Erdős Type Configuration Problems in Modules over Finite Rings

Transcription

1 Erdős Type Configuration Problems in Modules over Finite Rings by Esen Aksoy Yazici Submitted in Partial Fulfillment of the Requirements for the Degree Doctor of Philosophy Supervised by Professor Alex Iosevich Professor Jonathan Pakianathan Department of Mathematics Arts, Sciences and Engineering School of Arts and Sciences University of Rochester Rochester, New York 2014

2 ii Biographical Sketch Esen Aksoy Yazici was born on March 14, 1981, in Elazig, Turkey. She started her studies of mathematics in 1999 at Ankara University and received a B.S. In September 2004, she joined the Mathematics Program at Sabanci University and graduated with a master s degree in She started her Ph.D. studies at Sabanci University and took some doctorate level courses in She was then accepted to the University of Rochester in August She resumed her Ph.D. studies in the areas of geometric combinatorics and number theory under the guidance of her advisor, Alex Iosevich, and co-advisor, Jonathan Pakianathan, at the University of Rochester.

3 iii Acknowledgments First and foremost, I would like to express my deepest gratitude to my advisor, Alex Iosevich, for his excellent guidance during my graduate studies. His expertise and broad vision in mathematics has constantly inspired me and eventually led me to my current research. I thank him for sharing his knowledge and supporting me throughout this stage. I am also exceptionally grateful to my co-advisor, Jonathan Pakianathan, for his helpful suggestions and comments during this work. His excellent supervision contributed greatly to this dissertation and my mathematical insight. I am sincerely thankful to the faculty, staff and graduate students in the Department of Mathematics at the University of Rochester for their friendship and assistance. I, also, would like to thank Alev Topuzoğlu for her invaluable mentorship and encouragement. I appreciate my collaboration with Ayça Çesmelioğlu, Wilfried Meidl, and Alev Topuzoğlu from Sabanci University. Finally, I especially thank my husband for his understanding and helping me get through some difficult times. I am grateful to my family for their love and endless support.

4 iv Abstract Throughout this thesis we study two types of discrete problems over finite fields and rings. In the first part, we study Erdős type geometric problems for vector spaces over finite fields and modules over residue rings. The second part starts with some basic facts about permutation polynomials (PPs) over finite fields. We then introduce the Carlitz rank of a permutation polynomial and study related results.

5 v Contributors and Funding Sources This work was supervised by a dissertation committee consisting of Alex Iosevich (advisor), Jonathan Pakianathan (co-advisor) of the Department of Mathematics, and Alice Quillen of the Department of Physics and Astronomy. This work is partially funded by TÜBİTAK National Ph.D. Fellowship Programme.

6 vi Table of Contents Biographical Sketch ii Acknowledgments iii Abstract iv Contributors and Funding Sources v 1 Introduction 1 2 Background on Fourier Analysis 5 3 Distinct Distances-Congruence classes of k-simplices determined by subsets of F d q and Z d q Preliminaries F d q Setting: Z d q Setting: Product-Volume set of subsets of F d q and Z d q Preliminaries F d q Setting: Z d q Setting:

7 vii 5 Permutation Polynomials over F q Carlitz rank of a permutation polynomial Bibliography 48

8 1 1 Introduction In this work we consider geometric and arithmetical configurations in discrete settings arising in additive combinatorics, discrete geometry, and number theory. In particular, we study various Erdős type geometric problems over finite fields and residue rings Z q, where q = p l is an odd prime power. We also work on some enumerative problems on permutation polynomials over finite fields. The main focus of Erdős type problems is to determine the size of a discrete set so that the configuration of a given type is obtained. In this setting, it is well known that there is a rich analogy between discrete and continuous problems. More specifically, the techniques used for the geometry in the Euclidean plane are applicable to those in vector spaces over finite fields and conversely, the results coming from finite field geometry shed some light on the Erdős type problems in Euclidean settings. Recently, Erdős type geometric problems are of great interest and several problems have been studied by various authors in the context of finite fields, see for example [4], [3], [7], [8],[13], [14], [16] and the references therein. The results in them are mostly achieved via Fourier analytic methods for large enough sets and some other combinatorial and probabilistic approaches for relatively small ones. Our main focus here is to extend the results for Erdős type geometric prob-

9 2 lems to the setting of finite cyclic rings Z p l = Z/p l Z where p is a fixed odd prime. Among the Erdős type geometric problems, we study Erdős-Falconer distance problem, congruence classes of k-simplices up to orthogonal transformations, distribution of r-hinges, dot product problem, and distribution of areas of triangles determined by a given discrete set. It is worth to note here that the new problems in the study of these questions in these rings are due to the non unit elements that must be taken into arithmetical consideration separately. In Chapter 2 of this thesis, we present Fourier analytic tools defined on the finite cyclic groups Z q that are used throughout this work. For a further exploration we refer the reader to [18]. In Chapter 3 and 4, we first present various Erdős-Falconer type configuration questions in the context of finite fields F q together with a statement of the best known results. We then introduce our results for finite cyclic groups Z q, where q = p l. Chapter 3 starts with a complete description of the distance problem. Letting G = F q or Z q, the distance map λ on G d is given by the map, λ : G d G d G (x, y) x y = (x 1 y 1 ) (x d y d ) 2. This map does not induce a metric on G d. However, it is invariant under orthogonal transformations of G d. For subsets E G d, we shall consider the restriction map λ E E : E E G and ask for a threshold on the size of E to guarantee that the image of the map λ E E, (E) = { x y : (x, y) E E}, is about the size G = q. In other words, almost all distances are achieved.

10 3 An extension of this notion is the congruence classes of triangles, more generally, distribution of k-simplices with the vertices from the points of E. In Section 3.2, we introduce these problems on the base field G = F q, with a brief summary on the progress in this context. In Section 3.3, we focus on analogous problems for G = Z q. An asymptotically sharp bound for the distance set (E) for E Z d q is given in [9] and we first note that result. Then in Theorem we prove a sufficient lower bound on E to get a positive proportion of all triangle classes up to orthogonal transformations. In the rest of this section we study the Hinges problem. For a subset E Z d q, and a distance set α = {α i } r 1 i=1 (Z q) r 1, we define the r-hinges determined by the points of E as H r,α = {(x, x 1,..., x r 1 ) E... E : x x i = α i }, where r > 2 is an integer. In Theorem we determine a sufficient condition on the size of E so that the distribution of r-hinges among the points of E is uniform. This result is similar to the result obtained in [8, Theorem 3.1] in the context of finite fields. Chapter 4 focuses on the problems of dot product and volume set determined by subsets E of G d where G = F q or Z q. Given a subset E of G d, the dot product is defined as (E) = {x.y : x, y E}, where dot product of two vectors is defined by the usual formula x.y = x 1 y x d y d, for x = (x 1,..., x d ) and y = (y 1,..., y d ). We define the d- dimensional non-zero volumes of (d + 1)-simplices whose vertices are in E by V d (E) = {det(x 1 x d+1,..., x d x d+1 ) : x j E} \ {0}.

11 4 Section 4.2 is dedicated to recapping known results for dot products and volume sets in F d q. We then move to the residue ring G = Z q in Section 4.3. First, in Theorem for the sets of the form E = } A {{... A } with A Z q, we find a d-fold condition on the size of E, so that dot product set of E, (E), contains a positive proportion of the elements of Z q. This result can be seen as a variant of the dot product result for an arbitrary subset E of Z d q given in [9]. Later in this section, we prove a triangle area result in Theorem for subsets of Z 2 q. In Chapter 5, permutation polynomials (PPs) over finite fields F q are studied. A polynomial f(x) F q [x] is said to be a PP of F q if the induced map F q F q α f(α) is bijective. These polynomials play an important role in the study of secure transmission of data and combinatorial designs. In this chapter we first study the cycle structure of a PPs. Then in Section 5.1 we turn our attention to the problem of enumerating permutation polynomials of a given Carlitz rank. From a well known result of L. Carlitz (see [6]), it immediately follows that any permutation polynomial f(x) of F q can be represented by a polynomial P n (x) = (... ((a 0 x + a 1 ) q 2 + a 2 ) q a n ) q 2 + a n+1, n 0, (1.1) where a 1, a n+1 F q, a i F q = F q \ {0} for i = 0, 2,..., n. This representation is not necessarily unique and we define the Carlitz rank of a permutation polynomial f(x) to be to smallest n, i.e, smallest number of inversions, such that P n (x) represents f(x). In Theorem 5.1.1, we give a formula for the number of permutations of F q with Carlitz rank n, B(n), where 2 n < q 1. The results in Chapter 5 are published in [2]. 2

12 5 2 Background on Fourier Analysis For background on Fourier analysis on finite groups we refer the reader [18]. Here we present the necessary background for the modules Z d q over Z q. Let f, g : Z d q C. The Fourier transform of f is defined as f(m) = q d x Z d q χ( x.m)f(x), where χ(z) = exp(2πiz/q). We note the following properties: q 1, if m = 0 d χ(x.m) = (Orthogonality) x Z 0, otherwise d q f(x) = m Z d q χ(x.m) f(m) (Inversion) f(m)ĝ(m) = q d f(x)g(x) (Parseval) m Z d q x Z d q In particular, taking g = f in Parseval s identity, we have m Z d q f(m) 2 = q d x Z d q f(x) 2. Finally the average value of of f(x) is given by Avg(f) = f(0,..., 0) = q d x Z d q (Plancherel) f(x).

13 6 3 Distinct Distances-Congruence classes of k-simplices determined by subsets of F d q and Z d q The classical Erdős distance problem in discrete geometry asks for the number of distinct distances determined by n points in Euclidean space. In [11] Erdős conjectured that the minimum number of distinct distances determined by n points in the Euclidean plane is C n logn. Recently, Guth and Katz [12] settled the conjecture, up to a square log factor showing that n points determine at least C n logn distances. We should note that in higher dimensions the conjecture is still open, and the best results are due to Solymosi and Vu which can be found in [20]. Before stating the best known result in the context of finite fields and integer rings let us first introduce the necessary background. 3.1 Preliminaries Throughout, Tk d (E) will denote the the set of congruence classes of k-simplices determined by E F d q or Z d q. We simply write (E) for T d 1 (E), the set of distinct distances determined by the points of E.

14 7 3.2 F d q Setting: The geometry of subsets of vector spaces over finite fields is analogous to that of discrete points in the Euclidean setting. Iosevich and Rudnev (see [15]) proved that for E F d q, if E > 2q d+1 2 then the points of E determine all possible distances. They also showed that one cannot in general get Cq distinct distances, i.e a positive proportion of all distances, if E q d 2. In [14], the authors showed the following: (i) Let d > 2. If E > Cq d 2 exist c > 0 such that (E) cq. with a sufficiently large constant C, then there (ii) If d is odd, there exist c > 0 and E F d q such that E cq d+1 2 and (E) F q. Which implies that the exponent d+1 2 in [15] is sharp to get all q distances in odd dimensions. The question of whether the exponent d 2 in (i) can be reduced to get a positive proportion of all distances is still open in odd dimensions. (iii) If d > 2 is even, E > Cq d 2 (E) = F q. Also there exists c > 0 and E > cq d 2 exponent d 2 with a sufficiently large constant C gives with (E) F q. That is the is sharp to get all q distances in even dimensions greater than 2. The question of whether the exponent d 2 all distances is open. can be smaller to get a positive proportion of For d = 2, in [7, Theorem 2.2], it was shown that if E F 2 q with E > q 4 3, where q = 3 mod 4, then (E) > cq. In [3] this result has been generalized to an arbitrary odd q. The question of whether the exponent 4 3 is still open. here can be reduced A natural generalization of distance problem is the problem of distribution of k-simplices and in [3], it is proved that for E F 2 q, q odd, if E Cq 8 5, then T 2 2 (E) cq 3, which improves the previous known result of E q 7 4 in [4].

15 8 In more general setting, the following non-trivial exponent for congruence classes of k-simplices Tk d (E) is examined in [3]. Theorem Let Q be a non-degenerate quadratic form on F d q, where q is odd and d 2. Let E F d q. There exist constants C, c, depending only on 1 k d, d 1 d such that: if E Cq k+1, then T d k,q(e) cq (k+1 2 ). 3.3 Z d q Setting: In this section we concentrate on the distance relevant problems in context of finite cyclic rings Z p l, where p is an odd prime. Compared to configurations in vector spaces over finite fields, to overcome the difficulties arising from the zero divisors in these cyclic rings, an extra arithmetical machinery is developed. The distance problem for subsets of residue rings, namely for Z d q where q is an odd prime power, is studied by Covert, Iosevich and Pakianathan in [9]. Here is the related theorem. Theorem Let E Z d q, where q = p l and p is odd. Suppose E l(l + 1)q (2l 1)d + 1 2l 2l. Then, (E) Z q where Z q denotes the the set of unit elements in Z q. Given E Z 2 q, using a group theoretical approach analogous to [3], for T2 2 (E), the congruence classes of triangles determined by the points of E, we will prove the following result. Theorem Suppose E Z 2 q with q = p l and p 3 mod 4. If E 3 3p 2l 1 3, then T2 2 (E) q 3.

16 9 For the proof we will need the following lemmas. Let us first denote by SO 2 (Z q ) = {A M 2 (Z q ) : AA T = I, det(a) = 1} the special orthogonal group. Lemma Let ξ = (ξ 1, ξ 2 ) Z 2 q, where q = p l and p is an odd prime. If ξ = ξ1 2 + ξ2 2 0, then Stab(ξ) p l 1, where Stab is the stabilizer under the action of the special orthogonal group. Proof. Let ξ = (x, y) Z 2 q. Since ξ 0, we can write ξ = x 2 + y 2 = p i u, 0 i l 1, u Z q. Now if A = a b SO 2 (Z q ) fixes ξ, then from the identity b a a b x = x b a y y we get (a 1)x + by = 0 bx + (a 1)y = 0, and hence (a 1)(x 2 + y 2 ) 0 mod p l, b(x 2 + y 2 ) 0 mod p l. (3.1) Putting x 2 + y 2 = p i u in (3.1), we have a = p l i k + 1, 0 k < p i, and b = p l i m, 0 m < p i, (3.2) where a 2 + b 2 1 mod p l. (3.3) Now to conclude the argument, we claim that for a 0 a if b 0 and b are satisfying a b mod p l and a 2 + b 2 1 mod p l and condition (3.2),

17 10 respectively, then b 0 b. This will prove the lemma, for then the number of pairs (a, b) satisfying the conditions (3.2) and (3.3) is at most the number of possibilities of b which is p i. This is at most p l 1 as the valuation of a nonzero element is at most l 1. It remains to prove the claim and we will prove its contrapositive here. Suppose that b 0 = b and a b mod p l, a 2 + b 2 1 mod p l, so that a 2 0 a 2 mod p l. Writing a 0 = p l i k and a = p l i k + 1, It follows that (p l i k 0 + 1) 2 (p l i k + 1) 2 mod p l, p 2l 2i k p l i k 0 p 2l 2i k 2 + 2p l i k mod p l, therefore, p 2l 2i (k0 2 k 2 ) + 2p l i (k 0 k) 0 mod p l, p l i (k 0 k)(p l i (k 0 + k) + 2) 0 mod p l. Thus p l p l i (k 0 k)(p l i (k 0 + k) + 2), and since p p l i (k 0 + k) + 2 as p is odd, we must have p i k 0 k < p i. Hence we have k 0 k = 0, i.e., k 0 = k and therefore a 0 = a. Lemma Let ξ Z 2 q \ (0, 0), where q = p l and p 3 mod 4. If ξ = 0, then Stab(ξ) p l 1. For the proof of Lemma we will use Hensel s Lemma. Lemma (Hensel s Lemma). Let f(x) Z[x], f(r) 0 mod p and f (r) 0 mod p so that r is a simple root of f modulo p. Then for any k 2, there exists a unique ˆr in Z p k such that f(ˆr) 0 mod p k with ˆr r mod p. Proof of Lemma We first note that as p 3 mod 4, for ξ Z 2 q \ (0, 0), ξ = 0 implies that ξ = (p m u, p m v) for some m l and u, v 2 Z q. Now if A SO 2 (Z q ) fixes ξ = (p m u, p m v), it can be readily shown that it also fixes η = ( p m v, p m u). Hence A also fixes Span{ξ, η} = p m (Z p l Z p l) = Z p l m Z p l m.

18 11 Since ξ (0, 0), we have m l 1. We shall note that Z p l m Z p l m is smallest, and hence the number of matrices A that fixes Z p l m Z p l m is largest, when m = l 1. Therefore it is sufficient to consider the case m = l 1. In this case A fixes p l 1 (Z p l Z p l) = Z p Z p. We now write A = I 2 + B, where I 2 denotes the 2 2 identity matrix and B M 2 (Z q ). Then for any y Z 2 q we have Ap l 1 y = p l 1 y + Bp l 1 y, so that Bp l 1 y = 0 as A fixes p l 1 y. This implies that B = pb for some B M 2 (Z q ), and A = I 2 + pb Γ 1 SO 2 (Z q ). where Γ 1 denotes the matrices in M 2 (Z q ) congruent to I 2 mod p. It follows that A a b : a, b Z q, a 2 + b 2 1 mod q, a 1 mod p, b 0 mod p b a. (3.4) Now we count the number of matrices in (3.4). We first fix b. Since b 0 mod p, we have p l 1 choices for b. Then we consider the polynomial f(x) = x 2 (1 b 2 ) Z[x]. Note that f(x) = x 2 1 in Z p as b 0 mod p. Hence 1 is a root of f(x) and f (1) = 2 0 in Z p as p is odd. Hence by Hensel s Lemma there exists a unique a in Z q such that f(a) = a 2 (1 b 2 ) = 0 in Z q with a 1 mod p. Therefore the number matrices of the form in (3.4) is p l 1. This completes the proof. We make use of the following lemma from [3].

19 12 Lemma For any finite space F, any function f : F R 0, and any n 2 we have f n (z) F z F ( ) n f 1 + F n(n 1) 2 f n 2 z F where f 1 = z F f(z), and f = max z F f(z). ( f(z) f ) 2 1, F Lastly, we state the following lemma from [9] and use Remark for the proof of Theorem Lemma Let d 2 and j Z q, where q is odd. Set x = x x 2 d. Denote by S j = {x Z d q : x = j} the sphere of radius j. Then, S j = q d 1 (1 + o(1)). Remark Note that SO 2 (Z q ) = a b b : a 2 + b 2 = 1 a (3.5) and hence if we denote by S 1 the sphere of radius 1 in Z 2 q, then SO 2 (Z q ) = S 1 q, by Lemma Proof of Theorem We first recall that SO 2 (Z q ) = {A M 2 (Z q ) : AA T = I, det(a) = 1} and define an equivalence relation on (Z 2 q) 3 as (a, b, c) (a, b, c ) if θ SO 2 (Z q ) with a = θa, b = θb, c = θc. For E Z 2 q and a, b, c Z 2 q, let µ(a, b, c) = {(x, y, z) E 3 : θ SO 2 (Z q ) such that x y = θa, y z = θb, x z = θc.}. Note that µ(θa, θb, θc) = µ(a, b, c) for all θ SO 2 (Z q ), so µ can be viewed as a function µ : (Z 2 q) 3 / Z 0.

20 13 Then by the Cauchy-Schwarz inequality, E 6 = where µ(a, b, c) 2 T2 2 (E) (a,b,c) (Z 2 q )3 / (a,b,c) (Z 2 q )3 / µ 2 (a, b, c), T 2 2 (E) = {(a, b, c) (Z 2 q) 3 / : µ(a, b, c) 0}, which is equal to {(a, b, c) (Z 2 q) 3 / : (x, y, z) E 3 and θ SO 2 (Z q ) such that x y = θa, y z = θb, x z = θc}. We have, µ 2 (a, b, c) = {(x, y, z, x, y, z ) E 6 : θ 1, θ 2 SO 2 (Z q ) such that x y = θ 1 a, x y = θ 2 a y z = θ 1 b, y z = θ 2 b x z = θ 1 c, x z = θ 2 c} = {(x, y, z, x, y, z ) E 6 : θ 1, θ 2 SO 2 (Z q ) such that θ 1 1 (x y) = θ 1 2 (x y ) = a θ 1 1 (y z) = θ 1 2 (y z ) = b θ 1 1 (x z) = θ 1 2 (x z ) = c}

21 14 so that (a,b,c) (Z 2 q) 3 / µ 2 (a, b, c) = {(x, y, z, x, y, z ) E 6 : θ 1, θ 2 SO 2 (Z q ) such that For a fixed θ SO 2 (Z q ), let Then we have and therefore θ 1 1 (x y) = θ 1 2 (x y ) θ 1 1 (y z) = θ 1 2 (y z ) θ 1 1 (x z) = θ 1 2 (x z )} = {(x, y, z, x, y, z ) E 6 : θ SO 2 (Z q ) such that θ(x y) = x y, θ(y z) = y z, θ(x z) = x z } = {(x, y, z, x, y, z ) E 6 : θ SO 2 (Z q ) such that x θx = y θy = z θz}. ν θ (t) = {(u, v) E E : u θ(v) = t}. (3.6) ν 3 θ (t) = {(x, y, z, x, y, z ) E 6 : x θx = y θy = z θz = t}, (a,b,c) (Z 2 q )3 / µ 2 (a, b, c) θ SO 2 (Z q) t Z 2 q By Lemma 3.3.4, ( ) 3 νθ 3 (t) q 2 νθ ν q 2 θ t Z 2 q t Z 2 q ν 3 θ (t). (3.7) ( ν θ (t) ν ) 2 θ 1 q 2 where ν θ 1 = t Z 2 q ν θ(t) = E 2 and ν θ = sup t ν θ (t) E as when we first fix v in (3.6), u is uniquely determined. It follows that t Z 2 q ν 3 θ (t) q 4 E E t Z 2 q q 4 E 6 + 3q 2 E ( ν θ (t) ν θ 1 ξ Z 2 q\(0,0) q 2 ) 2 ν θ (ξ) 2 (by Plancherel Theorem)

22 15 and thus θ SO 2 (Z q) t Z 2 q ν 3 θ (t) SO 2 (Z q ) q 4 E 6 + 3q 2 E By Remark 3.3.1, SO 2 (Z q ) q and hence Noting that and θ SO 2 (Z q) t Z 2 q ν 3 θ (t) q 3 E 6 + 3q 2 E ν θ (t) = v Z 2 q = v,α Z 2 q = v,α Z 2 q E(v)E(t + θv) θ SO 2 (Z q) ξ Z 2 q\(0,0) E(v)χ(α.(t + θv))ê(α) Ê(α)χ(t.α)E(v)χ(α.θv) = Ê(α)χ(t.α) α Z 2 q v Z 2 q = Ê(α)χ(t.α) α Z 2 q v Z 2 q = q 2 α Z 2 q ν θ (ξ) = q 2 t Z 2 q θ SO 2 (Z q) ξ Z 2 q \(0,0) χ(α.θv)e(v) χ(θ T (α).v)e(v) Ê(α)χ(t.α)Ê( θt (α)), χ( t.ξ)ν θ (t) = q 2 χ( t.ξ)q 2 t Z 2 q α Z 2 q = Ê(α)Ê( θt (α)) α Z 2 q t Z 2 q = q 2 Ê(ξ)Ê( θt (ξ)), ν θ (ξ) 2 Ê(α)χ(t.α)Ê( θt (α)) χ(t.(α ξ)) ν θ (ξ) 2 (3.8)

23 16 we have θ SO 2 (Z q) ξ Z 2 q \(0,0) ν θ (ξ) 2 = q 4 θ SO 2 (Z q) ξ Z 2 q \(0,0) = q 4 θ SO 2 (Z q) ξ Z 2 q\(0,0) Ê(ξ) 2 Ê( θt (ξ)) 2 Ê(ξ) 2 Ê(θT (ξ)) 2 ( ) q 4 max Stab(ξ) ξ Z 2 q \(0,0) ξ (0,0) Plugging this value in (3.8) and using (3.7) we get (a,b,c) (Z 2 q )3 / where µ 2 (a, b, c) θ SO 2 (Z q) t Z 2 q ν 3 θ (t) ( q 3 E 6 + 3q 6 E Ê(ξ) 2 max Stab(ξ) ξ Z 2 q \(0,0) η (0,0) η = ξ ) ξ (0,0) Ê(η) 2 Ê(ξ) 2 η (0,0) η = ξ Ê(η) 2 = q 3 E 6 + 3q 6 E I (3.9) ( ) I = max ξ Z 2 q\(0,0) Stab(ξ) ξ (0,0) Ê(ξ) 2 η (0,0) η = ξ Ê(η) 2 We first note that Stab(ξ) p l 1 for ξ (0, 0) by Lemma and Extending the summation in η over all η and using Plancherel twice in I, we get Plugging this value in (3.9) gives so that (a,b,c) (Z 2 q) 3 / I p l 1 q 4 E 2. µ 2 (a, b, c) q 3 E 6 + 3q 2 E 3 p l 1, (3.10) T 2 2 (E) E 6 q 3 E 6 + 3q 2 E 3 p l 1 E 6 2q 3 E = q3 6 2

24 17 whenever E 3 3p 2l 1 3, which completes the proof. Another distance related problem is the hinges problem in the discrete setting. Employing the methods of [8], we have the following related result in the context of Z p l. Theorem Let r > 2 be an integer. For E Z d q, where q = p l, p is an odd prime, and a distance set α = {α i } r 1 i=1 (Z q) r 1, let H r,α = {(x, x 1,..., x r 1 ) E... E : x x i = α i } denote the r-hinges with distances α determined by the points of E. Then, H r,α = E r (1 + o(1)) qr 1 whenever E l(l + 1)q d(1 1 2l(r 2))+ 1 2l(r 2). For the proof we shall make use of Lemma and the following lemma from [9]. Lemma For j Z q and q = p l, we have sup m 0 Ŝ j (m) l(l + 1)q d+2l 1 2l. Proof of Theorem The proof will proceed by induction on r. We first consider the basis step r = 3: H 3,α = {(x, x 1, x 2 ) E E E : x x 1 = α 1, x x 2 = α 2 } = S α1 (x x 1 )S α2 (x x 2 )E(x)E(x 1 )E(x 2 ) x,x 1,x 2 = ( ) ( ) S α1 (x y)e(y) S α2 (x y )E(y ). (3.11) x E y y

25 18 Note that S αi (x y)e(y) = y m = m χ(m.(x y))ŝα i (m)e(y) y χ(m.x)ŝα i (m) χ( m.y)e(y) y = q d m χ(m.x)ŝα i (m)ê(m) = q d Ŝ αi (0)Ê(0) + qd m 0 χ(m.x)ŝα i (m)ê(m) = q d E S αi + q d m 0 χ(m.x)ŝα i (m)ê(m) By Lemma 3.3.5, S αi = q d 1 (1 + o(1)) for any α i Z q. Hence S αi (x y)e(y) = q 1 E (1 + o(1)) + q d χ(m.x)ŝα (m)ê(m), i m 0 y and (3.11) is equal to x E(q 1 E (1+o(1))+q d m 0 χ(m.x)ŝα 1 (m)ê(m))(q 1 E (1+o(1))+q d n 0 χ(n.x)ŝα 2 (n)ê(n)) = x E q 2 E 2 (1 + o(1)) + x E q d 1 E (1 + o(1)) + q 2d x E ( χ(m.x)ŝα (m)ê(m) + 1 m 0 n 0 χ(m.x)χ(n.x)ŝα 1 (m)ê(m)ŝα (n)ê(n) 2 m 0 n 0 = E 3 (1 + o(1)) + I + II, q 2 where I = x E q d 1 E (1 + o(1)) ( m 0 χ(m.x)ŝα 1 (m)ê(m) + n 0 χ(n.x)ŝα 2 (n)ê(n) ) χ(n.x)ŝα 2 (n)ê(n) ),

26 19 and II = q 2d x E χ(m.x)χ(n.x)ŝα 1 (m)ê(m)ŝα (n)ê(n). 2 m 0 n 0 Observe that I 2 q d 1 E (1 + o(1)) χ(m.x)ŝα (m)ê(m) i x E m 0 = 2 E q d 1 (1 + o(1)) χ(m.x)ŝα (m)ê(m)e(x) i m 0 x = 2 E q 2d 1 (1 + o(1)) Ê( m)ê(m)ŝα i (m) m 0 = 2 E q 2d 1 (1 + o(1)) Ê(m) 2 Ŝ αi (m) m 0 so that by Lemma and Plancherel equality I 2 E q 2d 1 (1 + o(1))l(l + 1)q d+2l 1 2l q d E = 2 E 2 q d 1 (1 + o(1))l(l + 1)q d+2l 1 2l. Also, II q 2d χ(m.x)ŝα (m)ê(m) 1 χ(n.x)ŝα (n)ê(n) 2 x E m 0 n 0 q 2d x E m 0 χ(m.x)ŝα i (m)ê(m) 2 q 2d x Z d q = q 2d m 0 n 0 = q 3d m 0 χ(m.x)ŝα i (m)ê(m)χ( n.x)ŝα i (n) Ê(n) m 0 n 0 Ŝ αi (m)ê(m)ŝα i (n) Ê(n) x Z d q Ŝα i (m) 2 Ê(m) 2 q 3d l 2 (l + 1) 2 q d+2l 1 l χ((m n).x) Ê(m) 2 (by Lemma 3.3.6) m 0 q 2d l 2 (l + 1) 2 q d+2l 1 l E.

27 20 Therefore, H 3,α = E 3 (1 + o(1)) + I + II q 2 = E 3 q (1 + o(1)) + 2 O(2 E 2 q d 1 (1 + o(1))l(l + 1)q d+2l 1 2l It follows that if E l(l + 1)q d(1 1 2l )+ 1 2l, then H 3,α = E 3 (1 + o(1)). q 2 + q 2d l 2 (l + 1) 2 q d+2l 1 l E ). Now for the inductive step we assume that H r,α = E r (1 + o(1)) when E q r 1 l(l + 1)q d(1 2l(r 1)) l(r 1). Setting S := Sαr, we can write H r+1,α = H r,α (x, x 1,..., x r 1 )E(x r )S(x x r ) x,x 1,...x r = H r,α (x, x 1,..., x r 1 )E(x r ) χ(m.(x x r ))Ŝ(m) x,x 1,...x r m = H r,α (x, x 1,..., x r 1 ) Ŝ(m)χ(m.x) χ( m.x r )E(x r ) x,x 1,...x r 1 m x r = q d H r,α (x, x 1,..., x r 1 ) Ŝ(m)χ(m.x)Ê(m) x,x 1,...x r 1 m = q d Ŝ(m)Ê(m) H r,α (x, x 1,..., x r 1 )χ(m.x) m x,x 1,...x r 1 = q d(r+1) m Ŝ(m)Ê(m)Ĥr,α( m, 0,..., 0) = q d S E H r,α + q d(r+1) m 0 Ŝ(m)Ê(m)Ĥr,α( m, 0,..., 0) = E r+1 (1 + o(1)) + R (by induction hypotesis) q r where R = q d(r+1) m 0 Ŝ(m)Ê(m)Ĥr,α( m, 0,..., 0).

28 21 Now using the Cauchy-Schwarz inequality, R 2 = R 2 = q 2d(r+1) m 0 Ŝ(m)Ê(m)Ĥr,α( m, 0,..., 0) 2 q 2d(r+1) ( m 0 q 2d(r+1) m 0 Ŝ(m) Ê(m) Ĥr,α( m, 0,..., 0) Ŝ(m) 2 Ê(m) 2 m 0 ) 2 Ĥr,α( m, 0,..., 0) 2 q 2d(r+1) l 2 (l + 1) 2 q d+2l 1 l q d E Ĥr,α( m, 0,..., 0) 2 m 0 q d+2l 1 2dr+d l l 2 (l + 1) 2 E m Ĥr,α(m, 0,..., 0) 2. (3.12) Note that the third inequality above follows from Lemma and Plancherel theorem. Let A := m Ĥr,α(m, 0,..., 0) 2. We first see that Ĥ r,α (m, 0,..., 0) = q dr where f(x) = E(x) x,x 1,...x r 1 χ( x.m)h r,α (x, x 1,..., x r 1 ) = q dr = q dr+d f(m), x,x 1,...x r 1 χ( x.m)e(x)... E(x r 1 )S α1 (x x 1 )... S αr 1 (x x r 1 ) x 1,...,x r 1 E(x 1 )... E(x r 1 )S α1 (x x 1 )... S αr 1 (x x r 1 ) = E(x) x 1 E(x 1 )S α1 (x x 1 ) x 2 E(x 2 )S α2 (x x 2 )... x r 1 E(x r 1 )S αr 1 (x x r 1 ) = E(x) x 1 E(x 1 )S α1 (x x 1 ) E (x S α2 )... E (x S αr 1 ) q (r 2)(d 1) (1 + o(1))e(x) x 1 E(x 1 )S α1 (x x 1 ) where in the last step we used the fact that E (x S) S = q d 1 (1 + o(1)).

29 22 It follows that A = m Ĥr,α(m, 0,..., 0) 2 = q 2dr+2d m = q 2dr+d x f(m) 2 f(x) 2 q 2dr+d q 2(r 2)(d 1) (1 + o(1)) x E(x) x 1,x 2 E(x 1 )E(x 2 )S α1 (x x 1 )S α1 (x x 2 ) q 2dr+d q 2(r 2)(d 1) (1 + o(1)) H 3,α q 2dr+d (q d 1 ) 2(r 2) E 3 q 2 (1 + o(1)) Now plugging this value of A in (3.12) we get R 2 q 2d 2 (q d 1 ) 2(r 2) q d+2l 1 l E 4 l 2 (l + 1) 2 (1 + o(1)), so that R q d 1 (q d 1 ) r 2 q d+2l 1 2l E 2 l(l + 1)(1 + o(1)) = q (d 1)(r 1) q d+2l 1 2l E 2 l(l + 1)(1 + o(1)). We conclude that, ( H r+1,α = E r+1 (1 + o(1)) + O q r and hence q (d 1)(r 1) q d+2l 1 2l H r+1,α = E r+1 (1 + o(1)) q r whenever E l(l + 1)q d(1 1 2l(r 1))+ 1 2l(r 1). ) E 2 l(l + 1)(1 + o(1)),

30 23 4 Product-Volume set of subsets of F d q and Z d q An important problem in additive combinatorics is the sum-product problem for finite subsets of abelian groups, e.g. given A F q or Z q, how large does A need to be to ensure that da 2 = A.A } +. {{.. + A.A } contains all q elements or a positive d-fold proportion of them. It is an easy observation that sum-product problem is closely related to the dot product problem in the context of finite fields and rings. Also, it turns out that dot product problem has immediate implications for the volume of simplices in d-dimensional spaces, which we will discuss in detail in the subsequent sections. 4.1 Preliminaries Given a subset E of F d q or Z d q, the dot product is defined as (E) = {x.y : x, y E}, where dot product of two vectors is defined by the usual formula x.y = x 1 y x d y d, for x = (x 1,..., x d ) and y = (y 1,..., y d ).

31 24 Let A(E) = {x.y : x, y E} = (E, E ) denote the set of triangle areas determined by two arbitrary points of E with the third vertex pinned at the origin. More generally, throughout V d (E) = {det(x 1 x d+1,..., x d x d+1 ) : x j E} \ {0}. will denote the set of d-dimensional nonzero volumes, defined by (d + 1)-simplices whose vertices are in E. 4.2 F d q Setting: Dot Product It is shown in [13] that for d 2, E F d q if E > q d+1 2, then (E) F q. Later in [14], it is shown that the exponent d+1 2 above is sharp. More precisely, in quadratic extension of prime fields they construct a subset E of F d q of size q d+1 2 ɛ for which (E) = o(q) Volume Set For d = 2 the dot product result in [13] in particular implies that if E F 2 q has size greater than q 3 2, then the points of E with the third vertex at the origin determines all possible nonzero distinct triangle areas. In [16], using a point-line incidence approach to the area problem, Iosevich et al. prove that if E F 2 q with E > q, then V 2 (E) q 1 2 and the triangles giving at least q 1 2 distinct areas can be chosen such that they share the same base.

32 Z d q Setting: Dot Product In [9], the authors prove the following. Theorem Let E Z d q, where q = p l. Suppose E lq (2l 1)d + 1 2l 2l. Then, (E) Z q. In particular for product sets E = A } {{... A } Z d q, we prove the following: d-fold Theorem Let E = } A. {{.. A } be a subset Z d q, where A Z q, q = p l. d-fold 2l 1 d( )+ Suppose E q 1 2l 2l. Then (E) q. Proof. Let ν(t) = {(x, y) E E : x.y = t}. Then by Cauchy-Schwarz inequality, 2 E 4 = ν(t) (E) ν(t) 2. (4.1) t Zq t Z q We can write ν(t) = E(x)E(y) x.y=t = E(y). x E x.y=t From the Cauchy-Schwarz inequality, it follows that ν 2 (t) E E(y)E(y ) x E x.y=x.y =t

33 26 so that ν 2 (t) E t = E q 1 s = E 4 q E(x)E(y)E(y ) x.y=x.y x,y,y χ(s(x.y x.y ))E(x)E(y)E(y ) l 1 + e i, (4.2) i=0 where e i = E q 1 χ(sx(y y ))E(x)E(y)E(y ) s x,y,y υ(s)=i = E q 2d 1 E(x) Ê(sx) 2 s x υ(s)=i = E q 2d 1 E(x) Ê(pi s x) 2 s Z p l i = E q 2d 1 s Z p l i x E(s x) Ê(pi x) 2. Now denoting l i x = {s x : s Z p l i }, we can write x e i = E q 2d 1 x E l i x Ê(pi x) 2. We note here that E l i x A = p α. To see this, let x = (x 1,..., x d ) and l i x := {sx : s Z p l i} {s x : s Z p l i } = l i x From the definition, it is clear that E l i x E l i x and we will show that E lx i A where E = A }. {{.. A }. d-fold Note that we can assume that at least one of the coordinates of x = (x 1,..., x d ) is unit. Since otherwise if val p (x j ) = n > 0 is among the smallest, we can write x = p n x where x = ( x 1,..., x j,..., x d ) with x j is unit and l ĩ x = li x gives E l ĩ x = E li x.

34 27 So we assume that x j is unit. If {s 1 x,..., s t x} = E lx, i then the jth coordinates s k x j of the vectors s k x, 1 k t, are all different in A. Hence t A. Thus, E lx i E lx i = t A. Clearly, we also have E lx i p l i. Therefore E lx i min{p l i, A }, and e i E q 2d 1 x min{p l i, p α } Ê(pi x) 2. (4.3) Now we estimate Ê(pi x) 2 = q 2d χ((u v)p i x)e(u)e(v) x x u,v = q d E(u)E(v) (u v)p i =0 For u = (u 1,..., u d ), v = (v 1,..., v d ) } A. {{.. A } if d-fold p i (u v) = p i (u 1 v 1,..., u d v d ) = (0,..., 0), then p i (u j v j ) = 0 for all 1 j d. That implies u j v j {p l i, 2p l i,..., p i p l i } and we have p i choices for u j v j. If we first fix v j in A different ways, then u j is uniquely determined. So we have A p i choices for each (u j, v j ), giving A d p id choices for (u, v). It follows that Ê(pi x) 2 q d p id A d x = E q d p id

35 28 plugging this value in (4.3) and summing over all e is for i = 0,..., l 1, we get l 1 l 1 e i E 2 q d 1 min{p l i, p α }p id i=0 i=0 ( l α = E 2 q d 1 p α p id + i=0 l 1 i=l α+1 E 2 q d 1 (p α p (l α)d + pp (l 1)d ) = E 2 q d 1 (p α+ld αd + p 1+ld d ) = E 2 q d 1 p ld (p α(1 d) + p 1 d ) p l i p id ) E 2 q d 1 p ld p 1 d = E 2 q d 1 q d q 1 d l = E 2 q 2d 1 q 1 d l. Plugging this value in (4.2), the inequality (4.1) yields (E) whenever E 2 1 d 2d+ q l 2l 1 d( )+ i.e. E q 1 2l 2l. E 4 E 2 q 2d 1 q 1 d l q, Volume Set We first note the dot product result, Theorem 4.3.1, for d = 2 implies that if E Z 2 q with E lq 2 1 2l then A(E) Z q. Applying an F 2 q analogous point-line incidence approach, for the subsets of Z 2 q we have the following result. Theorem Let E Z 2 q V 2 (E) q 1+p 1. 4 p where q = p l. Suppose that E > p 2l 1 2. Then Now before giving the proof, let us introduce the necessary background. Let q = p l and (a, b) Z 2 q. Let (a, b) = {t(a, b): t Z q }be the submodule of Z 2 q generated by (a, b), which gives the line through origin and the point (a, b) in Z 2 q. Now, consider the set Λ n = {(a, b) Z 2 q : p n a, b but (a, b) (0, 0) mod p n+1 },

36 29 and denote Λ n = λ n for n = 0, 1,, l 1. Lemma λ n = p 2(l n) p 2(l n 1). Proof. Since p n a, b in Z q we have p l n choices for a and b each and hence p 2(l n) choices for (a, b). Now we need to subtract p 2(l n 1) cases where p n+1 divides both a and b to get the desired result. Lemma Let L n = { (a, b) : (a, b) Λ n } denote the set of lines generated by the points of Λ n. Then L n = p l n + p l n 1. Proof. For (a, b) Λ n, note that (a, b) is cyclic and (a, b) = p l n. Hence there exist φ(p l n ) = p l n p l n 1 generators of the group which lies in Λ n. So that for each n, we have p 2(l n) p 2(l n 1) p l n p l n 1 = p l n + p l n 1 many lines in L n each containing p l n points. We now conclude that the average number of points in a line in Z 2 q is = l 1 n=0 (pl n + p l n 1 )p l n l 1 n=0 pl n + p l n 1 = p2l + p 2l 1 + p 2l p 2 + p p l + 2p l 1 + 2p l p + 1 p l In what follows we will only consider L 0, i.e. the set of all lines of full length q in Z 2 q. Lemma For any (a, b) Z 2 q if (a, b) Λ n, then (a, b) appears in p n distinct lines in L 0.

37 30 Proof. Say (a, b) Λ n and p n+1 a. Then (a, b) belongs to the lines generated by ( ) a, ip l n + b for i = 0,..., p n 1. Note that i p n p n 0 p l n + b = i p n 1 p l n + b mod p l p n would imply i 0 p l n = i 1 p l n mod p l i 0 = i 1 mod p n which is not the case. Hence the given points are all distinct. Since a p n is a unit in Z q, it follows that the lines determined by the given generators are all distinct. Lemma Let R i = {(x, y) Z 2 q Z 2 q : x y Λ i } and r i = R i, for i = 1,..., l 1. Then r := r 1 p + r 2 p r l 1 p l 1 2p 4l 1. Proof. Let x = (x 1, x 2 ), y = (y 1, y 2 ) in Zq 2. Now if x y = (x 1 y 1, x 2 y 2 ) Λ i then p i x 1 y 1 and x 2 y 2 but p i+1 x 1 y 1 or x 2 y 2. p i x 1 y 1 gives p l i choices for x 1 y 1 in Z q, we have q choices for y 1 and y 1 determines x 1 uniquely. Hence we have qp l i choices for x 1 and y 1. Same argument applies for x 2 and y 2. Altogether the condition p i x 1 y 1 and x 2 y 2 gives qp l i qp l i choices for x = (x 1, x 2 ), y = (y 1, y 2 ). To exclude the cases where p i+1 divides both x 1 y 1 and x 2 y 2 we need to subtract qp l (i+1) qp l (i+1) cases of x and y. Hence, r i = qp l i qp l i qp l i 1 qp l i 1. Now summing r i p i s over all i = 1,..., l 1 we get r = (qp l 1 qp l 1 qp l 2 qp l 2 )p + (qp l 2 qp l 2 qp l 3 qp l 3 )p 2. + (qpqp q 2 )p l 1 = q 2 (p 2l 1 + p 2l 2 p l p l 1 ) 2q 2 p 2l 1 = 2p 4l 1

38 31 Proof of Theorem Let L be a line in L 0 and consider the sum set E + L = {e + l : e E, l L} Since L E > q 2 e 1 + l 1 = e 2 + l 2 for some l 1 l 2 so that l 2 l 1 = e 1 e 2. (4.4) Here we aim to average the solutions of the equation (4.4) over p l + p l 1 lines in L 0. To start with, we count the number of solutions of (4.4) over all lines in L 0 in two cases: In the case e 1 = e 2, there are E and q 2 choices for e 1 = e 2 and l 1 = l 2, respectively. In the case e 1 e 2, we can choose e 1 and e 2 in E ( E 1) different ways, and once we fix them, we look at the difference e 1 e 2. At that point let S i = {(e 1, e 2 ) E E : e 1 e 2 Λ i } and s i = S i for i = 0,...l 1. We know from Lemma that if (e 1, e 2 ) S i, then e 1 e 2 lies on p i lines in L 0 and when we fix the line, l 2 l 1 can be written q different ways on that line. In other words, for all s i pairs (e 1, e 2 ) S i, l 1, l 2 is chosen p i q different ways over the lines in L 0. So altogether we have {(e 1, e 2, l 1, l 2 ) E E L L : (4.4) holds for some L L 0 } = E q 2 + s 0 q + s 1 pq + s 2 p 2 q s l 1 p l 1 q E 2 q + (s 0 + s 1 p + s 2 p s l 1 p l 1 )q = E 2 q + ( E 2 + s)q

39 32 where s = s 1 p + s 2 p s l 1 p l 1. Note that s r 2p 4l 1 2 E 2 by Lemma and the assumption on the size of E. Hence we get, {(e 1, e 2, l 1, l 2 ) E E L L : (4.4) holds for some L L 0 } 4 E 2 q It follows that there exists a L L 0 such that {(e 1, e 2, l 1, l 2 ) E E L L : (4.4) holds} 4 E 2 p l p l + p l 1 = 4 E 2 p 1 + p. If we let ν(n) denote the number of representation of n as e+l for some e E, l L, then by Cauchy-Schwarz, for this particular L, ( ) 2 E 2 L 2 = ν(n) n E+L E + L n E+L ν 2 (n) = E + L {(e 1, e 2, l 1, l 2 ) E E L L : (4.4) holds}. Hence, E + L E 2 L 2 {(e 1, e 2, l 1, l 2 ) E E L L : (4.4) holds} E 2 p 2l 4 E 2 p 1+p = q p p. We conclude that there exist points of E in at least q 1+p 4 p parallel lines. Here we shall note that there are totally q parallel lines to L, including itself, and one of them must contain two points of E with a unit distance in between. For otherwise, on each of these parallel lines there would be at most p l 1 points of E yielding E qp l 1 = p 2l 1 which is not the case. It follows that points of E determines at least q 1+p 1 distinct triangle areas with the same unit base on one of the 4 p parallel lines and with the third vertex being on q 1+p 1 different parallel lines. 4 p

40 33 5 Permutation Polynomials over F q Permutation polynomials (PPs) over finite fields play an important role in combinatorics, cryptograph and coding theory. More specifically, they are used for construction of several combinatorial designs, generating pseudorandom sequences by recursive procedures, and enhancing secure transmission of data. Although permutation polynomials have been studied for long, there are still many open problems regarding the subject. For a detailed literature on this subject we refer to [17] and [19]. In this chapter, we intend to introduce some basic facts about PPs first, and then present the Carlitz rank construction described in [2]. Definition Let F q be a finite field of q elements, where q = p n, p is a prime and n is a positive integer. A polynomial f(x) F q [x] is said to be a PP of F q if the induced map α f(α) from F q to itself is bijective. Given a permutation σ of the elements of F q, there exists a unique polynomial f σ F q [x] with deg(f σ ) < q and f σ (c) = σ(c) for all c F q. The polynomial f σ can be given by the formula f σ (x) = σ(c)(1 (x c) q 1 ). (5.1) c F q or by the Lagrange interpolation formula, see for instance [17]. From (5.1), we note that deg(f σ ) q 2, since all elements of F q sum up to zero.

41 34 Consider an (arbitrary) polynomial f F q [x]. One can associate f to the reduction polynomial g F q [x] by taking f mod (x q x), since g and f induce the same map over F q, as stated in the following lemma. Lemma For f, g F q [x], f(c) = g(c) for all c F q if and only if f(x) g(x) mod (x q x). Proof. Using division algorithm we have, f(x) g(x) = h(x)(x q x) + r(x) for some h, r F q [x] with deg(r) < q. Substituting c for x, we get f(c) = g(c) for all c F q if and only if r(c) = 0 for all c F q, which is equivalent to r = 0. Definition Let n be a positive integer. The set of all one-to-one mappings, i.e. permutations, from the set {1, 2,..., n} onto {1, 2,..., n} forms a group under the composition. This group is called the symmetric group of degree n, and denoted by S n. Let S = {f(x) f(x) is a PP of F q }. Define an operation "." on the set S in such a way that g(x).f(x) = h(x) whenever f(g(x)) h(x) mod (x q x). Under this operation (S,.) is a group and it is isomorphic to the symmetric group S q. Theorem For q > 2, S q is generated by x q 2 and all (non-constant) linear polynomials over F q. Proof. Note that the polynomial f a (x) = a 2 (((x a) q 2 + a 1 ) q 2 a) q 2 represents the transposition (0a), a F q. Since every permutation of F q is a product of transpositions and every transposition (bc) can be written as a product (0b)(0c)(0b), we conclude the proof. Hence, as pointed out in [10], any permutation (or permutation polynomial) of a finite field F q can be represented by a polynomial of the form P n (x) = (... ((a 0 x + a 1 ) q 2 + a 2 ) q a n ) q 2 + a n+1, n 0, (5.2)

42 35 where a 1, a n+1 F q, a i F q for i = 0, 2,..., n. Here, we make the following convention for P n (x). We put P n = P n or P n in the cases a n+1 = 0 or a n+1 0, respectively. Obviously, in both cases n is the number of times x q 2 occurs in the representation (5.2). For the polynomial P n (x), we consider the rational function r n (x) = (... ((a 0 x + a 1 ) 1 + a 2 ) a n ) 1 + a n+1, and we have the following lemma accordingly. Lemma r n (x) with its continued fraction expansion a n+1 + 1/(a n + 1/(... + a 2 + 1/(a 0 x + a 1 )...)), can be represented by the n th convergent where the α k and the β k are given recursively by R n (x) = α n+1x + β n+1 α n x + β n, (5.3) α k = a k α k 1 + α k 2 and β k = a k β k 1 + β k 2, (5.4) for k 2, with initial values α 0 = 0, α 1 = a 0, β 0 = 1, β 1 = a 1. Note that α k and β k cannot both be zero. Proof. The proof proceeds by induction. First note that R 1 (x) = a a 0 x + a 1 = α 2x + β 2 α 1 x + β 1, where α 2 = a 2 a 0, β 2 = a 2 a 1 + 1, α 1 = a 0, β 1 = a 1. Now assuming the assertion for R k 2, we have R k 1 = a k + 1 R k 2 = a k + = a k + α k 2x + β k 2 α k 1 x + β k 1 = 1 α k 1 x+β k 1 α k 2 x+β k 2 where α k = a k α k 1 + α k 2 and β k = a k β k 1 + β k 2. α kx + β k α k 1 x + β k 1,

43 36 Note that when a n+1 = 0, R n reduces to (α n 1 x+β n 1 )/(α n x+β n ), as α n+1 = α n 1 and β n+1 = β n 1 in this case. We define the set of poles, O n, as O n = {x i : x i = β i α i, i = 1,..., n} P 1 (F q ) = F q { }. (5.5) and the string of poles as O n : O n = (x i ) n i=1 = (x 1, x 2,..., x n ). Observe that a rational function R(x) = ax+b cx+d, c 0, is onto from F q \ { d/c} to F q \ {a/c} so for every rational function R n (x) in the form (5.3) we can define a corresponding permutation F n (x) via F n (x) = R n (x) for x x n = βn α n F n (x n ) = α n+1 α n when x n F q. and As for P n, we distinguish two cases for R n and F n and write R n (x) = R n (x), F n (x) = F n (x) for a n+1 = 0, and R n (x) = R n (x), F n (x) = F n (x) for a n+1 0 An easy observation yields that P n (x) = F n (x) for all x F q \ O n. We will explore how P n and F n differs from each other for the elements in O n subsequently. We will first state the following observation. Lemma Let P m (x) = P n (x) on F q, with associated rational functions R m (x) and R n (x), respectively. If m + n < q 2, then R m (x) = R n (x). Proof. Let F m (x) and F n (x) be the permutations corresponding to P m (x) and P n (x) respectively. Then F m (x) differs from P m (x) for at most m elements, and F n (x) differs from P n (x) for at most n elements of F q. Since P m (x) = P n (x) on F q, F m (x) = F n (x) for at least q m n elements of F q. We know that if F m (x) F n (x), then F m (x) = F n (x) has at most two solutions on F q. Thus, q m n > 2 implies F m = F n.

44 37 In sequel, for a cycle τ S q, the length of τ will be denoted by l(τ), and supp(τ) := {a : τ(a) a} will denote the elements of F q that are not fixed by τ. Suppose that the permutation P n (x) can be decomposed into F n (x) and m disjoint cycles τ (n) 1... τ m (n), that is, P n (x) = τ (n) 1... τ (n) m F n (x), (5.6) where l(τ (n) i ) 2, and τ (n) i = (F n (x i 1)... F n (x i l i )), for 1 i m. We define the set O n as O n = {y O n 1 : F n (y) supp(τ (n) i ) for some 1 i m} {x n } if x n F q. O n does not contain x n if x n =. Theorem Let P n 1 (x) be given by where P n 1 (x) = τ (n 1) 1... τ (n 1) m F n 1 (x), (5.7) 1. τ (n 1) 1,..., τ m (n 1) are disjoint cycles. 2. If P n 1 (x n 1 ) = F n 1 (x n 1 ) we assume that τ (n 1) 1 = (F n 1 (x n 1 )) and l(τ (n 1) i ) = l i 2, for 2 i m. 3. If P n 1 (x n 1 ) F n 1 (x n 1 ) by permuting the cycles if necessary, we assume that F n 1 (x n 1 ) supp(τ (n 1) 1 ) with x 1 1 = x n 1, and l(τ (n 1) ) = l i 2, for 1 i m. i Assuming x n 1, x n F q, we have Proof. We consider three cases: P n (x) = (F n (x n )F n (x n 1 ))τ (n) 1... τ (n) F n (x). m

45 38 Case 1: Suppose that F n 1 (x n ) is not in supp(τ (n 1) i ) for any i = 1,... m in the decomposition (5.7), i.e. x n O n 1. we need to show the following three properties: (i) If x / O n 1 and x x n, then P n (x) = F n (x). (ii) If y O n 1 satisfies y x 1 l 1, and P n 1 (y) = F n 1 (y ), then P n (y) = F n (y ). (iii) P n (x 1 l 1 ) = F n (x n ), and P n (x n ) = F n (x n 1 ). (i) If x / O n 1, then P n 1 (x) = F n 1 (x). Since x x n 1, x n, we have ( P n (x) = (P n 1 (x) + a n ) q 2 αn 2 x + β n 2 = + a n α n 1 x + β n 1 ( ) q 2 (an α n 1 + α n 2 )x + a n β n 1 + β n 2 = α n 1 x + β n 1 ) q 2 = This proves (i). ( αn x + β n ) q 2 = α n 1x + β n 1 α n 1 x + β n 1 α n x + β n = F n (x). (ii) Let P n 1 (y) = F n 1 (y ). Then (iii) P n (y) = (P n 1 (y) + a n ) q 2 = (F n 1 (y ) + a n ) q 2 ( ) αn 2 y q 2 ( + β n 2 αn y + β n = + a α n 1 y n = + β n 1 α n 1 y + β n 1 = α n 1y + β n 1 α n y + β n = F n (y ). ) q 2 P n (x 1 l 1 ) = (P n 1 (x 1 l 1 ) + a n ) q 2 = (F n 1 (x n 1 ) + a n ) q 2 ( ) q 2 ( ) q 2 αn 2 an α n 1 + α n 2 = + a n = α n 1 = α n 1 α n = F n (x n ) and α n 1

46 39 P n (x n ) = (P n 1 (x n ) + a n ) q 2 = (F n 1 (x n ) + a n ) q 2 ( ) q 2 αn 2 x n + β n 2 = + a n α n 1 x n + β n 1 ( (an α n 1 + α n 2 )x n + a n β n 1 + β n 2 = α n 1 x n + β n 1 ) q 2 = ( αn x n + β n α n 1 x n + β n 1 ) q 2 = 0 = F n (x n 1), where in the last step we used F n (x n 1 ) = α n 1x n 1 + β n 1 α n x n 1 + β n = 0. Case 2: Suppose that F n 1 (x n ) supp(τ (n 1) 1 ), say x n = x 1 r, i.e. τ (n 1) 1 = (F n 1 (x n 1 )... F n 1 (x 1 r 1) F n 1 (x 1 n)... F n 1 (x 1 l 1 )). In this case we need to show the following: (i) If x / O n 1 then P n (x) = F n (x). (ii) If y O n 1 P n (y) = F n (y ). satisfies y x 1 l 1, x 1 r 1, and P n 1 (y) = F n 1 (y ), then (iii) P n (x 1 l 1 ) = F n (x n ), and P n (x 1 r 1) = F n (x n 1 ). Note that (i), (ii) and the first statement of (iii) can be shown as in Case (1), so we will only show P n (x 1 r 1) = F n (x n 1 ): P n (x 1 r 1) = (P n 1 (x 1 r 1) + a n ) q 2 = (F n 1 (x n ) + a n ) q 2 ( ) q 2 ( αn 2 x n + β n 2 αn x n + β n = + a n = α n 1 x n + β n 1 α n 1 x n + β n 1 = 0 = F n (x n 1 ). ) q 2 Case 3: Lastly, suppose that F n 1 (x n ) is in the support of a cycle in (5.7), other than the first one, say in supp(τ n 1 2 ) and x n = x 2 s. We need to show: (i) If x / O n 1, then P n (x) = F n (x).

47 40 (ii) If y O n 1 satisfies y x 1 l 1, x 2 s 1, and P n 1 (y) = F n 1 (y ), then P n (y) = F n (y ). (iii) P n (x 1 l 1 ) = F n (x n ), and P n (x 2 s 1) = F n (x n 1 ). The proofs are same as that of (i), (ii) in Case 1, and (iii) in Case 2. Corollary Let x n 1, x n F q. Let where 1. τ (n 1) 1,..., τ m (n 1) P n 1 (x) = τ (n 1) 1... τ (n 1) m F n 1 (x) (5.8) are disjoint cycles. 2. If P n 1 (x n 1 ) F n 1 (x n 1 ) we assume without loss of generality that F n 1 (x n 1 ) supp(τ (n 1) 1 ) with x 1 1 = x n 1, and l(τ (n 1) i ) = l i 2, for 1 i m. 3. If P n 1 (x n 1 ) = F n 1 (x n 1 ) we assume without loss of generality that τ (n 1) 1 = (F n 1 (x n 1 )) and l(τ (n 1) i ) = l i 2, for 2 i m. Then P n (x) can be decomposed as follows: (i) If x n / O n 1 then P n (x) = (F n (x n )F n (x n 1 )... F n (x 1 l 1 ))τ (n) 2... τ (n) m F n (x), (5.9) (ii) if F n 1 (x n ) supp(τ (n 1) 1 ), say x n = x 1 r, then P n (x) = (F n (x n )F n (x 1 r+1)... F n (x 1 l 1 )) (F n (x n 1 )F n (x 1 2)... F n (x 1 r 1))τ (n) 2... τ (n) F n (x), (iii) if F n 1 (x n ) supp(τ (n 1) i ), j 1, say in supp(τ (n 1) 2 ) and x n = x 2 s, then (5.10) P n (x) = (F n (x 2 s)f n (x 2 s+1)... F n (x 2 l 2 )F n (x 2 1)... F n (x 2 s 1)F n (x 1 1)... F n (x 1 l 1 )) 3... τ (n) F n (x). τ (n) m m

48 41 Theorem Suppose P n 1 can be decomposed as Let x n =. Then P n 1 (x) = τ (n 1) 1... τ (n 1) F n 1 (x). P n (x) = τ (n) 1... τ (n) F n (x) P n+1 (x) = τ (n+1) 1... τ (n+1) F n+1 (x), and m m P n+2 (x) = (F n+2 (x n+2 )F n+2 (x n+1 ))τ (n+2) 1... τ (n+2) F n+2 (x), m m where τ (k) i denotes the cycle (F k (x i 1)... F k (x i l i )) as before. Proof. To prove the first two equalities it is sufficient to show the following properties: (i) If x / O n 1, then P n (x) = F n (x) andp n+1 (x) = F n+1 (x). (ii) If y O n 1 satisfies P n 1 (y) = F n 1 (y ), then P n (y) = F n (y ) and P n+1 (y) = F n+1 (y ). (i) Suppose that x / O n 1. Then P n 1 (x) = F n 1 (x), and ( P n (x) = (P n 1 (x) + a n ) q 2 αn 2 x + β n 2 = + a n α n 1 x + β n 1 ( ) q 2 (an α n 1 + α n 2 )x + a n β n 1 + β n 2 = α n 1 x + β n 1 ) q 2 = ( αn x + β n ) q 2 = α n 1x + β n 1 α n 1 x + β n 1 α n x + β n = F n (x). For x / O n 1 and x x n+1 the equality P n+1 (x) = F n+1 (x) also follows as above. Finally ( P n+1 (x n+1 ) = (P n (x n+1 ) + a n ) q 2 αn 1 x n+1 + β n 1 = + a n α n x n+1 + β n ( ) q 2 (an α n + α n 1 )x n+1 + a n β n + β n 1 = α n x n+1 + β n ) q 2 = ( ) q 2 αn+1 x n+1 + β n+1 = 0 = F n+1(x n+1), α n x n+1 + β n