 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 #CT\#C_{T}, where CC is a torsion coset.

Definition 3.1.

Let 𝒜={f:}\mathcal{A}=\{f:\mathbb{N}\to\mathbb{C}\} be the set of arithmetic functions. We say that f𝒜f\in\mathcal{A} is completely multiplicative if f(nm)=f(n)f(m)f(nm)=f(n)f(m) for all n,mn,m\in\mathbb{N}. The Dirichlet convolution of f,g𝒜f,g\in\mathcal{A} is defined as (fg)(n)=d|nf(d)g(n/d).(f\star g)(n)=\sum_{d|n}f(d)g(n/d). We can associate to f𝒜f\in\mathcal{A} the formal Dirichlet series n=1f(n)ns\sum_{n=1}^{\infty}f(n)n^{-s}, where ss is an indeterminate.

Definition 3.2.

Define η𝒜\eta\in\mathcal{A} by η(1)=1\eta(1)=1 and η(n)=0\eta(n)=0 for all n2n\geq 2 and ϵ𝒜\epsilon\in\mathcal{A} by ϵ(n)=1\epsilon(n)=1 for all nn\in\mathbb{N}. The Möbius function μ𝒜\mu\in\mathcal{A} is nonzero only on squarefree numbers and if nn is a product of rr distinct primes we have μ(n)=(1)r\mu(n)=(-1)^{r}.
For dd\in\mathbb{N} we define Id𝒜I_{d}\in\mathcal{A} as nndn\mapsto n^{d} and Jordan’s totient function Jd𝒜J_{d}\in\mathcal{A} given by Jd(n)=μIdJ_{d}(n)=\mu\star I_{d}, a generalisation of Euler’s totient function φ=J1\varphi=J_{1}.
Any character χ:(/m)\chi:\left(\mathbb{Z}/m\mathbb{Z}\right)^{*}\to\mathbb{C}^{*} gives rise to fχ𝒜f_{\chi}\in\mathcal{A} defined by fχ(n)=χ(n+m)f_{\chi}(n)=\chi(n+m\mathbb{Z}) if (n,m)=1(n,m)=1 and zero otherwise. We will sometimes abuse notation and write χ\chi for fχf_{\chi}.
The Riemann zeta function ζ(s)=n=1ns\zeta(s)=\sum_{n=1}^{\infty}n^{-s} is the series corresponding to ϵ\epsilon and the Dirichlet LL-functions are defined by L(s,χ)=n=1χ(n)nsL(s,\chi)=\sum_{n=1}^{\infty}\chi(n)n^{-s}.

Remark 3.3.

Together with the pointwise addition the Dirichlet convolution endows 𝒜\mathcal{A} with the structure of a commutative unitary ring with multiplicative identity η\eta. If f𝒜f\in\mathcal{A} is completely multiplicative we have f(gh)=(fg)(fh)f(g\star h)=(fg)\star(fh) for all g,h𝒜g,h\in\mathcal{A}. Here fgfg denotes the pointwise multiplication. For all f,g𝒜f,g\in\mathcal{A} we have

n=1f(n)nsm=1g(m)ms=n=1(fg)(n)ns\sum_{n=1}^{\infty}f(n)n^{-s}\sum_{m=1}^{\infty}g(m)m^{-s}=\sum_{n=1}^{\infty}(f\star g)(n)n^{-s}

as formal series. In particular if all three series converge absolutely for some ss\in\mathbb{C} the equality above holds in \mathbb{C}.

Remark 3.4.

Möbius inversion is equivalent to μϵ=η\mu\star\epsilon=\eta. The functions IdI_{d} and fχf_{\chi} are completely multiplicative for all dd\in\mathbb{N} and characters χ\chi respectively. The explicit formula for Jordan’s totient function reads Jd(n)=ndp|n(1pd)J_{d}(n)=n^{d}\prod_{p|n}(1-p^{-d}) where the product is taken over all primes dividing nn and as usual the empty product equals one. The Riemann ζ\zeta-function as well as the Dirichlet LL-functions converge absolutely if the real part of ss is strictly larger than 11.

Lemma 3.5.

Let d,md,m\in\mathbb{N} and aa\in\mathbb{Z}. Then

nxna(m)nd=xd+1(d+1)m+Od,m(xd) as x.\sum_{\begin{subarray}{c}n\leq x\\ n\equiv a(m)\end{subarray}}n^{d}=\frac{x^{d+1}}{(d+1)m}+O_{d,m}\left(x^{d}\right)\text{ as }x\to\infty.
Proof.

Let b{1,,m}b\in\{1,\dots,m\} such that ba(m)b\equiv a(m). Then every nxn\leq x which is congruent to aa modulo mm is of the form n=b+lmn=b+lm for some 0l(xb)/m0\leq l\leq(x-b)/m. Thus we have

nxna(m)nd=0l(xb)/m(b+lm)d=0l(xb)/mi=0d(di)bdi(lm)i=i=0d(di)bdimi0l(xb)/mli.\sum_{\begin{subarray}{c}n\leq x\\ n\equiv a(m)\end{subarray}}n^{d}=\sum_{0\leq l\leq(x-b)/m}(b+lm)^{d}=\sum_{0\leq l\leq(x-b)/m}\sum_{i=0}^{d}\binom{d}{i}b^{d-i}(lm)^{i}=\sum_{i=0}^{d}\binom{d}{i}b^{d-i}m^{i}\sum_{0\leq l\leq(x-b)/m}l^{i}.

By exercise 6.9 in [McC86] we know that

0l(xb)/mli=(xbm)i+1i+1+Od((xbm)i)=(x/m)i+1i+1+Od(xi).\sum_{0\leq l\leq(x-b)/m}l^{i}=\frac{\left(\frac{x-b}{m}\right)^{i+1}}{i+1}+O_{d}\left(\left(\frac{x-b}{m}\right)^{i}\right)=\frac{(x/m)^{i+1}}{i+1}+O_{d}\left(x^{i}\right).

So if id1i\leq d-1 then 0l(xb)/mli=Od(xd)\sum_{0\leq l\leq(x-b)/m}l^{i}=O_{d}(x^{d}) and hence

nxna(m)nd=mdxd+1(d+1)md+1+Od,m(xd)=xd+1(d+1)m+Od,m(xd).\sum_{\begin{subarray}{c}n\leq x\\ n\equiv a(m)\end{subarray}}n^{d}=m^{d}\frac{x^{d+1}}{(d+1)m^{d+1}}+O_{d,m}\left(x^{d}\right)=\frac{x^{d+1}}{(d+1)m}+O_{d,m}\left(x^{d}\right).\qed
Lemma 3.6.

Let d,md,m\in\mathbb{N}. Then

nx(n,m)=1μ(n)nd+1=md+1ζ(d+1)Jd+1(m)+Od(xd) as x.\sum_{\begin{subarray}{c}n\leq x\\ (n,m)=1\end{subarray}}\frac{\mu(n)}{n^{d+1}}=\frac{m^{d+1}}{\zeta(d+1)J_{d+1}(m)}+O_{d}\left(x^{-d}\right)\text{ as }x\to\infty.
Proof.

Let χ0\chi_{0} be the principal character modulo mm, and recall that μϵ=η\mu\star\epsilon=\eta. Since χ0\chi_{0} is completely multiplicative we have (μχ0)χ0=(μχ0)(ϵχ0)=χ0(μϵ)=χ0η=η(\mu\chi_{0})\star\chi_{0}=(\mu\chi_{0})\star(\epsilon\chi_{0})=\chi_{0}(\mu\star\epsilon)=\chi_{0}\eta=\eta. Therefore the infinite sum S=(n,m)=1μ(n)n(d+1)S=\sum_{(n,m)=1}\mu(n)n^{-(d+1)} satisfies SL(d+1,χ0)=1SL(d+1,\chi_{0})=1. We have L(d+1,χ0)=ζ(d+1)p|m(1p(d+1))=ζ(d+1)Jd+1(m)m(d+1)L(d+1,\chi_{0})=\zeta(d+1)\prod_{p|m}(1-p^{-(d+1)})=\zeta(d+1)J_{d+1}(m)m^{-(d+1)} and hence S=md+1ζ(d+1)Jd+1(m)S=\frac{m^{d+1}}{\zeta(d+1)J_{d+1}(m)}. For the error term we find

|n>x(n,m)=1μ(n)nd+1|\displaystyle\left|\sum_{\begin{subarray}{c}n>x\\ (n,m)=1\end{subarray}}\frac{\mu(n)}{n^{d+1}}\right| n>xn(d+1)l=x1ll+1t(d+1)𝑑t\displaystyle\leq\sum_{n>x}n^{-(d+1)}\leq\sum_{l=\lceil x\rceil-1}^{\infty}\int_{l}^{l+1}t^{-(d+1)}dt
=x1t(d+1)𝑑t=1/d(x1)d=Od(xd).\displaystyle=\int_{\lceil x\rceil-1}^{\infty}t^{-(d+1)}dt=1/d(\lceil x\rceil-1)^{-d}=O_{d}(x^{-d}).\qed
Theorem 3.7.

Let d,md,m\in\mathbb{N}, aa\in\mathbb{Z} and g=(a,m)g=(a,m). Then

nxna(m)Jd(n)=mdJd(g)xd+1(d+1)ζ(d+1)gdJd+1(m)+{Om(xlog(x))if d=1Od,m(xd)if d2. as x.\sum_{\begin{subarray}{c}n\leq x\\ n\equiv a(m)\end{subarray}}J_{d}(n)=\frac{m^{d}J_{d}(g)x^{d+1}}{(d+1)\zeta(d+1)g^{d}J_{d+1}(m)}+\begin{cases}O_{m}(x\log(x))&\text{if }d=1\\ O_{d,m}(x^{d})&\text{if }d\geq 2.\end{cases}\text{ as }x\to\infty.
Proof.

We claim that for natural numbers k,lk,l the following statements are equivalent: (i):klx(i):kl\leq x and kla(m)kl\equiv a(m) and (ii):kx,lx/k,(k,m)|g(ii):k\leq x,l\leq x/k,(k,m)|g and lAD¯(M)l\equiv A\bar{D}(M), where a=(k,m)A,k=(k,m)D,m=(k,m)Ma=(k,m)A,k=(k,m)D,m=(k,m)M and DD¯1(M)D\bar{D}\equiv 1(M). So assume that there are k,lk,l\in\mathbb{N} such that klxkl\leq x and kla(m)kl\equiv a(m). Since l1l\geq 1 we find kxk\leq x and lx/kl\leq x/k is equivalent to klxkl\leq x. Since (k,m)(k,m) divides both kk and mm, it also has to divide aa and hence also (a,m)=g(a,m)=g. Dividing kla=cmkl-a=cm through (k,m)(k,m) we find DlA(M)Dl\equiv A(M). Since (D,M)=1(D,M)=1 there is D¯\bar{D} such that DD¯1(M)D\bar{D}\equiv 1(M). If we multiply the congruence by D¯\bar{D} we get the congruence of (ii)(ii). For the other direction we multiply the congruence by DD and find DlA=cMDl-A=cM for some integer cc. Multiplying this equation by (k,m)(k,m) we find that klamodmkl\equiv a\mod m.
Recall that Jd=μIdJ_{d}=\mu\star I_{d}. Hence we find

nxna(m)Jd(n)=nxna(m)k|nμ(k)(n/k)d=klxkla(m)μ(k)ld=kx(k,m)|gμ(k)lx/klAD¯(M)ld,\sum_{\begin{subarray}{c}n\leq x\\ n\equiv a(m)\end{subarray}}J_{d}(n)=\sum_{\begin{subarray}{c}n\leq x\\ n\equiv a(m)\end{subarray}}\sum_{k|n}\mu(k)(n/k)^{d}=\sum_{\begin{subarray}{c}kl\leq x\\ kl\equiv a(m)\end{subarray}}\mu(k)l^{d}=\sum_{\begin{subarray}{c}k\leq x\\ (k,m)|g\end{subarray}}\mu(k)\sum_{\begin{subarray}{c}l\leq x/k\\ l\equiv A\bar{D}(M)\end{subarray}}l^{d},

where A,D¯A,\bar{D} and MM are defined as in the claim. By Lemma 3.5 the inner sum equals xd+1(d+1)kd+1M+Od,M(xd/kd)=xd+1(k,m)(d+1)kd+1m+Od,M(xd/kd)\frac{x^{d+1}}{(d+1)k^{d+1}M}+O_{d,M}(x^{d}/k^{d})=\frac{x^{d+1}(k,m)}{(d+1)k^{d+1}m}+O_{d,M}(x^{d}/k^{d}). Since MM is a divisor of mm, we can find an absolute constant which holds for all kk and therefore the sum of the error terms is Od,m(xdkxkd)=Od,m(xdlog(x))O_{d,m}(x^{d}\sum_{k\leq x}k^{-d})=O_{d,m}(x^{d}\log(x)) where we need the log term only if d=1d=1 since the sum is O(log(x))O(\log(x)) if d=1d=1 and O(1)O(1) if d2d\geq 2. Therefore we find

nxna(m)Jd(n)=kx(k,m)|gμ(k)xk+1(k,m)(d+1)kd+1m+R(x)=xd+1(d+1)mg|ggkx(k,m)=gμ(k)kd+1+R(x)\sum_{\begin{subarray}{c}n\leq x\\ n\equiv a(m)\end{subarray}}J_{d}(n)=\sum_{\begin{subarray}{c}k\leq x\\ (k,m)|g\end{subarray}}\mu(k)\frac{x^{k+1}(k,m)}{(d+1)k^{d+1}m}+R(x)=\frac{x^{d+1}}{(d+1)m}\sum_{g^{\prime}|g}g^{\prime}\sum_{\begin{subarray}{c}k\leq x\\ (k,m)=g^{\prime}\end{subarray}}\frac{\mu(k)}{k^{d+1}}+R(x)

where R(x)=Om(xlog(x))R(x)=O_{m}(x\log(x)) if d=1d=1 and R(x)=Od,m(xd)R(x)=O_{d,m}(x^{d}) if d2d\geq 2.
Observe that for all g|mg^{\prime}|m we have {k:kx,(k,m)=g}={gD:Dx/g,(D,m/g)=1}\{k\in\mathbb{N}:k\leq x,(k,m)=g^{\prime}\}=\{g^{\prime}D:D\leq x/g^{\prime},(D,m/g^{\prime})=1\}. This allows us to write

kx(k,m)=gμ(k)kd+1=1gd+1Dx/g(D,m/g)=1μ(gD)Dd+1=1gd+1Dx/g(D,m/g)=1(D,g)=1μ(gD)Dd+1=μ(g)gd+1Dx/g(D,m)=1μ(D)Dk+1\sum_{\begin{subarray}{c}k\leq x\\ (k,m)=g^{\prime}\end{subarray}}\frac{\mu(k)}{k^{d+1}}=\frac{1}{g^{\prime d+1}}\sum_{\begin{subarray}{c}D\leq x/g^{\prime}\\ (D,m/g^{\prime})=1\end{subarray}}\frac{\mu(g^{\prime}D)}{D^{d+1}}=\frac{1}{g^{\prime d+1}}\sum_{\begin{subarray}{c}D\leq x/g^{\prime}\\ (D,m/g^{\prime})=1\\ (D,g^{\prime})=1\end{subarray}}\frac{\mu(g^{\prime}D)}{D^{d+1}}=\frac{\mu(g^{\prime})}{g^{\prime d+1}}\sum_{\begin{subarray}{c}D\leq x/g^{\prime}\\ (D,m)=1\end{subarray}}\frac{\mu(D)}{D^{k+1}}

since μ(mn)=0\mu(mn)=0 if (m,n)>1(m,n)>1. By Lemma 3.6 the sum is md+1ζ(d+1)Jd+1(m)+Od(gd/xd).\frac{m^{d+1}}{\zeta(d+1)J_{d+1}(m)}+O_{d}(g^{\prime d}/x^{d}). Hence

kx(k,m)=gμ(k)kd+1=μ(g)md+1ζ(d+1)gd+1Jd+1(m)+Od(1).\sum_{\begin{subarray}{c}k\leq x\\ (k,m)=g^{\prime}\end{subarray}}\frac{\mu(k)}{k^{d+1}}=\frac{\mu(g^{\prime})m^{d+1}}{\zeta(d+1)g^{\prime d+1}J_{d+1}(m)}+O_{d}(1).

Finally we plug this into the equation above and find

nxna(m)Jd(n)\displaystyle\sum_{\begin{subarray}{c}n\leq x\\ n\equiv a(m)\end{subarray}}J_{d}(n) =mdxd+1(d+1)ζ(d+1)Jd+1(m)g|gμ(g)gd+Od(g)+R(x)\displaystyle=\frac{m^{d}x^{d+1}}{(d+1)\zeta(d+1)J_{d+1}(m)}\sum_{g^{\prime}|g}\frac{\mu(g^{\prime})}{g^{\prime d}}+O_{d}(g)+R(x)
=mdJd(g)xd+1(d+1)ζ(d+1)gdJd+1(m)+Od,m(1)+R(x)\displaystyle=\frac{m^{d}J_{d}(g)x^{d+1}}{(d+1)\zeta(d+1)g^{d}J_{d+1}(m)}+O_{d,m}(1)+R(x)

since Jd(g)=g|gμ(g)(g/g)dJ_{d}(g)=\sum_{g^{\prime}|g}\mu(g^{\prime})(g/g^{\prime})^{d}. The error term is Om(xlog(x))O_{m}(x\log(x)) if d=1d=1 and Od,m(xd)O_{d,m}(x^{d}) if d2d\geq 2. ∎

Lemma 3.8.

Let d,nd,n\in\mathbb{N}. Then a=1n(a,n)Jd((a,n))=Jd+1(n)\sum_{a=1}^{n}(a,n)J_{d}((a,n))=J_{d+1}(n).

Proof.

Fix gg a divisor of nn. If 1an1\leq a\leq n satisfies (a,n)=g(a,n)=g, then a=bga=bg for some bn/gb\leq n/g coprime to n/gn/g and hence the number of such aa equals φ(n/g)\varphi(n/g). Thus we can write the sum in question as g|nφ(n/g)gJd(g)\sum_{g|n}\varphi(n/g)gJ_{d}(g), which is nothing else than ((I1Jd)φ)(n)((I_{1}J_{d})\star\varphi)(n). Thus we want to prove (I1Jd)φ=Jd+1(I_{1}J_{d})\star\varphi=J_{d+1}. Since IlI_{l} is completely multiplicative we have

(I1Jd)φ\displaystyle(I_{1}J_{d})\star\varphi =(I1(Idμ))(I1μ)=Id+1(I1μ)(I1ϵ)μ\displaystyle=(I_{1}(I_{d}\star\mu))\star(I_{1}\star\mu)=I_{d+1}\star(I_{1}\mu)\star(I_{1}\epsilon)\star\mu
=Id+1μ(I1(μϵ))=Jd+1(I1η)=Jd+1η=Jd+1.\displaystyle=I_{d+1}\star\mu\star(I_{1}(\mu\star\epsilon))=J_{d+1}\star(I_{1}\eta)=J_{d+1}\star\eta=J_{d+1}.\qed
Lemma 3.9.

Let KK be an algebraically closed field of characteristic pp and d,nd,n\in\mathbb{N}. Then the number of 𝛇𝔾md(K)\boldsymbol{\zeta}\in\mathbb{G}_{m}^{d}(K) such that ord(𝛇)=n\mathrm{ord}(\boldsymbol{\zeta})=n is Jd(n)J_{d}(n) if (pn(p\nmid n or p=0)p=0) and is 0 if p|np|n and p>0p>0.

Proof.

Note that for all 𝜻𝔾md\boldsymbol{\zeta}\in\mathbb{G}_{m}^{d} we have 𝜻n=𝟏\boldsymbol{\zeta}^{n}=\mathbf{1} if and only if ord(𝜻)|n\mathrm{ord}(\boldsymbol{\zeta})|n. Let Ad(n)A_{d}(n) be the number of elements in 𝔾md\mathbb{G}_{m}^{d} of order nn. The number of solutions 𝜻𝔾md\boldsymbol{\zeta}\in\mathbb{G}_{m}^{d} of 𝜻n=𝟏\boldsymbol{\zeta}^{n}=\mathbf{1} equals (#μn)d(\#\mu_{n})^{d}. If p=0p=0 we have #μn=n\#\mu_{n}=n and if p>0p>0 and n=pαm,pmn=p^{\alpha}m,p\nmid m we have ζn1=(ζm1)pα\zeta^{n}-1=(\zeta^{m}-1)^{p^{\alpha}} and thus ζn=1\zeta^{n}=1 if and only if ζm=1\zeta^{m}=1. Since ζm=1\zeta^{m}=1 and its formal derivative mζm1m\zeta^{m-1} have no common roots, the equation ζm=1\zeta^{m}=1 has mm distinct roots. Therefore #μpαm=m\#\mu_{p^{\alpha}m}=m. Let Bd(n)=(#μn)dB_{d}(n)=(\#\mu_{n})^{d}. Then counting solutions in two ways yields Bd(n)=m|nAd(m).B_{d}(n)=\sum_{m|n}A_{d}(m). So by Möbius inversion we have Ad(n)=m|nμ(m)Bd(n/m)A_{d}(n)=\sum_{m|n}\mu(m)B_{d}(n/m). If p=0p=0 we have Bd(n)=nd=Id(n)B_{d}(n)=n^{d}=I_{d}(n) and therefore Ad=μId=JdA_{d}=\mu\star I_{d}=J_{d} and we are left to prove the lemma in positive characteristic.
So let us assume p>0p>0 and write n=pαmn=p^{\alpha}m where pmp\nmid m. Since pp and mm are coprime we can factor any divisor of nn uniquely in a product of coprime divisors of pαp^{\alpha} and mm respectively. Thus we find

Ad(n)=k|pαl|mμ(kl)Bd(n/(kl))=(k|pαμ(k))(l|mμ(l)(m/l)d)=η(pα)Jd(m)A_{d}(n)=\sum_{k|p^{\alpha}}\sum_{l|m}\mu(kl)B_{d}(n/(kl))=\left(\sum_{k|p^{\alpha}}\mu(k)\right)\left(\sum_{l|m}\mu(l)(m/l)^{d}\right)=\eta(p^{\alpha})J_{d}(m)

using that μϵ=η\mu\star\epsilon=\eta. If p|np|n then pα1p^{\alpha}\neq 1 and hence Ad(n)=0A_{d}(n)=0 since η(pα)=0\eta(p^{\alpha})=0. And if pnp\nmid n we have pα=1,m=np^{\alpha}=1,m=n and hence Ad(n)=Jd(n)A_{d}(n)=J_{d}(n).∎

Definition 3.10.

Let 𝛇(𝔾mn)tors\boldsymbol{\zeta}\in(\mathbb{G}_{m}^{n})_{\mathrm{tors}} and G<𝔾mnG<\mathbb{G}_{m}^{n} an algebraic subgroup. Then we define

ord(𝜻G)=min{ord(𝝃):𝝃𝜻G of finite order}.\mathrm{ord}(\boldsymbol{\zeta}G)=\min\{\mathrm{ord}(\boldsymbol{\xi}):\boldsymbol{\xi}\in\boldsymbol{\zeta}G\text{ of finite order}\}.
Proof of Theorem 1.4.

Since a monoidal transform preserves orders, we can assume that the coset is of the form {𝜼}×𝔾md\{\boldsymbol{\eta}\}\times\mathbb{G}_{m}^{d} for some 𝜼𝔾mnd\boldsymbol{\eta}\in\mathbb{G}_{m}^{n-d} of order gg by Lemma A.17. Then the order of (𝜼,𝜻)(\boldsymbol{\eta,\zeta}) equals lcm(g,ord(𝜻))=g(g,ord(𝜻))ord(𝜻)\mathrm{lcm}(g,\mathrm{ord}(\boldsymbol{\zeta}))=\frac{g}{(g,\mathrm{ord}(\boldsymbol{\zeta}))}\mathrm{ord}(\boldsymbol{\zeta}) for all 𝜻𝔾md\boldsymbol{\zeta}\in\mathbb{G}_{m}^{d}. Fix 1ag1\leq a\leq g and assume that ord(𝜻)amodg\mathrm{ord}(\boldsymbol{\zeta})\equiv a\mod g and (𝜼,𝜻)(\boldsymbol{\eta,\zeta}) is of order bounded by TT. Thus g/(g,a)ord(𝜻)Tg/(g,a)\mathrm{ord}(\boldsymbol{\zeta})\leq T and hence ord(𝜻)T(g,a)/g\mathrm{ord}(\boldsymbol{\zeta})\leq T(g,a)/g.
We first assume that p=0p=0. Then by Lemma 3.9 the number of elements of order nn in 𝔾md\mathbb{G}_{m}^{d} is given by Jd(n)J_{d}(n). We use Theorem 3.7 and Lemma 3.8 to find

(𝜼G)T\displaystyle(\boldsymbol{\eta}G)_{T} =a=1ghT(g,a)/gha(g)Jd(h)=a=1gJd((g,a))(g,a)Td+1(d+1)ζ(d+1)gJd+1(g)+R(T)\displaystyle=\sum_{a=1}^{g}\sum_{\begin{subarray}{c}h\leq T(g,a)/g\\ h\equiv a(g)\end{subarray}}J_{d}(h)=\sum_{a=1}^{g}\frac{J_{d}((g,a))(g,a)T^{d+1}}{(d+1)\zeta(d+1)gJ_{d+1}(g)}+R(T)
=Td+1(d+1)ζ(d+1)gJd+1(g)a=1g(g,a)Jk((g,a))+R(T)=Td+1(d+1)ζ(d+1)g+R(T),\displaystyle=\frac{T^{d+1}}{(d+1)\zeta(d+1)gJ_{d+1}(g)}\sum_{a=1}^{g}(g,a)J_{k}((g,a))+R(T)=\frac{T^{d+1}}{(d+1)\zeta(d+1)g}+R(T),

where R(T)=Og(Tlog(T))R(T)=O_{g}(T\log(T)) if d=1d=1 and R(T)=Od,g(Td)R(T)=O_{d,g}(T^{d}) if d2d\geq 2.
For the case p>0p>0 let us start with some remarks. Since gg is the order of 𝜼\boldsymbol{\eta} it is coprime to pp. Thus there exists an integer p¯\bar{p} such that pp¯1(modg)p\bar{p}\equiv 1\;(\mathrm{mod}\;g). Then for natural numbers hh we have ha(modg)h\equiv a\;(\mathrm{mod}\;g) and h0(modp)h\equiv 0\;(\mathrm{mod}\;p) if and only if happ¯(modgp)h\equiv ap\bar{p}\;(\mathrm{mod}\;gp). Let us also compute (ap¯p,gp)=p(ap¯,g)=p(a,g)(a\bar{p}p,gp)=p(a\bar{p},g)=p(a,g) since (p¯,g)=1(\bar{p},g)=1. By Lemma 3.9 there are also Jd(n)J_{d}(n) torsion points of order nn if (n,p)=1(n,p)=1. Otherwise there are none. Thus we have to subtract

a=1ghT(g,a)/gha(g)h0(p)Jk(h)\displaystyle\sum_{a=1}^{g}\sum_{\begin{subarray}{c}h\leq T(g,a)/g\\ h\equiv a(g)\\ h\equiv 0(p)\end{subarray}}J_{k}(h) =a=1ghT(g,a)/ghapp¯(pg)Jd(h)=a=1gJd(p(g,a))(g,a)Td+1(d+1)ζ(d+1)gJd+1(gp)+R(T)\displaystyle=\sum_{a=1}^{g}\sum_{\begin{subarray}{c}h\leq T(g,a)/g\\ h\equiv ap\bar{p}(pg)\end{subarray}}J_{d}(h)=\sum_{a=1}^{g}\frac{J_{d}(p(g,a))(g,a)T^{d+1}}{(d+1)\zeta(d+1)gJ_{d+1}(gp)}+R(T)
=Jd(p)Td+1Jd+1(p)(d+1)ζ(d+1)gJd+1(g)a=1g(g,a)Jd((g,a))+R(T)\displaystyle=\frac{J_{d}(p)T^{d+1}}{J_{d+1}(p)(d+1)\zeta(d+1)gJ_{d+1}(g)}\sum_{a=1}^{g}(g,a)J_{d}((g,a))+R(T)
=Jd(p)Td+1Jd+1(p)(d+1)ζ(d+1)g+R(T)\displaystyle=\frac{J_{d}(p)T^{d+1}}{J_{d+1}(p)(d+1)\zeta(d+1)g}+R(T)

from the term in characteristic 0 above. Here R(T)R(T) is as above and we used again Theorem 3.7 for the second and Lemma 3.8 for the last equation. Since 1Jd(p)Jd+1(p)=pd+11(pd1)pd+11=p1ppd1-\frac{J_{d}(p)}{J_{d+1}(p)}=\frac{p^{d+1}-1-(p^{d}-1)}{p^{d+1}-1}=\frac{p-1}{p-p^{-d}} the claimed result follows.∎

Remark 3.11.

Let Sa,m,d,T=nTna(m)Jd(n)S_{a,m,d,T}=\sum_{\begin{subarray}{c}n\leq T\\ n\equiv a(m)\end{subarray}}J_{d}(n). Then we have

limTSa,m,d,TSb,m,d,T=p|(a,m)(1pd)p|(b,m)(1pd).\lim_{T\to\infty}\frac{S_{a,m,d,T}}{S_{b,m,d,T}}=\frac{\prod_{p|(a,m)}(1-p^{-d})}{\prod_{p|(b,m)}(1-p^{-d})}.

In particular if we compare the contribution in congruence classes mod pp, the proportions are 1::1:(1pd)1:\dots:1:(1-p^{-d}) and hence the proportion of all the coprime congruence classes is (p1)/(ppd)(p-1)/(p-p^{-d}), 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 KK always denotes an algebraically closed field.

Definition 4.1.

Let n,mn,m\in\mathbb{N} and AMatn×m()A\in\mathrm{Mat}_{n\times m}(\mathbb{Z}). Then we call a factorisation A=TDSA=TDS Smith normal form of AA if SGLn(),TGLm()S\in\mathrm{GL}_{n}(\mathbb{Z}),T\in\mathrm{GL}_{m}(\mathbb{Z}) and D=diagn×m(𝛂)D=\mathrm{diag}_{n\times m}(\boldsymbol{\alpha}) for some 𝛂0n\boldsymbol{\alpha}\in\mathbb{Z}_{\geq 0}^{n} such that αi+1αi\alpha_{i+1}\mathbb{Z}\subset\alpha_{i}\mathbb{Z} for all 1i<min{n,m}1\leq i<\min\{n,m\}.

Remark 4.2.

That every matrix AMatn×m()A\in\mathrm{Mat}_{n\times m}(\mathbb{Z}) has a Smith normal form is a classical result established in [Smi61].

Definition 4.3.

Let PK[X1±1,,Xn±1]{0}P\in K[X_{1}^{\pm 1},\dots,X_{n}^{\pm 1}]\setminus\{0\} be a Laurent polynomial. There exists a unique polynomial QK[X1,Xn]Q\in K[X_{1},\dots X_{n}] which is coprime to X1XnX_{1}\cdots X_{n} and 𝐯n\mathbf{v}\in\mathbb{Z}^{n} such that P=(X1,,Xn)𝐯QP=(X_{1},\dots,X_{n})^{\mathbf{v}}Q. We define the Laurent degree as Ldeg(P)=deg(Q)\mathrm{Ldeg}(P)=\deg(Q).

Remark 4.4.

Since the zero locus of PP and QQ above in 𝔾mn\mathbb{G}_{m}^{n} are equal, a Laurent polynomial PK[X±1]{0}P\in K[X^{\pm 1}]\setminus\{0\} of degree d=Ldeg(P)d=\mathrm{Ldeg}(P) has at most dd roots.

Lemma 4.5.

Let PK[X1±1,,Xn±1]P\in K[X_{1}^{\pm 1},\dots,X_{n}^{\pm 1}], 𝐲𝔾mn\mathbf{y}\in\mathbb{G}_{m}^{n} and 𝐮n\mathbf{u}\in\mathbb{Z}^{n}. Suppose that P(𝐲Y𝐮)0P\left(\mathbf{y}Y^{\mathbf{u}^{\top}}\right)\neq 0. Then the Laurent degree of P(𝐲Y𝐮)K[Y±1]P\left(\mathbf{y}Y^{\mathbf{u}^{\top}}\right)\in K[Y^{\pm 1}] is bounded by 2|𝐮|Ldeg(P)2|\mathbf{u}|\mathrm{Ldeg}(P).

Proof.

Let RK[Y]{0}R\in K[Y]\setminus\{0\}, s0=minSupp(R)s_{0}=\min\mathrm{Supp}(R) and s1=maxSupp(R)s_{1}=\max\mathrm{Supp}(R). Then we have Ldeg(R)=s1s0\mathrm{Ldeg}(R)=s_{1}-s_{0}. By definition there exists 𝐯n\mathbf{v}\in\mathbb{Z}^{n} and QK[X1,,Xn]Q\in K[X_{1},\dots,X_{n}] of degree Ldeg(P)\mathrm{Ldeg}(P) such that P=(X1,,Xn)𝐯QP=(X_{1},\dots,X_{n})^{\mathbf{v}}Q. Let R=P(𝐲Y𝐮)R=P\left(\mathbf{y}Y^{\mathbf{u}^{\top}}\right) and suppose that sSupp(R)s\in\mathrm{Supp}(R). Then there exists 𝐢Supp(P)\mathbf{i}\in\mathrm{Supp}(P) such that s=𝐮𝐢s=\mathbf{u^{\top}i} and hence 𝐣Supp(Q)\mathbf{j}\in\mathrm{Supp}(Q) such that s=𝐮𝐯+𝐮𝐣s=\mathbf{u^{\top}v+u^{\top}j}. We can bound |𝐮𝐣|i=1n|ui|ji|𝐮|i=1nji|𝐮|deg(Q)|\mathbf{u^{\top}j}|\leq\sum_{i=1}^{n}|u_{i}|j_{i}\leq|\mathbf{u}|\sum_{i=1}^{n}j_{i}\leq|\mathbf{u}|\deg(Q). Thus there exist r0,r1r_{0},r_{1}\in\mathbb{Z} bounded by |𝐮|deg(Q)|\mathbf{u}|\deg(Q) such that si=𝐮𝐯+ri,i=0,1s_{i}=\mathbf{u^{\top}v}+r_{i},i=0,1. Then we find Ldeg(R)=s1s0=r1r02|𝐮|Ldeg(P)\mathrm{Ldeg}(R)=s_{1}-s_{0}=r_{1}-r_{0}\leq 2|\mathbf{u}|\mathrm{Ldeg}(P).∎

Definition 4.6.

We call a set X𝔾mnX\subset\mathbb{G}_{m}^{n} algebraic over KK if there exists polynomials P1,,PmK[X1±1,,Xn±1]P_{1},\dots,P_{m}\in K[X_{1}^{\pm 1},\dots,X_{n}^{\pm 1}] such that

X={𝐱𝔾mn:P1(𝐱)==Pm(𝐱)=0}.X=\{\mathbf{x}\in\mathbb{G}_{m}^{n}:P_{1}(\mathbf{x})=\dots=P_{m}(\mathbf{x})=0\}.

Let X𝔾mnX\subset\mathbb{G}_{m}^{n} be an algebraic subset of dimension dd. We call it admissible if it is finite or d1d\geq 1 and it does not contain a torsion coset of dimension dd. We define

X=XCX torsion cosetdim(C)1C.X^{*}=X\setminus\bigcup_{\begin{subarray}{c}C\subset X\text{ torsion coset}\\ \dim(C)\geq 1\end{subarray}}C.

For PK[X1±1,,Xn±1]P\in K[X_{1}^{\pm 1},\dots,X_{n}^{\pm 1}] we let Z(P)={𝐱𝔾mn(K):P(𝐱)=0}Z(P)=\{\mathbf{x}\in\mathbb{G}_{m}^{n}(K):P(\mathbf{x})=0\} be the zero locus of PP in 𝔾mn.\mathbb{G}_{m}^{n}.

Lemma 4.7.

Let n2n\geq 2 and 𝐚1,𝐚n1n\mathbf{a}_{1},\dots\mathbf{a}_{n-1}\in\mathbb{Z}^{n} be linearly independent and PK[X1±1,,Xn±1]{0}P\in K[X_{1}^{\pm 1},\dots,X_{n}^{\pm 1}]\setminus\{0\} and X=Z(P)X=Z(P). Then

#(Xi=1n1{𝐱𝔾mn:𝐱𝐚i=1})2(n1)!Ldeg(P)|𝐚1||𝐚n1|.\#\left(X^{*}\cap\bigcap_{i=1}^{n-1}\left\{\mathbf{x}\in\mathbb{G}_{m}^{n}:\mathbf{x}^{\mathbf{a}_{i}}=1\right\}\right)\leq 2(n-1)!\mathrm{Ldeg}(P)|\mathbf{a}_{1}|\cdots|\mathbf{a}_{n-1}|.
Proof.

A suitable version of Bézouts Theorem would suffice for our purposes. However we follow a more elementary and self-contained approach. Let A=(𝐚1,,𝐚n1)Matn×(n1)()A=\begin{pmatrix}\mathbf{a}_{1},\dots,\mathbf{a}_{n-1}\end{pmatrix}\in\mathrm{Mat}_{n\times(n-1)}(\mathbb{Z}). Let A=SDTA=SDT be a Smith normal form of AA with D=diagn×(n1)(α1,,αn1)D=\mathrm{diag}_{n\times(n-1)}(\alpha_{1},\dots,\alpha_{n-1}) and note that α1,,αn11\alpha_{1},\dots,\alpha_{n-1}\geq 1. Let 𝐮1,,𝐮nn\mathbf{u}_{1},\dots,\mathbf{u}_{n}\in\mathbb{Z}^{n} such that S1=(𝐮1𝐮n)S^{-1}=\begin{pmatrix}\mathbf{u}_{1}^{\top}\\ \vdots\\ \mathbf{u}_{n}^{\top}\end{pmatrix}. We claim that

i=1n1{𝐱𝔾mn:𝐱𝐚i=1}={y1𝐮1yn𝐮n:yiαi=1 for i=1,,n1 and yn𝔾mn}.\bigcap_{i=1}^{n-1}\left\{\mathbf{x}\in\mathbb{G}_{m}^{n}:\mathbf{x}^{\mathbf{a}_{i}}=1\right\}=\left\{y_{1}^{\mathbf{u}_{1}^{\top}}\cdots y_{n}^{\mathbf{u}_{n}^{\top}}:y_{i}^{\alpha_{i}}=1\text{ for }i=1,\dots,n-1\text{ and }y_{n}\in\mathbb{G}_{m}^{n}\right\}.

Note that 𝐱𝔾mn\mathbf{x}\in\mathbb{G}_{m}^{n} is in the intersection on the left if and only if 𝐱A=𝟏\mathbf{x}^{A}=\mathbf{1} and since TT has an inverse in Matn1()\mathrm{Mat}_{n-1}(\mathbb{Z}) this is equivalent to 𝐱SD=𝟏\mathbf{x}^{SD}=\mathbf{1}. Let 𝐲=𝐱S\mathbf{y=x}^{S}. Then 𝐲D=𝟏\mathbf{y}^{D}=\mathbf{1} if and only if yiαi=1y_{i}^{\alpha_{i}}=1 for i=1,,n1i=1,\dots,n-1. There is no condition on yny_{n}. Therefore any 𝐱𝔾mn\mathbf{x}\in\mathbb{G}_{m}^{n} such that 𝐱SD=𝟏\mathbf{x}^{SD}=\mathbf{1} is of the form 𝐱=𝐲S1=y1𝐮1yn𝐮n\mathbf{x=y}^{S^{-1}}=y_{1}^{\mathbf{u}_{1}^{\top}}\cdots y_{n}^{\mathbf{u}_{n}^{\top}} where yiαi=1,i<ny_{i}^{\alpha_{i}}=1,i<n and yn𝔾m.y_{n}\in\mathbb{G}_{m}. Note that 𝐮n\mathbf{u}_{n} is primitive since it is a row of S1GLn()S^{-1}\in\mathrm{GL}_{n}(\mathbb{Z}). We also find

𝐮nA=𝐞nS1A=𝐞nDT=𝟎.\mathbf{u}_{n}^{\top}A=\mathbf{e}_{n}^{\top}S^{-1}A=\mathbf{e}_{n}^{\top}DT=\mathbf{0}.

Since the kernel of AA^{\top} is one dimensional this shows that primitivity and being in the kernel of AA^{\top} determines 𝐮n\mathbf{u}_{n} up to sign.
Let g=dn1(A)g=d_{n-1}(A) and vi=det(𝐞i,A)/g.v_{i}=\det(\mathbf{e}_{i},A)/g\in\mathbb{Z}. Since gcd(v1,,vn)=dn1(A)/g=1\gcd(v_{1},\dots,v_{n})=d_{n-1}(A)/g=1, the vector 𝐯=(v1,,vn)\mathbf{v}=(v_{1},\dots,v_{n}) is primitive. Then for any j=1,,n1j=1,\dots,n-1 we have

𝐚j𝐯=a1,jdet(𝐞1,A)/g++an,jdet(𝐞n,A)/g=det(𝐚j,A)/g=0\mathbf{a}_{j}^{\top}\mathbf{v}=a_{1,j}\det(\mathbf{e}_{1},A)/g+\dots+a_{n,j}\det(\mathbf{e}_{n},A)/g=\det(\mathbf{a}_{j},A)/g=0

and hence |𝐮n|=|𝐯|(n1)!|𝐚1||𝐚n1|/g|\mathbf{u}_{n}|=|\mathbf{v}|\leq(n-1)!|\mathbf{a}_{1}|\cdots|\mathbf{a}_{n-1}|/g by the Leibniz formula. Lemmas B.5 and B.7 imply that g=dn1(D)=α1αn1g=d_{n-1}(D)=\alpha_{1}\cdots\alpha_{n-1}.
Now fix (y1,,yn1)μα1××μαn1(y_{1},\dots,y_{n-1})\in\mu_{\alpha_{1}}\times\dots\times\mu_{\alpha_{n-1}}. If P(y1𝐮1yn1𝐮n1Y𝐮n)=0P\left(y_{1}^{\mathbf{u}_{1}^{\top}}\cdots y_{n-1}^{\mathbf{u}_{n-1}^{\top}}Y^{\mathbf{u}_{n}^{\top}}\right)=0, then for all yn𝔾my_{n}\in\mathbb{G}_{m} the element (y1𝐮1yn𝐮n)(y_{1}^{\mathbf{u}_{1}^{\top}}\cdots y_{n}^{\mathbf{u}_{n}^{\top}}) is contained in a torsion coset of positive dimension and hence {y1𝐮1yn𝐮n:yn𝔾m}X=\{y_{1}^{\mathbf{u}_{1}^{\top}}\cdots y_{n}^{\mathbf{u}_{n}^{\top}}:y_{n}\in\mathbb{G}_{m}\}\cap X^{*}=\emptyset.
In the other case an element 𝐱{y1𝐮1yn𝐮n:yn𝔾m}X\mathbf{x}\in\{y_{1}^{\mathbf{u}_{1}^{\top}}\cdots y_{n}^{\mathbf{u}_{n}^{\top}}:y_{n}\in\mathbb{G}_{m}\}\cap X is of the form y1𝐮1yn𝐮ny_{1}^{\mathbf{u}_{1}^{\top}}\cdots y_{n}^{\mathbf{u}_{n}^{\top}} for some root yn𝔾my_{n}\in\mathbb{G}_{m} of the Laurent polynomial Q(Y)=P(y1𝐮1yn1𝐮n1Y𝐮n)Q(Y)=P\left(y_{1}^{\mathbf{u}_{1}^{\top}}\cdots y_{n-1}^{\mathbf{u}_{n-1}^{\top}}Y^{\mathbf{u}_{n}^{\top}}\right). By Lemma 4.5 we have Ldeg(Q)2|𝐮n|Ldeg(P)\mathrm{Ldeg}(Q)\leq 2|\mathbf{u}_{n}|\mathrm{Ldeg}(P). Since |𝐮n|(n1)!|𝐚1||𝐚n1|/g|\mathbf{u}_{n}|\leq(n-1)!|\mathbf{a}_{1}|\cdots|\mathbf{a}_{n-1}|/g we have

#({y1𝐮1yn𝐮n:yn𝔾mn}X)2(n1)!Ldeg(P)|𝐚1||𝐚n1|/g.\#\left(\left\{y_{1}^{\mathbf{u}_{1}^{\top}}\cdots y_{n}^{\mathbf{u}_{n}^{\top}}:y_{n}\in\mathbb{G}_{m}^{n}\right\}\cap X\right)\leq 2(n-1)!\mathrm{Ldeg}(P)|\mathbf{a}_{1}|\cdots|\mathbf{a}_{n-1}|/g.

Summing over all (y1,,yn1)μα1××μαn1(y_{1},\dots,y_{n-1})\in\mu_{\alpha_{1}}\times\dots\times\mu_{\alpha_{n-1}} - which are at most gg distinct elements - we get the claimed bound.∎

Lemma 4.8.

Let n2,α1,,αn1[0,1),β(1,)n\geq 2,\alpha_{1},\dots,\alpha_{n-1}\in[0,1),\beta\in(-1,\infty) and A=i=1n1(1αi)A=\prod_{i=1}^{n-1}(1-\alpha_{i}). Then for all T1T\geq 1 we have

k1Tα1ki(T/(k1ki1))αikn1(T/(k1kn2))αn1(k1kn1)β𝜶,βT(β+1)(1A).\sum_{k_{1}\leq T^{\alpha_{1}}}\cdots\sum_{k_{i}\leq(T/(k_{1}\cdots k_{i-1}))^{\alpha_{i}}}\cdots\sum_{k_{n-1}\leq(T/(k_{1}\cdots k_{n-2}))^{\alpha_{n-1}}}(k_{1}\cdots k_{n-1})^{\beta}\ll_{\boldsymbol{\alpha},\beta}T^{(\beta+1)(1-A)}.
Proof.

Comparing the sum with the integral yields that kxkααxα+1\sum_{k\leq x}k^{\alpha}\ll_{\alpha}x^{\alpha+1} as xx\to\infty for all real α>1\alpha>-1. The proof is by induction on nn. If n=2n=2 we have

kTαkββTα(β+1)=T(β+1)(1(1α)).\sum_{k\leq T^{\alpha}}k^{\beta}\ll_{\beta}T^{\alpha(\beta+1)}=T^{(\beta+1)(1-(1-\alpha))}.

So assume that n3n\geq 3 and the lemma holds for all mn1m\leq n-1. Let A=i=2n1(1αi)A^{\prime}=\prod_{i=2}^{n-1}(1-\alpha_{i}). Then

k1Tα1ki(T/(k1ki1))αikn1(T/(k1kn2))αn1(k1kn1)β\displaystyle\sum_{k_{1}\leq T^{\alpha_{1}}}\cdots\sum_{k_{i}\leq(T/(k_{1}\cdots k_{i-1}))^{\alpha_{i}}}\cdots\sum_{k_{n-1}\leq(T/(k_{1}\cdots k_{n-2}))^{\alpha_{n-1}}}(k_{1}\cdots k_{n-1})^{\beta}
=k1Tα1k1βk2(T/k1)αikn1((T/k1)/(k2kn2))αn1(k2kn1)β\displaystyle\qquad=\sum_{k_{1}\leq T^{\alpha_{1}}}k_{1}^{\beta}\sum_{k_{2}\leq(T/k_{1})^{\alpha_{i}}}\cdots\sum_{k_{n-1}\leq((T/k_{1})/(k_{2}\cdots k_{n-2}))^{\alpha_{n-1}}}(k_{2}\cdots k_{n-1})^{\beta}
βk1Tα1k1β(T/k1)(β+1)(1A)=T(β+1)(1A)k1Tα1k1β(β+1)(1A)\displaystyle\qquad\ll_{\beta}\sum_{k_{1}\leq T^{\alpha_{1}}}k_{1}^{\beta}(T/k_{1})^{(\beta+1)(1-A^{\prime})}=T^{(\beta+1)(1-A^{\prime})}\sum_{k_{1}\leq T^{\alpha_{1}}}k_{1}^{\beta-(\beta+1)(1-A^{\prime})}

by induction since T/k1T1α11T/k_{1}\geq T^{1-\alpha_{1}}\geq 1.
We claim that the exponent β(β+1)(1A)>1\beta-(\beta+1)(1-A^{\prime})>-1. This is equivalent to β+1>(β+1)(1A)\beta+1>(\beta+1)(1-A^{\prime}) and since β>1\beta>-1 also to A>0A^{\prime}>0 which is true since the αi\alpha_{i} are in [0,1)[0,1). Thus we can bound

k1Tα1k1β(β+1)(1A)𝜶,βTα1(β+1(β+1)(1A))=Tα1(β+1)A.\sum_{k_{1}\leq T^{\alpha_{1}}}k_{1}^{\beta-(\beta+1)(1-A^{\prime})}\ll_{\boldsymbol{\alpha},\beta}T^{\alpha_{1}(\beta+1-(\beta+1)(1-A^{\prime}))}=T^{\alpha_{1}(\beta+1)A^{\prime}}.

So the exponent in the bound equals

(β+1)(1A)+(β+1)α1A=(β+1)(1A(1α1))=(β+1)(1A).(\beta+1)(1-A^{\prime})+(\beta+1)\alpha_{1}A^{\prime}=(\beta+1)(1-A^{\prime}(1-\alpha_{1}))=(\beta+1)(1-A).\qed
Definition 4.9.

Let Fn=[1,1]nF_{n}=[-1,1]^{n} and Λn\Lambda\subset\mathbb{Z}^{n} a subgroup of rank nn. Then the successive minima of Λ\Lambda with respect to FnF_{n} are defined as

λi(Λ)\displaystyle\lambda_{i}(\Lambda) =inf{λ0:λFn contains i linear independent lattice points of Λ}\displaystyle=\inf\{\lambda\in\mathbb{R}_{\geq 0}:\lambda F_{n}\text{ contains }i\text{ linear independent lattice points of }\Lambda\}
=inf{λ0: there exist i linear indepedent points of Λ of norm at most λ}\displaystyle=\inf\{\lambda\in\mathbb{R}_{\geq 0}:\text{ there exist }i\text{ linear indepedent points of }\Lambda\text{ of norm at most }\lambda\}

where the norm of a vector 𝐯n\mathbf{v}\in\mathbb{Z}^{n} is given by |𝐯||\mathbf{v}|.

Lemma 4.10.

Let nn\in\mathbb{N} and Λn\Lambda\subset\mathbb{Z}^{n} be a subgroup of rank nn. Then λ1(Λ)λn(Λ)[n:Λ]\lambda_{1}(\Lambda)\cdots\lambda_{n}(\Lambda)\leq[\mathbb{Z}^{n}:\Lambda].

Proof.

This follows from Minkowski’s second theorem (Theorem V on page 218 in [Cas97]) and the fact that the volume of FnF_{n} equals 2n2^{n}. Furter the determinant of Λ\Lambda is given by the absolute value of the determinant of a matrix BB whose columns form a basis of Λ\Lambda. Since det(Λ)=|det(B)|=[n:Bn]=[n:Λ]\det(\Lambda)=|\det(B)|=[\mathbb{Z}^{n}:B\mathbb{Z}^{n}]=[\mathbb{Z}^{n}:\Lambda] the determinant is given by [n:Λ][\mathbb{Z}^{n}:\Lambda].∎

Definition 4.11.

For 𝛇(𝔾mn)tors\boldsymbol{\zeta}\in(\mathbb{G}_{m}^{n})_{\mathrm{tors}} we define Λ𝛇={𝐚n:𝛇𝐚=1}\Lambda_{\boldsymbol{\zeta}}=\left\{\mathbf{a}\in\mathbb{Z}^{n}:\boldsymbol{\zeta}^{\mathbf{a}}=1\right\}.

Lemma 4.12.

Let 𝛇(𝔾mn)tors\boldsymbol{\zeta}\in(\mathbb{G}_{m}^{n})_{\mathrm{tors}} be of order NN. Then Λ𝛇\Lambda_{\boldsymbol{\zeta}} is a subgroup of n\mathbb{Z}^{n} of rank nn and [n:Λ𝛇]=N[\mathbb{Z}^{n}:\Lambda_{\boldsymbol{\zeta}}]=N.

Proof.

Since Λ𝜻\Lambda_{\boldsymbol{\zeta}} contains NnN\mathbb{Z}^{n}, it has full rank. Consider the group homomorphism ϕ:n𝔾m\phi:\mathbb{Z}^{n}\mapsto\mathbb{G}_{m}, given by ϕ(𝐚)=𝜻𝐚\phi(\mathbf{a})=\boldsymbol{\zeta}^{\mathbf{a}}. Then ker(ϕ)=Λ𝜻\ker(\phi)=\Lambda_{\boldsymbol{\zeta}} and Im(ϕ)=μN\mathrm{Im}(\phi)=\mu_{N}. By the isomorphism theorem we find [n:Λ𝜻]=#μN=N.[\mathbb{Z}^{n}:\Lambda_{\boldsymbol{\zeta}}]=\#\mu_{N}=N. The last equality holds clearly in characteristic zero. It is also true in positive characteristic p>0p>0 since the order of an element in (𝔾mn)tors(\mathbb{G}_{m}^{n})_{\mathrm{tors}} is always coprime to pp.∎

Definition 4.13.

For a set V𝔾mnV\subset\mathbb{G}_{m}^{n} and T1T\geq 1 we define

VT={𝜻V(𝔾mn)tors:ord(𝜻)T}.V_{T}=\{\boldsymbol{\zeta}\in V\cap(\mathbb{G}_{m}^{n})_{\mathrm{tors}}:\mathrm{ord}(\boldsymbol{\zeta})\leq T\}.
Definition 4.14.

A hypersurface X𝔾mnX\subset\mathbb{G}_{m}^{n} is the zero set Z(P)Z(P) of PK[X1±1,,Xn±1](K[X1±1,,Xn±1]{0})P\in K[X_{1}^{\pm 1},\dots,X_{n}^{\pm 1}]\setminus(K[X_{1}^{\pm 1},\dots,X_{n}^{\pm 1}]^{*}\cup\{0\}) in 𝔾mn\mathbb{G}_{m}^{n}.

Remark 4.15.

Let PK[X1±1,,Xn±1](K[X1±1,,Xn±1]{0}).P\in K[X_{1}^{\pm 1},\dots,X_{n}^{\pm 1}]\setminus(K[X_{1}^{\pm 1},\dots,X_{n}^{\pm 1}]^{*}\cup\{0\}). Then the dimension of Z(P)Z(P) equals n1n-1.

Lemma 4.16.

Let X=Z(P)X=Z(P) be a hypersurface in 𝔾mn\mathbb{G}_{m}^{n}. Then

#XTnLdeg(P)Tn1/n.\#X^{*}_{T}\ll_{n}\mathrm{Ldeg}(P)T^{n-1/n}.
Proof.

Let 𝜻𝔾mn\boldsymbol{\zeta}\in\mathbb{G}_{m}^{n} be of order NTN\leq T. Since there are only finitely many points in n\mathbb{Z}^{n} of finite norm, the infimum in the definition of λi\lambda_{i} is a minimum. Thus there exists a system of nn linear independent vectors 𝐚1,,𝐚nΛ𝜻\mathbf{a}_{1},\dots,\mathbf{a}_{n}\in\Lambda_{\boldsymbol{\zeta}} such that |𝐚i|=λi(Λ𝜻)|\mathbf{a}_{i}|=\lambda_{i}(\Lambda_{\boldsymbol{\zeta}}) for all i=1,,ni=1,\dots,n. Then by Lemmas 4.10 and 4.12 we have |𝐚1||𝐚n|[n:Λ𝜻]=NT|\mathbf{a}_{1}|\cdots|\mathbf{a}_{n}|\leq[\mathbb{Z}^{n}:\Lambda_{\boldsymbol{\zeta}}]=N\leq T. Since |𝐚1||𝐚n||\mathbf{a}_{1}|\leq\dots\leq|\mathbf{a}_{n}| we have |𝐚i|n(i1)T/(|𝐚1||𝐚i1|)|\mathbf{a}_{i}|^{n-(i-1)}\leq T/(|\mathbf{a}_{1}|\cdots|\mathbf{a}_{i-1}|) for all i=1,,ni=1,\dots,n which yields |𝐚i|(T/(|𝐚1||𝐚i1|))1/(n(i1))|\mathbf{a}_{i}|\leq\left(T/(|\mathbf{a}_{1}|\cdots|\mathbf{a}_{i-1}|)\right)^{1/(n-(i-1))}. Therefore

(𝔾mn)T\displaystyle(\mathbb{G}_{m}^{n})_{T} {𝜻𝔾mn: there exist linearly independent 𝐚1,,𝐚n such that 𝜻𝐚i=1,\displaystyle\subset\{\boldsymbol{\zeta}\in\mathbb{G}_{m}^{n}:\text{ there exist linearly independent }\mathbf{a}_{1},\dots,\mathbf{a}_{n}\text{ such that }\boldsymbol{\zeta}^{\mathbf{a}_{i}}=1,
 and |𝐚i|(T/(|𝐚1||𝐚i1|))1/(ni+1) for all i=1,,n}.\displaystyle\quad\quad\text{ and }|\mathbf{a}_{i}|\leq\left(T/(|\mathbf{a}_{1}|\cdots|\mathbf{a}_{i-1}|)\right)^{1/(n-i+1)}\text{ for all }i=1,\dots,n\}.

The number of vectors 𝐚n\mathbf{a}\in\mathbb{Z}^{n} satisfying |𝐚|=k|\mathbf{a}|=k is bounded by a constant depending on nn times kn1k^{n-1} . By Lemma 4.7 have

#XT\displaystyle\#X^{*}_{T} 𝐚1,,𝐚n1#(Xi=1n1{𝐱𝔾mn:𝐱𝐚i=1})\displaystyle\leq\sum_{\mathbf{a}_{1},\dots,\mathbf{a}_{n-1}}\#\left(X^{*}\cap\bigcap_{i=1}^{n-1}\left\{\mathbf{x}\in\mathbb{G}_{m}^{n}:\mathbf{x}^{\mathbf{a}_{i}}=1\right\}\right)
nLdeg(P)𝐚1,,𝐚n1|𝐚1||𝐚n1|\displaystyle\ll_{n}\mathrm{Ldeg}(P)\sum_{\mathbf{a}_{1},\dots,\mathbf{a}_{n-1}}|\mathbf{a}_{1}|\cdots|\mathbf{a}_{n-1}|
nLdeg(P)k1,,kn1(k1kn1)n;\displaystyle\ll_{n}\mathrm{Ldeg}(P)\sum_{k_{1},\dots,k_{n-1}}(k_{1}\cdots k_{n-1})^{n};

in the first two sums the 𝐚i\mathbf{a}_{i} satisfy |𝐚i|(T/(|𝐚1||𝐚i1|))1/(ni+1)|\mathbf{a}_{i}|\leq\left(T/(|\mathbf{a}_{1}|\cdots|\mathbf{a}_{i-1}|)\right)^{1/(n-i+1)}, in the last step we sum over ki=|𝐚i|k_{i}=|\mathbf{a}_{i}| such that ki(T/(k1ki1))1/(ni+1)k_{i}\leq\left(T/(k_{1}\cdots k_{i-1})\right)^{1/(n-i+1)}. The last sum can be bounded by Lemma 4.8 with αi=1/(ni+1)\alpha_{i}=1/(n-i+1) and β=n\beta=n. We have i=1n1(1αi)=i=2ni1i=1/n\prod_{i=1}^{n-1}(1-\alpha_{i})=\prod_{i=2}^{n}\frac{i-1}{i}=1/n and thus the exponent equals (n+1)(11/n)=n1/n(n+1)(1-1/n)=n-1/n.∎

5 Common roots and admissibility

This section contains technical statements on algebraic subgroups of 𝔾mn\mathbb{G}_{m}^{n}. As we work in arbitrary characteristic and as the literature (cf. [BG06]) concerns mainly characteristic zero we give full proofs.

Lemma 5.1.

Let m,nm,n\in\mathbb{N}, UMatn×m()U\in\mathrm{Mat}_{n\times m}(\mathbb{Z}) of rank rr and 𝛇(𝔾mm)tors\boldsymbol{\zeta}\in(\mathbb{G}_{m}^{m})_{\mathrm{tors}} be a torsion point. Then the set of 𝐱𝔾mn\mathbf{x}\in\mathbb{G}_{m}^{n} such that 𝐱U=𝛇\mathbf{x}^{U}=\boldsymbol{\zeta} is either empty or a finite union of torsion cosets of dimension nrn-r. Moreover the union is not empty if r=mr=m.

Proof.

Suppose that there exists 𝐲𝔾mn\mathbf{y}\in\mathbb{G}_{m}^{n} satisfies 𝐲U=𝜻\mathbf{y}^{U}=\boldsymbol{\zeta}. Let Λ=Um\Lambda=U\mathbb{Z}^{m}. Then 𝐱U=𝜻\mathbf{x}^{U}=\boldsymbol{\zeta} if and only if 𝐱𝐲HΛ\mathbf{x}\in\mathbf{y}H_{\Lambda} and we can conclude with Lemma A.4. For the second statement let U=SDTU=SDT be a Smith normal form of UU, where D=diagn×m(α1,,αm)D=\mathrm{diag}_{n\times m}(\alpha_{1},\dots,\alpha_{m}). Let a=α1αma=\alpha_{1}\cdots\alpha_{m} and βi=a/αi,i=1,,m\beta_{i}=a/\alpha_{i},i=1,\dots,m. Let V=T1diagm×n(β1,,βm)S1Matm×n()V=T^{-1}\mathrm{diag}_{m\times n}(\beta_{1},\dots,\beta_{m})S^{-1}\in\mathrm{Mat}_{m\times n}(\mathbb{Z}). Then VU=aEmVU=aE_{m}. Since 𝔾m\mathbb{G}_{m} is algebraically closed we can find 𝜼𝔾mn\boldsymbol{\eta}\in\mathbb{G}_{m}^{n} such that 𝜼a=𝜻\boldsymbol{\eta}^{a}=\boldsymbol{\zeta}. Then 𝐲=𝜼V\mathbf{y}=\boldsymbol{\eta}^{V} satisfies 𝐲U=𝜼a=𝜻.\mathbf{y}^{U}=\boldsymbol{\eta}^{a}=\boldsymbol{\zeta}.

Definition 5.2.

We call 𝐮n{0}\mathbf{u}\in\mathbb{Z}^{n}\setminus\{0\} primitive if its entries are coprime.

Lemma 5.3.

Let 𝐮n\mathbf{u}\in\mathbb{Z}^{n} and z𝔾mz\in\mathbb{G}_{m}. Then (X1,,Xn)𝐮zK[X1±1,,Xn±1](X_{1},\dots,X_{n})^{\mathbf{u}}-z\in K[X_{1}^{\pm 1},\dots,X_{n}^{\pm 1}] is irreducible if and only if 𝐮\mathbf{u} is primitive.

Proof.

Note that for all AGLn()A\in\mathrm{GL}_{n}(\mathbb{Z}) the map ϕA:𝐢np𝐢(X1,,Xn)𝐢𝐢np𝐢(X1,,Xn)A𝐢\phi_{A}:\sum_{\mathbf{i}\in\mathbb{Z}^{n}}p_{\mathbf{i}}(X_{1},\dots,X_{n})^{\mathbf{i}}\mapsto\sum_{\mathbf{i}\in\mathbb{Z}^{n}}p_{\mathbf{i}}(X_{1},\dots,X_{n})^{A\mathbf{i}} is an automorphism of K[X1±1,Xn±1]K[X_{1}^{\pm 1},\dots X_{n}^{\pm 1}]. By the theorem of elementary divisors (Theorem III.7.8 in [Lan02]) there exists λ{0}\lambda\in\mathbb{Z}\setminus\{0\} and a basis 𝐛1,,𝐛n\mathbf{b}_{1},\dots,\mathbf{b}_{n} of n\mathbb{Z}^{n} such that 𝐮=λ𝐛1\mathbf{u}=\lambda\mathbf{b}_{1}. Let B=(𝐛1,,𝐛n)B=(\mathbf{b}_{1},\dots,\mathbf{b}_{n}). Then 𝐗𝐮z\mathbf{X^{u}}-z is irreducible if and only if ϕB1(𝐗𝐮z)=𝐗λB1𝐛1z=X1λz\phi_{B^{-1}}(\mathbf{X^{u}}-z)=\mathbf{X}^{\lambda B^{-1}\mathbf{b}_{1}}-z=X_{1}^{\lambda}-z is irreducible too. It is not hard to show that this is the case if and only if 1=|λ|=gcd(𝐮)1=|\lambda|=\gcd(\mathbf{u}).∎

Lemma 5.4.

Let R1,,RmK[X1±1,,Xn±1]{0}R_{1},\dots,R_{m}\in K[X_{1}^{\pm 1},\dots,X_{n}^{\pm 1}]\setminus\{0\} each be a product of factors of the form 𝐗𝐮ζ\mathbf{X^{u}}-\zeta, where ζ\zeta is a root of unity and 𝐮n\mathbf{u}\in\mathbb{Z}^{n} is primitive. Assume that m2m\geq 2 and R1,,RmR_{1},\dots,R_{m} are coprime. Then the set of their common roots is a finite union of torsion cosets of dimension at most n2n-2.

Proof.

If there are empty products there are no common roots, therefore we can assume that there are no RiR_{i} which equal 11. Suppose that 𝐱𝔾mn\mathbf{x}\in\mathbb{G}_{m}^{n} is a common root. Then for each i{1,m}i\in\{1,\dots m\} there exist a primitive 𝐮in\mathbf{u}_{i}\in\mathbb{Z}^{n} and ζi\zeta_{i} a root of unity such that 𝐗𝐮iζi|Ri\mathbf{X}^{\mathbf{u}_{i}}-\zeta_{i}|R_{i} and 𝐱𝐮i=ζi\mathbf{x}^{\mathbf{u}_{i}}=\zeta_{i}. Let U=(𝐮1,,𝐮m)Matn×m()U=(\mathbf{u}_{1},\dots,\mathbf{u}_{m})\in\mathrm{Mat}_{n\times m}(\mathbb{Z}) and 𝜻=(ζ1,,ζm)\boldsymbol{\zeta}=(\zeta_{1},\dots,\zeta_{m}). Then 𝐱U=𝜻\mathbf{x}^{U}=\boldsymbol{\zeta}. Let U=SDTU=SDT be a smith normal form of UU. The rank jj of UU is the largest index such that αj0\alpha_{j}\neq 0. Since the 𝐮i\mathbf{u}_{i} are primitive we have U𝟎U\neq\mathbf{0} and therefore there exists such a jj. Assume that j=1j=1. Then we find

𝐮i=U𝐞i=SDT𝐞i=SD𝐭i=α1t1,iS𝐞1=α1t1,i𝐬1.\mathbf{u}_{i}=U\mathbf{e}_{i}=SDT\mathbf{e}_{i}=SD\mathbf{t}_{i}=\alpha_{1}t_{1,i}S\mathbf{e}_{1}=\alpha_{1}t_{1,i}\mathbf{s}_{1}.

Since the 𝐮i\mathbf{u}_{i} are primitive, we have α1t1,i=±1\alpha_{1}t_{1,i}=\pm 1. Thus 𝐮i=±𝐮1\mathbf{u}_{i}=\pm\mathbf{u}_{1}. So for all ii there is some root of unity ξi\xi_{i} such that 𝐗𝐮1ξi|Ri\mathbf{X}^{\mathbf{u}_{1}}-\xi_{i}|R_{i} and 𝐱𝐮1=ξi\mathbf{x}^{\mathbf{u}_{1}}=\xi_{i}, since 𝐗𝐮ζ=ζ𝐗𝐮(𝐗𝐮ζ1)\mathbf{X^{u}}-\zeta=-\zeta\mathbf{X^{u}}(\mathbf{X^{-u}}-\zeta^{-1}). Since the RiR_{i} are coprime, not all ξi\xi_{i} are the same and hence there is no common root. So we can assume that j2j\geq 2. Then the set of 𝐱𝔾mn\mathbf{x}\in\mathbb{G}_{m}^{n} such that 𝐱U=𝜻\mathbf{x}^{U}=\boldsymbol{\zeta} is a finite union of torsion cosets of dimension njn2n-j\leq n-2 by Lemma 5.1. Since UU is from a finite set, this shows the claim.∎

Lemma 5.5.

Let 𝐮n{0},z𝔾m\mathbf{u}\in\mathbb{Z}^{n}\setminus\{0\},z\in\mathbb{G}_{m} and PK[X1±1,,Xn±1]{0}P\in K[X_{1}^{\pm 1},\dots,X_{n}^{\pm 1}]\setminus\{0\}. Assume that 𝐗𝐮z|P\mathbf{X^{u}}-z|P. Then there exist 𝐢,𝐣Supp(P)\mathbf{i,j}\in\mathrm{Supp}(P) and kk\in\mathbb{N} such that k𝐮=𝐢𝐣.k\mathbf{u=i-j}.

Proof.

Let QK[X1±1,,Xn±1]{0}Q\in K[X_{1}^{\pm 1},\dots,X_{n}^{\pm 1}]\setminus\{0\} be the Laurent polynomial such that P=(𝐗𝐮z)QP=(\mathbf{X^{u}}-z)Q. Let p𝐢p_{\mathbf{i}} and q𝐢q_{\mathbf{i}} be the coefficients of PP and QQ. Then we have p𝐢=q𝐢𝐮zq𝐢p_{\mathbf{i}}=q_{\mathbf{i-u}}-zq_{\mathbf{i}} for all 𝐢n\mathbf{i}\in\mathbb{Z}^{n}. Fix any 𝐢0Supp(Q)\mathbf{i}_{0}\in\mathrm{Supp}(Q) and for all kk\in\mathbb{Z} let 𝐢k=𝐢0+k𝐮\mathbf{i}_{k}=\mathbf{i}_{0}+k\mathbf{u}. Let l=max{k:𝐢kSupp(Q)}l=\max\{k\in\mathbb{Z}:\mathbf{i}_{k}\in\mathrm{Supp}(Q)\} and 𝐢=𝐢l+1\mathbf{i=i}_{l+1}. Then p𝐢=q𝐢𝐮zq𝐢=q𝐢𝐮0p_{\mathbf{i}}=q_{\mathbf{i-u}}-zq_{\mathbf{i}}=q_{\mathbf{i-u}}\neq 0 and therefore 𝐢Supp(P)\mathbf{i}\in\mathrm{Supp}(P). Similarly let l=min{k:𝐢kSupp(Q)}l^{\prime}=\min\{k\in\mathbb{Z}:\mathbf{i}_{k}\in\mathrm{Supp}(Q)\} and 𝐣=𝐢l\mathbf{j=i}_{l^{\prime}}. Then p𝐣=q𝐣𝐮zq𝐣=zq𝐣0p_{\mathbf{j}}=q_{\mathbf{j}-\mathbf{u}}-zq_{\mathbf{j}}=-zq_{\mathbf{j}}\neq 0 and therefore 𝐣Supp(P)\mathbf{j}\in\mathrm{Supp}(P). Then 𝐢𝐣=𝐢l+1𝐢l=(ll+1)𝐮\mathbf{i-j}=\mathbf{i}_{l+1}-\mathbf{i}_{l^{\prime}}=(l-l^{\prime}+1)\mathbf{u}. Since lll\geq l^{\prime} the scalar k=ll+1k=l-l^{\prime}+1 is a natural number.∎

Lemma 5.6.

Let G<𝔾mnG<\mathbb{G}_{m}^{n} be a subgroup of codimension rr, LMatn×r()L\in\mathrm{Mat}_{n\times r}(\mathbb{Z}) a basis of ΛG\Lambda_{G} and P=𝐢np𝐢(X1,,Xn)𝐢K[X1±1,,Xn±1]P=\sum_{\mathbf{i}\in\mathbb{Z}^{n}}p_{\mathbf{i}}(X_{1},\dots,X_{n})^{\mathbf{i}}\in K[X_{1}^{\pm 1},\dots,X_{n}^{\pm 1}]. For 𝐚n\mathbf{a}\in\mathbb{Z}^{n} the map χ𝐚:G𝔾mn,𝐱𝐱𝐚\chi_{\mathbf{a}}:G\to\mathbb{G}_{m}^{n},\mathbf{x}\mapsto\mathbf{x^{a}} is a character on GG. For a character χ:G𝔾m\chi:G\to\mathbb{G}_{m} we put Lχ={𝐢Supp(P):χ𝐢=χ}L_{\chi}=\{\mathbf{i}\in\mathrm{Supp}(P):\chi_{\mathbf{i}}=\chi\}. For those χ\chi such that LχL_{\chi}\neq\emptyset we fix 𝐢χLχ\mathbf{i}_{\chi}\in L_{\chi}. For all 𝐢Supp(P)\mathbf{i}\in\mathrm{Supp}(P) we have 𝐢𝐢χ𝐢ΛG\mathbf{i-i}_{\chi_{\mathbf{i}}}\in\Lambda_{G}. Thus there is a unique 𝐮𝐢𝔾mr\mathbf{u_{i}}\in\mathbb{G}_{m}^{r} such that 𝐢=𝐢χ𝐢+L𝐮𝐢\mathbf{i=i}_{\chi_{\mathbf{i}}}+L\mathbf{u_{i}}. Finally we define Pχ=𝐢Lχp𝐢(Y1,,Yr)𝐮𝐢K[Y1±1,,Yr±1]P_{\chi}=\sum_{\mathbf{i}\in L_{\chi}}p_{\mathbf{i}}(Y_{1},\dots,Y_{r})^{\mathbf{u_{i}}}\in K[Y_{1}^{\pm 1},\dots,Y_{r}^{\pm 1}].
Then we have P=χ:Lχ(X1,,Xn)𝐢χPχ((X1,,Xn)L)P=\sum_{\chi:L_{\chi}\neq\emptyset}(X_{1},\dots,X_{n})^{\mathbf{i}_{\chi}}P_{\chi}((X_{1},\dots,X_{n})^{L}) and if 𝐱𝔾mn\mathbf{x}\in\mathbb{G}_{m}^{n} is such that P(𝐱G)=0P(\mathbf{x}G)=0, then 𝐢Lχp𝐢𝐱𝐢=0=Pχ(𝐱L)\sum_{\mathbf{i}\in L_{\chi}}p_{\mathbf{i}}\mathbf{x}^{\mathbf{i}}=0=P_{\chi}(\mathbf{x}^{L}) for all characters χ\chi such that LχL_{\chi}\neq\emptyset.

Proof.

The argument has similarities with the proof of Proposition 3.2.14 in [BG06]. First we show that the 𝐮𝐢\mathbf{u_{i}} are well defined. Let 𝐢Supp(P)\mathbf{i}\in\mathrm{Supp}(P). Since 𝐢χ𝐢Lχ𝐢\mathbf{i}_{\chi_{\mathbf{i}}}\in L_{\chi_{\mathbf{i}}} we have 𝐱𝐢χ𝐢=𝐱𝐢\mathbf{x}^{\mathbf{i}_{\chi_{\mathbf{i}}}}=\mathbf{x^{i}} for all 𝐱G\mathbf{x}\in G and hence 𝐢𝐢χ𝐢ΛG\mathbf{i}-\mathbf{i}_{\chi_{\mathbf{i}}}\in\Lambda_{G}.

Let us abbreviate 𝐗=(X1,,Xn)\mathbf{X}=(X_{1},\dots,X_{n}). Then we have

P=𝐢Supp(P)p𝐢𝐗𝐢=χ:Lχ𝐢Lχp𝐢𝐗𝐢χ𝐢+L𝐮i=χ:Lχ𝐗𝐢χ𝐢Lχp𝐢(𝐗L)𝐮i=χ:Lχ𝐗𝐢χPχ(𝐗L).P=\sum_{\mathbf{i}\in\mathrm{Supp}(P)}p_{\mathbf{i}}\mathbf{X^{i}}=\sum_{\chi:L_{\chi}\neq\emptyset}\sum_{\mathbf{i}\in L_{\chi}}p_{\mathbf{i}}\mathbf{X}^{\mathbf{i}_{\chi_{\mathbf{i}}}+L\mathbf{u}_{i}}=\sum_{\chi:L_{\chi}\neq\emptyset}\mathbf{X}^{\mathbf{i}_{\chi}}\sum_{\mathbf{i}\in L_{\chi}}p_{\mathbf{i}}(\mathbf{X}^{L})^{\mathbf{u}_{i}}=\sum_{\chi:L_{\chi}\neq\emptyset}\mathbf{X}^{\mathbf{i}_{\chi}}P_{\chi}(\mathbf{X}^{L}).

Suppose that 𝐱𝔾mn\mathbf{x}\in\mathbb{G}_{m}^{n} is such that P(𝐱G)=0P(\mathbf{x}G)=0. Let 𝐠G\mathbf{g}\in G. Then we have

0=P(𝐱𝐠)=χ:Lχ𝐢Lχp𝐢𝐱𝐢χ𝐢(𝐠)=χ:Lχ(𝐢Lχp𝐢𝐱𝐢)χ(𝐠).0=P(\mathbf{xg})=\sum_{\chi:L_{\chi}\neq\emptyset}\sum_{\mathbf{i}\in L_{\chi}}p_{\mathbf{i}}\mathbf{x^{i}}\chi_{\mathbf{i}}(\mathbf{g})=\sum_{\chi:L_{\chi}\neq\emptyset}\left(\sum_{\mathbf{i}\in L_{\chi}}p_{\mathbf{i}}\mathbf{x}^{\mathbf{i}}\right)\chi(\mathbf{g}).

By Artin’s Lemma (Theorem 4.1 in chapter VI in [Lan02]) we must have 𝐢Lχp𝐢𝐱𝐢=0\sum_{\mathbf{i}\in L_{\chi}}p_{\mathbf{i}}\mathbf{x}^{\mathbf{i}}=0 for all χ\chi.

We compute

Pχ(𝐱L)=𝐢Lχp𝐢𝐱L𝐮𝐢=𝐢Lχp𝐢𝐱𝐢𝐢χi=𝐱𝐢χ(𝐢Lχp𝐢𝐱𝐢)=0.P_{\chi}(\mathbf{x}^{L})=\sum_{\mathbf{i}\in L_{\chi}}p_{\mathbf{i}}\mathbf{x}^{L\mathbf{u_{i}}}=\sum_{\mathbf{i}\in L_{\chi}}p_{\mathbf{i}}\mathbf{x}^{\mathbf{i-i}_{\chi_{i}}}=\mathbf{x}^{-\mathbf{i}_{\chi}}\left(\sum_{\mathbf{i}\in L_{\chi}}p_{\mathbf{i}}\mathbf{x}^{\mathbf{i}}\right)=0.

Lemma 5.7.

Let n2n\geq 2, PK[X1±1,,Xn±1]P\in K[X_{1}^{\pm 1},\dots,X_{n}^{\pm 1}] and X=Z(P)X=Z(P). Then the following are equivalent

  1. 1.

    XX is admissible.

  2. 2.

    There are no primitive 𝐮n{0}\mathbf{u}\in\mathbb{Z}^{n}\setminus\{0\} and root of unity ζ𝔾m\zeta\in\mathbb{G}_{m} such that 𝐗𝐮ζ|P\mathbf{X^{u}}-\zeta|P.

  3. 3.

    There are no 𝐮n{0}\mathbf{u}\in\mathbb{Z}^{n}\setminus\{0\} and root of unity ζ𝔾m\zeta\in\mathbb{G}_{m} such that 𝐗𝐮ζ|P\mathbf{X^{u}}-\zeta|P.

Proof.

If P=0P=0, X=𝔾mnX=\mathbb{G}_{m}^{n} is not admissible and every QK[X1±1,,Xn±1]Q\in K[X_{1}^{\pm 1},\dots,X_{n}^{\pm 1}] divides PP. And if PK[X1±1,,Xn±1]P\in K[X_{1}^{\pm 1},\dots,X_{n}^{\pm 1}]^{*} we have X=X=\emptyset which is admissible. Since 𝐗𝐮ζ\mathbf{X^{u}}-\zeta is never a unit if 𝐮n{0}\mathbf{u}\in\mathbb{Z}^{n}\setminus\{0\} and ζ𝔾mn\zeta\in\mathbb{G}_{m}^{n}, the second and the third statement are also true. Thus we can assume that PK[X1,,Xn]{0}P\not\in K[X_{1},\dots,X_{n}]^{*}\cup\{0\} and hence the dimension of XX is n11.n-1\geq 1. The implication 1)2)1)\Rightarrow 2) follows directly from Lemma 5.1 by contraposition.

2)3):2)\Rightarrow 3): We prove the contraposition and assume that there exists 𝝀n{0}\boldsymbol{\lambda}\in\mathbb{Z}^{n}\setminus\{0\} such that 𝐗𝝀ζ|P\mathbf{X}^{\boldsymbol{\lambda}}-\zeta|P. Write 𝝀=g𝐮\boldsymbol{\lambda}=g\mathbf{u} for a primitive 𝐮n\mathbf{u}\in\mathbb{Z}^{n} and a g1g\geq 1. Since XY|XgYg[X,Y]X-Y|X^{g}-Y^{g}\in\mathbb{Z}[X,Y] and we can write ζ=ηg\zeta=\eta^{g} for some η𝔾m\eta\in\mathbb{G}_{m} we find 𝐗𝐮η|𝐗𝝀ζ\mathbf{X^{u}}-\eta|\mathbf{X}^{\boldsymbol{\lambda}}-\zeta and hence there exists a primitive 𝐮n{0}\mathbf{u}\in\mathbb{Z}^{n}\setminus\{0\} and a root of unity η𝔾m\eta\in\mathbb{G}_{m} such that 𝐗𝐮η|P.\mathbf{X^{u}}-\eta|P.

3)1)3)\Rightarrow 1) We show the contraposition and assume that there is an (n1)(n-1)-dimensional algebraic subgroup G𝔾mnG\subset\mathbb{G}_{m}^{n} and a torsion point 𝜻𝔾mn\boldsymbol{\zeta}\in\mathbb{G}_{m}^{n} such that 𝜻GX\boldsymbol{\zeta}G\subset X. By Theorem A.12 the rank of ΛG\Lambda_{G} equals one. Thus there exists 𝝀n{0}\boldsymbol{\lambda}\in\mathbb{Z}^{n}\setminus\{0\} which is a basis of ΛG\Lambda_{G}. We can apply Lemma 5.6 and find that 𝜻𝝀\boldsymbol{\zeta^{\lambda}} is a root of PχK[Y]P_{\chi}\in K[Y] and hence Y𝜻𝝀|PχY-\boldsymbol{\zeta^{\lambda}}|P_{\chi} for all characters χ:G𝔾mn\chi:G\to\mathbb{G}_{m}^{n} such that LχL_{\chi}\neq\emptyset. So let QχK[Y±1]Q_{\chi}\in K[Y^{\pm 1}] be such that Pχ=(Y𝜻𝝀)QχP_{\chi}=(Y-\boldsymbol{\zeta^{\lambda}})Q_{\chi}. Then again by Lemma 5.6 we find

P=χ:Lχ𝐗𝐢χPχ(𝐗𝝀)=(𝐗𝝀𝜻𝝀)χ:Lχ𝐗𝐢χQχ(𝐗𝝀).P=\sum_{\chi:L_{\chi}\neq\emptyset}\mathbf{X}^{\mathbf{i}_{\chi}}P_{\chi}(\mathbf{X}^{\boldsymbol{\lambda}})=(\mathbf{X}^{\boldsymbol{\lambda}}-\boldsymbol{\zeta^{\lambda}})\sum_{\chi:L_{\chi}\neq\emptyset}\mathbf{X}^{\mathbf{i}_{\chi}}Q_{\chi}(\mathbf{X}^{\boldsymbol{\lambda}}).

Therefore 𝐗𝝀𝜻𝝀|P\mathbf{X}^{\boldsymbol{\lambda}}-\boldsymbol{\zeta^{\lambda}}|P, where 𝝀n{0}\boldsymbol{\lambda}\in\mathbb{Z}^{n}\setminus\{0\} and 𝜻𝝀𝔾m\boldsymbol{\zeta^{\lambda}}\in\mathbb{G}_{m} is a torsion point.∎

6 Bound for cosets

Although we already esablished an asymptotic formula for #CT\#C_{T} if CC 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 G<𝔾mnG<\mathbb{G}_{m}^{n} we denote by G0G^{0} the linear torus equal to the connected component of GG which contains the identity.

Lemma 6.2.

Let 𝛇(𝔾mn)tors\boldsymbol{\zeta}\in(\mathbb{G}_{m}^{n})_{\mathrm{tors}} and let G<𝔾mnG<\mathbb{G}_{m}^{n} be an algebraic subgroup of dimension dd. Then we have (𝛇G)T[G:G0]Td+1/ord(𝛇G)(\boldsymbol{\zeta}G)_{T}\leq[G:G^{0}]T^{d+1}/\mathrm{ord}(\boldsymbol{\zeta}G).

Proof.

Let us first assume that GG is a linear torus. In this case we can assume that 𝜻G={𝜼}×𝔾md\boldsymbol{\zeta}G=\{\boldsymbol{\eta}\}\times\mathbb{G}_{m}^{d} for some 𝜼(𝔾mnd)tors\boldsymbol{\eta}\in(\mathbb{G}_{m}^{n-d})_{\mathrm{tors}} of order g:=ord(𝜻G)g:=\mathrm{ord}(\boldsymbol{\zeta}G) by Lemma A.17. Thus the order of (𝜼,𝝃)(\boldsymbol{\eta},\boldsymbol{\xi}) is lcm(g,ord(𝝃))\mathrm{lcm}(g,\mathrm{ord}(\boldsymbol{\xi})) for any 𝝃(𝔾md)tors\boldsymbol{\xi}\in(\mathbb{G}_{m}^{d})_{\mathrm{tors}}. Hence (𝜼,𝝃)(𝜻G)T(\boldsymbol{\eta},\boldsymbol{\xi})\in(\boldsymbol{\zeta}G)_{T} if and only if ord(𝝃)gcd(g,ord(𝝃))T/g(\boldsymbol{\xi})\leq\gcd(g,\mathrm{ord}(\boldsymbol{\xi}))T/g. By Lemma 3.9 the number of elements of order NN in 𝔾md\mathbb{G}_{m}^{d} is Jd(N)J_{d}(N) if p=char(K)=0p=\mathrm{char}(K)=0 or (N,p)=1(N,p)=1 and zero otherwise. In particular Jd(N)J_{d}(N) is always an upper bound. Hence we find

#(𝜼G)T1mgcd(g,m)T/gJd(m)=e|gm:gcd(g,m)=e,meT/gJd(m).\#(\boldsymbol{\eta}G)_{T}\leq\sum_{1\leq m\leq\gcd(g,m)T/g}J_{d}(m)=\sum_{e|g}\sum_{m:\gcd(g,m)=e,m\leq eT/g}J_{d}(m).

With the formula Jd(n)=ndp|n(1pd)J_{d}(n)=n^{d}\prod_{p|n}(1-p^{-d}) one can see that Jd(ab)adJd(b)J_{d}(ab)\leq a^{d}J_{d}(b) for all a,ba,b\in\mathbb{N}. And since Jd=μIdJ_{d}=\mu\star I_{d}, we have m|nJd(m)=nd\sum_{m|n}J_{d}(m)=n^{d} for all nn\in\mathbb{N} by the Möbius inversion formula. If d|gd|g and mm is as in the inner sum, then we can write m=kem=ke for some kT/gk\leq T/g. Thus we find

#(𝜼G)Te|g1kT/gJd(ke)e|gJd(e)1kT/gkdgd(T/g)d+1Td+1/g.\#(\boldsymbol{\eta}G)_{T}\leq\sum_{e|g}\sum_{1\leq k\leq T/g}J_{d}(ke)\leq\sum_{e|g}J_{d}(e)\sum_{1\leq k\leq T/g}k^{d}\leq g^{d}(T/g)^{d+1}\leq T^{d+1}/g.

For the general case we decompose 𝜻G\boldsymbol{\zeta}G into a disjoint union of [G:G0][G:G^{0}] torsion cosets 𝜻iG0\boldsymbol{\zeta}_{i}G^{0} and observe that ord(𝜻iG0)ord(𝜻G)\mathrm{ord}(\boldsymbol{\zeta}_{i}G^{0})\geq\mathrm{ord}(\boldsymbol{\zeta}G) since 𝜻iG0𝜻G\boldsymbol{\zeta}_{i}G^{0}\subset\boldsymbol{\zeta}G. Therefore

#(𝜻G)T=i=1[G:G0](𝜻iG0)Ti=1[G:G0]Td+1/ord(𝜻iG0)[G:G0]Td+1/ord(𝜻G).\#(\boldsymbol{\zeta}G)_{T}=\sum_{i=1}^{[G:G^{0}]}(\boldsymbol{\zeta}_{i}G^{0})_{T}\leq\sum_{i=1}^{[G:G^{0}]}T^{d+1}/\mathrm{ord}(\boldsymbol{\zeta}_{i}G^{0})\leq[G:G^{0}]T^{d+1}/\mathrm{ord}(\boldsymbol{\zeta}G).\qed

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 𝜻G\boldsymbol{\zeta}G be a maximal coset. There are only finitely many possibilities for GG. But it may happen that there are infinitely many torsion cosets 𝜻G\boldsymbol{\zeta}G which are contained in the hypersurface. We construct a map from the torsion cosets of GG to torsion points of a variety of lower dimension and get a bound by induction.

Lemma 7.1.

Let n2n\geq 2 and XX an admissible hypersurface. Let G𝔾mnG\subset\mathbb{G}_{m}^{n} be a linear torus of dimension dd. Let r=ndr=n-d and LMatn×r()L\in\mathrm{Mat}_{n\times r}(\mathbb{Z}) such that Lr=ΛGL\mathbb{Z}^{r}=\Lambda_{G}. Then there exist an admissible hypersurface Y𝔾mrY\subset\mathbb{G}_{m}^{r} and finitely many torsion cosets C1,,Cm𝔾mrC_{1},\dots,C_{m}\subset\mathbb{G}_{m}^{r} of dimension at most r2r-2 with the property that for every T1T\geq 1 the map 𝛇G𝛇L\boldsymbol{\zeta}G\mapsto\boldsymbol{\zeta}^{L} is an injection from the torsion cosets 𝛇GX\boldsymbol{\zeta}G\subset X of order bounded by TT to YTi=1m(Ci)TY_{T}\cup\bigcup_{i=1}^{m}(C_{i})_{T}.

Proof.

Without loss of generality we can assume that there exist torsion cosets of GG which are contained in XX. Thus by admissibility we have dn2d\leq n-2 and hence r2r\geq 2. Let PK[X1±1,,Xn±1]P\in K[X_{1}^{\pm 1},\dots,X_{n}^{\pm 1}] such that X=Z(P)X=Z(P). Let PχK[Y1±1,,Yr±1]P_{\chi}\in K[Y_{1}^{\pm 1},\dots,Y_{r}^{\pm 1}] be defined as in Lemma 5.6 for all χ:G𝔾mn\chi:G\to\mathbb{G}_{m}^{n} such that LχL_{\chi}\neq\emptyset. For any such χ\chi let RχR_{\chi} be the product of all factors of PχP_{\chi} of the form 𝐘𝐮ζ\mathbf{Y^{u}}-\zeta for some primitive 𝐮r{0}\mathbf{u}\in\mathbb{Z}^{r}\setminus\{0\} and ζ𝔾m\zeta\in\mathbb{G}_{m} of finite order, which are irreducible by Lemma 5.3, counted with multiplicity. Let QχK[Y1±1,,Yr±1]Q_{\chi}\in K[Y_{1}^{\pm 1},\dots,Y_{r}^{\pm 1}] such that Pχ=QχRχ.P_{\chi}=Q_{\chi}R_{\chi}. Define Q=χ:LχQχK[Y1±1,,Yr±1]Q=\prod_{\chi:L_{\chi}\neq\emptyset}Q_{\chi}\in K[Y_{1}^{\pm 1},\dots,Y_{r}^{\pm 1}] and let Y=Z(Q)𝔾mrY=Z(Q)\subset\mathbb{G}_{m}^{r}. Since r2r\geq 2 we can apply Lemma 5.7 and see that YY is admissible by construction of QQ.
We claim that the RχR_{\chi} are coprime. Assume by contradiction that they have a common factor F=𝐘𝐮ζF=\mathbf{Y^{u}}-\zeta, where 𝐮r{0}\mathbf{u}\in\mathbb{Z}^{r}\setminus\{0\} and ζ\zeta is a root of unity and write Rχ=FRχR_{\chi}=FR_{\chi}^{\prime}. Then we find by Lemma 5.6 that

P=χ:Lχ𝐗𝐢χPχ(𝐗L)=F(𝐗L)χ:Lχ𝐗𝐢χQχ(𝐗L)Rχ(𝐗L).P=\sum_{\chi:L_{\chi}\neq\emptyset}\mathbf{X}^{\mathbf{i}_{\chi}}P_{\chi}(\mathbf{X}^{L})=F(\mathbf{X}^{L})\sum_{\chi:L_{\chi}\neq\emptyset}\mathbf{X}^{\mathbf{i}_{\chi}}Q_{\chi}(\mathbf{X}^{L})R^{\prime}_{\chi}(\mathbf{X}^{L}).

Thus 𝐗L𝐮ζ\mathbf{X}^{L\mathbf{u}}-\zeta divides PP. We now show that L𝐮0L\mathbf{u}\neq 0. Let χ\chi be such that LχL_{\chi}\neq\emptyset. Then Pχ0P_{\chi}\neq 0. Since 𝐮0,ζ0\mathbf{u}\neq 0,\zeta\neq 0 and F|PχF|P_{\chi} there exist 𝐢𝐣Lχ\mathbf{i\neq j}\in L_{\chi} and a natural number kk such that k𝐮=𝐮𝐢𝐮𝐣k\mathbf{u=u_{i}-u_{j}} by Lemma 5.5. We have kL𝐮=L(𝐮𝐢𝐮𝐣)=𝐢𝐣0kL\mathbf{u}=L(\mathbf{u_{i}-u_{j})=i-j}\neq 0 since χ𝐢=χ=χ𝐣\chi_{\mathbf{i}}=\chi=\chi_{\mathbf{j}}. Therefore L𝐮n{0}L\mathbf{u}\in\mathbb{Z}^{n}\setminus\{0\} and 𝐗L𝐮ζ|P\mathbf{X}^{L\mathbf{u}}-\zeta|P which contradicts the fact that XX is admissible by Lemma 5.7. So we showed that the RχR_{\chi} are coprime. We can therefore apply Lemma 5.4 and see that the common zeros in 𝔾mr\mathbb{G}_{m}^{r} is a finite union of torsion cosets C1,,CmC_{1},\dots,C_{m} of dimension at most r2r-2.
Let T1T\geq 1 and 𝜻G\boldsymbol{\zeta}G be such that 𝜻GX\boldsymbol{\zeta}G\subset X and ord(𝜻G)T.\mathrm{ord}(\boldsymbol{\zeta}G)\leq T. We claim that 𝜻G𝜻L\boldsymbol{\zeta}G\mapsto\boldsymbol{\zeta}^{L} is well defined and injective. This is a well defined map, because for 𝐱G\mathbf{x}\in G we have (𝜻𝐱)L=𝜻L(\boldsymbol{\zeta}\mathbf{x})^{L}=\boldsymbol{\zeta}^{L} since the columns of LL are in ΛG\Lambda_{G}. To prove injectivity assume that 𝜻L=𝝃L\boldsymbol{\zeta}^{L}=\boldsymbol{\xi}^{L}. Let 𝐱=𝜻𝝃1\mathbf{x}=\boldsymbol{\zeta\xi}^{-1} and 𝝀ΛG\boldsymbol{\lambda}\in\Lambda_{G}. Then there exists 𝐮r\mathbf{u}\in\mathbb{Z}^{r} such that 𝝀=L𝐮\boldsymbol{\lambda}=L\mathbf{u}. Therefore 𝐱𝝀=𝜻L𝐮𝝃L𝐮=𝟏𝐮=1\mathbf{x}^{\boldsymbol{\lambda}}=\boldsymbol{\zeta}^{L\mathbf{u}}\boldsymbol{\xi}^{-L\mathbf{u}}=\mathbf{1}^{\mathbf{u}}=1 which shows that 𝐱G\mathbf{x}\in G by Theorem A.12 and hence 𝜻G=𝝃G\boldsymbol{\zeta}G=\boldsymbol{\xi}G. Furthermore the order of the image is bounded by the order of the coset.

By Lemma 5.6 we have Pχ(𝜻L)=0P_{\chi}(\boldsymbol{\zeta}^{L})=0. If there is χ\chi such that Qχ(𝜻L)=0Q_{\chi}(\boldsymbol{\zeta}^{L})=0 the element 𝜻L\boldsymbol{\zeta}^{L} is a root of QQ of order bounded by TT and hence in YTY_{T}. If not, it is a common root of the RχR_{\chi} and hence contained in the finite union of the torsion cosets of dimension at most r2r-2.∎

Lemma 7.2.

Let (an)0(a_{n})\in\mathbb{R}_{\geq 0}^{\mathbb{N}} be a sequence, α>1\alpha>1 a real number and suppose that there is c>0c>0 such that n=1TancTα\sum_{n=1}^{T}a_{n}\leq cT^{\alpha} for all TT\in\mathbb{N}. Then there is C=C(c,α)>0C=C(c,\alpha)>0 such that n=1Tan/nCTα1\sum_{n=1}^{T}a_{n}/n\leq CT^{\alpha-1} for all TT\in\mathbb{N}.

Proof.

Note that for all n<Tn<T we have 1n=1T+k=nT1(1k1k+1)=1T+k=nT11k(k+1)\frac{1}{n}=\frac{1}{T}+\sum_{k=n}^{T-1}\left(\frac{1}{k}-\frac{1}{k+1}\right)=\frac{1}{T}+\sum_{k=n}^{T-1}\frac{1}{k(k+1)}. Comparing the sum with the integral and using α2>1\alpha-2>-1 we can write

n=1Tann\displaystyle\sum_{n=1}^{T}\frac{a_{n}}{n} =n=1T1k=nT1ank(k+1)+1Tn=1Tan=k=1T11k(k+1)n=1kan+1Tn=1Tan\displaystyle=\sum_{n=1}^{T-1}\sum_{k=n}^{T-1}\frac{a_{n}}{k(k+1)}+\frac{1}{T}\sum_{n=1}^{T}a_{n}=\sum_{k=1}^{T-1}\frac{1}{k(k+1)}\sum_{n=1}^{k}a_{n}+\frac{1}{T}\sum_{n=1}^{T}a_{n}
ck=1T1kα2+cTα1αcTα1.\displaystyle\leq c\sum_{k=1}^{T-1}{k}^{\alpha-2}+cT^{\alpha-1}\ll_{\alpha}cT^{\alpha-1}.\qed
Lemma 7.3.

Let n3,X=Z(P)n\geq 3,X=Z(P) an admissible hypersurface in 𝔾mn\mathbb{G}_{m}^{n} and 1dn21\leq d\leq n-2. Assume that we know that #YTYTr1/r\#Y_{T}\ll_{Y}T^{r-1/r} for all admissible hypersurfaces YY in 𝔾mr\mathbb{G}_{m}^{r}, where 2rn12\leq r\leq n-1. Let G<𝔾mnG<\mathbb{G}_{m}^{n} be a linear torus of dimension dd. Then

#(𝜼(𝔾mn)tors𝜼GX(𝜼G)T)X,GTn1/(nd) for all T1.\#\left(\bigcup_{\begin{subarray}{c}\boldsymbol{\eta}\in(\mathbb{G}_{m}^{n})_{\text{tors}}\\ \boldsymbol{\eta}G\subset X\end{subarray}}(\boldsymbol{\eta}G)_{T}\right)\ll_{X,G}T^{n-1/(n-d)}\text{ for all }T\geq 1.
Proof.

Let us denote by MM the quantity on the left. Note that ord(𝜼G)>T\mathrm{ord}(\boldsymbol{\eta}G)>T implies that (𝜼G)T=(\boldsymbol{\eta}G)_{T}=\emptyset. Thus we have only to consider cosets of order bounded by TT. Let aNa_{N} be the number of torsion cosets 𝜼G\boldsymbol{\eta}G such that 𝜼GX\boldsymbol{\eta}G\subset X and ord(𝜼G)=N.\mathrm{ord}(\boldsymbol{\eta}G)=N. Then we can use Lemma 6.2 to find the bound MGTd+1N=1TaN/NM\ll_{G}T^{d+1}\sum_{N=1}^{T}a_{N}/N. Let r=ndn1r=n-d\leq n-1 and Y𝔾mrY\subset\mathbb{G}_{m}^{r} and C1,,Cm𝔾mrC_{1},\dots,C_{m}\subset\mathbb{G}_{m}^{r} be the admissible hypersurface and the torsion cosets from Lemma 7.1. By hypothesis there exists a constant c=c(Y)c=c(Y) such that #YTcTr1/r\#Y_{T}\leq cT^{r-1/r} and by Lemma 6.2 we have #(Ci)TTr1\#(C_{i})_{T}\leq T^{r-1} for all T1T\geq 1. So by Lemma 7.1 for all N1N\geq 1 we have

N=1TaN#(Y)T+i=1m#(Ci)T(c+m)Tr1/r.\sum_{N=1}^{T}a_{N}\leq\#(Y)_{T}+\sum_{i=1}^{m}\#(C_{i})_{T}\leq(c+m)T^{r-1/r}.

Since r2r\geq 2 we have r1/r3/2>1r-1/r\geq 3/2>1 and therefore N=1TaN/NYTr1/r1\sum_{N=1}^{T}a_{N}/N\ll_{Y}T^{r-1/r-1} by Lemma 7.2. So we get MX,GTd+1+r11/r=Tn1/(nd).M\ll_{X,G}T^{d+1+r-1-1/r}=T^{n-1/{(n-d)}}.

Lemma 7.4.

Let nn\in\mathbb{N} and X=Z(P)X=Z(P) a hypersurface in 𝔾mn\mathbb{G}_{m}^{n}. Then #XTXTn.\#X_{T}\ll_{X}T^{n}. If n=1n=1 we even have #XTX1\#X_{T}\ll_{X}1.

Proof.

The proof is by induction on nn. If n=1n=1 the set Z(P)Z(P) is finite and hence the lemma is true in this case. So assume that n2n\geq 2 and the lemma is true in dimension smaller than nn. Write P=i=DDRi(X1,,Xn1)XniP=\sum_{i=-D}^{D}R_{i}(X_{1},\dots,X_{n-1})X_{n}^{i}. Let R{RD,RD}K[X1±1,,Xn1±1]R\in\{R_{-D},R_{D}\}\subset K[X_{1}^{\pm 1},\dots,X_{n-1}^{\pm 1}] be a nonzero Laurent polynomial which exists since P0P\neq 0. To estimate #Z(P)T\#Z(P)_{T} we divide XTX_{T} into two sets: Y1={(x1,,xn)XT:R(x1,,xn1)0}Y_{1}=\{(x_{1},\dots,x_{n})\in X_{T}:R(x_{1},\dots,x_{n-1})\neq 0\} and Y2=XTY1Y_{2}=X_{T}\setminus Y_{1}. First we bound #Y1\#Y_{1}. Fix 𝐱=(x1,,xn1)𝔾mn1\mathbf{x}=(x_{1},\dots,x_{n-1})\in\mathbb{G}_{m}^{n-1} and assume that R(𝐱)0R(\mathbf{x})\neq 0. Then there are at most 2D2D solutions x𝔾mx\in\mathbb{G}_{m} satisfying P(𝐱,x)=0P(\mathbf{x},x)=0. We can bound #Y12D(𝔾mn1)T2DTn\#Y_{1}\leq 2D(\mathbb{G}_{m}^{n-1})_{T}\leq 2DT^{n} by Lemma 6.2. If RK[X1±1,,Xn±1]R\in K[X_{1}^{\pm 1},\dots,X_{n}^{\pm 1}]^{*}, then we have XT=Y1X_{T}=Y_{1} and the claimed bound holds.
Otherwise we also have to bound #Y2\#Y_{2} and V=Z(R)V=Z(R) is a hypersurface. Let G={1}××{1}×𝔾mG=\{1\}\times\dots\times\{1\}\times\mathbb{G}_{m}. Then

Y2𝐱VT((𝐱,1)G)T.Y_{2}\subset\bigcup_{\mathbf{x}\in V_{T}}((\mathbf{x},1)G)_{T}.

The order of a coset (𝐱,1)G(\mathbf{x},1)G equals ord(𝐱)\mathrm{ord}(\mathbf{x}). Let cc be a constant such that #VTcTn1\#V_{T}\leq cT^{n-1} for all TT which exists by induction. Let aNa_{N} be the number of points of order NN in VV. Thus by Lemma 6.2 we have to bound the sum

#Y2𝐱VTT2/ord(𝐱)=T2N=1TaN/N.\#Y_{2}\ll\sum_{\mathbf{x}\in V_{T}}T^{2}/\mathrm{ord}(\mathbf{x})=T^{2}\sum_{N=1}^{T}a_{N}/N.

By induction we know that N=1TaN=#VTV1\sum_{N=1}^{T}a_{N}=\#V_{T}\ll_{V}1 if n=2n=2 and N=1TaN=#VTVTn1\sum_{N=1}^{T}a_{N}=\#V_{T}\ll_{V}T^{n-1} if n>2n>2. So if n=2n=2 we find #Y2T2N=1TaN/NRT2\#Y_{2}\ll T^{2}\sum_{N=1}^{T}a_{N}/N\ll_{R}T^{2}. If n>2n>2 we have n1>1n-1>1, so we can apply Lemma 7.2 and find #Y2T2N=1TaN/Nn,RT2+(n1)1=Tn\#Y_{2}\ll T^{2}\sum_{N=1}^{T}a_{N}/N\ll_{n,R}T^{2+(n-1)-1}=T^{n}. Thus we have #Y2n,PTn\#Y_{2}\ll_{n,P}T^{n} and hence #XT=#Y1+#Y2XTn\#X_{T}=\#Y_{1}+\#Y_{2}\ll_{X}T^{n}.∎

Theorem 7.5.

Let n2n\geq 2, XX be an admissible hypersurface in 𝔾mn\mathbb{G}_{m}^{n}. Then #XTXTn1/n.\#X_{T}\ll_{X}T^{n-1/n}.

Proof.

Since any torsion coset in XX is contained in a maximal one, we find

X(𝔾mn)torsX𝜻GX maximal torsion coset,dim(G)1𝜻G.X\cap(\mathbb{G}_{m}^{n})_{\mathrm{tors}}\subset X^{*}\cup\bigcup_{\begin{subarray}{c}\boldsymbol{\zeta}G\subset X\text{ maximal torsion coset},\\ \dim(G)\geq 1\end{subarray}}\boldsymbol{\zeta}G.

Let 𝒢={G<𝔾mn:G linear torus, there is 𝜻(𝔾mn)tors:𝜻GX is maximal and dim(G)1}\mathcal{G}=\{G<\mathbb{G}_{m}^{n}:G\text{ linear torus, there is }\boldsymbol{\zeta}\in(\mathbb{G}_{m}^{n})_{\mathrm{tors}}:\boldsymbol{\zeta}G\subset X\text{ is maximal and }\dim(G)\geq 1\} which is finite by Corollary A.8. Then we have

#XT#XT+G𝒢#(𝜻(𝔾mn)tors𝜻GX(𝜻G)T)\#X_{T}\leq\#X^{*}_{T}+\sum_{G\in\mathcal{G}}\#\left(\bigcup_{\begin{subarray}{c}\boldsymbol{\zeta}\in(\mathbb{G}_{m}^{n})_{\text{tors}}\\ \boldsymbol{\zeta}G\subset X\end{subarray}}(\boldsymbol{\zeta}G)_{T}\right)

and since 𝒢\mathcal{G} is finite it is enough to show the bound for each summand separately. The proof is by induction on nn. If n=2n=2 we have XT=XTX_{T}=X^{*}_{T} and the theorem follows from Lemma 4.16. Let n3n\geq 3 and G𝒢G\in\mathcal{G} be of dimension dd. We have 1dn21\leq d\leq n-2 by admissibility. Thus the hypothesis of Lemma 7.3 is satisfied by induction. We can apply it and find #(𝜼(𝔾mn)tors𝜼GX(𝜼G)T)X,GTn1/(nd)\#\left(\bigcup_{\begin{subarray}{c}\boldsymbol{\eta}\in(\mathbb{G}_{m}^{n})_{\text{tors}}\\ \boldsymbol{\eta}G\subset X\end{subarray}}(\boldsymbol{\eta}G)_{T}\right)\ll_{X,G}T^{n-1/(n-d)}. Thus we have #XTXTn1/(nd)\#X_{T}\ll_{X}T^{n-1/(n-d)}.∎

8 Generalisation

We deduce the general theorem from the case where the subvariety is a hypersurface using algebraic geometry.

Lemma 8.1.

Let n2n\geq 2, X𝔾mnX\subset\mathbb{G}_{m}^{n} be an algebraic set of dimension dd and π:X𝔾mn1\pi:X\to\mathbb{G}_{m}^{n-1} the projection to the first n1n-1 coordinates. Let Z={𝐩𝔾mn1:dim(π1(𝐩))1}Z=\{\mathbf{p}\in\mathbb{G}_{m}^{n-1}:\dim(\pi^{-1}(\mathbf{p}))\geq 1\}. Then there exists D>0D>0 such that for all 𝐩π(X)\mathbf{p}\in\pi(X) we have #π1(𝐩)D\#\pi^{-1}(\mathbf{p})\leq D or 𝐩Z\mathbf{p}\in Z. The set ZZ is an algebraic set of dimension at most d1d-1. The Zariski closure Y=π(X)¯Y=\overline{\pi(X)} of the image is of dimension at most dd.

Proof.

Let 𝐩π(X)\mathbf{p}\in\pi(X). The number of irreducible components in the fibres is uniformly bounded and hence there exists D>0D>0 such that all finite fibres, contain at most DD points. Note that the fibres of positive dimension are of the form 𝐪×𝔾m\mathbf{q}\times\mathbb{G}_{m} and therefore V={𝐱X:dim𝐱π1(π(𝐱))1}=A×𝔾mV=\{\mathbf{x}\in X:\dim_{\mathbf{x}}\pi^{-1}(\pi(\mathbf{x}))\geq 1\}=A\times\mathbb{G}_{m} for some A𝔾mn1A\subset\mathbb{G}_{m}^{n-1}. By Theorem 14.112 in [GW20] we have that VV and hence also AA is closed. Thus Z=π(V)=AZ=\pi(V)=A is closed. Since Z×𝔾mXZ\times\mathbb{G}_{m}\subset X we have dim(Z)d1\dim(Z)\leq d-1. 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 #XT1mT#{𝐱X:𝐱m=𝟏}XTd+1\#X_{T}\leq\sum_{1\leq m\leq T}\#\{\mathbf{x}\in X:\mathbf{x}^{m}=\mathbf{1}\}\ll_{X}T^{d+1}. The first inequality follows immediately. It suffices to show the second one.
Let X1,,XrX_{1},\dots,X_{r} be the irreducible components of XX. Then max1irdim(Xi)=d\max_{1\leq i\leq r}\dim(X_{i})=d and XT=i=1r(Xi)TX_{T}=\bigcup_{i=1}^{r}(X_{i})_{T} for all T1T\geq 1. Therefore we can assume that XX is irreducible.
If d=0d=0 then XX is finite and the claim is immediate. If n=1n=1 and d=1d=1, then X=𝔾mX=\mathbb{G}_{m} and the sum is bounded by m=1TmT2\sum_{m=1}^{T}m\leq T^{2}. So we may assume d1d\geq 1 and n2n\geq 2
We continue by induction on n+dn+d, the case n+d=1n+d=1 is already done.
Let π:𝔾mn𝔾mn1\pi:\mathbb{G}_{m}^{n}\to\mathbb{G}_{m}^{n-1} denote the projection to the first n1n-1 coordinates. The set

Z={𝐩𝔾mn1:dim(π|X1(𝐩))1}Z=\{\mathbf{p}\in\mathbb{G}_{m}^{n-1}:\dim(\pi|_{X}^{-1}(\mathbf{p}))\geq 1\}

is Zariski closed and of dimension at most d1d-1 by Lemma 8.1. Let TT\in\mathbb{N} and observe that the sum in question is at most aT+bTa_{T}+b_{T}, where

aT=m=1T#{(𝐩,t)X:𝐩Z,𝐩m=𝟏,tm=1} and bT=m=1T#{(𝐩,t)X:𝐩Z,𝐩m=𝟏,tm=1}.a_{T}=\sum_{m=1}^{T}\#\{(\mathbf{p},t)\in X:\mathbf{p}\in Z,\mathbf{p}^{m}=\mathbf{1},t^{m}=1\}\text{ and }b_{T}=\sum_{m=1}^{T}\#\{(\mathbf{p},t)\in X:\mathbf{p}\not\in Z,\mathbf{p}^{m}=\mathbf{1},t^{m}=1\}.

Then

aTm=1T#{𝐩Z:𝐩m=𝟏}mTm=1T#{𝐩Z:𝐩m=𝟏}.a_{T}\leq\sum_{m=1}^{T}\#\{\mathbf{p}\in Z:\mathbf{p}^{m}=\mathbf{1}\}m\leq T\sum_{m=1}^{T}\#\{\mathbf{p}\in Z:\mathbf{p}^{m}=\mathbf{1}\}.

By induction applied to Z𝔾mn1Z\subset\mathbb{G}_{m}^{n-1} we have aTZTdim(Z)+1Td+1a_{T}\ll_{Z}T^{\dim(Z)+1}\leq T^{d+1}.
If 𝐩𝔾mn1Z\mathbf{p}\in\mathbb{G}_{m}^{n-1}\setminus Z there are at most DD possibilities for t𝔾mt\in\mathbb{G}_{m} such that (𝐩,t)X(\mathbf{p},t)\in X by Lemma 8.1. We apply the current lemma by induction to the Zariski closure π(X)¯𝔾mn1\overline{\pi(X)}\subset\mathbb{G}_{m}^{n-1} which has dimension at most dd. So bTDm=1T#{𝐩π(X)¯:𝐩m=𝟏}XTd+1b_{T}\leq D\sum_{m=1}^{T}\#\{\mathbf{p}\in\overline{\pi(X)}:\mathbf{p}^{m}=\mathbf{1}\}\ll_{X}T^{d+1}. ∎

Definition 8.2.

Let XX be a topological space. Then we denote by Irr(X)\mathrm{Irr}(X) the set of irreducible components of XX.

Remark 8.3.

If an algebraic subset X𝔾mnX\subset\mathbb{G}_{m}^{n} is irreducible and of dimension n1n-1, it is a hypersurface. This can be deduced from the same fact for subvarieties of 𝔸n\mathbb{A}^{n}, which is stated in Theorem 1.21 in [Sha13].

Theorem 8.4.

Let KK be an algebraically closed field and X𝔾mnX\subset\mathbb{G}_{m}^{n} irreducible and admissible of dimension dd. Then we have #XTTd+11/(d+1)\#X_{T}\ll T^{d+1-1/(d+1)} for all T1T\geq 1.

Proof.

As the conclusion of the theorem is invariant under permuting coordinates we may freely do so. As 𝔾mn\mathbb{G}_{m}^{n} itself is not admissible we may assume dn1d\leq n-1. If d=0d=0, the set XX is finite and hence we have XT#XX1X_{T}\leq\#X\ll_{X}1. So we can assume that d1d\geq 1.
Next we fix a good projection. To this end we consider the coordinate functions x1,,xnx_{1},\dots,x_{n} restricted to XX. They are denoted x1|X,,xn|Xx_{1}|_{X},\dots,x_{n}|_{X} and members of the function field of XX. By dimension theory, the function field K(x1|X,,xn|X)K(x_{1}|_{X},\dots,x_{n}|_{X}) has transcendence degree dd over KK. After permuting coordinates we may assume that x1|X,,xd|Xx_{1}|_{X},\dots,x_{d}|_{X} are algebraically independent.

We claim that there exists k{d+1,,n}k\in\{d+1,\dots,n\} such that x1|X,,xd|X,xk|Xx_{1}|_{X},\dots,x_{d}|_{X},x_{k}|_{X} are multiplicatively independent. Let us assume the contrary and deduce a contradiction. For each such kk there exist ak,1,,ak,d,ak,ka_{k,1},\dots,a_{k,d},a_{k,k}\in\mathbb{Z} not all zero such that

x1|Xak,1xd|Xak,dxk|Xak,k=1x_{1}|_{X}^{a_{k,1}}\cdots x_{d}|_{X}^{a_{k,d}}x_{k}|_{X}^{a_{k,k}}=1 (1)

identically on XX. By algebraic independence we must have ak,k0a_{k,k}\neq 0. So the ndn-d vectors (ak,1,,ak,d,0,,0,ak,k,0,,0)n(a_{k,1},\dots,a_{k,d},0,\dots,0,a_{k,k},0,\dots,0)\in\mathbb{Z}^{n} for k{d+1,,n}k\in\{d+1,\dots,n\} are linearly independent. They generate a subgroup Λ<n\Lambda<\mathbb{Z}^{n} of rank ndn-d. The algebraic subgroup H=HΛ<𝔾mnH=H_{\Lambda}<\mathbb{G}_{m}^{n} is of dimension dd. By (1) we have XHX\subset H. But since XX is irreducible and of the same dimension as HH, it is an irreducible component of HH and hence a torsion coset. This contradicts the admissibility of XX.

After permuting coordinates we may assume k=d+1k=d+1. Let ϕ:𝔾mn𝔾md+1\phi:\mathbb{G}_{m}^{n}\to\mathbb{G}_{m}^{d+1} denote the projection onto the first d+1d+1 coordinates. Then the Zariski closure Y=ϕ(X)¯Y=\overline{\phi(X)} of ϕ(X)\phi(X) in 𝔾md+1\mathbb{G}_{m}^{d+1} is irreducible and dim(Y)d\dim(Y)\leq d. For all (𝐱,𝐱)X(\mathbf{x,x}^{\prime})\in X we have xi|X(𝐱,𝐱)=xi|Y(𝐱)x_{i}|_{X}(\mathbf{x,x}^{\prime})=x_{i}|_{Y}(\mathbf{x}). Thus x1|Y,,xd|Yx_{1}|_{Y},\dots,x_{d}|_{Y} must be algebraically inedependent elements of the function field of YY and hence we have dim(Y)=d\dim(Y)=d. Next we show that YY is admissible. Assume that it is not. Then there exists a torsion coset 𝜻GY\boldsymbol{\zeta}G\subset Y of dimension dd. Since both 𝜻G\boldsymbol{\zeta}G and YY are irreducible we have Y=𝜻GY=\boldsymbol{\zeta}G. Then there exists 𝝀d+1{0}\boldsymbol{\lambda}\in\mathbb{Z}^{d+1}\setminus\{0\} such that 𝐲𝝀=𝜻𝝀\mathbf{y}^{\boldsymbol{\lambda}}=\boldsymbol{\zeta^{\lambda}} for all 𝐲Y\mathbf{y}\in Y. Let NN be the order of 𝜻\boldsymbol{\zeta}. Then since (x1,,xd+1)Y(x_{1},\dots,x_{d+1})\in Y for all (x1,,xn)X(x_{1},\dots,x_{n})\in X we find x1|XNλ1xd+1|XNλd+1=1x_{1}|_{X}^{N\lambda_{1}}\cdots x_{d+1}|_{X}^{N\lambda_{d+1}}=1 identically on XX, a contradiction. Hence YY is admissible. By Remark 8.3 we have that YY is a hypersurface.

Let

Z={𝐱X:dim𝐱(ϕ|X1(ϕ(𝐱)))1}.Z=\{\mathbf{x}\in X:\dim_{\mathbf{x}}\left(\phi|_{X}^{-1}(\phi(\mathbf{x}))\right)\geq 1\}.

It is an algebraic subset of 𝔾mn\mathbb{G}_{m}^{n} by Theorem 14.112 in [GW20]. The set V={𝐲𝔾md+1:dim(ϕ|X1(𝐲))1}V=\{\mathbf{y}\in\mathbb{G}_{m}^{d+1}:\dim\left(\phi|_{X}^{-1}(\mathbf{y})\right)\geq 1\} is algebraic and so is Z=ϕ|X1(V)Z=\phi|_{X}^{-1}(V). By Corollary 14.116 in [BG06] there is a nonempty open UU of YY such that dim(ϕ|X1(𝐮))=0\dim(\phi|_{X}^{-1}(\mathbf{u}))=0 for all 𝐮U\mathbf{u}\in U. Since ϕ(X)\phi(X) is dense in YY, its intersection with UU is nonempty and therefore ZZ is a proper subset of XX. In particular we find dim(Z)d1\dim(Z)\leq d-1.

Let T1T\geq 1. Then XTZTBTX_{T}\subset Z_{T}\cup B_{T} where

BT\displaystyle B_{T} ={𝐱X:dim𝐱(ϕ|X1(ϕ(𝐱)))=0,ord(ϕ(𝐱))T}.\displaystyle=\{\mathbf{x}\in X:\dim_{\mathbf{x}}(\phi|_{X}^{-1}(\phi(\mathbf{x})))=0,\mathrm{ord}(\phi(\mathbf{x}))\leq T\}.

Applying Proposition 1.3 we find #ZTZTdim(Z)+1Td\#Z_{T}\ll_{Z}T^{\dim(Z)+1}\leq T^{d} for all T1T\geq 1. Let CC be a uniform bound for the number of irreducible components in a fibre. For any 𝐱BT\mathbf{x}\in B_{T} we have 𝐲=ϕ(𝐱)YT\mathbf{y}=\phi(\mathbf{x})\in Y_{T} and {x}\{x\} is an irreducible component of ϕ|X1(𝐲)\phi|_{X}^{-1}(\mathbf{y}). Thus we find

#BT𝐲YTdim(ϕ|X1(𝐲))=0#ϕ|X1(𝐲)𝐲YT#Irr(ϕ|X1(𝐲))C#YT.\#B_{T}\leq\sum_{\begin{subarray}{c}\mathbf{y}\in Y_{T}\\ \dim(\phi|_{X}^{-1}(\mathbf{y}))=0\end{subarray}}\#\phi|_{X}^{-1}(\mathbf{y})\leq\sum_{\mathbf{y}\in Y_{T}}\#\mathrm{Irr}(\phi|_{X}^{-1}(\mathbf{y}))\leq C\#Y_{T}.

Since YY is an admissible hypersurface in 𝔾md+1\mathbb{G}_{m}^{d+1} we can apply Theorem 7.5 to bound #YTXTd+11/(d+1)\#Y_{T}\ll_{X}T^{d+1-1/(d+1)} for all T1T\geq 1. The theorem follows from #XT#ZT+#BT\#X_{T}\leq\#Z_{T}+\#B_{T}.∎

Lemma 8.5.

Let ϕ:GG\phi:G\to G^{\prime} be a surjective morphism of algebraic groups. Let H<GH^{\prime}<G^{\prime} be a subgroup and gGg^{\prime}\in G^{\prime}. Then dim(ϕ1(H))=dim(kerϕ)+dim(H)\dim(\phi^{-1}(H^{\prime}))=\dim(\ker\phi)+\dim(H^{\prime}).

Proof.

Let H=ϕ1(H)H=\phi^{-1}(H^{\prime}) and ψ:HH\psi:H\to H^{\prime} which maps gg to ϕ(g)\phi(g). Since ϕ\phi is surjective, ψ\psi is surjective too. We have dimH=dim(ker(ψ))+dimH\dim H=\dim(\ker(\psi))+\dim H^{\prime} and since ker(ϕ)=ker(ψ)\ker(\phi)=\ker(\psi) the lemma follows. ∎

Proof of Theorem 1.2.

If d=0d=0, XX and G=Stab(X)G=\mathrm{Stab}(X) are both finite. The inequality is satisfied, since the exponent equals 0. Thus we can assume that d1d\geq 1. If δ=d\delta=d let 𝐱X\mathbf{x}\in X and HH be the connected component of the identity of GG. The coset 𝐱HX\mathbf{x}H\subset X is irreducible and of dimension dd. Thus we find X=𝐱HX=\mathbf{x}H. Since XX is admissible 𝐱\mathbf{x} is not a torsion point and hence #XT=0\#X_{T}=0. Thus we can assume that δ<dn1\delta<d\leq n-1.

Let LMatn×(nδ)()L\in\mathrm{Mat}_{n\times(n-\delta)}(\mathbb{Z}) be a basis of ΛG\Lambda_{G}. Let ϕ:𝔾mn𝔾mnδ\phi:\mathbb{G}_{m}^{n}\to\mathbb{G}_{m}^{n-\delta} given by 𝐱𝐱L\mathbf{x}\mapsto\mathbf{x}^{L}. Note that ϕ\phi is surjective since dim(Im(ϕ))=ndim(G)=nδ\dim(\mathrm{Im}(\phi))=n-\dim(G)=n-\delta. Let Y𝔾mnδY\subset\mathbb{G}_{m}^{n-\delta} be the Zariski closure of ϕ(X)\phi(X). Then YY is irreducible and hence either admissible or a torsion coset. Since the fibres of ϕ\phi are cosets of GG we have dim(Y)=dδ\dim(Y)=d-\delta. Assume by contradiction that YY is a torsion coset. Then there exists a torsion point 𝜻𝔾mnδ\boldsymbol{\zeta}\in\mathbb{G}_{m}^{n-\delta} of order NN and a linear torus H<𝔾mnδH<\mathbb{G}_{m}^{n-\delta} such that Y=𝜻HY=\boldsymbol{\zeta}H. Then H=𝜻N=1𝜻HH^{\prime}=\bigcup_{\boldsymbol{\zeta}^{N}=1}\boldsymbol{\zeta}H is an algebraic subgroup of 𝔾mnδ\mathbb{G}_{m}^{n-\delta} of dimension dδd-\delta. By Lemma 8.5 we have dim(ϕ1(H))=δ+(dδ)=d\dim(\phi^{-1}(H^{\prime}))=\delta+(d-\delta)=d. Since ϕ(X)Y=𝜻HH\phi(X)\subset Y=\boldsymbol{\zeta}H\subset H^{\prime} and d=dimXd=\dim X we see that XX is an irreducible component of ϕ1(H)\phi^{-1}(H^{\prime}) which are torsion cosets by Lemma A.4. Since d1d\geq 1 this contradicts admissibility. Thus YY is admissible.

Let aka_{k} be the number of cosets of GG of order kk contained in XX. We have

XT=𝜻GX torsion cosetord(𝜻G)T(𝜻G)T and hence #XTGTδ+1k=1Tak/k for all T1X_{T}=\bigcup_{\begin{subarray}{c}\boldsymbol{\zeta}G\subset X\text{ torsion coset}\\ \mathrm{ord}(\boldsymbol{\zeta}G)\leq T\end{subarray}}(\boldsymbol{\zeta}G)_{T}\text{ and hence }\#X_{T}\ll_{G}T^{\delta+1}\sum_{k=1}^{T}a_{k}/k\text{ for all }T\geq 1

using Lemma 6.2. If 𝐱GX\mathbf{x}G\subset X is a coset of order kk, it is the preimage of 𝐱L\mathbf{x}^{L} which is of order at most kk. Therefore ϕ\phi induces an injection 𝐱Gϕ(𝐱)\mathbf{x}G\mapsto\phi(\mathbf{x}) from the cosets of GG which are contained in XX and of order bounded by TT into YTY_{T}. Thus we find k=1Tak#YTXTdδ+11/(dδ+1)\sum_{k=1}^{T}a_{k}\leq\#Y_{T}\ll_{X}T^{d-\delta+1-1/(d-\delta+1)} for all T1T\geq 1 by Theorem 8.4. Since dδ1d-\delta\geq 1 the exponent is strictly larger than 11 and using Lemma 7.2 we find

#XTGTδ+1k=1Tak/kXT(δ+1)+(dδ+11/(dδ+1))1=Td+11/(dδ+1).\#X_{T}\ll_{G}T^{\delta+1}\sum_{k=1}^{T}a_{k}/k\ll_{X}T^{(\delta+1)+(d-\delta+1-1/(d-\delta+1))-1}=T^{d+1-1/(d-\delta+1)}.\qed
Corollary 8.6.

Let KK be an algebraically closed field and X𝔾mnX\subset\mathbb{G}_{m}^{n} admissible of dimension dd. Let δ=minYIrr(X)dim(Stab(Y))\delta=\min_{Y\in\mathrm{Irr}(X)}\dim(\mathrm{Stab}(Y)). Then #XTT1+d1/(dδ+1)\#X_{T}\ll T^{1+d-1/(d-\delta+1)}.

Proof.

Note that #XTYIrr(X)#YT\#X_{T}\leq\sum_{Y\in\mathrm{Irr}(X)}\#Y_{T} and apply Theorem 1.2 to each summand.∎

Remark 8.7.

The number δ\delta in Corollary 8.6 is also equal to

min𝐱Xmax{dim(G):G linear torus, 𝐱GX}.\min_{\mathbf{x}\in X}\max\{\dim(G):G\text{ linear torus, }\mathbf{x}G\subset X\}.

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 KK denotes a field of characteristic 0.

The following theorem is sometimes called the Manin-Mumford- conjecture. In the current setting it was proved by Laurent.

Theorem 9.1.

Let X𝔾mnX\subset\mathbb{G}_{m}^{n} be an algebraic set. Then there are finitely many torsion cosets C1,,CmC_{1},\dots,C_{m} such that X(𝔾mn)tors=i=1m(Ci(𝔾mn)tors)X\cap(\mathbb{G}_{m}^{n})_{\mathrm{tors}}=\bigcup_{i=1}^{m}(C_{i}\cap(\mathbb{G}_{m}^{n})_{\mathrm{tors}}).

Proof.

This is Théorème 2 in [Lau84].∎

Proof of Corollary 1.5.

By Theorem 9.1 there exist finitely many torsion cosets C1,,CmC_{1},\dots,C_{m} in 𝔾mn\mathbb{G}_{m}^{n} such that X(𝔾mn)tors=i=1m(Ci(𝔾mn)tors)X\cap(\mathbb{G}_{m}^{n})_{\mathrm{tors}}=\bigcup_{i=1}^{m}(C_{i}\cap(\mathbb{G}_{m}^{n})_{\mathrm{tors}}). Intersecting both sides with (𝔾mn)T(\mathbb{G}_{m}^{n})_{T} we get XT=i=1m(Ci)TX_{T}=\bigcup_{i=1}^{m}(C_{i})_{T}. If d=0d=0, there are at most finitely many torsion points in XX and thus we have #XTX1\#X_{T}\ll_{X}1 in this case. If d1d\geq 1 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 𝔽p¯\overline{\mathbb{F}_{p}}

We use the Lang-Weil estimates to establish lower bounds if K=𝔽¯pK=\overline{\mathbb{F}}_{p}.

Theorem 10.1.

Let XnX\subset\mathbb{P}^{n} be a geometrically irreducible variety defined over a finite field KK of positive characteristic with qq elements. Assume that XX is of dimension rr. Then there exists C=C(X)C=C(X) such that |#X(K)qr|Cqr(1/2)|\#X(K)-q^{r}|\leq Cq^{r-(1/2)}.

Proof.

This is Theorem 1 in [LW54].∎

Proof of Theorem 1.6.

Considering an irreducible component of XX of dimension dd we can assume that XX is irreducible. Let ZZ be the Zariski closure of Y={(1:x1::xn)n:(x1,,xn)X}Y=\{(1:x_{1}:\dots:x_{n})\in\mathbb{P}^{n}:(x_{1},\dots,x_{n})\in X\} in n\mathbb{P}^{n}. Then ZZ is irreducible and of dimension dd. Since ZZ is irreducible, the proper and closed subset B=Z{(x0::xn)n:x0xn=0}B=Z\cap\{(x_{0}:\dots:x_{n})\in\mathbb{P}^{n}:x_{0}\cdots x_{n}=0\} is of dimension at most d1d-1 and we have Z=YBZ=Y\cup B.

Let kk\in\mathbb{N} be such that ZZ is defined over 𝔽pk\mathbb{F}_{p^{k}}. Let ll\in\mathbb{N} be a multiple of kk. Then 𝔽pk\mathbb{F}_{p^{k}} is a subfield of 𝔽pl\mathbb{F}_{p^{l}} and hence ZZ and BB are defined over 𝔽pl\mathbb{F}_{p^{l}} too. By Theorem 10.1 there exist constants C,D>0C,D>0 independent of ll such that |#Z(𝔽pl)pld|Cpl(d1/2)|\#Z(\mathbb{F}_{p^{l}})-p^{ld}|\leq Cp^{l(d-1/2)} and #B(𝔽pl)Dpldim(B)Dpl(d1)\#B(\mathbb{F}_{p^{l}})\leq Dp^{l\dim(B)}\leq Dp^{l(d-1)}. Let T1T\gg 1 be large enough and ll a multiple of kk such that pl1Tpl+k1p^{l}-1\leq T\leq p^{l+k}-1. Note that for all (x0::xn)Z(𝔽pl)B(𝔽pl)(x_{0}:\dots:x_{n})\in Z(\mathbb{F}_{p^{l}})\setminus B(\mathbb{F}_{p^{l}}) we have (x1/x0,,xn/x0)Xpl1(x_{1}/x_{0},\dots,x_{n}/x_{0})\in X_{p^{l}-1} since for all nonzero x𝔽plx\in\mathbb{F}_{p^{l}} we find xpl1=1x^{p^{l}-1}=1. Therefore we get

#XT\displaystyle\#X_{T} #Xpl1#Z(𝔽pl)#B(𝔽pl)\displaystyle\geq\#X_{p^{l}-1}\geq\#Z(\mathbb{F}_{p^{l}})-\#B(\mathbb{F}_{p^{l}})
pldCpl(d1/2)Dpl(d1)=pld(1Cpl/2Dpl).\displaystyle\geq p^{ld}-Cp^{l(d-1/2)}-Dp^{l(d-1)}=p^{ld}(1-Cp^{-l/2}-Dp^{-l}).

Thus if TT and hence ll is large enough, we have #XTpld212pkd(pk+l)d12pkdTd\#X_{T}\geq\frac{p^{ld}}{2}\geq\frac{1}{2p^{kd}}(p^{k+l})^{d}\geq\frac{1}{2p^{kd}}T^{d}.∎

Appendix A Algebraic subgroups and subgroups of n\mathbb{Z}^{n}

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 KK denotes an algebraically closed field.

Definition A.1.

For an algebraic subgroup GG we define

ΛG={𝐯n:𝐱𝐯=1 for all 𝐱G}\Lambda_{G}=\{\mathbf{v}\in\mathbb{Z}^{n}:\mathbf{x^{v}}=1\text{ for all }\mathbf{x}\in G\}

and for a subgroup Λ<n\Lambda<\mathbb{Z}^{n} we set

HΛ={𝐱𝔾mn:𝐱𝐯=1 for all 𝐯Λ}.H_{\Lambda}=\{\mathbf{x}\in\mathbb{G}_{m}^{n}:\mathbf{x^{v}}=1\text{ for all }\mathbf{v}\in\Lambda\}.
Definition A.2.

Let Λ<n\Lambda<\mathbb{Z}^{n} be a subgroup. The saturation Λ~\tilde{\Lambda} of Λ\Lambda is defined as

Λ~={𝐯n: there exists k:k𝐯Λ}.\tilde{\Lambda}=\{\mathbf{v}\in\mathbb{Z}^{n}:\text{ there exists }k\in\mathbb{N}:k\mathbf{v}\in\Lambda\}.

We call Λ\Lambda primitive if [Λ~:Λ]=1[\tilde{\Lambda}:\Lambda]=1.

Definition A.3.

Let pp be a prime or 0 and Λ<n\Lambda<\mathbb{Z}^{n} a subgroup. We say that Λ\Lambda is pp-full if p=0p=0 or if pp is prime and for all 𝐯n\mathbf{v}\in\mathbb{Z}^{n} we have that p𝐯Λp\mathbf{v}\in\Lambda implies that 𝐯Λ\mathbf{v}\in\Lambda.

Lemma A.4.

Let p=char(K)p=\mathrm{char}(K) and Λ\Lambda be a pp-full subgroup of n\mathbb{Z}^{n} of rank nrn-r. Then HΛH_{\Lambda} is an algebraic subgroup of 𝔾mn\mathbb{G}_{m}^{n} of dimension rr, which is the union of [Λ~:Λ][\tilde{\Lambda}:\Lambda] torsion cosets of dimension rr. We also have ΛHΛ=Λ\Lambda_{H_{\Lambda}}=\Lambda.

Proof.

By the theorem of elementary divisors (Theorem III.7.8 in [Lan02]) there exists a basis 𝐛1,,𝐛n\mathbf{b}_{1},\dots,\mathbf{b}_{n} of n\mathbb{Z}^{n} and λ1,,λnr{0}\lambda_{1},\dots,\lambda_{n-r}\in\mathbb{Z}\setminus\{0\} such that λ1𝐛1,,λnr𝐛nr\lambda_{1}\mathbf{b}_{1},\dots,\lambda_{n-r}\mathbf{b}_{n-r} is a basis of Λ\Lambda. Let B=(𝐛1,,𝐛n)GLn()B=(\mathbf{b}_{1},\dots,\mathbf{b}_{n})\in\mathrm{GL}_{n}(\mathbb{Z}). Then the monoidal transform 𝐱𝐱B\mathbf{x}\mapsto\mathbf{x}^{B} gives an isomorphism between HΛH_{\Lambda} and F×𝔾mrF\times\mathbb{G}_{m}^{r} where F={𝐱𝔾mnr:x1λ1=1,,xnrλnr=1}F=\{\mathbf{x}\in\mathbb{G}_{m}^{n-r}:x_{1}^{\lambda_{1}}=1,\dots,x_{n-r}^{\lambda_{n-r}}=1\} since 𝐱λi𝐛i=1\mathbf{x}^{\lambda_{i}\mathbf{b}_{i}}=1 if and only if 𝐱λiB𝐞i=1\mathbf{x}^{\lambda_{i}B\mathbf{e}_{i}}=1 and thus if and only if 𝐲=𝐱B\mathbf{y=x}^{B} satisfies 𝐲iλi=1\mathbf{y}_{i}^{\lambda_{i}}=1. Then we see that HΛH_{\Lambda} is a union of #F\#F linear tori of dimension rr.

We claim that span{𝐛1,,𝐛nr}=Λ~\mathrm{span}\{\mathbf{b}_{1},\dots,\mathbf{b}_{n-r}\}=\tilde{\Lambda}. Since λi𝐛iΛ\lambda_{i}\mathbf{b}_{i}\in\Lambda the inclusion \subset holds. For the other let 𝐯=α1𝐛1++αn𝐛nΛ~\mathbf{v}=\alpha_{1}\mathbf{b}_{1}+\dots+\alpha_{n}\mathbf{b}_{n}\in\tilde{\Lambda}. There exists mm\in\mathbb{N} such that m𝐯Λ.m\mathbf{v}\in\Lambda. Hence we can write m𝐯=β1λ1𝐛1++βnrλnr𝐛nrm\mathbf{v}=\beta_{1}\lambda_{1}\mathbf{b}_{1}+\dots+\beta_{n-r}\lambda_{n-r}\mathbf{b}_{n-r} and the comparison of coefficients yields αnr+1==αn=0\alpha_{n-r+1}=\dots=\alpha_{n}=0. Hence we have 𝐯span{𝐛1,,𝐛nr}\mathbf{v}\in\mathrm{span}\{\mathbf{b}_{1},\dots,\mathbf{b}_{n-r}\} and the claim is true.

Consider the surjective group homomorphism nrΛ~\mathbb{Z}^{n-r}\to\tilde{\Lambda} given by (α1,,αnr)α1𝐛1++αnr𝐛nr(\alpha_{1},\dots,\alpha_{n-r})\mapsto\alpha_{1}\mathbf{b}_{1}+\dots+\alpha_{n-r}\mathbf{b}_{n-r} and compose it with the reduction Λ~Λ~/Λ\tilde{\Lambda}\to\tilde{\Lambda}/\Lambda. Let D=diagnr,nr(𝝀)D=\mathrm{diag}_{n-r,n-r}(\boldsymbol{\lambda}). Then the kernel of the composition is λ1××λnr=Dnr\lambda_{1}\mathbb{Z}\times\dots\times\lambda_{n-r}\mathbb{Z}=D\mathbb{Z}^{n-r}. Thus we find [Λ~:Λ]=[nr:Dnr]=i=1nr|λi|[\tilde{\Lambda}:\Lambda]=[\mathbb{Z}^{n-r}:D\mathbb{Z}^{n-r}]=\prod_{i=1}^{n-r}|\lambda_{i}|. Since Λ\Lambda is pp-full the λi\lambda_{i} are coprime to pp, if p>0p>0 and therefore we have #μ|λi|=|λi|\#\mu_{|\lambda_{i}|}=|\lambda_{i}| in all cases. Thus we find #F=i=1nr#μ|λi|=i=1nr|λi|=[Λ~:Λ]\#F=\prod_{i=1}^{n-r}\#\mu_{|\lambda_{i}|}=\prod_{i=1}^{n-r}|\lambda_{i}|=[\tilde{\Lambda}:\Lambda].

Let Λ=λ1××λnr×{0}××{0}<n\Lambda^{\prime}=\lambda_{1}\mathbb{Z}\times\dots\times\lambda_{n-r}\mathbb{Z}\times\{0\}\times\dots\times\{0\}<\mathbb{Z}^{n}. In a first step we reduce the last statement to ΛHΛ=Λ\Lambda_{H_{\Lambda^{\prime}}}=\Lambda^{\prime}. We have seen above that 𝐱𝐱B\mathbf{x}\mapsto\mathbf{x}^{B} is an isomorphism between HΛH_{\Lambda} and HΛH_{\Lambda^{\prime}}. Therefore we have

ΛHΛ={𝐯n:𝐱𝐯=1 for all 𝐱HΛ}={BB1𝐯n:𝐱B1𝐯=1 for all 𝐱HΛ}=BΛHΛ.\Lambda_{H_{\Lambda}}=\{\mathbf{v}\in\mathbb{Z}^{n}:\mathbf{x^{v}}=1\text{ for all }\mathbf{x}\in H_{\Lambda}\}=\{BB^{-1}\mathbf{v}\in\mathbb{Z}^{n}:\mathbf{x}^{B^{-1}\mathbf{v}}=1\text{ for all }\mathbf{x}\in H_{\Lambda^{\prime}}\}=B\Lambda_{H_{\Lambda^{\prime}}}.

So if we assume that ΛHΛ=Λ\Lambda_{H_{\Lambda^{\prime}}}=\Lambda^{\prime} we find ΛHΛ=BΛ=Λ\Lambda_{H_{\Lambda}}=B\Lambda^{\prime}=\Lambda.
In the second step we prove ΛHΛ=Λ\Lambda_{H_{\Lambda^{\prime}}}=\Lambda^{\prime} and let 𝐯Λ\mathbf{v}\in\Lambda^{\prime}. Then we have 𝐱𝐯=1\mathbf{x^{v}}=1 for all 𝐱HΛ\mathbf{x}\in H_{\Lambda^{\prime}} and hence 𝐯ΛHΛ\mathbf{v}\in\Lambda_{H_{\Lambda^{\prime}}}.

For the other inclusion let 𝐯ΛHΛ\mathbf{v}\in\Lambda_{H_{\Lambda^{\prime}}}. Let i{1,,nr}i\in\{1,\dots,n-r\}. Since λi\lambda_{i} is coprime to pp if p>0p>0 there always exists a root of unity ξλi\xi_{\lambda_{i}} of order |λi||\lambda_{i}|. We have (1,,1,ξλi,1,,1)HΛ(1,\dots,1,\xi_{\lambda_{i}},1,\dots,1)\in H_{\Lambda^{\prime}} where ξλi\xi_{\lambda_{i}} is the ii-th coordinate. Hence ξλivi=1\xi_{\lambda_{i}}^{v_{i}}=1 which shows that viλiv_{i}\in\lambda_{i}\mathbb{Z}. Now let i{nr+1,,n}i\in\{n-r+1,\dots,n\}. Then for all x𝔾mx\in\mathbb{G}_{m} we have (1,,1,x,1,,1)HΛ(1,\dots,1,x,1,\dots,1)\in H_{\Lambda^{\prime}} where xx is the ii-th coordinate and hence xvi=1x^{v_{i}}=1. This implies that vi=0v_{i}=0, and thus 𝐯Λ\mathbf{v}\in\Lambda^{\prime} and the other inclusion holds too. ∎

Definition A.5.

A torus coset is a coset of a linear torus.

Lemma A.6.

Let X𝔾mnX\subset\mathbb{G}_{m}^{n}, be an algebraic set defined by Laurent polynomials Pi:=𝛌ai,𝛌𝐱𝛌=0P_{i}:=\sum_{\boldsymbol{\lambda}}a_{i,\boldsymbol{\lambda}}\mathbf{x}^{\boldsymbol{\lambda}}=0 and let LiL_{i} be the set of exponents appearing in the monomials in Pi,i=1,mP_{i},i=1,\dots m. Let 𝐳G\mathbf{z}G be a maximal coset in XX. Then G=HΛG=H_{\Lambda}, where Λ\Lambda is a subgroup generated by vectors of type 𝛌i𝛌i\boldsymbol{\lambda}^{\prime}_{i}-\boldsymbol{\lambda}_{i} with 𝛌i,𝛌iLi,\boldsymbol{\lambda}^{\prime}_{i},\boldsymbol{\lambda}_{i}\in L_{i}, for i=1,,mi=1,\dots,m.
Similarly suppose that a linear torus GG has a torus coset zGXzG\subset X which is maximal among torus cosets contained in XX. Then G=HΛ~G=H_{\tilde{\Lambda}}, where Λ\Lambda is a subgroup generated by vectors of type 𝛌i𝛌i\boldsymbol{\lambda}^{\prime}_{i}-\boldsymbol{\lambda}_{i} with 𝛌i,𝛌iLi,\boldsymbol{\lambda}^{\prime}_{i},\boldsymbol{\lambda}_{i}\in L_{i}, for i=1,,mi=1,\dots,m.

Proof.

Suppose that 𝐳GX\mathbf{z}G\subset X is maximal. Let us define Li,χ={𝝀Li:χ𝝀=χ}L_{i,\chi}=\{\boldsymbol{\lambda}\in L_{i}:\chi_{\boldsymbol{\lambda}}=\chi\}. By Lemma 5.6 we have

𝝀Li,χai,𝝀𝐳𝝀=0\sum_{\boldsymbol{\lambda}\in L_{i,\chi}}a_{i,\boldsymbol{\lambda}}\mathbf{z}^{\boldsymbol{\lambda}}=0

for every ii and χ\chi such that Li,χL_{i,\chi}\neq\emptyset.
Let Λ=χi=1m𝝀,𝝀Li,χ(𝝀𝝀)\Lambda=\sum_{\chi}\sum_{i=1}^{m}\sum_{\boldsymbol{\lambda,\lambda}^{\prime}\in L_{i,\chi}}\mathbb{Z}(\boldsymbol{\lambda}-\boldsymbol{\lambda}^{\prime}). By definition of Li,χL_{i,\chi} we find G<HΛG<H_{\Lambda}. On the other hand note that for all 𝐡HΛ\mathbf{h}\in H_{\Lambda} and all χ\chi and ii the term 𝐡𝝀\mathbf{h}^{\boldsymbol{\lambda}} is always the same, say hi,χh_{i,\chi}, for all 𝝀Li,χ\boldsymbol{\lambda}\in L_{i,\chi} by definition of Λ\Lambda. Therefore we can write

Pi(𝐳𝐡)=χ:Li,χ(𝝀Li,χai,𝝀𝐳𝝀)hi,χ=0P_{i}(\mathbf{zh})=\sum_{\chi:L_{i,\chi}\neq\emptyset}\left(\sum_{\boldsymbol{\lambda}\in L_{i,\chi}}a_{i,\boldsymbol{\lambda}}\mathbf{z}^{\boldsymbol{\lambda}}\right)h_{i,\chi}=0

and find 𝐳HΛX\mathbf{z}H_{\Lambda}\subset X. This holds in both cases. Since 𝐳G\mathbf{z}G is a maximal coset in XX we have G=HΛ.G=H_{\Lambda}. in the first case. In the second case we note that GG is an irreducible subset of HΛH_{\Lambda} which contains the identity. Thus the connected component of the identity H0H^{0} of HΛH_{\Lambda} contains GG. Thus we have 𝐳G𝐳H0X\mathbf{z}G\subset\mathbf{z}H^{0}\subset X. Since H0H^{0} is a linear torus and by the maximality of 𝐳G\mathbf{z}G among torus cosets in XX we find G=H0G=H^{0}. Since HΛ~H_{\tilde{\Lambda}} is also a linear torus inside HΛH_{\Lambda} with finite index, we have G=HΛ~G=H_{\tilde{\Lambda}}.∎

Corollary A.7.

Every algebraic subgroup GG of 𝔾mn\mathbb{G}_{m}^{n} is of type HΛH_{\Lambda} for some subgroup Λ\Lambda of n\mathbb{Z}^{n}.

Proof.

Apply Lemma A.6 to GG and its maximal coset GG.∎

Corollary A.8.

Let X𝔾mnX\subset\mathbb{G}_{m}^{n} be an algebraic subset. Then there are only finitely many linear tori G<𝔾mnG<\mathbb{G}_{m}^{n} with the property that there exists 𝐱𝔾mn\mathbf{x}\in\mathbb{G}_{m}^{n} such that 𝐱G\mathbf{x}G is a maximal torus coset in XX.

Proof.

Assume that 𝐱G\mathbf{x}G is a maximal torus coset in XX. By Lemma A.6 there exists Λ<n\Lambda<\mathbb{Z}^{n} generated by vectors of type 𝝀𝝀\boldsymbol{\lambda}^{\prime}-\boldsymbol{\lambda} with 𝝀,𝝀\boldsymbol{\lambda}^{\prime},\boldsymbol{\lambda} in a finite set LL such that G=HΛ~G=H_{\tilde{\Lambda}}. Since the generators come from a finite set, there are also only finitely many Λ\Lambda and hence GG is from a finite set.∎

Definition A.9.

Let Λ<n\Lambda<\mathbb{Z}^{n} be a subgroup and pp a prime or 0. The pp-saturation Λp\Lambda_{p} is defined by

Λp={𝐯n: there exists k0 in :pk𝐯Λ} if p is a prime, and Λ0=Λ.\Lambda_{p}=\{\mathbf{v}\in\mathbb{Z}^{n}:\text{ there exists }k\geq 0\text{ in }\mathbb{Z}:p^{k}\mathbf{v}\in\Lambda\}\text{ if }p\text{ is a prime, and }\Lambda_{0}=\Lambda.
Remark A.10.

The pp-saturation of a subgroup Λ\Lambda is pp-full.

Lemma A.11.

Let Λ<n\Lambda<\mathbb{Z}^{n} and char(K)=p0.(K)=p\geq 0. We have HΛ=HΛp.H_{\Lambda}=H_{\Lambda_{p}}.

Proof.

If p=0p=0 we have Λ0=Λ\Lambda_{0}=\Lambda and the equality obviously holds. Thus we may assume that p>0p>0. Since ΛΛp\Lambda\subset\Lambda_{p} it suffices to show HΛHΛpH_{\Lambda}\subset H_{\Lambda_{p}}. Let 𝐱HΛ\mathbf{x}\in H_{\Lambda} and 𝐯Λp\mathbf{v}\in\Lambda_{p}. There exists a natural number kk such that pk𝐯Λp^{k}\mathbf{v}\in\Lambda. Therefore 𝐱pk𝐯=1\mathbf{x}^{p^{k}\mathbf{v}}=1. Since the characteristic of the field is p>0p>0 we have (x+y)pk=xpk+ypk(x+y)^{p^{k}}=x^{p^{k}}+y^{p^{k}} and hence we find 0=𝐱𝐯pk1pk=(𝐱𝐯1)pk0=\mathbf{x^{v}}^{p^{k}}-1^{p^{k}}=(\mathbf{x^{v}}-1)^{p^{k}} which implies that 𝐱𝐯=1\mathbf{x^{v}}=1 and thus 𝐱HΛp\mathbf{x}\in H_{\Lambda_{p}}.∎

Theorem A.12.

The map ΛHΛ\Lambda\mapsto H_{\Lambda} is a bijection between pp-full subgroups of n\mathbb{Z}^{n} and the algebraic subgroups of 𝔾mn\mathbb{G}_{m}^{n}. The inverse is given by GΛGG\mapsto\Lambda_{G}. Moreover we have rk(Λ)+dim(HΛ)=n\mathrm{rk}(\Lambda)+\dim(H_{\Lambda})=n and HΛH_{\Lambda} is a linear torus if and only if Λ\Lambda is primitive.

Proof.

Corollary A.7 and Lemma A.11 show that the map is surjective. For injectivity assume that Λ\Lambda and Λ\Lambda^{\prime} are two pp-full subgroups of n\mathbb{Z}^{n} such that HΛ=HΛH_{\Lambda}=H_{\Lambda^{\prime}}. By Lemma A.4 we have Λ=ΛHΛ=ΛHΛ=Λ\Lambda=\Lambda_{H_{\Lambda}}=\Lambda_{H_{\Lambda^{\prime}}}=\Lambda^{\prime}. This proves bijectivity. The other statements follow directly from Lemma A.4.∎

Definition A.13.

For a prime pp let RpR_{p} be the localisation of \mathbb{Z} at the powers of pp.

Lemma A.14.

Let nn\in\mathbb{N} and pp be a prime. Then there is a bijection between the submodules MM of RpnR_{p}^{n} and the pp-full subgroups Λ\Lambda of n\mathbb{Z}^{n}. The maps are given by MMnM\mapsto M\cap\mathbb{Z}^{n} and ΛRΛ\Lambda\mapsto R\Lambda.

Proof.

We abbreviate R=RpR=R_{p}. First we show that both maps are well defined. Let MM be a submodule of RnR^{n}. Clearly MnM\cap\mathbb{Z}^{n} is a subgroup of n\mathbb{Z}^{n}. Assume that p𝐯Mnp\mathbf{v}\in M\cap\mathbb{Z}^{n} for some 𝐯n\mathbf{v}\in\mathbb{Z}^{n}. There exists 𝐦M\mathbf{m}\in M such that p𝐯=𝐦p\mathbf{v=m} and therefore 𝐯=𝐦/pM\mathbf{v}=\mathbf{m}/p\in M since 1/pR1/p\in R and hence 𝐯Mn\mathbf{v}\in M\cap\mathbb{Z}^{n}. Thus the first map is well defined. That RΛR\Lambda is a submodule of RnR^{n} is straight forward.
Now we look at the compositions of the maps and show first that R(Mn)=MR(M\cap\mathbb{Z}^{n})=M. The inclusion \subset is immediate since RM=MRM=M. For the other direction let 𝐦M\mathbf{m}\in M. Then there exists rr\in\mathbb{Z} and 𝐯n\mathbf{v}\in\mathbb{Z}^{n} such that 𝐦=pr𝐯\mathbf{m}=p^{r}\mathbf{v}. We have 𝐯=pr𝐦M\mathbf{v}=p^{-r}\mathbf{m}\in M, so 𝐯Mn\mathbf{v}\in M\cap\mathbb{Z}^{n} and hence 𝐦=pr𝐯R(Mn)\mathbf{m}=p^{r}\mathbf{v}\in R(M\cap\mathbb{Z}^{n}). Secondly we want to show that (RΛ)n=Λ(R\Lambda)\cap\mathbb{Z}^{n}=\Lambda. Here the inclusion \supset is direct. For the other direction let 𝐰=r𝐯(RΛ)n\mathbf{w}=r\mathbf{v}\in(R\Lambda)\cap\mathbb{Z}^{n} where rRr\in R and 𝐯Λ\mathbf{v}\in\Lambda. If rr\in\mathbb{Z}, then 𝐰Λ\mathbf{w}\in\Lambda. Else we write r=pskr=p^{-s}k for an integer kk and ss\in\mathbb{N}. Then we find that ps𝐰=k𝐯Λp^{s}\mathbf{w}=k\mathbf{v}\in\Lambda. Since Λ\Lambda is pp-full we find also in this case that 𝐰Λ\mathbf{w}\in\Lambda. ∎

Corollary A.15.

If char(K)=p>0(K)=p>0, the map MHMnM\mapsto H_{M\cap\mathbb{Z}^{n}} is a bijection between the submodules of RpnR_{p}^{n} and the algebraic subgroups of 𝔾mn\mathbb{G}_{m}^{n}.

Proof.

Combine Theorem A.12 and Lemma A.14.∎

Definition A.16.

A monoidal transform ψ:𝔾mn𝔾mn\psi:\mathbb{G}_{m}^{n}\to\mathbb{G}_{m}^{n} is an automorphism given by 𝐱𝐱A\mathbf{x}\mapsto\mathbf{x}^{A} for some AGLn().A\in\mathrm{GL}_{n}(\mathbb{Z}).

Lemma A.17.

Let C𝔾mnC\subset\mathbb{G}_{m}^{n} be a torsion coset of dimension dd. Then there exists a monoidal transformation ψ\psi and 𝛈(𝔾mnd)tors\boldsymbol{\eta}\in(\mathbb{G}_{m}^{n-d})_{\mathrm{tors}} such that ψ(C)={𝛈}×𝔾md.\psi(C)=\{\boldsymbol{\eta}\}\times\mathbb{G}_{m}^{d}.

Proof.

Let GG be the linear torus in 𝔾mn\mathbb{G}_{m}^{n} and 𝜻(𝔾mn)tors\boldsymbol{\zeta}\in(\mathbb{G}_{m}^{n})_{\mathrm{tors}} such that C=𝜻GC=\boldsymbol{\zeta}G. Let Λ=ΛG\Lambda=\Lambda_{G}. Let B=(𝐛1,,𝐛n)=(Bl,Br)missingGLn()B=(\mathbf{b}_{1},\dots,\mathbf{b}_{n})=(B_{l},B_{r})\in\mathbb{\mathrm{missing}}{GL}_{n}(\mathbb{Z}) such that Blnd=ΛB_{l}\mathbb{Z}^{n-d}=\Lambda. Let ψ:𝐱𝐱B\psi:\mathbf{x}\mapsto\mathbf{x}^{B} and 𝜼=𝜻Bl\boldsymbol{\eta=\zeta}^{B_{l}}. Then using B1B^{-1} one can show that ψ(𝜻G)={𝜼}×𝔾md\psi(\boldsymbol{\zeta}G)=\{\boldsymbol{\eta}\}\times\mathbb{G}_{m}^{d} 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 kk entries of the diagonal matrix in the Smith Normal Form of AA is equal to the gcd of the determinats of the kk-minors of AA. This is used to bound the number of isolated torsion points.

Definition B.1.

For n,mn,m\in\mathbb{N}, a matrix A=(ai,j)Matn×m()A=(a_{i,j})\in\mathrm{Mat}_{n\times m}(\mathbb{Z}), I{1,,n}\emptyset\neq I\subset\{1,\dots,n\} and J{1,,m}\emptyset\neq J\subset\{1,\dots,m\} we define AI{}_{I}A as the #I×m\#I\times m matrix with entries ai,ja_{i,j} where ii and jj range over II and {1,,n}\{1,…,n\}, respectively, and in increasing order. Similarly AJA_{J} denotes the n×#Jn\times\#J matrix with entries ai,ja_{i,j} where ii and jj range over {1,,n}\{1,…,n\} and JJ, respectively, and in increasing order. Finally we see that (AI)J=(AJ)I({}_{I}A)_{J}={}_{I}(A_{J}) and denote this matrix by AJI{}_{I}A_{J}.

Remark B.2.

For natural numbers k,m,n,k,m,n, matrices AMatk×m(),BMatm×n(),I{1,,k}A\in\mathrm{Mat}_{k\times m}(\mathbb{Z}),B\in\mathrm{Mat}_{m\times n}(\mathbb{Z}),\emptyset\neq I\subset\{1,\dots,k\} and J{1,,n}\emptyset\neq J\subset\{1,\dots,n\} we have (AB)I=AIB{}_{I}(AB)={}_{I}AB and (AB)J=ABJ(AB)_{J}=AB_{J}.

Definition B.3.

For n,mn,m\in\mathbb{N} and AMatn×m()A\in\mathrm{Mat}_{n\times m}(\mathbb{Z}) let us define for k=1,,min{n,m},k=1,\dots,\min\{n,m\},

dk(A)=gcd{det(AJI):I{1,n},J{1,m},#I=k=#J}0d_{k}(A)=\gcd\{\det({}_{I}A_{J}):I\subset\{1,\dots n\},J\subset\{1,\dots m\},\#I=k=\#J\}\in\mathbb{Z}_{\geq 0}

the gcd of the determinants of the k×kk\times k-minors of AA.

Theorem B.4.

Let m,n,AMatn×m()m,n\in\mathbb{N},A\in\mathrm{Mat}_{n\times m}(\mathbb{Z}) and BMatm×n()B\in\mathrm{Mat}_{m\times n}(\mathbb{Z}). Then we have

det(AB)=I{1,,m}#I=ndet(AI)det(BI).\det(AB)=\sum_{\begin{subarray}{c}I\subset\{1,\dots,m\}\\ \#I=n\end{subarray}}\det(A_{I})\det({}_{I}B).
Proof.

This is the Cauchy-Binet formula. A proof can be found in [Gan86], page 27-28.∎

Lemma B.5.

Let n,m,AMatn×m()n,m\in\mathbb{N},A\in\mathrm{Mat}_{n\times m}(\mathbb{Z}), PGLn()P\in\mathrm{GL}_{n}(\mathbb{Z}) and QGLm()Q\in\mathrm{GL}_{m}(\mathbb{Z}). Then for all k=1,,min{n,m}k=1,\dots,\min\{n,m\} we have dk(PAQ)=dk(A)d_{k}(PAQ)=d_{k}(A).

Proof.

Since P,QP,Q are invertible it is enough to show that dk(A)dk(PAQ)d_{k}(A)\mid d_{k}(PAQ). So let I{1,n},J{1,m}I\subset\{1,\dots n\},J\subset\{1,\dots m\} such that #I=k=#J\#I=k=\#J. By Theorem B.4 we have

det(PIAQJ)=S{1,,n}#S=kdet(PSI)det(ASQJ)=S{1,,n}#S=kdet(PSI)T{1,,m}#T=kdet(ATS)det(QJT).\det({}_{I}PAQ_{J})=\sum_{\begin{subarray}{c}S\subset\{1,\dots,n\}\\ \#S=k\end{subarray}}\det({}_{I}P_{S})\det({}_{S}AQ_{J})=\sum_{\begin{subarray}{c}S\subset\{1,\dots,n\}\\ \#S=k\end{subarray}}\det({}_{I}P_{S})\sum_{\begin{subarray}{c}T\subset\{1,\dots,m\}\\ \#T=k\end{subarray}}\det({}_{S}A_{T})\det({}_{T}Q_{J}).

Therefore dk(A)det(PIAQJ)d_{k}(A)\mid\det({}_{I}PAQ_{J}) and hence dk(A)d_{k}(A) divides their greatest common divisor dk(PAQ)d_{k}(PAQ). ∎

Definition B.6.

For n,mn,m\in\mathbb{N} and 𝛂=(α1,,αmin{n,m})min{n,m}\boldsymbol{\alpha}=(\alpha_{1},\dots,\alpha_{\min\{n,m\}})\in\mathbb{Z}^{\min\{n,m\}} we denote by diagn×m(𝛂)\mathrm{diag}_{n\times m}(\boldsymbol{\alpha}) the n×mn\times m matrix D=(di,j)D=(d_{i,j}) where di,j=0d_{i,j}=0 if iji\neq j and di,i=αid_{i,i}=\alpha_{i} for all i=1,,min{n,m}i=1,\dots,\min\{n,m\}.

Lemma B.7.

Let n,m,l=min{n,m},α1,,αl0n,m\in\mathbb{N},l=\min\{n,m\},\alpha_{1},\dots,\alpha_{l}\in\mathbb{Z}_{\geq 0} such that αi+1αi\alpha_{i+1}\mathbb{Z}\subset\alpha_{i}\mathbb{Z} for all i=1,,l1i=1,\dots,l-1. Then dk(diagn×m(α1,,αl))=α1αkd_{k}(\mathrm{diag}_{n\times m}(\alpha_{1},\dots,\alpha_{l}))=\alpha_{1}\cdots\alpha_{k} for all k=1,,lk=1,\dots,l.

Proof.

Let k{1,,l},D=diagn×m(α1,,αl)k\in\{1,\dots,l\},D=\mathrm{diag}_{n\times m}(\alpha_{1},\dots,\alpha_{l}) and I{1,,n},J{1,,m}I\subset\{1,\dots,n\},J\subset\{1,\dots,m\} such that #I=k=#J\#I=k=\#J. If IJI\neq J there is a row of zeros in DJI{}_{I}D_{J} since the entries are di,jd_{i,j} where iIi\in I and jJj\in J and hence the determinant vanishes. If I=JI=J the determinant of DJI{}_{I}D_{J} is given by iIαi\prod_{i\in I}\alpha_{i}. Therefore we find that dk(D)=gcd({iIαi:I{1,,l},#I=k})=α1αkd_{k}(D)=\gcd(\{\prod_{i\in I}\alpha_{i}:I\subset\{1,\dots,l\},\#I=k\})=\alpha_{1}\dots\alpha_{k}, where we use the divisibility properties of the αi\alpha_{i}.∎

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