Cover Page. The handle holds various files of this Leiden University dissertation

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1 Cover Page The handle holds various files of this Leiden University dissertation Author: Junjiang Liu Title: On p-adic decomposable form inequalities Issue Date:

2 Chapter 1 Preliminaries In this chapter, we have collected definitions, facts and basic lemmas that are used throughout the thesis. 1.1 Decomposable forms Definition A homogeneous polynomial F Q[X 1,..., X n ] of degree d is called a decomposable form if it factorizes into linear forms over some algebraic closure of Q. Lemma Let F Q[X 1,..., X n ] be any decomposable form. Then it factors into linear forms over some algebraic number field. Proof. [3, Chapter 2, Theorem 1] For a polynomial F in n variables and an n n matrix T = (a ij ) i,j=1,...,n, we define n n F T (X 1,..., X n ) := F ( a 1j X j,..., a nj X j ). j=1 j=1 Definition Two homogeneous polynomials F, G Z[X 1,..., X n ] of the same degree are called equivalent if there exists a matrix T GL n (Z) such that F = G T. 17

3 Definition A norm form over Q is a decomposable form of the shape F = AN L/Q (λ 1 X 1 + λ 2 X λ n X n ) = d ( ) σi (λ 1 )X 1 + σ i (λ 2 )X σ i (λ n )X n i=1 where A Q, L = Q(λ 1,..., λ n ) and σ 1,..., σ d are the embeddings of L in Q. Lemma Let F Q[X 1,..., X n ] be a decomposable form which is irreducible over Q. Then F is a norm form. Proof. [3, Chapter 2, Theorem 2] Let F Z[X 1,..., X n ] be a decomposable form of degree d. Let P be the set of all primes and M Q = P { }. We denote the set of places of Q by M Q. Let S = {, p 1,..., p r } be a fixed (once for all) finite subset of M Q, where p 1,.., p r are primes. Put S 0 = S \ { }. We are interested in the solutions to inequalities of the shape F (x) p m in x = (x 1, x 2,..., x n ) Z n with gcd(x 1, x 2,..., x n, p 1 p r ) = 1. (1.1.1) Define I F,S (m) to be the set of solutions to (1.1.1) and put N F,S (m) = I F,S (m). By Lemma 1.1.2, every decomposable form F Z[X 1,..., X n ] can be factored into linear forms over some number field. Let K be the smallest extension of Q in which F factorizes into linear forms. This field is called the splitting field of F. Write F = d i=1 L i with L i K[X 1,..., X n ] linear forms. For every p M Q, we choose an extension of p to K. For p M Q, let K p be the completion of K at p. Identify K as a subfield of K p. We use the same factorization of F for each p S. Henceforth, we fix a factorization F = g i=1 F i = d i=1 L i, where F 1,..., F g Z[X 1,..., X n ] are irreducible and L 1,..., L d K[X 1,..., X n ] are linear forms. We denote the coefficient vector of a linear form L(X) by L, that is, if L(X) = a 1 X a n X n, then L = (a 1,..., a n ). M = (b 1,..., b n ) is M(X) = b 1 X b n X n. Conversely, the linear form associated to Let I(F ) denote the set of all ordered n-tuples (L i1, L i2,..., L in ) which are linearly independent over K. 18

4 We define the L 2 -norm of x = (x 1,..., x n ) C n by x := x x n 2. For p P, x = (x 1,..., x n ) Q n p, we define the p-sup-norm of x by x p = max{ x 1 p,..., x n p }. For each p S, define H p (F ) := d L i p where denotes the L 2 -norm and p (p S 0 ) denotes the p-sup-norm. Then the height of F is given by i=1 H(F ) = H p (F ). (1.1.2) Note that H p (F ) is independent of the factorization of F for each p S and so is H(F ). Define the semi-discriminant S(F ) of F by S(F ) := (L i1,l i2,...,l in ) I(F ) It is known that S(F ) Q (see, e.g. [19], Lemma 2). det(l i1, L i2,..., L in ). For p S, define the normalized p-adic semi-discriminant by NS(F ) p := (L i1,l i2,...,l in ) I(F ) det(l i1, L i2,..., L in ) p n j=1 L i j p. Define b(l i ) as the number of times that L i appears in some element of I(F ) and put b(f ) := max 1 i d b(l i). Since any two linear forms dividing the same factor F l of F are conjugate to one another, we have b(l i1 ) = b(l i2 ) whenever L i1 and L i2 belong to the same factor F l of F. Put b(l) := b(l i ) for any linear factor L i of F l. Let K be an algebraic number field. For x = (x 1,..., x n ) K n, define x p = ( x 1 2 p + + x n 2 p) 1/2 if p is archimedean, x p = max ( x 1 p,..., x n p ) if p is nonarchimedean. 19

5 The field height H K (x) of x is defined as where d p = [K p : Q p ] is the local degree. H K (x) = x dp p p M K Lemma Let F = d i=1 L i Z[X 1,..., X n ] be a decomposable form. Then each L i is proportional to a linear form L i with algebraic coefficients in an algebraic number field of degree at most d. Furthermore, H(F ) H Q(L i )(L i ) 1 for i = 1,..., d. Proof. This lemma follows from [19, Lemma 2]. For the convenience of the readers, we give the proof. First, suppose that F is irreducible over Q. We know that F (X) = c N K/Q (L(X)) for some number field K and some c Q. Then any linear factor L i of F is proportional to some conjugate of L. Choose for each p S an extension of p to Q. Assume that F = a F where a Z and F is primitive, i.e., the coefficients of F have gcd 1. Then by Gauss s Lemma for each p S 0, we have H p (F ) = d L i p = af p = a p. i=1 Hence by the product formula H(F ) = H (F ) H p (F ) = H (F ) a 0 By [19, Lemma 2], we have H (F )/ a = H K (L) 1. Hence H(F ) H K (L) 1. a p H (F ). a Second, suppose that F = g l=1 F l where F l Z[X 1,..., X n ] is an irreducible form. Then any L i of F is a linear factor of F li (1 l i g) and hence proportional to some L i 20

6 with algebraic coefficients in a number field of degree at most deg(f li ) d and satisfying H(F li ) H Q(L i )(L i ) 1. Therefore H(F ) = g H(F l ) H(F li ) 1. l=1 We often use the following well known lemma, which is called Hadamard s inequality. Lemma Let p M Q and L i1, L i2,..., L in Q n p. Then we have det(l i1, L i2,..., L in ) p n j=1 L i j p 1. Lemma Assume that I(F ). Then we have (a) NS(F ) p H(F ) b(f ). (b) For each tuple (L p1, L p2,..., L pn ) I(F ) (p S), we have n j=1 1 L pj p H(F ) b(f ) n!. det(l p1, L p2,..., L pn ) p Proof. This lemma follows from [19, Lemma 3]. For the convenience of the readers, we give the proof. (a) First, there is no loss of generality to assume that F is primitive. Indeed, let F = af with F primitive and a Z, and assume that (a) is true for F. Then we have b(f ) = b(f ), and also NS(F ) p = NS(F ) p, H(F ) b(f ) = ( a p ) b(f ) H(F ) b(f ) H(F ) b(f ) = H(F ) b(f ). 21

7 Hence NS(F ) p = NS(F ) p H(F ) b(f ) H(F ) b(f ). Now assume that F is primitive. Recall that S(F ) Q. For p M Q \ S, we have S(F ) p = det(l i1, L i2,..., L in ) p (i 1,...,i n) I(F ) (i 1,...,i n) I(F ) L i1 p L i2 p L in p = F l b(l) p = 1 l and then by the product formula, S(F ) p = 1 p M Q \S S(F ) 1. p Hence have NS(F ) p = I(F ) S(F ) p n j=1 L = i j p S(F ) p d i=1 L i b(li) p 1 g l=1 H p(f l ) b(l) = 1 g l=1 H (F l ) b(l) 1 H (F ) b(f ) = 1 H(F ) b(f ) (b) Apply Hadamard s inequality. For each {(L p1, L p2,..., L pn ) : p S} in I(F ) we n j=1 1 L pj p ( det(l p1, L p2,..., L pn ) p NS(F ) p ) 1 n! H(F ) b(f ) n!. 1.2 Measures, geometry of numbers In this section, let F Z[X 1,..., X n ] be a decomposable form of degree d such that F (x) 0 for every x Q n \ {0}. Let S be a finite subset of M Q including. Define Z S := {x Q : x p 1 for p / S} 22

8 and A n S := Qn p. We identify Q n with a subset of A n S via the diagonal embedding. Let µ be the normalized Lebesgue measure on R = Q such that µ ([0, 1]) = 1. Let µ n be the product measure on R n. For p S 0, let µ p be the normalized Haar measure on Q p such that µ p (Z p ) = 1. Let µ n p be the product measure on Q n p. Define the product measure µ n = µn p on A n S. Define A F,S (m) := {(x p ) p A n S : F (x p ) p m, x p p = 1 for p S 0 }. We view Z n S as a subset of An S by identifying x Zn S with (x) A n S. identification, we may interpret the set of the solutions of (1.1.1) as A F,S (m) Z n. With this Counting lattice points We prove a general result on counting the number of lattice points in a symmetric convex body. Here, by a lattice we mean a free Z-module of rank n in Q n. For p P, we view Q as a subfield of Q p. For a lattice M, we define { M p = Z p M = ζ i m i : ζ i Z p, m i M, I finite i I }. Lemma of rank n. (a) Let M be a lattice and p be a prime, then M p is a free Z p -module (b) If a 1, a 2,..., a n is a Z p -basis of M p, then det(a 1, a 2,..., a n ) p = det M p. Proof. Obvious. Let A p GL(n, Q p ) for p S 0 and A q = Id for q P \ S 0 where Id is the n n unit matrix. Define M = {x Q n : A p x p 1 for p P}. The following lemma is well-known. For the convenience of the readers, we give the proof. 23

9 Lemma (a) M is a lattice. (b) We have M p = {x Q n p : A p x p 1} for p P. (c) det M = 0 det(a p ) p. Proof. (a) For p P, write A p = (a ij ) i,j and A 1 p = (b ij ) i,j. Define C p = max i,j { a ij p, b ij p }. Then C p > 0 for p S 0 and C p = 1 for p P \ S. Choose α Q such that α p = C 1 p for p P. For every x Q n p and p P, we have A p x p (max a ij p ) x p C p x p, i,j x p = (A 1 p A p )x p = A 1 p (A p x) p C p A p x p. Hence Cp 1 x p A p x p C p x p for p P. Then for m M, we have Hence αm Z n. So M α 1 Z n. αm p = C 1 p m p A p m p 1 for p P. On the other hand, for x αz n, we have α 1 x p 1 for p P. Hence A p x p C p x p α 1 x p 1 for p P. This implies αz n M. So αz n M α 1 Z n which implies that M is a lattice. (b) Let p P and define R p := {x Q n p : A p x p 1}. By definition { M p = Z p M = finite } ζ i m i : ζ i Z p, m i M. Since A p m i p 1 for m i M, we have A p m p 1 for m M p. Hence M p R p. We need to prove R p M p. Let x R p. There is x Q n such that x x p α p and A p (x x ) p 1. Hence A p x p 1 and x x αz n p M p. So it suffices to prove that x M p. By the Chinese Remainder Theorem, there is β Q such that β 1 p < 1, β q A q x q 1 for q p. 24

10 So β p = 1. Now we have A p (βx ) p = A p x p 1 and A q (βx ) q 1 for q p. Hence βx M. Since β Z p, it follows that x M p. Therefore, also x M p. (c) For each p P, M p = { x Q n p : A p x p 1 } = { x Q n p : A p x Z n p} = A 1 p Z n p. This implies that the columns of A 1 p serve as a Z p -basis of M p. Hence Since det M Q, we have det M = det M p = det A 1 p p = det A p 1 p. p P det M 1 p = 0 det M 1 p = 0 det A p p. The following two lemmas are Lemma 8 and Lemma 9 from Thunder [19]. They allow us to bound the number of lattice points inside a symmetric convex body. Lemma Let C R n be a symmetric convex body and let Λ R n be a lattice. Suppose there are n linearly independent lattice points in C Λ. Then there are y 1, y 2,..., y n C such that C Λ 3 n 2 n(n 1)/2 det(y 1, y 2,..., y n ). det Λ Lemma Let a 1, a 2,..., a n be positive reals and K 1, K 2,..., K n C[X 1,..., X n ] linearly independent linear forms with det(k 1,..., K n) = 1. Let C = {y R n : K i (y) a i, i = 1,..., n}. Then C Z n either lies in a proper linear subspace of Q n or C Z n 3 n 2 n(n 1)/2 n! The volume of C is at most 2 n n! n i=1 a i. 25 n a i. i=1

11 Apply these lemmas to the following situation. Let C := {(x p ) A n S : K pi (x p) p a pi for p S, i = 1,..., n}, where for p S K p1,..., K pn are linearly independent linear forms in Q p [X 1,..., X n ] with and the a pi are positive reals. det(k p1, K p2,..., K pn) p = 1, K p1 p = = K pn p = 1 Lemma Let C be as above. Then we have µ n (C) i=1 n a pi. Assume that there are n linearly independent points in C Q n. Then we have C Z n S i=1 n a pi. The implicit constants implied by the symbols depend only on n, S Proof. For each p S 0, we choose λ pi Q p such that a pi λ pi p < pa pi. For p =, we choose λ i = a i. Then C = {(x p ) A n S : λ 1 pi K pi (x p) p 1 for p S, i = 1,..., n}. For p S 0, let A p be a matrix with rows (λ 1 p1 K p1, λ 1 p2 K p2,..., λ 1 pn K pn). Let A q = Id, for q P \ S 0. Define O := {x R n : λ 1 i K i (x) 1, i = 1,..., n}, Λ := {x Z n S : A p(x) p 1, p S 0 }. Then O is a symmetric convex body. By Lemma 1.2.2, Λ is a lattice and clearly C Z n S = O Λ. 26

12 Suppose there are n linearly independent lattice points in O. Then by Lemma and Lemma 1.2.4, there are y 1, y 2,..., y n O such that O Λ 3 n 2 n(n 1)/2 det(y 1, y 2,..., y n ) det Λ 0 3 n 2 n(n 1)/2 n! n i=1 a i det Λ By Lemma 1.2.2, we have det Λ = A p p = det(λ 1 p1 K p1, λ 1 p2 K p2,..., λ 1 pn K pn) p 0 0 det(k p1 =, K p2,..., K pn) p 1 1 n i=1 λ = n pi p 0 i=1 λ n pi p 0 i=1 pa. pi Hence n C Z n S = O Λ 3n 2 n(n 1)/2 i=1 n! a i det Λ 3 n 2 n(n 1)/2 n! p n 0. (1.2.1) n a pi. (1.2.2) Now we compute the volume of C. For p S 0, define the Q p -linear map φ p : Q n p Q n p by y p = φ p (x p ) = A p x p. Further, let φ = Id. This gives a product map Φ = φ p : A n S An S. Then we have Φ(C) = {(y p ) A n S : K i (x ) a i, y p p 1 for p S 0, i = 1,..., n}. Hence, using (1.2.1) µ n (O) = µ n (Φ(C)) = det φ p p µ n (C) = det A p p µ n (C) 0 1 = n 0 i=1 λ µ n (C) = det Λ µ n (C). pi p i=1 Therefore n µ n (C) = µn (O) det Λ 2n n! i=1 a i det Λ 2 n n! p n 0 n a pi. i=1 27

13 1.2.2 Geometry of numbers In this section, we recall some results from the adelic Geometry of numbers. Definition An n-dimensional symmetric p-adic convex body is a set C p Q n p with the following properties: (a) C p is compact in the topology of Q n p and has 0 as an interior point, (b) for x C p, a Q p with a p 1 we have ax C p, (c) if p =, then for x, y C, λ R with 0 λ 1 we have (1 λ)x + λy C p, (d) if p is finite, then for x, y C p we have x + y C p. Definition An n-dimensional S-convex body is a Cartesian product C = C p Q n p = A n S where for p S, C p is an n-dimensional symmetric p-adic convex body. For λ > 0 set λc := λc 0 C p. For i = 1,..., n, we define the i-th successive minimum λ i of C p to be the minimum of all λ R 0 such that λc p Z n S contains at least i Q-linearly independent points. From the definition of p-adic convex body and from the fact that Z n S is discrete in An S, it follows that these minima exist and Define 0 < λ 1 λ n <. For p S, let M p1,..., M pn be linearly independent linear forms in Q p [X 1,..., X n ]. C p := { x Q n p : max 1 i n M pi(x) p 1 }. Then C p is a symmetric p-adic convex body and C p is a n-dimensional S-convex body. Let λ 1,..., λ n denote the successive minima of C p. 28

14 Lemma λ 1 λ n det(m p1,..., M pn ) p Proof. See Lemma 6 in [6]. For a linear form L = a 1 X a n X n with a 1,..., a n Q p and σ Gal(Q p /Q p ), define σ(l) := σ(a 1 )X σ(a n )X n. Definition For p M Q, a set L = {L 1,..., L d } of linear forms with coefficients in Q p (resp. a finite extension E p of Q p ) is called Q p -symmetric (resp. Gal(E p /Q p )- symmetric) if for every σ Gal(Q p /Q p ) (resp. σ Gal(E p /Q p )) σ(l) := {σ(l 1 ),..., σ(l d )} = L. We also need the following lemma. Let d n be an integer. For p S, let L p = {L p1,..., L pd } Q p [X 1,..., X n ] be a Q p -symmetric set of linear forms of maximal rank n. Define C p := { x Q n p : max 1 i d L pi(x) p 1 }. Let λ 1,..., λ n be the successive minima of C p. Further let R p = R p (L p ) := max 1 i 1,...,i n d det(l pi i,..., L pin ) p where the maximum is taken over all tuples i 1,..., i n from {1,..., d}. Lemma λ 1 λ n R p Proof. See [6, Lemma 6]. 29

15 1.3 Basic properties of the volume of A F,S (m) Recall that A F,S (m) = {(x p ) p A n S : F (x p ) p m, x p p = 1 for p S 0 }. In this section, we show how the volume of A F,S (m) changes under some actions defined below. In fact, we can consider a slightly more general situation by letting F vary at each p S. Let S = {, p 1,..., p r }. For each p S, take T p GL n (Q p ) and let F p Q p [X 1,..., X n ] be a decomposable form of degree d. Define A(F p : p S, m) := {(x p ) p A n S : F p (x p ) p m, x p p = 1 for p S 0 }. Write A(F p : p S) := A(F p : p S, 1). The first fact we notice is that the volume of A(F p : p S, m) is homogeneous in m. Lemma For m R 0, we have µ n( A(F p : p S, m) ) = m n/d µ n( A(F p : p S) ). Remark The statement is also true if µ n (A(F p : p S)) is infinite. Proof of Lemma We can write A(F p : p S, m) as a disjoint union { } A(F p : p S, m) = (x p ) p A n S : F (x ) mp z pzr r,. F p (x pi ) pi = p zi i (i = 1,..., r) Therefore µ n( A(F p : p S, m) ) = = z 1,...,z r Z 0 z 1 0,...,z r 0 z 1 0,...,z r 0 µ n { F (x ) mp z pzr r } r i=1 m n/d µ n { F (x ) p z pzr r } = m n/d µ n( A(F p : p S) ). 30 µ n p i { F p (x pi ) pi = p zi i } r i=1 µ n p i { F p (x pi ) pi = p zi i }

16 The next lemma tells us how µ n (A(F p : p S)) changes under the action of T p GL n (Q p ), p S. Lemma Let T p GL n (Q p ) for each p S. Then µ n( A((F p ) Tp : p S) ) ( ) = det(t p ) 1 p µ n (A(F p : p S). Proof. Let k = (k p : p S) Z r+1 with k = 0 and S 0 = r. Put k = (k p : p S 0 ). Define A(k ) = {(x p ) p A n S : Then we can write F p (x p ) p 1, T 1 p (x p ) p = p kp for p S 0 }. A(F p : p S) = k Z r A(k ). Let a = 0 p kp. For each p S, define a map φ kp : Q n p Q n p by φ kp (x p ) = atp 1 (x p ). This gives a product map φ k = φ k p : A n S An S. For each p S, put y p = atp 1 (x p ). Then we have } φ k (A(k )) = {(y p ) p A n S : F p (T p (y p )) p 1, y p p = a p p kp = 1 for p S 0 So we have Therefore = µ n( A((F p ) Tp : p S) ) = k Z r k Z r φ k (A(k )) = A((F p ) Tp : p S). k Z r µ n a n det T 1 p p µ n p(a(k )) = = ( det T p p ) 1 µ n (A(F p : p S)). ( φk (A(k ) ) = k Z r k Z r det φ kp p µ n p(a(k )) det T p 1 p µ n p(a(k )) 31

17 Corollary Let λ p Q p for p S. Then µ n (A(λ p F p : p S)) = ( λ p p ) n/d µ n (A(F p : p S)). Proof. Let T = (( λ p p ) 1/d ) Id GL n (R) and T p = Id GL n (Q p ) for p S 0. By Lemma 1.3.3, we have µ n (A(λ p F p : p S)) = ( det T p p ) 1 µ n (A((F p : p S)) = ( λ p p ) n/d µ n (A(F p : p S)). 1.4 Analysis of the small solutions In this section, we use the sup-norm of x = (x 1,..., x n ) R n defined by Given B 0 > 0, we define x := max{ x 1,..., x n }. A F,S (m, B 0 ) := {(x p ) p A F,S (m) : x B 0 }. We estimate the difference between the volume µ n (A F,S (m, B 0 )) and the number of integer points in A F,S (m, B 0 ), i.e., A F,S (m, B 0 ) Z n. First, we need the following generalization of [19, Lemma 14]. Lemma Let F 1,..., F r R[X 1,..., X n ]. Suppose that F i has total degree d i for i = 1,..., r. Let B 0, m 1,..., m r be positive reals and A := {x R n : F i (x) m i, i = 1,..., r}. Assume that A {x R n : x B 0 }. Then we have µ n (A) A Z n n2 r d 1 d r (2B 0 + 1) n 1. 32

18 Proof. The proof is by induction on n. Let n = 1. If F R[X] and deg F = d, we know that the set {x R : F (x) m} = {x R : F 2 (x) m 2 } is a disjoint union of at most 2d closed intervals. So A = {x R : F i (x) m i, i = 1,..., r} is a disjoint union of at most 2 r d 1 d r closed intervals. For each interval I we know that µ (I) I Z 1. Hence µ (A) A Z n 2 r d 1 d r. Let n 2. Let v n be the counting measure of Z n, that is v n (B) = B Z n for a subset B of Z n. For x = (x 1,..., x n ), let x = (x 1,..., x n 1 ) and define A(x ) := {x n R : (x, x n ) A}, A := {x R n 1 : A(x ) }, A(x n ) := {x R n 1 : (x, x n ) A}, A := {x n R : A(x n ) }. By our assumption A {x R n : x B 0 }, we have A(x ) [ B 0, B 0 ], A [ B 0, B 0 ] n 1, A(x n ) [ B 0, B 0 ] and A [ B 0, B 0 ]. Hence we can apply the induction hypothesis to A(x ) := {x n R : F i,x (x n ) m i, i = 1,..., r} with deg F i,x (X n ) deg F i = d i for i = 1,..., r and A(x n ) := { x R n 1 : F i,xn (x ) m i, i = 1,..., r } with deg F i,xn (X 1,..., X n 1 ) deg F i = d i for i = 1,..., r. 33

19 By the Fubini-Tonelli Theorem (see [2, 41.1]), we have µ n (A) A Z n = µ n (A) v n (A) { } { } = dµ n 1 (x ) dµ 1 (x n ) dv 1 (x n ) dv n 1 (x ) A x A(x n) A x n A(x ) { } = µ n 1 (A(x n ))dµ 1 (x n ) dv n 1 (x ) dµ 1 (x n ) A A x A(x n) { } + dµ 1 (x n ) dv n 1 (x ) v 1 (A(x ))dv n 1 (x ) A x n A(x ) A. This leads to µ n (A) A Z n µ n 1 (A(x n ))dµ 1 (x n ) v n 1 (A(x n ))dµ 1 (x n )+ A A + µ 1 (A(x ))dv n 1 (x ) v 1 (A(x ))dv n 1 (x ) A A ( µ n 1 (A(x n )) v n 1 (A(x n )) ) dµ 1 ( (x n ) + µ 1 (A(x )) v 1 (A(x )) ) dv n 1 (x ) A A µ n 1 (A(x n )) v n 1 (A(x n )) dµ 1 (x n ) + µ 1 (A(x )) v 1 (A(x )) dv n 1 (x ) A A (n 1)2 r d 1 d r (2B 0 + 1) n 2 dµ 1 (x n ) + (2 r d 1 d r )dv n 1 (x ) A A (n 1)2 r d 1 d r (2B 0 + 1) n 2 2B r d 1 d r (2B 0 + 1) n 1 n2 r d 1 d r (2B 0 + 1) n 1. Remark We obtain a slightly weaker version of Thunder s Lemma 14 in [19] by putting F 1 (X) = F (X), F 2 (X) = X 1, F 3 (X) = X 2,..., F n+1 = X n and m 1 = m, m 2 = = m r = B 0. We also need a lemma of Schlickewei. 34

20 Lemma Let Λ be any lattice in R n and a Euclidean norm on R n. Then Λ has a basis {a 1,..., a n } with the property that for any x = y 1 a y n a n with y 1,..., y n R, we have x 2 (3/2)n max{ y 1 a 1,..., y n a n }. Proof. See [13, Lemma 2.2 ]. Combining the previous two lemmas, we obtain the following: Lemma Let F R[X 1,..., X n ] be a polynomial of total degree d. Let m, B 0 and m i, i = 1,..., n be positive reals. Let A = {x R n : F (x) m, x i m i, i = 1,..., n}. Let Λ R n be a lattice of rank n and be the sup-norm on R n. Assume A {x R n : x B 0 } and inf{ x : 0 x Λ} = θ > 0. Then µ n (A) n4 n det Λ A Λ B 0 nd (2 + 1) n 1. θ Proof. Choose a basis {a 1,..., a n } of Λ as in Lemma and let A be the matrix with columns a 1,..., a n. For the set A = A 1 A = {y R n : Ay A}, we have A = {y R n : F (Ay) m, < a i, y > m i for i = 1,..., n} where <, > is the usual inner product of vectors. By Lemma 1.4.3, for each y = n i=1 y ia i A we have that y i 4n x a i n4 n x a i n4 n B 0. θ Hence A {y R n : y n4 n B 0 /θ}. We know that µ n (A ) = µ n (A)/ det Λ, A Z n = A Λ. 35

21 Now an application of Lemma to A gives µn (A) det Λ A Λ = µ n n4 (A ) A Z n n B 0 nd (2 θ + 1) n 1. We use Lemma to analyze the set A F,S (m, B 0 ). Without loss of generality, we assume that F is primitive. For p S 0, let E p be the splitting field of F over Q p. Choose a factorization F = L p1 L pd with L pi = a pi,1 x 1 + +a pi,n x n E p [X 1,..., X n ]. We can fix the factorization such that L pi p = max{ a pi,1 p,..., a pi,n p } = 1 (i = 1,.., d). Let e p be the ramification index of E p /Q p. Then e p [Ep : Q p ] and hence e p d!. For every x p Z n p with x p p = 1, there are z p1,..., z pd Z 0 such that L pi (x p ) p = p zpi/ep, i = 1,..., d. For a tuple z := (z pi : p S 0, i = 1,..., d) Z dr 0, we define { } A F,S (m, B 0, z) = (x p ) p A F,S (m, B 0 ) : L pi (x p ) p = p zpi/ep (p S 0, i = 1,..., d). Then A F,S (m, B 0 ) = z Z A dr F,S (m, B 0, z). 0 Define m(z) = m i=1 zpi)/ep. Then 0 p ( d (x p ) p A F,S (m, B 0, z) implies F (x ) m(z). For each x A F,S (m, B 0, z) Z n and q P \ S 0, we have F (x) q 1. By the product formula, we have p M Q F (x) p = 1, hence F (x) p 1. Therefore 0 p ( d i=1 zpi)/ep F (x) n d/2 H (F ) B d 0 = H(F ) B d 0. This implies 0 d i=1 z pi log ( ) n d/2 H(F ) B0 d / log 2, e p 36

22 hence 0 i=1 d z pi d! ( ) log n d/2 H(F ) B0 d log 2 (1.4.1) Let Z be the set of tuples z = (z pi : p S 0, 1 i d) Z dr 0 that satisfy condition (1.4.1). Then A F,S (m, B 0, z) Z n implies that z Z. Note that Z ( d! log ( ) ) dr n d/2 H(F ) B0 d / log 2 (1 + log ( ) ) dr H(F )B 0 (1.4.2) where the implicit constant depends only on n, d and S. So we can write A F,S (m, B 0 ) Z n as a finite disjoint union A F,S (m, B 0 ) Z n = (A F,S (m, B 0, z) Z n ). z Z Lemma With the notation above, we have µ n (A F,S (m, B 0, z)) A F,S (m, B 0, z) Z n (B0 + 1) n 1 where the implicit constant depends only on n, d and S. Proof. Write A F,S (m, B 0, z) as follows: A F,S (m, B 0, z) = (x p) p A n S : F (x ) m(z), x B 0, L pi (x p ) p = p zpi/ep, 1 i d, p S 0, x p p = 1, p S 0. Let L p,d+1 = X 1,..., L p,d+n = X n and z p,d+1 = = z p,d+n = 0 for p S. Then { } A F,S (m, B 0, z) = (x p ) p A n S : F (x ) m(z), x B 0, L pi (x p ) p = p zpi/ep, i = 1,..., d + n, p S 0. Put I 0 = {(p, i) : p S 0, 1 i d + n}. For I I 0, define F (x ) m(z), x B 0, A F,S (m, B 0, z, I) = (x p) p A n S : L p,i (x p ) p < p zpi/ep ((p, i) I), L p,i (x p ) p p zp,i/ep ((p, i) I 0 \ I). 37.

23 Then A F,S (m, B 0, z) = A F,S (m, B 0, z, ) I =1 A F,S(m, B 0, z, I). Hence by the rule of inclusion-exclusion and µ n (A F,S (m, B 0, z)) = µ n (A F,S (m, B 0, z, )) I =1 µn (A F,S (m, B 0, z, I)) + I =2 µn (A F,S (m, B 0, z, I)) A F,S (m, B 0, z) = A F,S (m, B 0, z, ) I =1 A F,S(m, B 0, z, I) + I =2 A F,S(m, B 0, z, I) where both sums are over all subsets of I 0. Therefore µ n (A F,S (m, B 0, z)) A F,S (m, B 0, z) Z n µ n (A F,S (m, B 0, z, I)) A F,S (m, B 0, z, I) Z n. I I 0 The set I 0 has at most 2 I0 = 2 (d+n)r subsets. Hence we are left to show that for each non-empty subset I I 0 µ n (A F,S (m, B 0, z, I)) A F,S (m, B 0, z, I) Z n (B 0 + 1) n 1. Note that the set x Zn : L pi (x) p < p zpi/ep, (pi) I; L pi (x) p p zpi/ep, (pi) I 0 \ I; x p 1, p P \ S defines a lattice Λ = Λ(z, I) Z n. So we have θ = inf{ x : 0 x Λ} 1. Define S = S(z, B 0 ) = {x R n : F (x) m(z), x B 0 }. Then A F,S (m, B 0, z, I) Z n = S Λ. Since µ n (A F,S (m, B 0, z, I)) = µ n (S)/ det Λ, we are left to show that for each lattice Λ Z n µn (S) det Λ S Λ (B0 + 1) n 1. But this follows from Lemma

24 With the lemma above, we finally arrive at the next Proposition which allows us to estimate the difference between the volume µ n (A F,S (m, B 0 )) and the cardinality of A F,S (m, B 0 ) Z n. Proposition With the notation above, we have µ n( A F,S (m, B 0 ) ) A F,S (m, B 0 ) Z n (B 0 + 1) n 1 ( 1 + log ( H(F )B 0 ) ) dr where the implicit constant depends only on n, d and S. Proof. Since A F,S (m, B 0 ) = z Z A dr F,S (m, B 0, z), we have 0 ( µ n A F,S (m, B 0 ) ) A F,S (m, B 0 ) Z n ( µ n A F,S (m, B 0, z) ) A F,S (m, B 0, z) Z n z Z ( (B 0 + 1) n 1 (B 0 + 1) n log ( ) ) dr H(F )B 0 z Z where in the estimates we have used (1.4.2) and Lemma

25