Evalutaion of certain exponential sums of quadratic functions over a finite fields of odd characteristic

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1 University of South Florida Scholar Commons Graduate Theses and Dissertations Graduate School 2006 Evalutaion of certain exonential sums of quadratic functions over a finite fields of odd characteristic Sandra D. Draer University of South Florida Follow this and additional works at: htt://scholarcommons.usf.edu/etd Part of the American Studies Commons Scholar Commons Citation Draer, Sandra D., "Evalutaion of certain exonential sums of quadratic functions over a finite fields of odd characteristic" (2006. Graduate Theses and Dissertations. htt://scholarcommons.usf.edu/etd/2508 This Thesis is brought to you for free and oen access by the Graduate School at Scholar Commons. It has been acceted for inclusion in Graduate Theses and Dissertations by an authorized administrator of Scholar Commons. For more information, lease contact scholarcommons@usf.edu.

2 Evaluation of Certain Exonential Sums of Quadratic Functions over a Finite Field of Odd Characteristic by Sandra D. Draer A thesis submitted in artial fulfillment of the requirements for the degree of Master of Arts Deartment of Mathematics College of Arts and Sciences University of South Florida Major Professor: Xiang Dong Hou, Ph.D. Brian Curtin, Ph.D. Stehen Suen, Ph.D. Date of Aroval: June 22, 2006 Keywords: Artin-Schreier Theorem, Gauss sum, law of quadratic recirocity, Legendre symbol, quadratic form c Coyright 2006, Sandra D. Draer

3 Contents List of Tables ii Abstract iii. Introduction 2. Background from Number Theory 3 3. Quadratic Forms on F n, the Multivariable Aroach 5 4. Quadratic Forms on F n, the Single Variable Aroach 0 5. Comutation of the Nullity 6. S(f + bx Follows from S(f 4 7. From S(f, n to S(f, q s n, q, From S(f, n to S(f, 2 s n, s > From S(f, n to S(f, n When ν 2 (α = ν 2 (α 2 = = ν 2 (α k 3. The Formula for S(ax +α Tables of Numerical Results 37 References 45 i

4 List of Tables Values of l m (f with n = 3, α 4 38 Continued 39 Continued 40 2 Values of l m (f with n = 5, α Continued 42 2 Continued 43 2 Continued 44 ii

5 Evaluation of Certain Exonential Sums of Quadratic Functions Over a Finite Field of Odd Characteristic Sandra Draer ABSTRACT Let be an odd rime, and define f(x as follows: k f(x = a i x α i + F n[x] where 0 α < α 2 < < α k = α. We consider the exonential sum S(f, n = ζ Trn(f(x, i= x F n where ζ = e 2πi/ and Tr n is the trace from F n to F. We rovide necessary background from number theory and review the basic facts about quadratic forms over a finite field F through both the multivariable and single variable aroach. Our main objective is to comute S(f, n exlicitly. The sum S(f, n is determined by two quantities: the nullity and the tye of the quadratic form Tr n (f(x, denoted by l n (f and t n (f, resectively. We give an effective algorithm for the comutation of l n (f. Tables of numerical values of l n (f are included. However, t n (f is more subtle and more difficult to determine. Most of our investigation concerns t n (f. We obtain relative formulas for S(f, mn in terms of S(f, n when ν (m min{ν (α i : i k}, where ν is the -adic order. The formulas are obtained in three searate cases, using different methods: (i m = q s, where q is a rime different from 2 and ; (ii m = 2 s ; and (iii m =. In case (i, we use a congruence relation resulting from a suitable action by the Galois grou Gal(F q s n/f n. For case (ii, in addition to the congruence in case (i, a secial artition of F 2n is needed. In case (iii, the congruence method does not work. However, the Artin-Schreier Theorem for the extension F n/f n allows us to comute the trace of F n/f n rather exlicitly. When ν 2 (α = ν 2 (α 2 = = ν 2 (α k < ν 2 (n, we are able to determine S(f, n exlicitly. As a secial case, we have exlicit formulas for S(ax +α. Most of the results of the thesis are new and generalize revious results by Carlitz, Baumert, McEliece, and Hou. iii

6 . Introduction Let be a rime and n a ositive integer. Let F n denote the finite field with n elements. When m n, the trace from F n to F m is written as Tr n/m. We denote the Tr n/ as Tr n. Let e n (y = e 2πiTrn(y/ for y F n. In 979, Carlitz wrote a aer [3] evaluating the sum x F q e n (ax 3 + bx, a, b F q, where q = 2 n. In 980, Carlitz wrote another aer [4] evaluating the sum x F q e n (ax + + bx, a, b F q, where q = n again, but is an odd rime. A similar sum e n (ax α +, a F n, α 0, x F n was also imlied by the results of Baumert and McEliece [2]. Also see [7]. In 2005, Hou wrote a aer [8] for the = 2 case, generalizing the result by Carlitz. More recisely, in [8], the olynomial ax 3 + bx in [3] was relaced by a olynomial of the form f(x = k i= a ix 2αi+ + bx F 2 n[x]. Possible extension of the results of [8] for the case of = 2 to the case of odd resented the oortunity for this thesis. This toic, evaluation of exonential sums, has alications in many areas within and outside mathematics. In number theory, exonential sums are a owerful tool to study the number of solutions of olynomial equations over finite fields, see [3, Chater 6]. Exonential sums are also widely used in coding theory, information theory, crytograhy and combinatorics. In fact, the sum x F e n n(ax α + arose in the study of weights of irreducible cyclic codes (see Baumert and McEliece [2] and the study of the cross-correlation function between two maximal linear sequences (see Helleseth [7]. In this thesis, we always let be a odd rime. We define f(x as follows: f(x = k a i x αi+ F n[x] where 0 α < α 2 < < α k = α are integers. We consider the sum S(f, n = e n (f(x i= x F n as Hou did in his aer [8], where he handles the case of = 2. The function Tr n (f(x is a quadratic form in the F -coordinates of x F n when F n is identified with F n as a F -vector sace. Thus, S(f, n is an exonential sum of a quadratic form over F. The exonential sum of a quadratic form over F is comletely determined by the nullity (n rank and the tye (discriminant of the quadratic form. In Section 5, we will see that there is an effective method for comuting the nullity. However, there is no direct way to identify the tye of Tr n (f(x from f(x. Most of our investigation is about the determination of the tye of Tr n (f(x.

7 Here is a briefly outline of the thesis. Section 2 contains some necessary theorems and information from number theory. Sections 3 and 4 discuss the quadratic forms on F n using both the multivariable and single-variable aroach. We may include a linear term to f(x; the resulting sum will be S(f + bx, n, b F n. In fact, S(f + bx, n, as a function of b, is the Fourier transform of Tr n (f(x. In Section 5, we show that S(f + bx, n follows from S(f, n in a straightforward way. In Section 6, the method for the comutation of the nullity is shown. In Sections 7 through 9, we derive relative formulas for S(f, mn in terms of S(f, n. Here, S(f, mn is the exonential sum of the same olynomial f F n[x] over an extension field F mn. Three cases require different methods. In Section 7, we assume m = q s where q is a rime different from and 2. We use a congruence relation resulting from a suitable action by the Galois grou Gal(F q s n/f n to determine the tye of Tr q s n(f. In Section 8, We assume m = 2 s. In addition to the congruence in Section 7, a secial artition of F 2n is needed. In Section 9, we assume m =. In this case, the congruence method in the revious two sections does not work. However, the Artin-Schreier Theorem for the extension F n/f n allows us to comute the trace of F n/f n rather exlicitly. We are able to exress S(f, n in terms of S(f, n under the condition that min{ν (α i : i k}, where ν is the -adic order. In Section 0, we look at the case where ν 2 (α = ν 2 (α 2 = = ν 2 (α k < ν 2 (n. An exlicit formula for S(f, n is obtained. Section handles the secial case of S(ax +α which can be evaluated as a result of the revious section. Tables of numerical values of certain nullities are included in Section 2. 2

8 2. Background from Number Theory In this section, we collect some well-known facts from number theory to be used later. Let be an odd rime and a Z. The Legendre symbol ( a is defined as ( a 0 if a, = if a is a square in F, if a is a nonresidue in F. Theorem 2.. (i We have ( = ( 2 (, ( 2 = ( 8 (2. (ii (The law of quadratic recirocity Let q be an odd rime with q. Then ( ( q = ( q 4 ( (q. For the roof of Theorem 2., see [, 5.2] Let q = n, where is an odd rime and n Z +. Let e n : F q C be the canonical additive character, i.e., e n (y = ζ Tr Fq/F (y, y F q, where ζ = e 2πi/. Let η : F q C be the quadratic character of F q, i.e., 0 if y = 0, η(y = if y is a square in F q, if y is a nonsquare in F q. When q =, η(y is the Legendre symbol ( y. The Gauss quadratic sum on F q, denoted by G(η, is defined as G(η = y F q η(ye(y. Lemma 2.2. (i We have G(η = y F q e(y 2. (ii For each a F q, Proof. (i: We have (2. G(η = η(0e n (0 + y F q e(ay 2 = η(a e(y 2. y F q y F q η(ye n (y = 3 y (F q 2 e n (y y F q \(F q 2 e n (y.

9 On the other hand, (2.2 = y F q Adding (2. and (2.2, we have e n (y = G(η = + 2 y (F q 2 e n (y + y (F q 2 e n (y = y F q \(F q 2 e n (y. y F q e n (y 2. (ii: If a is a square in F q, say, a = b 2, b F q, then e n (ay 2 ( = e n (by 2 = e n (y 2. y F q y F q y F q If a is a non square in F q, then e n (y 2 + e n (ay 2 = 2 e n (y = 0. y F q y F q y F q So, e n (y 2 = e n (ay 2 = η(a e n (ay 2. y F q y F q y F q The Gauss quadratic sum is comletely determined. quadratic sum over F, i.e., g = ζ y2. y F It is easy to show that g = { ± 2 if (mod 4, ± 2 i if (mod 4. We use g to denote the Gauss It took Gauss four years ( to determine that signs in both cases above are ositive, i.e., { } 2 if (mod 4, g = = i 4 ( i if (mod 4 The Gauss quadratic sum G(η over F q follows from the Davenort-Hasse Theorem on the Gauss sum of a lifted character [5]. We have where q = n. G(η = ( n g n, 4

10 3. Quadratic Forms on F n, the Multivariable Aroach A quadratic form on F n q is a function F : F n q F q defined by a homogeneous olynomial in (x,..., x n. When q is odd, a quadratic form on F n q can be written as F (x,..., x n = (x,..., x n A(x,..., x n T, where A is an n n symmetric matrix over F q. The matrix A is called the matrix of F. Two quadratic forms F (x = xax T and G(x = xbx T, where x = (x,..., x n, are called equivalent if there exists P GL(n, F q such that F (x = G(xP, or equivalently, A = P BP T. So, the classification of quadratic forms on F n q (q odd under equivalence is the same as the classification n n symmetric matrices over F q under congruence. Congruence of symmetric matrices is denoted by =. The classification of quadratic form over finite fields is well-known. When q is odd, the classification is simle and is given in the next theorem. When q is even, the classification is slightly involved, see [, 6]. For the rest of this section q is a ower of an odd rime. Theorem 3.. Let x = (x,..., x n and A be an n n symmetric matrix over F n q. Then every quadratic form F (x = xax T on F n is equivalent to x x 2 r + dx 2 r, where 0 r n and d F q. The integer r is called the rank of F and is denoted by rank F. The image of d in F q/(f q 2, i.e. d(f q 2, is called the discriminant of F. (Sometimes we simly say that the discriminant of F is d. If r = 0, the discriminant of F is defined to be. Two quadratic forms are equivalent if and only if they have the same rank and discriminant. The roof of Theorem 3. will follow two lemmas. Lemma 3.2. Every element of F q is a sum of two squares. Proof. The multilicative grou F q contains 2 (q squares and 2 (q nonsquares. So, F q has 2 (q + squares (including 0. We claim that the set of squares in F q is not closed under addition. Otherwise, the set of squares of F q would be a subgrou of (F q, +, which is imossible since 2 (q + does not divide q. Thus, there exists a nonsquare d F q such that d = a 2 + b 2 for some a, b F q. Now let x F q be arbitrary. If x is a square, say, x = y 2, then x = y If x is a nonsquare, since F q/(f q 2 = 2, we must have x = dy 2 for some y F q. Then x = (a 2 + b 2 y 2 = (ay 2 + (by 2. 5

11 Lemma 3.3. Every n n symmetric matrix of rank r over F q is congruent to... d Proof. Let a a 2 a n a 2 a 22 a 2n A =... a n a 2n a nn be an n n symmetric matrix over F q. We may assume A 0. By suitable ermutations of rows and columns of A, we may assume that the first row of A is not all 0. First, we show that A is congruent to a diagonal matrix. There are two cases for A, a 0 or a = 0. Case. Assume a 0. Let a 2 a 0 P =.... GL(n, F q. a n a 0 Then a 0 0 P AP T 0 =. A, 0 where A is an (n (n symmetric matrix over F q. Using induction on n, we may assume that A is congruent to a diagonal matrix. So, A is congruent to a diagonal matrix. Case 2. Assume a = 0. Then one of a 2,..., a n is nonzero, say, a 2 0. Let P =... GL(n, F q. Then 2a 2 P AP T =..., where 2a 2 0. Therefore, we are in case. So we have roved that A is congruent to a diagonal matrix. We can use row and column ermutations to move all non-zero diagonal entries of A to the first entries in the diagonal. 6

12 Then A has the form: a... a rr 0, a,..., a rr F q There are two cases for each a ii : it is either a square or a nonsquare in F q. Without loss of generality, we may assume that a,..., a ss are squares and a s+,s+,..., a rr are nonsquares. So, { b 2 i if i s, a ii = d b 2 i if s + i r, where b,..., b r F q and d F q is a nonsquare. Let b... b P = r.... Then (3. P AP T =... d... d By Lemma 3.2, d = a 2 + b 2, a, b F q. So, [ [ [ ] [ ] d a b a b = = d] b a] b a Therefore, the diagonal matrix in (3. is congruent to or [ ]. d deending on whether r s is even or odd. 7

13 Proof of Theorem 3.. Let F (x,..., x n = (x,..., x n A(x,..., x n T be a quadratic form in x,..., x n over F q, where A is an n n symmetric matrix over F q. By Lemma 3.2, there exists Q GL(n, F q such that QAQ T =... d r, d F q. Then F (x,..., x n = F ((x,..., x n Q = (x,..., x n QAQ T (x,..., x n T = x 2 + x x 2 r + dx 2 r. Let G(x,..., x n be another quadratic form over F q with rank r and discriminant d. If r = r and η(d = η(d, of course, G = F. On the other hand, if G = F, by the next theorem, we have η(d g r n r = = (x,...,x n F n (x,...,x n F n = η(d g r n r. ζ F (x,...,x n ζ G(x,...,x n It follows that r = r and η(d = η(d. Let F be a quadratic form on F n q tye of F. So, with discriminant d. Then η(d {±} is called the tye of F = { if the discriminant of F is a square, if the discriminant of F is a nonsquare. Theorem 3.4. Let F be a quadratic form on F n with tye t and rank r. Then (x,...,x n F n ζ F (x,...,x n = t g r n r. 8

14 Proof. We may assume F (x,..., x n = x x2 r + dx2 r, 0 r n, d F q. Then (x,...,x n F n = ( x F n = ( ζ x2 ζ x2 x F = ( ζ x2 x F ζ F (x,...,x n ( ( ζ x2 x F r ( x r F n ζ dx2 x F ( ζ x2 r ( ζ x2 x F n r x r F n ( ( ζ dx2 r ( x r+ F n x n F n n r ζ dx2 x F =g r η(dg n r (by Lemma 2.2 (ii. 9

15 4. Quadratic Forms on F n, the Single Variable Aroach In Section 3, quadratic forms on F n q are reresented by homogeneous olynomials of degree 2 in x,..., x n ; this way of reresenting quadratic forms is usually referred to as the multi-variable aroach. However, since F n q is identified with the finite field F q n, there is another way of reresenting quadratic forms of F n q which is called the single-variable aroach. Let a F q n and α 0. Let ɛ,..., ɛ n be a basis of F q n over F q and let x = x ɛ + +x n ɛ n, (x,..., x n F n q. Let Tr = Tr Fq n/f q. Then [ ( n q Tr(ax qα + α +] = Tr a x i ɛ i [ = Tr a [ = Tr a = i,j i= ( n i= n i= x i ɛ qα i ( n ] x j ɛ j j= ɛ qα i ɛ j x i x j ] Tr(aɛ qα i ɛ j x i x j is a quadratic form in x,..., x n over F q. Therefore, if a,..., a k F q and α,..., α k 0, then ( k (4. Tr a i x qα i+ i= is a quadratic form in x,..., x n over F q. On the other hand, every quadratic form on F n q (identified with F q n can be written in the form of (4.. In fact, if n is odd, a basis of the F q -sace of quadratic forms on F n q is given by Tr(ax qi +, 0 i n, a A, 2 where A is a basis of F q n over F q. If n is even, a basis of the F q -sace of quadratic forms on F n q is given by Tr(ax qi +, 0 i n 2, a A and Tr(ax qn/2 +, b B, where A is a basis of F q n over F q and B is a basis of F q n/2 over F q. See [9] for details. 0

16 5. Comutation of the Nullity Let F (x,..., x n = (x,..., x n A(x,..., x n T be a quadratic form on F n where A is an n n symmetric matrix over F. For z, x F n, we have F (x + z F (x = xaz T + zax T + zaz T = 2zAx T + zaz T. Thus, za = 0 if and only if F (x + z F (x is constant for all x F n. So n rank F = dim F {z F n : F (x + z F (x = constant for all x F n }. The number n rank F is called the nullity of F. Consider (5. f(x = k a i x αi+ + bx F n[x], i= where 0 α α 2 α k = α. In Section 4, we saw that every quadratic form on F n (identified with F n can be written as Tr n (f(x for some f F n[x] of the form (5.. Denote the nullity of Tr n (f(x by l n (f, i.e. l n (f = dim F L n (f, where L n (f = {z F n : Tr n (f(x + z f(x = constant for all x F n} which is called the null sace of f. Let t n (f denote the tye of Tr n (f(x. Then by Theorem 3.4, we have S(f, n = t n (fg n ln(f ln(f. Thus, S(f, n is comletely determined by l n (f and t n (f.

17 There is an effective algorithm to comute l n (f. We have (5.2 where α i Tr n (f(x + z f(x ( k = Tr n a i (x + z α i + i= k a i x + α i i= ( k = Tr n (a i (x + z α i + a i x α i + i= ( k = Tr n (a i (x α i + + x αi z + xz αi + z α i + a i x α i + i= = Tr n ( k i= ( k = Tr n a i z α i + + i= a i (x αi z + xz αi + z α i + k i= a i (x αi z + xz αi ( k = Tr n f(z + Tr n (a i x αi z + a i xz αi i= ( k = Tr n f(z + Tr n (a i x αi z αi + a i xz αi i= = Tr n f(z + Tr n ( k = Tr n f(z + Tr n (x i= i xz αi + a i xz αi (a α i k i= = Tr n f(z + Tr n ( x α (a α i i z αi + a i z αi k i= (a α α i i = Tr n f(z + Tr n (x α f (z, z α αi + a α i z α+αi is a ositive integer and α i α i (mod n, and k f (z = (a α α i i z α αi + a α i z α+αi. i= Note that f F n[x] is a -olynomial with no reeated roots. Thus, Tr n (f(x + z f(x is constant for all x F n if and only if z is a root of f. Therefore, and (5.3 L n (f = {z F n : f (z = 0} l n (f = dim F {z F n : f (z = 0} = log {z F n : f (z = 0} = log deg(f, x n x. Since α k = α, then the degree of f is always 2α. Let (5.4 s = min {m : n m, l m (f = 2α}. 2

18 Thus, F s is the slitting field of f over F n. It is then obvious that for all multiles m of n, (5.5 l m (f = l (m,s (f. Therefore, we can comute the nullity l m (f for all 0 < m 0 (mod n using the following method. First, use (5.3 to comute l in (f for i =, 2,... until l s (f = 2α. Then F s is the slitting field of f and l m (f is given by (5.5. Examle 5.. Let Then we find that f(x = 2x x x 33 + F 3 [x]. f (x = x x 3 + 2x x x 35 + x 36. We can then use Mathematica [5] to find that { deg(f 3 6 if m = 26,, x 3m x = if m < 26. Thus, the slitting field of F 3 is F So Examle 5.2. Let Then we find that l m (f = { 6 if 26 m, 0 otherwise. f(x = 3x x x 73 + F 7 [x]. f (x = x x 7 + 3x x x 75 + x 76. We can then use Mathematica [5] to find that 7 6 if m = 75, deg(f 7 5 if m < 75, 25 m,, x 7m x = 7 2 if m < 75, 7 m, 7 if m < 75, 25 m, 7 m. The slitting field of f over F 7 is F Write m = 5 a 7 b m, where a, b 0 and (5 7, m =. Then for every m > 0, 6 if a 2 and b, 5 if a 2 and b = 0, l m (f = 2 if a and b, if a and b = 0. In fact, we have comuted the values of l m (f with n = 3, α 4, and n = 5, α 3. The results are tabulated in Section 2. 3

19 6. S(f + bx Follows from S(f Let f F n[x] be given in (5.. Our main objective is to comute the sum S(f, n. Let b F n. The sum S(f + bx, n = e n (f(x + bx x F n is also imortant. In fact, when viewed as a function of b, S(f + bx, n is the Fourier transform of Tr n (f. However, we will see that the seemingly more general sum S(f +bx, n follows easily from S(f, n. Hence, it suffices to determine S(f, n. Theorem 6.. Let b F n. Then S(f, ns(f + bx, n = Proof. We have x F n { n+ln(f e n (f(x 0 if f (x = b α has a solution x 0 F n, 0 otherwise. S(f, ns(f + bx, n = e n (f(x e(f(y + by = x,y F n = x,y F n = x,y F n = x F n = n y F n e n (f(x f(y by e n (f(x + y f(y by e n (f(x + yf (x α by (by (5.2 e n (f(x x F n f (x=b α y F n e n (f(x. e n [y(f (x α b] If f (x = b α has no solution in F n, then S(f, ns(f + bx, n = 0. Thus, assume f (x = b α has a solution, x 0 F n. Then the solution set of f (x = b α is x 0 + {x F n : f (x = 0} since f (x is a -olynomial. The solution set can also be written as x 0 + L n (f where L n (f is the null sace of f. Since Tr n f(x is a quadratic form on F n, it is constant on 4

20 x 0 + L n (f. So Corollary 6.2. S(f + bx, n = S(f, ns(f + bx, n = n x F n f (x=b α = n x x 0 +L n(f e n (f(x = n ln(f e n (f(x 0 e n (f(x = n+ln(f e n (f(x 0. { e(f(x 0 S(f, n if f (x = b α has a solution x 0 F n, 0 otherwise. Proof. Since S(f, n 0, then by Theorem 6., we know that the only case we have to consider is when f (x = b α has a solution x 0 F n. Then by Theorem 6., S(f + bx, n = n+ln(f S(f, n e(f(x 0 = n+ln(f S(f, n 2 e(f(x 0S(f, n. Since S(f, n = t n (fg n ln(f ln(f, we have S(f, n 2 = n+ln(f. This yields the desired result. 5

21 7. From S(f, n to S(f, q s n, q, 2 In this section, assume that q is an odd rime such that q and s 0. Theorem 7.. In the ring Z[ζ ], we have S(f, q s n Equivalently, t q s n(fg qs n l q s n (f x F n l q sn(f ( q e q s n(f(x (mod q. s(n ln(f tn (fg n ln(f ln(f (mod q. Proof. Let T (i = {x F q s n \ F n : Tr q s n(f(x = i}, i F. We can then write S(f, q s n = e q s n(f(x + e q s n(f(x x F q s n\f n i=0 x F n x F n = T (i ζ i + e q s n(f(x. The Galois grou Aut(F q s n/f n acts on T (i. Take x T (i and take σ Aut(F q s n/f n. Since f F n[x], we have σ(f(x = f(σ(x for all x F q s n. Thus, Tr q s n( f(σ(x = Trq s n( σ(f(x = Trq s n(f(x = i. Secondly, note that since x / F n, then σ(x / F n. (Otherwise, we would have x = σ σ(x F n. Therefore, σ(x T (i. T (i is a union of Aut(F q s n/f n-orbits of cardinality of q to ositive owers. Recall that the cardinality of the orbit is the size of the grou divided by the size of the stabilizer. Thus, since Aut(F q s n/f n = q s, then the cardinality of every orbit must be q to some ositive ower. If the cardinality of the orbit is, then the element of that orbit is fixed by every automorhism, and that orbit is contained in the base field F n. But, the orbit is in T (i, which is a contradiction as the orbit cannot be in F n. Thus, the cardinality of every orbit is equal to q h, where 0 < h s. Thus, T (i is a union of Aut(F q s n/f n-orbits of size q h where 0 < h s. So T (i 0 (mod q for all 0 i. Therefore, e q s n(f(x 0 (mod q which gives us and x F q s n\f n S(f, q s n x F n e q s n(f(x (mod q. For the equivalence in the second art of the theorem, since x F n e q s n(f(x = S(f, q s n = t q s n(fg qs n l q s n (f l q s n(f x F n ζ qs Tr n(f(x 6 = ( q s(n ln(f tn (fg n ln(f ln(f,

22 then t q s n(fg qs n l q s n (f Lemma 7.2. We have and. Proof. Let X = {x F qs n X and X is a union of Aut(F qs n l q sn(f ( q s(n ln(f tn (fg n ln(f ln(f (mod q. l q s n(f l n(f (mod q 2 [l q s n(f l n (f] Z \ F n : f (x = 0. Then the Galois grou Aut(F qs n /F n-orbits of cardinality >. Then, l q s n(f ln(f = X 0 (mod q ln(f ( l q s n(f l n(f 0 (mod q /F n acts on since is a rime different from q and l n (f is an integer, then l q s n(f l n(f 0 (mod q. Therefore, l q s n(f l n(f (mod q. To rove the second art of the lemma, begin with the equation from Theorem 7., (7. t q s n(fg qs n l q s n (f l q sn(f ( q s(n ln(f tn (fg n ln(f ln(f (mod q. Using the above equation and l q s n(f l n(f (mod q, we can simlify (7. to be (7.2 t q s n(fg qs n l q s n (f ( q s(n ln(f tn (fg n ln(f (mod q. Assume to the contrary of 2 [l q s n(f l n (f] / Z, i.e. l q s n(f l n (f is odd. Then exactly one of q s n l q s n(f or n l n (f is odd. This is true as q s n and n have the same arity, but l q s n(f and l n (f have different arities. Also, since g = 2 or i 2, then g 2 = ±. Thus, since q s n l q s n(f or n l n (f is odd, then (7.2 has g to an even ower on one side and g to an odd ower on the other side. This gives us to some ower on one side and to some ower multilied by g on the other side, resectively. Thus, g a b (mod q for some a, b Z, a 0. Let σ Aut(Q(g /Q such that σ(g = g. Since Q(g (ζ and Q(ζ /Q is Galois, we can extend σ to τ Aut(Q(ζ /Q. By alying τ to g a b (mod q, we have g a b (mod q. By subtracting the two equations, the result is 2g a 0 (mod q, which is a contradiction. Thus, 2 [l q s n(f l n (f] Z. Let o q ( denote the multilicative order of in Z/qZ. Then, using the above equation l q s n(f l n(f (mod q, we find that o q( (l q s n(f l n (f Z. Clearly 2 (l q s n(f l n(f ( oq( (l q s n(f l n(f (mod q. Theorem 7.3. We have (7.3 t q s n(f = t n (f ( q sln(f ( 4 ( [l q s n(f l n(f]+ oq( [l q s n(f l n(f]. 7

23 Proof. First, recall that 2 (q ( q (mod q and note that qs s(q (mod 4. We can start with (7.2 and simlify it to be t q s n(fg qs n l q s n (f t q s n(fg qs n n t q s n(fg n(qs ( q ( q ( q s(n ln(f tn (fg n ln(f (mod q s(n ln(f tn (fg l q s n(f l n(f (mod q s(n ln(f tn (fg l q s n(f l n(f Recall that the Gauss sum is g = i 4 ( 2 2. Then, (mod q. Using this fact, then g 2 = ( i 4 ( = i 2 ( 2 = ( 2. Also, simlify g n(qs = [( 2 ] 2 n(qs = ( 4 n( (qs 2 n(qs = ( 4 n( s(q 2 ns(q = ( 4 ns( (q 2 (q sn = ( ns( (q ( sn 4 (mod q. q g l q s n(f l n(f = [( 2 ] 2 [l q s n(f l n(f] = ( 4 ( [l q s n(f l n(f] 2 [l q s n(f l n(f] = ( 4 ( [l q s n(f l n(f] ( oq( (l q s n(f l n(f = ( 4 ( [l q s n(f l n(f]+ oq( (l q s n(f l n(f (mod q (mod q Then, the above values can be substituted into as t q s n(fg n(qs ( q s(n ln(f tn (fg l q s n(f l n(f (mod q sn t q s n(f( ns( (q ( 4 q ( q s(n ln(f tn (f( 4 ( [l q s n(f l n(f]+ oq( [l q s n(f l n(f] 8 (mod q

24 t q s n(f t n (f ( q s(n ln(f( sn q ( 4 ( [l q s n(f l n(f]+ oq( [l q s n(f l n(f] 4 ns( (q (mod q = t n (f ( q s(n ln(f( sn( 4 ( [l q s n(f l n(f] 4 ns( (q + oq( [l q s n(f l n(f] = t n (f ( q = t n (f ( q = t n (f ( q q s(n ln(f( sln(f(q sln(f ( (q q sn ( q ( q sn( ([l 4 ( q s n (f l n(f] sn(q + oq( [l q s n(f l n(f] sn( ([l 4 ( q s n (f l n(f] sn(q + oq( [l q s n(f l n(f] sn ( 4 ( ([l q s n (f l n(f] sn(q + oq( [l q s n(f l n(f]. In the above equation, both sides are ±. Thus, the two sides are equal. We can use the law of quadratic recirocity which states ((q = ( 4 ( (q. q This allows us to further simlify our equation. t q s n(f = t n (f ( q ( sln(f (q ( q = t n (f ( q = t n (f ( q = t n (f ( q = t n (f ( q Examle 7.4. Let sn ([l ( 4 ( q s n (f l n(f] sn(q + oq( [l q s n(f l n(f] sln(f ( ( 4 ( (q sn ( 4 ( ([l q s n (f l n(f] sn(q + oq( [l q s n(f l n(f] sln(f ([l ( 4 sn( (q + 4 ( q s n (f l n(f] sn(q + oq( [l q s n(f l n(f] sln(f (sn(q +[l ( 4 ( q s n (f l n(f] sn(q + oq( [l q s n(f l n(f] sln(f ( 4 ( [l q s n(f l n(f]+ oq( [l q s n(f l n(f]. f(x = 2x x x 33 + F 3 [x] as in Examle 5.. Then we found the nullity of f(x for every m to be { 6 if 26 m, l m (f = 0 otherwise. Since S(f, = x F 3 e (f(x = x F 3 e ( x 2 = η( x F 3 e (x 2 = g 3, we have t (f =. Now, let m be odd and not divisible by 3. Since m is not divisible by 26, then l m (f = 0 and it follows from (7.3 that t m (f =. 9

25 Examle 7.5. Let f(x = 3x x x 73 + F 7 [x] as in Examle 5.2. Then we found the nullity of f(x for every m = 5 a 7 b m, (m, 5 7 =, to be 6 if a 2 and b, 5 if a 2 and b = 0, l m (f = 2 if 0 a and b, if 0 a and b = 0. Since S(f, = e (f(x = = 7, x F 7 x F 7 we have t (f =. Now, let m be odd and not divisible by 7. Write m = 5 a m where (m, =. Note that o 5 (7 = 4 and ( 5 7 =. It follows from (7.3 that {( a( m t m (f = 7 if a, ( a+( m 7 if a 2. In the next section, we will revisit these examles and determine t m (f for all even m not divisible by 7. 20

26 8. From S(f, n to S(f, 2 s n, s > 0 To find S(f, 2 s n, if we use the congruence we used in the revious section, it would yield an answer for t 2 s n(f with congruence modulo 2. Since and - are congruent (mod 2, that does not hel us to determine the sign. In this case, instead of looking at t 2 s n(f modulo 2, we will look at t 2 s n(f modulo 4. Let T (i = {x F 2 s n \ F 2n : Tr 2 s n(f(x = i}. We can write S(f, 2 s n = e 2 s n(f(x + e 2 s n(f(x x F 2 s n\f 2n x F 2n = T (i ζ i + i=0 x F 2n e 2 s n(f(x. The Galois grou Aut(F 2 s n/f n acts on T (i, and T (i is a union of Aut(F 2 s n/f n- orbits of cardinality of 2 to some ower greater than or equal to 2. Recall that Aut(F 2 s n/f n is cyclic with an order of 2 s. If x F 2 s n \ F 2n, then the stabilizer of x in Aut(F 2 s n/f n does not contain Aut(F 2 s n/f 2n so the stabilizer of x must be roerly contained in Aut(F 2 s n/f 2n, i.e. contained in Aut(F 2 s n/f 4n. (Note that all subgrous of a cyclic grou of order 2 s form a chain. Thus, the Aut(F 2 s n/f n-orbits have cardinality that is divisible by 4. So T (i 0 (mod 4 for all 0 i. Therefore, e 2 s n(f(x 0 (mod 4. x F 2 s n\f 2n This gives us (8. S(f, 2 s n = e 2 s n(f(x (mod 4. x F 2n Consider the elements of F 2n. For every x F 2n, x 2n = if and only if x F n. So, we can artition F 2n into F 2n = F n A B, where A = {x F 2n : x n = } B = {x F 2n : x n ±}. The set B can be artitioned even further into four-element subsets of the form {±x, ±x n }. Since f( x = f(x and f(x n = f(x n, then e 2 s n(f(x is constant on {±x, ±x n }. We can then write (8.2 e 2 s n(f(x 0 (mod 4. x B Now, consider A. Choose β F n such that β is a nonsquare. Also, let x 2 0 = β where x 0 F 2n. We then find that x n 0 =. Also, for every y A, y can be written as 2

27 y = x 0 x where x F n, so A can be rewritten as A = x 0F n. Recall that So, (8.3 f(x = x A e 2 s n(f(x = x A = k a i x αi+. i= e 2 s n x F n e 2 s n = x F n e 2 s n = x F n ( k a i x α i+ i= ( k i= a i x α i+ 0 x α i+ ( k a i β 2 (αi+ x α i+ i= e 2 s n( f(x, where f(x = k a i β 2 (αi+ x αi+ F n[x]. i= Notice that f(x = f(βx when β is fixed. Combine (8. through (8.3. We have S(f, 2 s n e 2 s n(f(x (mod 4 x F 2n ( = x F n e 2 s n(f(x + + x F n x A x B = e 2 s n(f(x + e 2 s n( f(x (mod 4 x F n ( 2 s(n ln(f ( 2 s(n ln( f = S(f, n + S( f, n. Since S(f, n = t n (fg n ln(f ln(f, then S( f, n ln( f n = t n ( fg ln( f and S(f, 2 s n = t 2 s n(fg 2s n l 2 s n (f l 2 sn(f. By substituting these three values into the above equation, we have t 2 s n(fg 2s n l 2 s n (f l 2 s n(f ( 2 s(n ln(f ( 2 s(n ln( tn (fg n ln(f ln(f f n ln( f + tn ( fg ln( f (mod 4. ( Since 2 = ( 8 (2 and ( 2 ( (mod 4, we can further simlify the above to t 2 s n(f( 2 ( l 2 s n(f g 2s n l 2 s n (f ( 8 (2 s(n l n(f+ 2 ( ln(f t n (fg n ln(f + ( 8 (2 s(n l n( f+ 2 ( ln( f n ln( f t n ( fg (mod 4. 22

28 Theorem 8.. Let and f(x = f(x = k a i x α i+ i= k a i β 2 (α i + x α i+ and let s > 0. Then l n (f + l n ( f + l 2 s n(f is even. Moreover, { t n (ft n ( (8.4 t 2 s n(f = f if l n (f l n ( f (mod 2, t n (ft n ( f( 8 (2 s if l n (f l n ( f (mod 2. i= Proof. We begin the roof by showing that l n (f + l n ( f + l 2 s n(f is even. Suose to the contrary that l n (f + l n ( f + l 2 s n(f is odd. Then, either one or all three terms in l n (f + l n ( f + l 2 s n(f must be odd. This means that exactly one or three of n l n (f, n l n ( f, or 2 s n l 2 s n(f must be odd. Recall that g 2 = i 2 ( 2 and that ( 2 ( (mod 4, so g 2 i 2 ( 2 ( 2 ( (mod 4 = i 2 ( 2 i ( = i 2 ( 2 +( = i 2 (2 =. We use the above facts, in the equation (8.5 t 2 s n(f( 2 ( l 2 s n(f g 2s n l 2 s n (f ( 8 (2 s(n l n(f+ 2 ( ln(f t n (fg n ln(f + ( 8 (2 s(n l n( f+ 2 ( ln( f n ln( f t n ( fg (mod 4. The above equation becomes g u (mod 4 for some u Z if only one of the three numbers n l n (f, n l n ( f, or 2 s n l 2 s n(f is odd; the above equation becomes vg (mod 4 for some v Z if all three numbers n l n (f, n l n ( f, and 2 s n l 2 s n(f are odd. Let τ Aut(Q(ζ /Q such that τ(g = g. We can aly τ to both of the above cases. In the first case, we get { g u (mod 4, g u (mod 4. This imlies 2g 0 (mod 4, which is a contradiction. For the second case, we have { vg (mod 4, vg (mod 4. This leads to 2 0 (mod 4 which is a contradiction as well. Therefore, l n (f + l n ( f + l 2 s n(f is even. 23

29 To rove the second art of the theorem, we begin by assuming that l n (f l n ( f (mod 2. Since l n (f and l n ( f have the same arity, then l 2 s n(f is even. Consider t 2 s n(f( 2 ( l 2 s n(f g 2s n l 2 s n (f ( 8 (2 s(n l n(f+ 2 ( ln(f t n (fg n ln(f + ( 8 (2 s(n l n( f+ 2 ( ln( f n ln( f t n ( fg (mod 4. Thus, since g 2 (mod 4, then g 2s n l 2 s n (f (mod 4. We can use this and the assumtion that l n (f l n ( f (mod 2 to continue with the above. t 2 s n(f( 2 ( l 2 s n(f ( 8 (2 s(n l n(f+ 2 ( ln(f t n (fg n ln(f + ( 8 (2 s(n l n(f+ 2 ( ln(f n ln(f t n ( fg (mod 4 n ln(f (t n (f + t n ( fg ( 8 (2 s(n l n(f+ 2 ( ln(f (mod 4 (t n (f + t n ( fδ (mod 4 where δ {±, ±g }. Since 2 (g is integral over Q, then g (mod 2. We can rewrite this to be 2g 2 (mod 4 and generalize it further to 2δ 2 (mod 4. Returning to t 2 s n(f (t n (f + t n ( fδ (mod 4, recall that as t n (f is the tye of f, it is either or. Then t n (f + t n ( f can be either 0, 2, or 2. If t n (f + t n ( f = 0, then t 2 s n(f (mod 4. If t n (f + t n ( f = 2, then t 2 s n(f 2δ (mod 4. If t n (f + t n ( f = 2, then t 2 s n(f 2δ (mod 4. Thus, { if tn (f + t t 2 s n(f = n ( f } = 0 if t n (f + t n ( f = t n (ft n ( = ±2 f. Now assume that l n (f l n ( f (mod 2. Since they have different arities but the sum is even, then l 2 s n(f is odd. Without loss of generality, assume that n l n (f is odd and n l n ( f is even. Then, we can simlify (8.5. t 2 s n(f( 2 ( l 2 s n(f g 2s n l 2 s n (f ( 8 (2 s(n l n(f+ 2 ( ln(f t n (fg n ln(f + ( 8 (2 s(n l n( f+ 2 ( ln( f n ln( f t n ( fg (mod 4 t 2 s n(f( 2 ( g ( 8 (2 s+ 2 ( ln(f t n (fg + ( 2 ( ln( f t n ( f (mod 4 = ( 8 (2 s+ 2 ( (ln( f+ 2 ( t n (fg + ( 2 ( ln( f t n ( f t 2 s n(f( 2 ( g ( 8 (2 s+ 2 ( ln( f+ 2 ( t n (fg ( 2 ( ln( f t n ( f (mod 4 g ( 2 ( [ t 2 s n(f t n (f( 8 (2 s+ 2 ( ln( f ] t n ( f( 2 ( ln( f (mod 4 24

30 So, Thus, result. { t n (f( 8 (2 s+ 2 ( ln( f if t n ( t 2 s n(f = f( 2 ( ln( f =, t n (f( 8 (2 s+ ( ln( f 2 if t n ( f( 2 ( ln( f = = t n (ft n ( f( 8 (2 s. Examle 8.2. Let f(x = 2x x x 33 + F 3 [x] as in Examle 7.4 where t (f = and l (f = 0. Choose β = 2 as the nonsquare in F 3. Then f(x = 2x x x Since S( f, = x F 3 e ( f(x = x F 3 e ( x 2 = η( x F 3 = g 3 then t ( f =. Also, l ( f = 0. By (8.4, t 2 s(f =, for s > 0. Let m = 2 s 3 b m where (m, =. Note that o 3 (3 = 3. Then, by (7.3, { if b, t m (f = otherwise. Examle 8.3. Let f(x = 3x x x 73 + F 7 [x] as in Examle 7.5 where t (f = and l (f =. Choose β = 3 as the nonsquare in F 7. Then f(x = 4x x x Since then t ( f =. Also, l ( f = 0. By (8.4, S( f, = x F 7 e ( f(x = x F 7 e (4x 2 = g 7 t 2 s(f =, for s > 0. Let m = 2 s 5 a m where s > 0 and (m, =. Then, by (7.3, {( a( m t m (f = 7 if a, ( a+( m 7 if a

31 9. From S(f, n to S(f, n Note that both S(f, n and S(f, n contain a ower of. If we used the congruence method from the two revious sections, we would only get 0 0 (mod which is useless. Therefore, to find a relative formula for S(f, n in terms of S(f, n, we have to use a different method. Let b F n such that Tr n (b 0. Using the Artin-Schreier Theorem [0, Ch. 5, Pro. 7.8], [2, Ch. VI, Thm 6.4], we have F m = F n(ɛ where ɛ = ɛ + b, and the roots of the irreducible olynomial x x b are ɛ + j, j F. Then, for every integer t 0, we have According to [3, Lemma 7.3], So we have j F j s = Tr n/n (ɛ t = j F (ɛ + j t = j F t = s=0 s=0 ( t j s ɛ t s s t ( t ɛ t s j s s j F { if s > 0 and s 0 (mod (, 0 otherwise. (9. Tr n/n (ɛ t = ( t ɛ t i(. i( i>0 Lemma 9.. Let u, v be integers such that 0 u, v. Then Tr n/n (ɛ u+v = { 0 if u + v, 2(, if u + v =, 2(. If α is a ositive integer, then ( v Tr n/n (ɛ u+vα = (b b α u+v ( u + v ( { 0 if u + v 2( + if u + v = 2(. 26

32 Proof. Write u + v = u + v, where 0 u, v. We now have Tr n/n (ɛ u+v = ( u + v ɛ u+v i( i( i>0 ( ( u + v u + v = ɛ u+v ( 2( ( u + v ( u = ɛ u+v ( + v 2( ( u + v ( u = ɛ u+v ( + v ( 2 + ( ( ( u u = ɛ u+v ( v 2 = ɛ u+v 2( ɛ u+v 2( ɛ u+v 2( ɛ u+v 2( { if (u, v = (, 0 or ( 2,, i.e. u + v = or 2( 0 otherwise. Using (9. again as we rove the second art of the theorem, Tr n/n (ɛ u+vα = ( u + v α ɛ u+vα i(. i( i>0 Write i( q = s + s + + s α α + t α where 0 s, s,..., s α,. By the Lucas Theorem [4], ( {( u + v α u v s( t (mod if s = = s α = 0, i( 0 (mod otherwise. So, Tr n/n (ɛ u+vα = = = 0 s,t 0<s+t α 0 (mod ( 0 s u 0 t v 0<s+t 0 (mod ( ( u s 0 s u 0 t v u+v>s+t u+v (mod ( ( u + v α s + t α ( v t ( u s ɛ u+vα (s+t α ɛ u s+(v tα ( v t 27 ɛ s+tα (s u s, t v t.

33 Since ɛ = ɛ + b, then, by induction, we have ɛ α = ɛ + b b α. Thus, Tr n/n (ɛ u+vα = = 0 s u 0 t v u+v>s+t u+v (mod ( 0 s u 0 t v u+v>s+t u+v (mod ( ( u s ( u s ( v t ( v t ɛ s (ɛ + b b α t t τ=0 ( t (ɛ + b b α t τ ɛ s+τ τ In the above sum, s + τ s + t u + v (. Since Tr n/n (ɛ u+vα F n, then we only have to sum the terms where s + τ = 0 which is when s = τ = 0. So, Tr n/n (ɛ u+vα = ( = 0 t v u+v>t u+v (mod ( v u + v ( ( v (ɛ + b b α t t (b b α u+v ( + { 0 if u + v 2(, if u + v = 2(. Thus, result. Theorem 9.2. Let f be as reviously stated. Assume Then, Moreover, ν (n < min{ν (α i : i k}. S(f, n = 2 ( 3(n+ln(f S(f, n 2 S(f, n. l n (f = l n (f, t n (f = t n (f. Proof. Let v (n = v and write n = v n where n. Choose b F v such that Tr v(b 0. Then Tr n (b = Tr v n (b = n Tr v(b 0. Since v (α i > v, we have (9.2 b b α i = Tr α i(b = 0. (Note that if α i = 0 then b b α Tr n/n (ɛ u+vαi = is an emty sum. Recall that { 0 if u + v, 2( if u + v =, 2(. 28

34 for all i k and all 0 u, v. Let x = x 0 ɛ x ɛ F n, where x u F n, 0 u. Then Tr n ( f(x = Tr n ( k i= = Tr n ( k = Tr n [ k = Tr n [ k a i (x 0 ɛ x ɛ +α i a i i= 0 u,v a i i= 0 u,v i= = Tr n [ k i= = Tr n [ k + 2 ( 3 u= i= = Tr n [ k + 2 ( 3 u= a i x u x α i v ɛ u+vα i ] x u x α i v Tr n/n (ɛ u+vαi 0 u,v u+v=, 2( ] x u x α i v ( a i x +α i + x 0 x α i + x x α i 0 + x +α i 2 ( + 2 ( 3 u= ( a i x +α i + x 0 x α i + x x α i 0 + x +α i 0 x +α i 0 + x +α i (x u x α i u + x ux α i u i= ] ( a i (x 0 + x +αi x +α i 0 + x +α i k i= ] a i (x u x α i u + x ux α i u = Tr n [f(x 0 + x f(x 0 + f(x 2 ( + = Tr n [f(x 0 + x f(x 0 + f(x 2 ( + = Tr n [f(x 0 + x f(x 0 + f(x 2 ( + = Tr n [f(x 0 + x f(x 0 + f(x 2 ( + = Tr n [f(x 0 + x f(x 0 + f(x 2 ( ( 2 ( 3 u= 2 ( 3 u= 2 ( 3 u= 2 ( 3 u= 2 ( 3 u= k i= (x u x α i u + x ux α i u 2 ( ] (a i x u x α i u + a ix u x α i u k (a i x u x α i i= k x u (a i x α i i= x u i= k (a i x α i u + i a α i u + i a α i u + i a α i x u f (x u α]. ] x α i u x u x α i u ] x α i u ] ]

35 Then, = S(f, n (x 0,...,x F n e n = S(f, n 2 S(f, n = S(f, n 2 S(f, n [ f(x 0 + x + f(x 0 f(x 2 ( 2 ( 3 n= 2 ( 3 n= ( x u,x u F n ln(f+n = 2 ( 3(n+ln(f S(f, n 2 S(f, n. Since S(f, n = t n g n ln(f ln(f, then 2 ( 3 u= e n ( xu f (x u α t n (fg n ln(f ln(f = 2 ( 3(n+ln(f S(f, n 2 S(f, n Consider the two sides searately. The left-hand side equals x u f (x u α] t n (f(i 4 ( 2 2 n l n(f l n(f = t n (fi 4 ( 2 (n l n(f 2 (n ln(f+ln(f = t n (fi 4 ( 2 (n l n(f 2 (n+ln(f. When we look at the right-hand side, we get This yields 2 ( 3(n+ln(f S(f, n 2 S(f, n = 2 ( 3(n+ln(f t n (fg n ln(f ln(f 2 t n (fg n ln(f ln(f = 2 ( 3(n+ln(f n+ln(f t n (f(i 4 ( 2 2 n l n(f l n(f = t n (f 2 ( 3(n+ln(f n+ln(f i 4 ( 2 (n l n(f 2 (n ln(f+ln(f = t n (f 2 (n+ln(f 3 2 (n+ln(f n+ln(f i 4 ( 2 (n l n(f 2 (n+ln(f = t n (f 2 (n+ln(f i 4 ( 2 (n l n(f. t n (fi 4 ( 2 (n l n(f 2 (n+ln(f = t n (f 2 (n+ln(f i 4 ( 2 (n l n(f. Thus, l n (f = l n (f. We can then simlify further Thus, result. t n (f = t n (fi 4 ( 2 [(n l n(f+(n l n(f] = t n (fi 4 ( 2 [(n l n(f+(n l n(f] = t n (fi 4 ( 2 [(+(n l n(f] = t n (f. 30

36 0. When ν 2 (α = ν 2 (α 2 = = ν 2 (α k Lemma 0.. Let α,..., α k 0 be integers. Then gcd( α +,..., α k + > 2 ν 2 (α = = ν 2 (α k <. When ν 2 (α = = ν 2 (α k <, gcd( α +,..., α k + = gcd(α,...,α k +. Proof. It is sufficient to rove the lemma with k = 2. α ( Since i (α,α 2, i =, 2, are odd, (α,α 2 + α i + for i =, 2. Thus, (α,α 2 + ( α +, α 2 +. Also, since 2 (α +, α 2 + ( 2 2 (2α, 2 (2α 2 = 4 (2(α,α 2 = (α,α (α,α and since ( α + 2, (α,α 2 2 =, we have 2 (α +, α ((α,α 2 + which is ( α +, α 2 + (α,α 2 +. Thus, ( α +, α 2 + = (α,α 2 +. ( Clearly, α i > 0 for every i k. Assume to the contrary that ν(α > ν(α 2. Suose that α = 2 i α and α 2 = 2 j α 2 where i > j and α and α 2 are odd. Then, ( α +, α 2 + ( 2i α α 2 +, 2α 2 ( 2i α α 2 +, 2 i α α 2 = 2 which is a contradiction. Thus, we have roven the lemma. Lemma 0.2. Let α, β 0 be integers. Then { ( α +, β (α,β + if ν 2 (β > ν 2 (α, = 2 if ν 2 (β ν 2 (α. Proof. Since (0, m = m, then if any α or β is 0, the conclusion is obvious. Thus, assume α, β > 0. 3

37 α First assume ν 2 (β > ν 2 (α. Recall that ν 2 is the 2-adic order function. Since (α, β is odd, then (α, β + α α +. Since (α, β is even, then (α, β + β. Thus, (α, β + ( α +, β. Note that 2 (α +, β 2 (2α, β = 2 ((2α,β = 2 (2(α,β = 2 ((α,β ( (α,β +. Since ( 2 (α +, 2 ((α,β = this imlies that 2 (α +, β (α,β +. For each x Z and an odd integer k > 0, we have ν 2 ( + x k = ν 2 ( + x. By this we get ν 2 ( α + = ν 2 ( (α,β + and ν 2 ( β ν 2 ( α +. Then ν 2 ( α +, β = ν 2 ( (α,β +. Then, using 2 (α +, β (α,β +, we get ( α +, β (α,β +. Thus, ( α +, β (α,β +. For the second art, assume that ν 2 (β ν 2 (α. Then, ( α + 2 β 2, β 2 2 (2α, β = 2 ((2α,β = 2 ((α,β = ( α Since ( α + 2, α 2 =, then ( α + 2, β 2 = which gives us This comletes the roof. ( α +, β = 2. 2, β. 2 Theorem 0.3. Let f be as reviously stated. Assume that ν 2 (α = = ν 2 (α k = ν and that ν 2 (n > ν. Then 2 ν+ l n (f and t n (f = ( ( 4 ( 2 2 ν + n ln(f ( ( 4 ( 2 + n ln(f 2 ν+ 2 = ν+ if ν = 0, ( n ln(f 2 ν+ if ν > 0. Proof. Using the two lemmas above, we have (0. gcd( α +,..., α k +, n = ( gcd(α,...,α k +, n = gcd(α,...,α k,n + 0 (mod ( 2ν +. Let q = 2ν +. Then 2 ν+ is the multilicative order of (mod q, namely, o q ( = 2 ν+. Since 2 ν+ n, then q n. First we need to show that 2 ν+ l n (f. To do that, we show that {x F n : f (x = 0} is a vector sace over F 2 ν+. Let x F n such that f (x = 0. Also, let y F 2ν+. Then, 32

38 we want to show that f (xy = 0. Using the definition of f, we have f (yx α = k i= (a i y αi x αi + a α i i y αi x αi. We claim that y ±α i = y α for all i k. By equation (0., α i (mod q which imlies that ±α i (mod q. Then, α ±α i (mod q. This leads to α α i (mod q for all i k. Since o q ( = 2 ν+, then α α i 0 (mod 2 ν+. Thus, y α α i = y which imlies that y α = y ±α. Then f (yx α becomes f (yx α = k i= (a i y α x αi + a α i i y α x αi ( k = y α (a i x αi + a α i i x αi i= = y α f (x α = 0. Thus, f (yx = 0. So, we have roved that 2 ν+ l n (f. Now, choose z F n such that o(z = q. Since α i + 0 (mod q for all i k, we have f(yx = f(x for all y z. So, t n (fg n ln(f ln(f = S(f, n = + = + x F n e n (f(x x F n / z y z = + = + q x F n / z x F n / z (mod q. e n (f(xy e n (f(x q e n (f(x In the above, ln(f (mod q since o q ( = 2 ν+ l n (f. Also, [ ] g 2ν+ = i 4 ( 2 2 ν+ 2 = i 4 ( 2 2 ν2 2 (2ν 2 hence, This leads to g n ln(f = ( 4 ( 2 2 ν 2ν ( 4 ( 2 2 ν + = g 2ν+ n ln(f 2 ν+ (mod q; ( ( 4 ( 2 2 ν + n ln(f 2 ν+ (mod q. t n (f = ( ( 4 ( 2 2 ν + n ln(f 2 ν+. Corollary 0.4. Assume (mod 4, α,..., α k are all odd and n is even. Then t n (f =. Proof. This follows immediately from Theorem

39 Examle 0.5. Let f(x = 2x x 33 + F 3 [x]. Then f (x = x x x 34 + x 36. The slitting field of f over F 3 is F 3 2 and 6 if a 2, b l 2 a 3 b m (f = 4 if a =, b or a 2, b = 0, 2 if a =, b = 0 or a = 0, b, if a, b = 0. where (m, 2 3 =. By Corollary 0.4, for even n, t n (f =. 34

40 . The Formula for S(ax +α Theorem 0.3 hels to rovide a roof for the evaluation of the sum of S(ax α +, n, a secial case of the sum in Theorem 0.3 Corollary.. Let a F n and let α 0. (i If ν 2 (n ν 2 (α, (ii If ν 2 (n = ν 2 (α +, S(ax α +, n = 2 n S(ax α +, n = η(a( n i 4 ( 2n 2 n. 2 [n+(2α,n] ( α ( n if a (2α,n =, otherwise. (iii If ν 2 (n > ν 2 (α +, S(ax α +, n = 2 [n+(2α,n] 2 n ( α ( n if a (2α,n =, otherwise. Proof. We first need to determine l n (f. We claim that ( α ( n (. l n (f = (2α, n if a (2α,n = ( 0 otherwise. n (2α,n Note that f (x = a α x 2α + ax = a α x(x 2α + a α. So, if f (x = 0 has a solution in F n, the number of solutions will be (2α, n = (2α,n. So, { (2α, n if f (x = 0 has a solution in F l n (f = n, 0 otherwise. Notice that f (x = 0 has a solution in F n if and only if aα = x 2α for some x F n. This haens if and only if ( a α n ( 2α, n =, ( a α n (2α,n =, ( ( n (2α,n n (2α,n a (a α n (2α,n =, ( α ( n (2α,n =, This yields (.. ( α ( n a (2α,n = ( 35 n (2α,n.

41 (i Since ν 2 (n ν 2 (α, we can use Lemma 0.2 which says ( α +, n = 2. Then, x x α + is a 2-to- ma from F n to (F n2. Thus, S(ax α +, n = + e n (ax α+ = + 2 e n (ax x F n x (F n 2 = + e n (ax 2 = e n (ax 2 = η(a e n (x 2 x F n x F n x F n = η(a( n g n (by the Davenort-Hasse theorem [5], [3, 5.2] = η(a( n i 4 ( 2n 2 n. n (2α,n (ii Since ν 2 (n = ν 2 ((2α, n, then is odd. By (., ( α ( n l n (f = (2α, n if a (2α,n =, 0 otherwise. We can then use Theorem 0.3, t n (f = ( ( 4 ( 2 α+ n ln(f 2 ν 2 (α+. If l n (f = (2α, n, then n ln(f 2 ν 2 (α+ is even since ν 2 (n (2α, n > ν 2 (n = ν 2 (α +. So, t n (f = ; hence S(ax α +, n = g n ln(f ln(f = i 4 ( 2(n (2α,n 2 [n+(2α,n] = 2 [n+(2α,n] since ν 2 (n + (2α, n 2. If l n (f = 0, then n ln(f 2 ν 2 (α+ is odd and t n (f = ( 4 ( 2α. So, S(ax α +, n = ( 4 ( 2α g n = ( 4 ( 2α i 4 ( 2n 2 n = ( 4 ( 2 (α n 2 2 n = 2 n since ν 2 (α + n 2 > ν 2(α 0. To summarize, we have S(ax α +, n = 2 [n+(2α,n] ( α ( n if a (2α,n =, 2 n otherwise. (iii In this case, By Theorem 0.3, n (2α,n is even. By (., l n (f = (2α, n 0 otherwise. ( α ( n if a (2α,n =, t n (f = ( ( 4 ( 2 α+ n ln(f 2 ν 2 (α+. The conclusion follows the same way as in (ii. 36

42 Let f(x = 2. Tables of Numerical Results k a i x αi+ F n[x], 0 α < < α k, i= where a k F n and let 0 < m 0 (mod n. Since Tr m(a k f = a k Tr m(f, then l m (f = l m (a k f, where a k f is monic. Thus, when we comute l m(f with a k F, we can assume that a k =. We have two tables for l m (f in this section. Table gives the values for l m (f where n = 3 and α 4. Table 2 gives the values for l m (f where n = 5 and α 3. Both tables were comuted using Mathematica [5]. In both tables, the first column contains the values for a i where 0 i k of f(x as defined above. The next column is the integer s such that F s is the slitting field for f (x. The last column lists all the airs (m, l m (f such that m s. The values of l m (f for arbitrary m s follows from (5.5. The tables begin on the next age. 37

43 Table. Values of l m (f with n = 3, α 4 a 0,..., a k s (m, l m (f, m s (,0 0 4 (,0 (2,0 (4,2 6 (,0 (2, (3,0 (6,2 2 3 (, (3, (,0 (2,0 (4,0 (8,4 0 2 (,0 (2,0 (3,0 (4,2 (6,0 (2, (, (2,2 (3,2 (6,4 0 8 (,0 (2, (3,0 (6,3 (9,0 (8,4 2 (, (2, (3,2 (4,3 (6,2 (2,4 2 5 (,0 (5, (, (3,3 (9,4 2 2 (,0 (2, (3,0 (4,3 (6,2 (2, (,0 (2,0 (5,0 (0, (,0 (2,0 (3,0 (4,2 (6,0 (2, (,0 (2, (3,0 (6,3 (9,0 (8, (, (3,3 (9, (,0 (2,0 (4,2 (8, (, (2, (3,2 (5, (6,2 (0,5 (5,2 (30, (,0 (2, (3,0 (5,4 (6,2 (0,5 (5,4 (30, (, (2,2 (3,2 (4,4 (6,4 (2, (,0 (3, (,0 (2,0 (3,0 (26, (,0 (2, (3,0 (5,0 (6,2 (0,5 (5,0 (30,6 0 2 (, (2, (3,2 (4,3 (6,2 (2, (,0 (2,0 (4,0 (7,0 (4,0 (28, (, (2, (3,2 (4, (6,2 (8,5 (2,2 (24,6 36 (,0 (2, (3,0 (4,3 (6,3 (9,0 (2,5 (8,4 (36,6 2 7 (,0 (7, (,0 (3, (,0 (2,0 (4,2 (5,4 (0,4 (20, (, (2,2 (3,3 (6,5 (9,4 (8, (, (3,2 (5,5 (5, (,0 (2,0 (4,0 (7,0 (4,0 (28, (,0 (2, (3,0 (4,3 (6,2 (2, (,0 (2, (3,0 (4, (6,2 (8,5 (2,2 (24,6 2 4 (,0 (2,0 (7,0 (4, (, (2, (3,3 (4,3 (6,3 (9,4 (2,5 (8,4 (36, (,0 (2,0 (3,0 (26, (, (2,2 (3,2 (6,5 (9,2 (8, (,0 (2,0 (4,2 (5,0 (0,4 (20,6 38

44 Table. Continued a 0,..., a k s (m, l m (f, m s (,0 (2,0 (4,0 (8,0 (6, (,0 (2,0 (3,0 (4,0 (6,0 (8,4 (2,0 (24, (, (2,2 (3,2 (4,4 (6,4 (2, (,0 (2, (3,0 (5,0 (6,3 (9,0 (0,5 (5,0 (8,4 (30,7 (45, 0 (90, (, (2, (3,2 (4, (6,2 (7, (2,2 (4, (2,2 (28,7 (42, 2 (84, (,0 (2,0 (4,2 (3,0 (26,6 (52, (, (3,3 (5,5 (9,4 (5,7 (45, (,0 (2, (3,0 (4, (6,2 (7,0 (2,2 (4, (2,0 (28,7 (42, 2 (84, (,0 (2,0 (4,2 (3,6 (26,6 (52, (,0 (2,0 (3,0 (4,2 (6,0 (9,0 (2,6 (8,0 (36, (, (2,2 (3,2 (4,2 (6,4 (8,6 (2,4 (24, (,0 (2,0 (5,4 (0, (, (2, (3,3 (4,3 (6,3 (9,6 (2,5 (8,6 (36,8 0 4 (,0 (4, (,0 (2, (3,0 (6,2 (7,6 (4,7 (2,6 (42, (,0 (2, (3,0 (4,3 (6,3 (9,0 (2,5 (8,6 (36, (,0 (2,0 (4,0 (82, (, (2, (3,2 (6,2 (7, (4,7 (2,2 (42, (, (2,2 (3,3 (6,6 (9,4 (8, (,0 (2,0 (3,0 (4,2 (6,0 (8,6 (2,4 (24, (,0 (2,0 (4,0 (5,0 (0,0 (20, (,0 (4, (,0 (2, (3,0 (4,3 (5,4 (6,2 (0,5 (2,4 (5,4 (20,7 (30, 6 (60, (, (2, (3,2 (6,2 (3, (26,7 (39,2 (78, (,0 (2,0 (4,0 (82, (, (2, (3,2 (4,3 (5, (6,2 (0,5 (2,4 (5,2 (20,7 (30, 6 (60, (,0 (2, (3,0 (6,2 (3,6 (26,7 (39,6 (78, (,0 (2, (3,0 (6,2 (7,0 (4,7 (2,0 (42, (, (2, (3,2 (6,2 (3, (26,7 (39,2 (78, (,0 (2,0 (3,0 (4,2 (5,4 (6,0 (0,4 (2,4 (5,4 (20,6 (30, 4 (60, (, (3,2 (3,7 (39, (,0 (2, (3,0 (4, (6,3 (8,5 (9,0 (2,3 (8,4 (24,7 (36, 4 (72, (,0 (2,0 (4,2 (3,6 (26,6 (52, (,0 (2,0 (4,0 (5,0 (8,0 (0,0 (6,0 (20,0 (40,0 (80, (,0 (2,0 (3,0 (5,0 (6,0 (0,4 (5,0 (30, (, (2,2 (3,3 (4,4 (6,5 (9,4 (2,7 (8,6 (36, (, (2, (3,2 (4,3 (6,2 (2,8 0 4 (,0 (4, (,0 (2, (3,0 (6,2 (3,0 (26,7 (39,0 (78, (,0 (2, (3,0 (4,3 (5,0 (6,2 (0,5 (2,4 (5,0 (20,7 (30, 6 (60,8 40 (,0 (2,0 (4,0 (5,4 (8,4 (0,4 (20,4 (40,8 39

45 Table. Continued a 0,..., a k s (m, l m (f, m s 2 9 (, (3,3 (9, (,0 (2,0 (4,2 (7,0 (4,0 (28,8 2 8 (, (2,2 (3,2 (6,5 (9,2 (8, (,0 (2,0 (4,0 (5,0 (8,0 (0,0 (6,0 (20,0 (40,0 (80, (,0 (4, (,0 (2, (3,0 (4,3 (6,3 (9,0 (2,7 (8,4 (36, (, (2, (3,3 (5, (6,3 (9,4 (0,5 (5,3 (8,4 (30,7 (45, 4 (90, (,0 (2,0 (4,0 (82, (, (2, (3,2 (4,3 (6,2 (8,7 (2,4 (24, (,0 (2, (3,0 (4, (6,2 (7,0 (2,2 (4, (2,0 (28,7 (42, 2 (84, (, (2,2 (3,2 (5,5 (6,4 (0,6 (5,6 (30, (,0 (2,0 (4,2 (7,6 (4,6 (28, (,0 (2,0 (4,0 (5,0 (8,0 (0,0 (20,0 (40, (, (3,2 (7,7 (2, (,0 (2, (3,0 (6,2 (3,6 (26,7 (39,6 (78, (,0 (2,0 (3,0 (4,2 (5,0 (6,0 (0,4 (2,4 (5,0 (20,6 (30, 4 (60, (,0 (2,0 (4,0 (5,0 (8,0 (0,0 (6,0 (20,0 (40,0 (80, (,0 (3,0 (5,4 (5, (, (2,2 (3,2 (4,4 (6,5 (9,2 (2,7 (8,6 (36, (,0 (2, (3,0 (6,2 (3,0 (26,7 (39,0 (78, (, (2, (3,3 (4, (6,3 (8,5 (9,4 (2,3 (8,4 (24,7 (36, 4 (72, (,0 (2,0 (4,2 (3,0 (26,6 (52, (,0 (2, (3,0 (4,3 (6,2 (2, (,0 (2,0 (4,0 (82, (, (3,2 (3,7 (39, (,0 (2,0 (4,2 (7,0 (4,0 (28,8 2 8 (, (2,2 (3,3 (6,5 (9,6 (8, (,0 (2,0 (4,0 (5,0 (8,0 (0,0 (6,0 (20,0 (40,0 (80, (, (2, (3,2 (4,3 (5,5 (6,2 (0,5 (2,4 (5,6 (20,7 (30, 6 (60, (,0 (2,0 (4,0 (5,0 (8,4 (0,4 (20,4 (40, (,0 (2, (3,0 (6,3 (9,0 (8, (,0 (2,0 (4,0 (82, (, (2, (3,3 (4,3 (6,3 (9,4 (2,7 (8,4 (36, (,0 (2, (3,0 (5,4 (6,3 (9,0 (0,5 (5,4 (8,4 (30,7 (45, 4 (90, (, (2,2 (3,2 (5, (6,4 (0,6 (5,2 (30, (,0 (2,0 (4,2 (7,0 (4,6 (28, (,0 (2,0 (4,0 (5,0 (8,0 (0,0 (20,0 (40, (,0 (4, (,0 (2, (3,0 (4,3 (6,2 (8,7 (2,4 (24, (, (2, (3,2 (4, (6,2 (7, (2,2 (4, (2,2 (28,7 (42, 2 (84,8 40