On Kronecker s Theorem

Transcription

1 Thijs Vorselen On Kronecker s Theorem over the adèles Master s thesis, defended on April 27, 2010 Thesis advisor: Jan-Hendrik Evertse Mathematisch Instituut Universiteit Leiden

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3 Contents Introduction iv 1 Kronecker s Theorem An introduction to the geometry of numbers Kronecker s approximation theorem Systems of linear equations over principal ideal domains A more general Kronecker s Theorem Valuations Some algebraic preliminaries An introduction to absolute values Completions Absolute values on number fields An effective Kronecker s Theorem Uniform distribution An effective Kronecker s Theorem Subspace Theorem Geometry of numbers over the adèles Adèles Strong Approximation Theorem Fundamental domain The Haar measure on the adèle space Minkowski s Theorem for adèle spaces Kronecker s Theorem over the adèles Kronecker s theorem for adèle spaces An effective adèlic Kronecker s Theorem A more general adèlic Kronecker s Theorem Bibliography List of symbols Index iii

4 Introduction In this thesis we state and prove effective versions and adèlic generalizations of Kronecker s Theorem. Kronecker s Theorem takes an important place in the field of mathematics called Diophantine approximation. This field of mathematics is concerned with approximating real numbers by rational numbers. Kronecker s Theorem deals with inhomogeneous Diophantine inequalities and is published in 1884 by Kronecker in a paper [10] called Näherungsweise ganzzahlige Auflösung linearer Gleichungen. Let L i (q) = α i1 q α in q n (i = 1,..., m) be m linear forms with real coefficients α ij and let A be the m n matrix with elements α ij. The following theorem is a special case of Kronecker s Theorem. Theorem 1. (Kronecker) Assume that {z Q m : A T z Q n } = {0}. (1) Then for every ε > 0, b = (b 1,..., b m ) R m, there exist p = (p 1,..., p m ) Z m, q Z n such that L i (q) p i b i ε for i = 1,..., m. Condition (1) of this theorem is a necessary condition. Another important theorem from Diophantine approximation is Dirichlet s Theorem. This theorem is proven by Dirichlet in 1840 and deals with homogeneous Diophantine inequalities. If we compare Kronecker s Theorem with Dirichlet s Theorem, then we come across an interesting difference. Theorem 2. (Dirichlet) For every ε with 0 < ε < 1, there exist p Z m, q Z n with q 0 such that L i (q) p i ε for i = 1,..., m, q ε m/n. In contrast to Kronecker s Theorem this theorem gives an upper bound for q that is easy to calculate in terms of ε. We call such an upper bound an effective upper bound. Motivated by this observation we ask ourselves if there also exists an effective upper bound in the case of Kronecker s Theorem. In this thesis we answer this question in the affirmative for a matrix A with algebraic elements α ij. In 1966 Mahler [15] published a proof of Kronecker s Theorem with methods from the geometry of numbers. Geometry of numbers is concerned with convex bodies and integer vectors in n-dimensional Euclidean space. In 1988 R. Kannan and L. Lovász iv

5 also use geometry of numbers to prove a quantitative version of Theorem 1 for the case n = 1, published in [9]. Their theorem gives an ineffective upper bound for q and is the starting point of our proof of an effective Kronecker s Theorem. In the first chapter we give an introduction to geometry of numbers and generalize the proof of R. Kannan and L. Lovász to arbitrary m. We also give a more general quantitative version of Kronecker s Theorem. In the second chapter we recall some results from algebraic number theory and more specifically the theory of valuations. These results are used in Chapter 3 to prove an effective version of Kronecker s Theorem in the case that the elements of A are algebraic. In Chapter 4 we introduce the adèle space. This space was introduced by Chevalley in 1940 and is useful to solve number theoretic problems. Many of the concepts and theorems in geometry of numbers have been generalized over the adèles. An important result is the adèlic version of the Second Theorem of Minkowki. This theorem is published by R. McFeat [16] in Unaware of McFeat s result E. Bombieri and J. Vaaler [3] proved the same theorem in Motivated by the results in geometry of numbers over the adèles we wondered if it would be possible to generalize a version of Kronecker s Theorem over the adèles. In Chapter 5 we prove some adèlic versions of Kronecker s Theorem. Again the article by R. Kannan and L. Lovász forms the starting point of our proof. The adèlic Second Theorem of Minkowski also plays a crucial role. I would like to thank Jan-Hendrik Evertse for the idea of this thesis and for many suggestions and corrections. v

6 Chapter 1 Kronecker s Theorem In this chapter we give an introduction to the geometry of numbers and prove some versions of Kronecker s Theorem using geometry of numbers. 1.1 An introduction to the geometry of numbers Let n be an integer with n 1. We denote by R n the vector space of n-dimensional column vectors. We use shorthand x, y R n by (x 1,..., x n ) T, (y 1,..., y n ) T R n, respectively. A body is a closed, bounded, connected subset of R n with inner points. A body C is called convex if for all x, y C and all t [0, 1] the point (1 t)x + ty lies in C. A body C in R n is called symmetric, if x C implies that x C. Henceforth, let C be a bounded symmetric convex body in R n. Such a body C is Lebesgue measurable, so it has a finite volume, which we denote by V (C). For λ R we define λc = {x R n : x = λy, y C}. Let, denote the standard inner product on R n, given by x, y = n x i y i for x, y R n. i=1 The polar of C, denoted by C, is given by C := {x R n : x, y 1 for all y C}. This is again a bounded symmetric convex body. See Gruber and Lekkerkerker [7], Chapter 2, Section 14, Theorem 1 for a proof of this fact. A lattice in R n is a discrete subgroup of R n that spans R n as a real vector space. If L is a lattice and {x 1,..., x n } is a basis of L, then det(x 1,..., x n ) is called the determinant of L and is denoted by det L. Define the n n matrix A := [x 1,..., x n ], where x 1,..., x n are the columns of A. This matrix A is invertible, because x 1,..., x n are linearly independent. We have L = {Ax : x Z n }. 1

7 The dual of L is given by L = {x R n : x, y Z for all y L}. It is a corollary of the next lemma that L is again a lattice in R n. We use the following notation. Let GL n (R) denote the set of invertible n n matrices over a ring R. Lemma 1.1. Let A GL n (R) and let L = {Ax : x Z n } be the associated lattice in R n. Then its dual is given by L = {(A 1 ) T x : x Z n }. Proof. We have (A 1 ) T x, Ay = x T A 1 Ay = x T y = x, y. Hence, (A 1 ) T x, Ay Z if and only if x, y Z. Further, it is easy to see that {x R n : x, y Z for all y Z n } = Z n. We conclude that This proves the lemma. L = {(A 1 ) T x : x Z n }. by We define the n successive minima λ 1 (C, L),..., λ n (C, L) of C with respect to L λ i (C, L) = inf{λ R >0 : dim(λc L) i}. If it is not necessary to specify C and L, we will denote λ i (C, L) by λ i. It is easy to see that 0 < λ 1 λ n <. We defined successive minima as infima, but they are in fact minima, as suggested by their name. Theorem 1.2. (Second Minkowski s Convex Body Theorem) Let C be a bounded symmetric convex body and L a lattice in R n. Then the successive minima λ 1,..., λ n of C with respect to L satisfy 2 n n! λ 1... λ n V (C) det L 2n. Proof. See Minkowski [17], Kapitel V, for the original proof. We refer to Gruber and Lekkerkerker [7], Chapter 2, Section 9, Theorem 1 for a shorter proof by Estermann. 2

8 Let C be a bounded symmetric convex body and L a lattice in R n. For any u L we define the set λc + u = {x + u : x λc}. Then µ(c, L), defined by µ(c, L) = inf{λ R >0 : u L(λC + u) = R n }, is called the covering radius or inhomogeneous minimum of C with respect to L. For many problems in the geometry of numbers, it is useful to have an upper bound for the product of µ(c, L) and λ 1 (C, L ). Lemma 1.3. Let C be a symmetric convex body in R n and L a lattice in R n. Then µ(c, L)λ 1 (C, L ) 1 2 n2. Proof. See Lagarias, Lenstra, and Schnorr [12], Theorem 2.9. Theorem 1.4. Let C be a convex body in R n. Its successive minima λ 1,..., λ n and covering radius µ satisfy 1 λ 2 n µ 1(λ λ n ). Proof. First we prove the lower bound by contradiction. Let t be the number of linearly independent vectors in (µ + 1λ 2 n)c L. Suppose that µ + 1λ 2 n < λ n, implying that t < n. Let x 1,..., x t (µ + 1λ 2 n)c L be linearly independent. Choose x λ n C L such that x Span{x 1,..., x t }. There exists u L such that 1x u µc 2 by the definition of the covering radius. By symmetry and convexity of C we find that both u = u 1x + 1x and x u = 1x u + 1x are in (µ + 1λ n)x. Hence, u, x u Span{x 1,..., x t }, which contradicts x Span{x 1,..., x t }. This proves the lower bound. Now, we prove the upper bound. Choose linearly independent x 1,..., x n L such that x i λ i C for i = 1,..., n. Let us take an arbitrary x R n. There exist α 1,..., α n R such that x = α 1 x α n x n, because x 1,..., x n span R n. Choose a 1,..., a n Z such that α i a i 1 for i = 1,..., n. Then x (a 2 1x a n x n ) 1 (λ λ n )C. This proves the theorem. 3

9 1.2 Kronecker s approximation theorem The aim of this section is to obtain a quantitative version of Kronecker s approximation theorem for linear forms. The proof of this version is based on geometry of numbers. This idea comes originally from K. Mahler. He published a paper [15] in 1966 in which he applies geometry of numbers to prove Kronecker s Theorem. Let R be a ring. We denote the space of n-dimensional column vectors by R n and the ring of m n matrices over R by R m,n. We define the following two norms on R n : 1. x 1 := n i=1 x i for all x R n, called the sum norm, and 2. x := max{ x 1,..., x n } for all x R n called the maximum norm. Let L i (q) = α i1 q α in q n (i = 1,..., m) be m linear forms with real coefficients and define α 11 α 1n A := α m1 α mn The following theorem is a special case of Kronecker s approximation theorem for linear forms. Theorem 1.5. (Kronecker) Let A R m,n and assume that {z Q m : A T z Q n } = {0}. (1.1) Then for every ε > 0, b 1,..., b m R, there exist p 1,..., p m Z, q Z n such that L i (q) p i b i ε for i = 1,..., m. (1.2) Proof. A proof is given in this section. See [10] for Kronecker s original paper. The next lemma states that condition (1.1) is necessary for this theorem. Lemma 1.6. If for every ε > 0, b = (b 1,..., b m ) R m there exist p = (p 1,..., p m ) Z m, q Z n, such that L i (q) p i b i < ε then condition (1.1) is satisfied. Proof. This is a proof by contradiction. Suppose there exists z Q m such that A T z Q n and z 0. Let z be the vector obtained by multiplying z with the lowest common multiple of the denominators of the coordinates in z and A T z. Note that 4

10 z Z m and A T z Z n. There exists b R m with z, b Z since z 0. For all vectors p Z m, q Z n we have z, Aq p b = (Aq) T z z, p z, b = q T A T z z, p z, b. Note that q T A T z z, p Z for all p Z m, q Z n and z, b Z. Hence, z, Aq p b z, b > 0 independent of p and q. This proves the lemma. Theorem 1.5 does not give an upper bound for q. The following theorem by Dirichlet proves an effective upper bound for homogeneous linear forms. Theorem 1.7. (Dirichlet) For every ε with 0 < ε < 1, there exist p Z m, q Z n with q 0 such that L i (q) p i ε for i = 1,..., m, q ε m/n. Proof. Dirichlet proved this in See Cassels [5] for a proof. We wonder if we can calculate such an effective upper bound for q in the case of a system of inhomogenous linear forms. The following theorem implies a noneffective upper bound. The inequalities in (1.2) can be stated in the following more efficient way Aq p b ε. We denote the distance from a real number x to the nearest integer by x. Theorem 1.8. Let Q, ε be positive reals such that for all integers a 1,..., a m, not all zero, Q n a 1 α j1 + + a m α jm + ε j=1 Then for all b R n there exist p Z m, q Z n such that m a i 1(m + 2 n)2. (1.3) i=1 Aq p b ε, q Q. Kannan and Lovász proved this theorem for n = 1. The following proof expands theirs ([9], Theorem 5.5) to arbitrary n. 5

11 Proof. Let C = {x R m+n : x 1} be the unit cube and let L be the lattice generated by the columns of the matrix ( ) Im A B = ε 0 I, Q n where I n denotes the n n identity matrix. It is easy to see that C = {x R m+n : x 1 1} and that the dual lattice L is generated by the columns of the inverse transpose of B ( (B 1 ) T Im 0 = Q ε AT That is, lattice points in L are of the form ( ) a Q ε (AT a + c) with a Z m, c Z n. Condition (1.3) in Theorem 1.8 implies that for all (a, c) (0, 0) one has ( ) a Q 1 ε (AT a + c) (m + 2 n)2 /ε. 1 Hence, by definition of λ 1 (C, L ) we have Q ε I n ). λ 1 (C, L ) 1 2 (m + n)2 /ε. For all b R m there exist vectors p Z m, q Z n such that B ( p q) + ( b 0) µ(c, L) by definition of µ(c, L). Since µ(c, L) ε by Lemma 1.3 we have This proves the theorem. p Aq + b ε q Q. Lemma 1.9. Theorem 1.8 implies Theorem 1.5. Proof. Choose an arbitrary ε > 0. For every a = (a 1,..., a m ) Z m with a 0 we find n j=1 a 1α 1j + + a m α mj > 0, because of condition (1.1). Now take 1 Q = 1 n (m + n)2 min a 1 α 1j + + a n α mj. 2 a (m+n)2 /ε a 0 With this ε and Q, condition (1.3) is satisfied. The lemma follows. j=1 6

12 1.3 Systems of linear equations over principal ideal domains In this section we prove a criterion for the solvability of systems of linear equations over principal ideal domains. Recall that a principal ideal domain is a domain, of which all ideals are principal. This result is used for the proof of a more general Kronecker s Theorem in the next section. A diagonal matrix is a square matrix in which all elements outside the main diagonal are zero. Let R be a principal ideal domain. For a, b R, a 0 we denote a b if a divides b. Theorem Let A R m,n. Then there exist V 1 GL m (R), V 2 GL n (R) such that A given by A := V 1 AV 2 is a diagonal matrix δ 1... A δ = t, where δ 1,..., δ t R with δ 1... δ t and t = rank(a). Proof. See Bourbaki [4], Livre II, Chapitre VII, Section 4.5, Corollaire 1 of Proposition 4. For a proof in the special case that R = Z, see Smith [21]. The matrix A is called the Smith normal form of A. Bachem and Kannan, [1], published a polynomial time algorithm to calculate the Smith normal form in the special case that R = Z. Theorem Let R be a principal ideal domain with field of fractions K and let A R m,n, b R m. Then the following two assertions are equivalent. (i) There exists x R n such that Ax = b. (ii) {z K m : A T z R n } {z K m : b T z R}. Proof. First we prove (i) (ii). Let z K m, such that A T z R n. Then This proves (i) (ii). b T z = x T A T z x T R n R. 7

13 Now, we prove (ii) (i). Let A R m,n and b R m satisfy assertion (ii). Let A be the Smith normal form of A and let V 1, V 2 be the matrices such that A = V 1 AV 2 as in Theorem If A T z R n then we have V2 T A T V1 T z R n and as V 2 GL n (R) we get A T V1 T z R n. By assertion (ii) we see that (V 1 b) T z = b T V1 T z R. We conclude that A and b := (b 1,..., b m) T = V 1 b also satisfy assertion (ii). For every z = (z 1,..., z m ) K m we have δ 1... A T δ z = t z = (δ 0 1 z 1,..., δ t z t, 0,..., 0) T It is easy to see that δ 1 b 1,..., δ t b t and that b t+1 = 0,..., b m = 0. Define x := (b 1/δ 1,..., b t/δ t, 0,..., 0) and note that x R n and A x = b. Define x := V 1 2 x, then we have x R m and Ax = b. This proves the theorem. 1.4 A more general Kronecker s Theorem We already proved a special case of Kronecker s Theorem for linear forms. In this section we use that result to prove the following more general form of Kronecker s theorem. We use the following notation in this proof. Let again A R r,s be the r s matrix Define a norm on R r,s by A := α 11 α 1s..... α r1 α rs A := max 1 i r s α ij. j=1. Note that Ax A x for all A R r,s, x R s. 8

14 Theorem Let A R r,s and b R r. Then the following two assertions are equivalent. (i) For every ε > 0 there exists x Z s such that Ax b ε. (ii) For all z R r with A T z Z s we have b T z Z. In the proof of this theorem we need some lemmas, in which we refer repeatedly to assertions (i) and (ii). Lemma Let U GL r (R), V GL s (Z) and put à := UAV, b := Ub. Then: (i) is equivalent to the assertion that for every ε > 0 there is x Z s such that Ãx b ε, (ii) is equivalent to the assertion that {z R r : ÃT z Z s } {z R r : b T z Z}. Proof. Suppose assertion (i) holds. Then for every ε > 0 there exists x Z s such that Ax b ε U. Define x := V 1 x. Then à x b = UAx Ub ε. The proof of the reverse implication is entirely similar. Suppose assertion (ii) holds. If there exists z R r, w Z such that V T A T (U T z) = w then A T (U T z) = (V T ) 1 w Z. By assertion (ii) we have b T U T z Z. We conclude that {z R r : ÃT z Z s } {z R r, b T z Z}. Again, the reverse implication is proved in the same manner. Lemma Assume (ii) holds. Then there exist U GL r (R), V GL s (Z) such that I t 0 A 1 b 1 UAV = 0 I m t A 2, Ub = b where 0 t m r, A 1 Q t,s m, A 2 R m t,s m, b 1 Q t, b 2 R m t, and {z 1 Z t : A T 1 z 1 Z s m } {z 1 Z t : b T 1 z 1 Z}, {z 2 Z m t : A T 2 z 2 Z s m } = {0}. 9

15 Proof. Let m := rank(a) and suppose without loss of generality that the first m rows of A are linearly independent. Then by elementary linear algebra, there exists U 1 GL r (R) such that ( ) Im A U 1 A = with A R m,s m. 0 0 By Lemma 1.13, the validity of (ii) is unaffected if we replace A by U 1 A. That is, (ii) is equivalent to the assertion that ( ) {z R r Im A : z Z s } {z R r : b T U T z Z}. (1.4) 0 0 For this to hold, we must have U 1 b = (b, 0) T with b R m. Thus, (1.4) becomes {z Z m : A T z Z s m } {z Z m : b T z Z}. (1.5) The left-hand side is a sub-z-module M of Z s m, hence free of rank t m. If t = 0 we are done. Suppose t > 0. Then by Theorem 1.11 there is a basis {d 1,..., d m } of Z m and there are positive integers δ 1,..., δ t, such that {δ 1 d 1,..., δ t d t } is a Z-basis of M. Now, let D := [d 1,..., d m ] be the matrix with columns d 1,..., d m. Then D GL m (Z). Define  := DT A, ˆb := D T b. Then ( D T 0 0 I r t ) ( Im A 0 0 ) ( (D T ) I s m ) = ( Im  0 0 ). (1.6) The right-hand side is clearly of the shape UAV with U GL r (R), V GL s (Z). Writing ẑ := D 1 z we see that (1.5) is equivalent to {ẑ Z m : ÂT ẑ Z s m } {ẑ Z m : ˆb T ẑ Z}. (1.7) Let A 1 consist of the first t rows of  and A 2 of the last m t rows. Further, let b 1 and b 2 consist of respectively the first t and the last m t coordinates of ˆb. Notice that the left-hand side of (1.7) consists of all vectors of the shape (δ 1 x 1,..., δ r x t, 0,..., 0) T with x 1,..., x t Z. Hence, A 1 Q t,s m, b 1 Q t and {z 1 Z t : A T 1 z 1 Z s m } {z 1 Z t : b T 1 z 1 Z}. Further, by applying (1.7) with vectors (0,..., 0, z t+1,..., z m ) T, we see that {z 2 Z t : A T 2 z 2 Z s m } = {0}. 10

16 This proves the lemma. Proof of Theorem The proof of (i) (ii) is very similar to that of (i) (ii) of Theorem 1.11, and is therefore left to the reader. Now, we prove (ii) (i). By Lemmas 1.13 and 1.14, we may assume without loss of generality, that I t 0 A 1 b 1 A = 0 I m t A 2, b = b 2, with A 1, A 2, b 1, b 2 as in Lemma Writing x T = (q T, p T 1, p T 2 ), We can rewrite (i) as A 1 q p 1 b 1 ε, A 2 q p 2 b 2 ε, (1.8) to be solved in q Z s m, p 1 Z t, p 2 Z m t. By Theorem 1.11 there exist q Z s m, p 1 Z t such that A 1 q p 1 = b 1. Recall that A 1 Q t,s m. Let d be a positive integer, such that da 1 Z t,s m. By Theorem 1.8, there exist q Z s m, p 2 Z t, such that Hence, A 2 q p 2 1 d (b 2 A 2 q ) ε d. (1.9) A 1 (q + dq ) (p 1 + da 1 q ) b 1 = 0, A 2 (q + dq ) dp 2 b 2 < ε, which implies that (1.8) is satisfied with q = q + dq, p 1 = p 1 + da 1 q, p 2 = dp 2. This proves the theorem. 11

17 Chapter 2 Valuations In Chapter 3 we give an effective version of Kronecker s Theorem for a matrix A R m,n with algebraic entries. For this we need some algebraic number theory and more specifically, the theory of valuations. 2.1 Some algebraic preliminaries We refer to Algebraic Number Theory by J. Neukirch [18] for algebraic number theory. In this section we have stated some definitions and results. A number field K is a finite extension of Q. Define O K = {x K : monic f Z[X] such that f(x) = 0} This set is a subring of K and is called the ring of integers of K. This ring is a Dedekind domain, which means that it is a noetherian, integrally closed, integral domain in which every non-zero prime ideal is a maximal ideal. Ideals in Dedekind domains can be factorized into prime ideals in a unique way. Let K L be an extension of number fields and let p be a non-zero prime ideal of O K. Then the set po L = {xy : x p, y O L } is an ideal of O L. Using the unique ideal factorization in O L we get po L = P e 1 1 P eg g for certain prime ideals P 1,..., P g of O L and positive integers e 1,..., e g. We call e i the ramification index of P i over p and f i := [O L /P i : O K /p] the residue class degree of P i over p. Theorem 2.1. (Fundamental identity) Let K L be an extension of number fields of degree d. Then one has g e i f i = d. i=1 12

18 Proof. See Neukirch [18], Chapter I, Proposition 8.2. A fractional ideal of O K is a finitely generated non-zero sub-o K -module a (0) of K. Every fractional ideal a of O K can be expressed uniquely as a product a = p k 1 1 p kn n, where p 1,..., p n are prime ideals and k 1,..., k n Z. For a non-zero prime ideal p of O K we define ord pi (a) := k i for i = 0,..., n and ord p (a) := 0 if p {p 1,..., p n }. We have ord p (a) 0 for only finitely many prime ideals p. For an element x K, we define ord p (x) := ord p (b), where b = (x) is the fractional O K -ideal generated by x. The set of fractional ideals of O K is a group under multiplication, denoted by I K. Let P K denote the group of principal fractional ideals of O K, i.e., the set of fractional ideals generated by one element of K. The factor group I K /P K is the ideal class group of K and its order is called the class number of K. Theorem 2.2. The ideal class group I K /P K is finite. Proof. See Janusz [8], Chapter I, Theorem Note that O K is a principal ideal domain if and only if the ideal class group of K is trivial. Proposition 2.3. Let K be a number field of degree d = [K : Q] over Q and let a be a fractional ideal of O K. Then as a Z-module, a is free of rank d. Proof. This follows directly from Neukirch [18], Chapter I, Proposition Let K L be an extension of number fields. The map Tr L/K : L K x σ σ(x) is called the trace and the map N L/K : L K x σ σ(x) is called the norm. Here the sum and product range over all K-isomorphic embeddings σ of L in C. 13

19 We can also define a norm on the group of fractional ideals. Any fractional ideal a can be expressed uniquely as a = bc 1, where b, c are integral ideals such that b + c = (1). We then define N(a) by N(a) := (#O K /b) (#O K /c) 1. Let K L an extension of number fields of degree d. We define the discriminant of a basis {x 1,..., x d } of L over K by ( 2 D L/K (x 1,, x d ) = det(x (j) i ) i,j=1) d, where x (1) i,..., x (d) i are the d images of x i under the K-isomorphic embeddings of L in C. By some elementary calculations, we can rewrite this as ( ) d D L/K (x 1,, x d ) = det Tr K/Q (x i x j ). i,j=1 We have the following important proposition regarding the discriminant. Proposition 2.4. Let {x 1,..., x d } be a Q-basis of K. Then D K/Q (x 1,..., x d ) 0. If {x 1,..., x d } and {y 1,..., y d } are Z-bases of the same fractional ideal of O K, then we have D K/Q (x 1,..., x d ) = D K/Q (y 1,..., y d ). Proof. See Lang [13], Chapter I, Section 2, Proposition 8 and Proposition 12. Let again K be a number field and a a fractional ideal of O K. By Proposition 2.4, we can define the discriminant D K/Q (a) of a fractional ideal a by D K/Q (a) = D K/Q (x 1,..., x d ), where {x 1,..., x d } is a Z-basis of a as a Z-module. In particular, we define the discriminant D K of a number field K by D K = D K/Q (O K ). Hence, if we choose a basis {ω 1,..., ω d } of O K, then we get D K = det(tr K/Q (ω i ω j )) d i,j=1. As ω i ω j O K for i, j = 1,, d, we find that Tr K/Q (ω i ω j ) Z. Now, it follows that D K Z. 14

20 Proposition 2.5. Let a be a fractional ideal of O K. Then D K/Q (a) = N(a) 2 D K. Proof. This follows directly by expressing a as a product of two integral ideals a = bc 1 and applying Lang [13], Chapter I, Section 2, Proposition 12 to these ideals. 2.2 An introduction to absolute values Definition 2.6. An absolute value on a field K is a function v : K R 0 such that 1. x v = 0 if and only if x = 0 for all x K; 2. xy v = x v y v for all x, y K; 3. there exists C R >0, such that x + y v C max{ x v, y v } for all x, y K. Definition 2.7. A valuation on a field K is a function v : K R { } such that 1. v(x) = + if and only if x = 0 for all x K; 2. v(xy) = v(x) + v(y) for all x, y K; 3. v(x + y) min {v(x), v(y)} for all x, y K. An example of a valuation on a number field is given by ord p. It is easy to see that a valuation v gives rise to an absolute value v : K R 0 defined by x v = c v(x) with c R >1. In other literature, sometimes the term valuation is used for absolute value. Example 2.8. The most trivial example of an absolute value, which can be defined on any field K, is the absolute value that sends all x K with x 0 to 1. This absolute value is called trivial. Henceforth, all absolute values we consider are assumed to be non-trivial. Example 2.9. The standard absolute value on Q,, is defined by We call the infinite prime. : Q R 0 x x. 15

21 Example Let x Q\{0}. We can write x as x = b with b, c Z. If we c subsequently extract the highest possible power of p from b and c, we get x = p m b, gcd(bc, p) = 1. c The p-adic absolute value x p of x is then defined by x p := 1 p m. Further, we put 0 p := 0. Definition Let K be a field. Two absolute values v and w on K are called equivalent if and only if there exists a real number c > 0 such that x v = x c w for all x K. An equivalence class of absolute values on K is called a place of K. We denote the collection of all places of K by M K. Proposition Let v and w be two absolute values on a field K. Then the following two assertions are equivalent: (i) v and w are equivalent, (ii) x v < 1 x w < 1 for all x K. Proof. See Neukirch [18], Chapter II, Proposition 3.3. Definition An absolute value v on a field K is called non-archimedean if it satisfies the ultrametric inequality Otherwise it is called archimedean. x + y v max{ x v, y v } for all x, y K. The p-adic absolute values are non-archimedean and the standard absolute value is archimedean. The following lemma shows that we can define this property for places. Lemma Equivalent absolute values are either both archimedean or both nonarchimedean. Proof. Let v and w be equivalent absolute values on a field K. Hence, there exists an c > 0 such that x v = x c w for all x K. Suppose v is non-archimedean, then x + y w = x + y c v max{ x v, y v } c = max{ x c v, y c v} = max{ x w, y w } 16

22 for all x, y K. Hence, w is archimedean too. By symmetry we get the other implication. For a field K, denote by M K the set of places of K, by MK the set of archimedean places of K, and by MK fin the set of non-archimedean places of K. By the following theorem we can identify these places in the case that K = Q. Theorem (Ostrowski) Every non-trivial absolute value on Q is equivalent either to or to p for some prime number p. Proof. For a proof see Neukirch [18], Chapter II, Proposition 3.7. Hence, we may identify a non-archimedean place of Q with its corresponding prime number. Thus, we may write M Q = { } {primes}. Theorem (Product formula) We have p M Q x p = 1 for all x Q. Proof. Take an arbitrary x Q. The theorem follows directly from writing and calculating p M Q x p. x = ±p k 1 1 p k 2 2 p kt t 2.3 Completions We are already familiar with the construction of the real numbers as completion of the rational numbers. This is done with respect to the standard absolute value. We can adjust the same construction to arbitrary fields K with respect to any absolute value v on this field K. A field K is called complete with respect to an absolute value v, if every Cauchy sequence in K converges in K with respect to v. Let R be the ring of Cauchy sequences in K with respect to v. This is a ring under pointwise addition and multiplication of the sequences. The subset of R consisting of all sequences in R converging to 0 with respect to v form a maximal ideal of R which we denote by m. The field K v := R/m is called the completion of K with respect to v. It is easy to see that K v contains K. We can extend v to K v in the following way. For every 17

23 x K v there exists a Cauchy sequence (x k ) k=1 in K with lim x k = x. We define v k on K v by x v := lim x k v. The field K v is complete with respect to this valuation. k The completion K v of a field K with respect to an archimedean absolute value is equal to R or C. This follows directly from the following result regarding fields that are complete with respect to an archimedean absolute value. Theorem (Ostrowski) Every field that is complete with respect to an archimedean absolute value is isomorphic to either R or C. Proof. See Neukirch [18], Chapter II, Theorem 4.2. The next definition is that of an extension of an absolute value. This is very important in the next section, where we look at absolute values on number fields. Definition Let K L be a field extension and let v and w be absolute values on K and L respectively. We say that w extends v if the restriction of w to K is equal to v. Now, it is natural to speak of an extension of a place as in the following definition. Definition Let K L be a field extension and let v, w be places on respectively K and L. A place w extends v, denoted w v, if the restriction of a representative of w to K is equivalent to a representative of v. We say that w lies above v. We have the following important theorem regarding extensions of absolute values. Theorem Let K be a complete field with respect to an absolute value v and let K L be a field extension of finite degree d. Then there exists an unique extension of v to L given by and L is complete with respect to w. x w := d N L/K (x) v for x L Proof. See Neukirch [18], Chapter II, Theorem 4.8. Theorem Let v be an absolute value on K. There exists a unique extension of v from K v to K v also denoted by v, given by x v := d N Kv(x)/K v (x) v for x K v. 18

24 Proof. Let x K v. Let L 1 and L 2 be any two finite field extions of K v containing x. There exist two unique absolute values by Theorem 2.20, on respectively L 1 and L 2, given by x v1 := d N L1 /K v (x) v for x L 1 x v2 := d N L2 /K v (x) v for x L 2 on respectively L 1 and L 2. These absolute values agree on L 1 L 2, because the extension of v to L 1 L 2 is unique by Theorem We conclude that there exists an unique extension of v to K v. 2.4 Absolute values on number fields In this section we state and prove some results of absolute values in the more specific case of number fields. Henceforth, let K be a number field and let K denote the algebraic closure of K. Let σ 1,..., σ r be the real embeddings of K, and let σ r+1,..., σ r+2s be the complex embeddings of K ordered such that σ r+s+i = σ r+i for i = 1,..., s. Define d := [K : Q]. Note that r + 2s = d. Let σ : K C be an embedding of K in C. We define an absolute value σ associated to an embedding σ by x σ := σ(x) x σ := σ(x) 2 for all x K if σ is real, for all x K if σ is complex. Let σ be a complex embedding. We denote its complex conjugate by σ. Note that we have σ(x) = σ(x) for all x K. So, conjugate embeddings give rise to the same absolute value. We call a place associated to a real embedding a real place and a place associated to a complex embedding a complex place. There are r real places and s complex places of K. Lemma Every archimedean place on K is induced by an embedding σ : K C. Proof. Let v be an archimedean abolute value on K. The completion K v of K with respect to v is either isomorphic to R or C as stated in Theorem Hence, the 19

25 natural embedding composed with this isomorphism is an embedding of K in either R or C. Every archimedean absolute value on R or C is equivalent to the standard absolute value. Hence, v is equivalent to σ. This proves the lemma. Henceforth, we identify the embeddings {σ 1,..., σ r+s } with the corresponding archimedean places and we choose σi as standard representative for the place σ i. This is possible by Lemma Lemma The ring of integers O K of K is equal to {x K : x v 1 for all v M fin K }. Proof. First we prove that O K {x K : x v 1 for all v MK fin }. Take an arbitrary x O K and v MK fin. Note that v p for some prime number p, because v is non-archimedean. As x is integral over Z, we can find a monic polynomial f = X n + a n 1 X n a 0 Z[X] such that x n + a n 1 x n a 0 = 0. There is a c R >0 such that v = c p on Q. Hence, a i v 1 for i = 0,..., n 1. If we take the absolute value and use the ultrametric inequality we get x n v = a n 1 x n a 1 x + a 0 v max 0 i n 1 a ix i. We get a contradiction if x v > 1. We conclude that x v 1 for all v M fin We still have to show that {x K : x v 1 for all v MK fin} O K. Choose an x K such that x v 1 for all v MK fin. Let f be the monic minimal polynomial of x over Q. It is given by f(x) = X n + a n 1 X n a 1 X + a 0 with a 0,..., a n 1 Q. We have to prove that a 0,..., a n 1 Z. Let α 1,..., α n be the roots of f in K v. Sending x to one of the values α 1,..., α n induces an embedding of K(x) in K v. Let σ 1,..., σ n be those embeddings. Extend v to K v and define an absolute value w by y w := σ i (y) v for y K(x). The restriction to K of this absolute value is a non-archimedean, hence x w 1. It follows that α i v 1 for i = 1,..., n. The coefficients a 0,..., a n 1 of f are up to sign sums of products of subsets of {α 1,..., α n }. We conclude that a 1 v 1,..., a n 1 v 1 for all v MK fin. Hence, a i p 1 for all p MQ fin and so a i Z 20 K.

26 for i = 0,..., n 1. This proves the lemma. Every prime ideal p 0 of O K gives rise to a non-archimedean absolute value p by defining x p = N(p) ordp(x) for x K, 0 p = 0. to the set of non- Lemma The map f from the set of prime ideals of O K archimedean places of K given by p place of p is a bijection. Proof. Note that the set {x O K : x p < 1} is equal to p. So, different prime ideals give representatives for different places by Proposition This proves the injectivity of f. To prove surjectivity of f take an arbitrary non-archimedean absolute value v on K. The set {x O K : x v < 1} is an ideal. By Lemma 2.23 and the property xy v = x v y v we get primality of this ideal. The image of this prime ideal under f gives back an absolute value equivalent to v by Proposition Henceforth, for an algebraic number field K, we identify every prime ideal of O K with the corresponding non-archimedean place. Thus, M K = {prime ideals of O K } {σ 1,..., σ r+s }. Lemma If x K then x v 1 for almost all v M K. Proof. The fractional ideal (x) of O K generated by x K has a unique prime factorization (x) = p k p kg g. Hence, x v 1 for all v M fin K with v p 1,..., v p n. Lemma Let K L be an extension of number fields and let v M K, w M L with w v. Then x w = N Lw/Kv (x) [Lw:Kv] v for all x L. Proof. Note that there exists a c R >0 such that x w = N Lw/K v (x) c v for all x L 21

27 by Theorem Now, we calculate c for v, w non-archimedean. We have v = p for a prime ideal p in O K. The unique prime factorization in O L gives po L = P e P eg g. As {x K : x w < 1} = p we know that w = P i for some 1 i g. Expressing x Pi in terms of x p for x K, we get x Pi = (#O L /P i ) ord P i (x) = (#O L /P i ) e iord p(x) = (#O K /p) e if i ord p(x) = x e if i p. By Neukirch [18], Chapter II, Proposition 6.8 (with Remark) and the explanation under Proposition 8.5 we have [L w : K v ] = e i f i. We conclude that x w = N Lw/K v (x) v for all x L. It is easy to derive the same relation for v, w archimedean. Theorem Let K L be an extension of number fields of degree d. Let v M K. For the places w M L extending v we have the following formula: x w = N L/K (x) v for all x L. w v Proof. Let v M K. We have x w = N Lw/Kv (x) v w v w v and by Neukirch [18], Chapter II, Corollary 8.4 N Lw/Kv (x) v = N L/K (x) v. This proves the theorem. w v Theorem (Product formula) For a number field K we have the following product formula: v M K x v = 1 for all x K. 22

28 Proof. Using Theorem 2.27 for a field extension Q K, we get x v = x v = N K/Q (x) p = 1 for all x K. v M K p M Q p M Q This proves the theorem. v p Now, we are able to define the height for x Q n. Definition Let x = (x 1,..., x n ) K n with x 0, then H K (x) = v M K max( x 1 v,..., x n v ) is called the height of x with respect to K and denoted by H K (x). Let L be a field extension of K and let x K n. Then we have H L (x) = max( x 1 w,..., x n w ) w M L = max( x 1 w,..., x n w ) v M K w v = v M K w v = max( x 1 v,..., x n v ) [Lw:Kv] v M K max( x 1 v,..., x n v ) [L:K] = H K (x) d. Now, we define the absolute height for x Q n by choosing a number field K such that x K n, and putting H(x) = H K (x) 1/[K:Q]. By what we just observed, this is independent of the choice of K. 23

29 Chapter 3 An effective Kronecker s Theorem 3.1 Uniform distribution Let A be m n matrix with real coefficients satisfying the condition (1.1) of Kronecker s Theorem. Then by Kronecker s Theorem there exist p Z m, q Z n such that Aq p b ε. (3.1) In Theorem 1.8 we proved a non-effective upper bound for q. In the next sections we also prove an effective upper bound and in this section we will use the theory of uniform distribution to get some heuristics on the order of the upper bound for q in terms of ε. Define x as the largest integer not greater than x. Definition 3.1. Let (x k ) k=1 be a sequence of reals. For a, b R with 0 a < b < 1 let T ([a, b); K) denote the number of x k with k K such that x k x k [a, b). The sequence (x k ) k=1 is said to be uniformly distributed (modulo 1) if for all intervals [a, b) in [0, 1). T ([a, b); K) lim K K = b a Theorem 3.2. (Weyl s criterion) A sequence (x k ) k=1 and only if for all integers h 0 is uniformly distributed if lim K 1 K K e 2πihx k = 0. k=1 Proof. See Kuipers and Niederreiter [11], Chapter 1, Theorem 2.1. See Weyl [22] for Weyl s original proof. Corollary 3.3. Let α R\Q. Then the sequence (x k ) k=1 defined by x k = kα is uniformly distributed. 24

30 Proof. If x 1, we have the following identity K k=0 x k = 1 xk+1 1 x. Since α Q, we have e 2πihα 1 for all h 0. Hence, lim K 1 K K k=1 ( ) e 2πihαk 1 1 (e 2πiα ) K = lim 1 = 0, K K 1 e 2πiα since the numerator is bounded. Corollary 3.3 follows directly from Weyl s criterion, Theorem 3.2. Corollary 3.4. Let (q k ) k=1 be a sequence in Rn, ordered such that if s, t are any positive integers with q s < q t then s < t. If α = (α 1,..., α n ) Q n then the sequence (x k ) k=1 defined by x k = q k, α is uniformly distributed. Proof. By Weyl s criterion, it suffices to prove lim K 1 K K e 2πihx k = 0 for all integers h 0. k=1 Choose an integer h 0. For every K N there is a non-negative integer L such that (2L + 1) n K < (2L + 3) n. We split 1 (2L+1) n K k=1 e 2πihx k into S1 + S 2, where S 1 := 1 K S 2 := 1 K (2L+1) n k=1 K e 2πihx k k=(2l+1) n +1 and e 2πihx k. 25

31 Without loss of generality, assume α 1 Q. Then 1 L L S 1 (2L + 1) n l 1 = L l n= L ( 1 L L L e 2πihα 1l 1 (2L + 1) n l 1 = L l 2 = L 1 L = (2L + 1)n 1 e 2πihα 1l 1 (2L + 1) n l 1 = L 1 L = e 2πihα 1l 1. 2L + 1 Further, l 1 = L e 2πih(α 1l 1 + +α nl n) l n= L S 2 (2L + 3)n (2L + 1) n (2L + 1) ( n = ) n 1. 2L + 1 e 2πih(α 2l 2 + +α nl n)) Note that S 1 0 as K and S 2 0 as K. We conclude 1 K lim e 2πihx k = 0. K K Corollary 3.4 follows. k=1 The theory of uniform distribution can be extended to higher dimensions. But before stating the definition of uniform distribution in n-dimensional space, we need some new notation. Let a = (a 1,..., a n ), b = (b 1,..., b n ) be two real vectors in R n with a i < b i for i = 1,..., n. We denote the n-dimensional interval n i=1 [a i, b i ) by [a, b). For x R n we denote the vector (x 1 x 1,, x n x n ) by x x. By 0 and 1 we denote the two n-dimensional vectors defined by 0 = (0,, 0) and 1 = (1,, 1). Definition 3.5. Let (x k ) k=1 be a sequence in Rn. Let [a, b) be an n-dimensional interval and let T ([a, b); K) denote the number of x k with k K such that x k x k [a, b). A sequence (x k ) k=1 is said to be uniformly distributed (modulo 1) in Rn if T ([a, b); K) lim K K 26 = n (b j a j ) j=1

32 for all intervals [a, b) [0, 1). Theorem 3.6. (Weyl s criterion) A sequence (x k ) k=1 in Rn is uniformly distributed if and only if for every h Z n, h 0, lim K 1 K K e 2πi h,xk = 0. k=1 Proof. See Kuipers and Niederreiter [11], Chapter 1, Theorem 6.2. See Weyl [22] for Weyl s original proof. Corollary 3.7. A sequence (x k ) k=1 in Rn is uniformly distributed if and only if for every h Z n, h 0, the sequence of real numbers ( h, x k ) k=1, is uniformly distributed. Proof. See Kuipers and Niederreiter [11], Chapter 1, Theorem 6.3. See Weyl [22] for Weyl s original proof. Corollary 3.8. Define α = (α 1,..., α m ) with α 1,..., α m Q-linearly independent. The sequence (x) k=1 defined by x k = kα is uniformly distributed in R m. Proof. For every h Z n with h 0, we have h, α Q by the Q-linear independence of α 1,..., α m. The result follows from Corollary 3.3. Lemma 3.9. Let (q k ) k=1 be a sequence in Zn, ordered such that if s, t are any positive integers with q s < q t then s < t. Further, let A be an m n-matrix with real entries, that fulfills the condition of Kronecker s theorem, {z Q m : A T z Q n } = {0}. (3.2) Finally, let x k = Aq k for k = 1,..., n. Then the sequence (x k ) k=1 distributed in R m. is uniformly Proof. For any h Z n we have h, x k = h, Aq k = A T h, q k. 27

33 As A satisfies equation (3.2) one has A T h Q n for all h 0. The sequence ( A T h, q k ) k=1 is a sequence as in Corollary 3.4. This implies that it is uniformly distributed. Now Theorem 3.6 gives the required result. For x R n define x = max( x 1,..., x n ). Let A be a matrix as in Lemma 3.9. This lemma implies that for all b R m and 0 < ε < 1 we have the limit 2 #{q Z n : q Q, Aq b ε} lim = (2ε) m. Q Q n Note that Aq b < ε if and only if Aq b Aq b [0, ε1) (1 ε1, 1). This explains the factor 2 m. Kronecker s theorem states that for every ε there exist q Z n and p Z m such that Aq p b ε. This also follows from the limit just stated. For an effective Kronecker s theorem we want to give an upper bound Q for q in terms of ε. The limit suggests that for most A and b there should be a solution q Z n of Aq b ε with q of the order ε m/n. 3.2 An effective Kronecker s Theorem At this point we know enough algebraic number theory to prove an effective version of Kronecker s approximation theorem for linear forms with algebraic coefficients. Proposition Let K = Q(α 1,..., α n ) R be a number field and let d = [K : Q]. Then we have q 0 + q 1 α q n α n q 1 d (n + 1) 1 d H(1, α 1,..., α n ) d for every q = (q 0,..., q n ) Z n+1 with q 0 + q 1 α q n α n 0. Proof. By assumption K R. Denote by v the place represented by. As q 0 + q 1 α q n α n 0 we may rewrite the product formula, Theorem 2.28, as q 0 + q 1 α q n α n = q 0 + q 1 α q n α n 1 w w M K w v q 1 d (n + 1) 1 d w M K w v max{1, α 1 w,..., α n w } 1 q 1 d (n + 1) 1 d H(1, α 1,..., α n ) d. 28

34 This proves the proposition. With this proposition we can make Theorem 1.8 effective for the case that A is an m n matrix with real algebraic elements satisfying condition (1.1). Let α α 1n A =..... α m1... α mn with real algebraic elements α ij. The extensions Q(α 1j,..., α mj ) over Q are finite for j = 1,..., n, because α 1j,..., α mj are algebraic over Q. Define d j := [Q(α 1j,..., α mj ) : Q] (j = 1,..., n) d := max(d 1,..., d n ). (3.3) We define the height of A as H (A) = (n + 1) d 1 max H(1, α 1j,..., α mj ) d. j=1,...,n This notation is not standard, but it turns out to be very useful in the next theorem. Theorem Let A = (α ij ) be an m n matrix with real algebraic elements such that {z Q m : A T z Q n } = {0}. (3.4) Let d be given by (3.3) and define for ε > 0 ( ) d 1 1 Q(ε) := d (m + n) 2d H (A). 2 d 1 ε Then for every ε > 0, b R m, there exist p Z m, q Z n with Aq p b ε, q Q(ε). Proof. Let a Z m with a 0. Condition (3.4) implies that there is a j {1,..., m}, such that α 1j a α mj a m 0. Proposition 3.10 gives Let ε > 0. If we choose Q such that Q α 1j a α mj a m H (A) 1 a 1 d j. n a 1 α 1j + + a m α mj + ε j=1 29 m a i 1(m + 2 n)2, i=1

35 then we can apply Theorem 1.8. In view of Proposition 3.10 this inequality is satisfied if QH (A) 1 a 1 d + ε a 1 2 (m + n)2. By elementary calculus the minimum of the function f defined by is assumed at Hence, we must choose Q such that f(x) = QH (A) 1 x 1 d + εx ( ) QH (A) 1 1 d (d 1) x min =. εn QH (A) 1 x 1 d min + εx min 1 2 (m + n)2. An easy calculation shows that this inequality is satisfied for Q = d (m + n) 2d H (A)ε 1 d. 2 d 1 Theorem 3.11 now follows by applying Theorem 1.8. Recall that the heuristic argument in Section 3.1 suggests that in general Kronecker s Theorem should hold with q ε m. We only get this expected best result if A is a m 1 matrix and the extension Q Q(α 11,..., α m1 ) is of degree m. In Section 3.3 below we derive in the case that A is a m 1 matrix and any δ > 0 an upper bound for q of order (ε 1 ) m+δ. This upper bound is independent of the degree of the extension, but is not effectively computable by the method of proof. It is also possible to apply this result to Kronecker s theorem in the more general form as in Section 1.4. Before stating this theorem we need the following definitions. Let A be a r s matrix with real algebraic elements α ij. Define K as the field extension of Q generated by the elements of A. We define c(a) := [K : Q]. Theorem Let A R r,s, b R r both with algebraic elements, such that for all z R r with A T z Z s we have b T z Z. Then there exists C R >0, effectively computable and depending on A and b, such that for all ε > 0, there exists x Z s with ( ) c(a) 1 1 Ax b ε, x C. ε 30

36 Proof. For the proof of this theorem we follow the steps of Theorem By Lemma 1.13 and Lemma 1.14 we can find U GL r (R) such that UA is in the shape as demanded at the start of the proof of Theorem Let A 1, A 2, b 1, and b 2 be defined as in Lemma These matrices and vectors are effectively computable, as there exists a polynomial time algorithm to calculate the Smith normal form. See Bachem and Kannan, [1], for this. A basis for the module M in the proof of Theorem 1.12 is also effectively computable by expressing the elements of A as Q-linear combinations of a chosen Q-basis of K, and then using linear algebra over Z. By Theorem 1.11 there exists q Z s m, p 1 Z t such that A 1 q p 1 = b 1. By Lemma 1.13 and inequality (1.9) of Theorem 1.12 we have A 2 q p 2 1 d (b 2 A 2 q ) ε d U. We want to give an upper bound for q in this inequality. By Theorem 3.11 there exist q Z n, p 2 Z m t such that  2q p 2 1 d (b 2 + A 2 q ) U ε d, q c(a) ( ) c(a) 1 d 2 (m + c(a)+1 n)2c(a) H (A 2 ). ε The proof of this theorem follows if we use this result and continue with the proof of Theorem Subspace Theorem In this section we use the Subspace Theorem, see below, to prove an effective version of Kronecker s Theorem for matrix A R m,1 with algebraic entries. First we need the following definition. We say that n linear forms L i = α i1 x α in x n, i = 1,..., n, are linearly independent if α 11 α 1n det(l 1,..., L n ) = α n1 α nn 31

37 Theorem (Subspace Theorem) Let L 1,..., L n be n linearly independent forms with algebraic coefficients in C and let δ > 0. Then the set of solutions of the equation 0 L 1 (x) L n (x) x δ with x Z n is contained in the union of finitely many proper linear subspaces of Q n. Proof. See Schmidt [20]. Corollary Let α 1,..., α n C be algebraic and linearly independent over Q and let δ > 0. Then there exist only finitely many x Z n with α 1 x α n x n x 1 n δ. Proof. This proof is by induction. If α 1 0 then there exist only finitely many x Z such that α 1 x x δ. This proves the theorem for the case n = 1. Assume that the theorem is true for n 1. Let α 1,..., α n C be linearly independent over Q. Take L 1 (x) = α 1 x α n x n and L i (x) = x i for 2 i n. Note that L 1,..., L n are chosen such that det(l 1,..., L n ) 0. The set of solutions of L 1 (x) x 1 n δ is contained in the set of solutions of L 1 (x) L n (x) x δ By the Subspace theorem we find that the solutions with x Z n are contained in the union of finitely many proper linear subspaces of Q n of dimension n 1. These subspaces are of the form c 1 x c n x n = 0 with (c 1,..., c n ) Q n \{0}. For such a subspace the solutions of L 1 (x) x 1 n δ are also solutions of (α 1 c 1 c n α n )x (α n 1 c n 1 c n α n )x n 1 x 1 n δ. Note that α 1 c 1 c n α n,, α n 1 c n 1 c n α n are linearly independent over Q. Hence, these equations have only finitely many solutions (x 1,..., x n 1 ) Z n 1 by the induction hypothesis. This proves the corollary. So, if 1, α 1,..., α n are linearly independent with α 1,..., α n C, then there exists a constant C R >0 such that all x Z n satisfy the inequality α 1 x α n x n C x n δ. Hence, we proved a lower bound for the sum α 1 x α n x n as in Proposition In contrast to Proposition 3.10 we do not have an effective lower bound and we have the stronger condition that 1, α 1,..., α n must be linearly independent. But for this result the lower bound is in many cases much stronger than the one depending on the degree of the field extension. As in Section 3.2 we would now like to formulate a version of Kronecker s Theorem using this result. This can not be done for Kronecker s Theorem for linear forms, because of the stronger condition that 1, α 1,..., α n must be linearly independent. This is why we only consider the problem of inhomogeneous simultaneous approximation of real numbers by rational numbers. 32.