Subfield value sets over finite fields

Transcription

1 Subfield value sets over finite fields Wun-Seng Chou Institute of Mathematics Academia Sinica, and/or Department of Mathematical Sciences National Chengchi Universit Taipei, Taiwan, ROC Javier Gomez-Calderon, Gar L Mullen Department of Mathematics The Pennslvania State Universit Universit Park, PA 1680, U S A {jxg11,mullen}@mathpsuedu Daniel Panario, and David Thomson School of Mathematics and Statistics Carleton Universit 115 Colonel B Dr, Ottawa ON Canada, K1S 5B6 {daniel,dthomson}@mathcarletonca Abstract Let F q e be a finite field, and let F q d be a subfield of F q e The value set of a polnomial f ling within F q d is defined as the set of images {f(c F q d : c F q e} This work is concerned with the cardinalit of value sets of polnomials ling within subfields In particular, we give the cardinalit of subfield value sets of linearized polnomials, of power polnomials and of Dickson polnomials D n (x, a of degree n and parameter a, where a n F q d This author would like to thank the National Science Council for partial support of this work under grant number NSC M These authors would like to sincerel thank Gar McGuire and the Claude Shannon Institute of Universit College Dublin, Dublin, Ireland, for their support during the Jan - Mar, 011 period, where part of this work took place These authors are partiall supported b NSERC of Canada 010 Mathematics Subject Classification: Primar 11T06; Secondar 1F99 Ke words and phrases: Dickson polnomial, linearized polnomial, power polnomial, value set, permutation polnomial, König-Rados Theorem 1

2 W-S Chou et al 1 Introduction Let q be a prime power, let F q denote the finite field of order q, and let F q denote the (cclic multiplicative group of F q For integers 1 d e, F q d is a subfield of the finite field F q e if and onl if d divides e In this paper we consider the value set of a polnomial f F q e[x] ling within a subfield F q d of F q e, or simpl the subfield value set The subfield value set is defined as the set of images f(c F q d, where c runs over F q e When the subfield is omitted, the value set of f is simpl the set of images of f Das and Mullen [] stud value sets of polnomials over finite fields; in particular, the obtain a lower bound for the cardinalit of the value set of a polnomial over F q The idea of studing functions on extension fields with their images in subfields is a ver natural one For example, the absolute trace function defined for α F q e b Tr(α = α + α q + + α qe 1 maps onto the subfield F q uniforml in the sense that it maps onto each element of the subfield F q equall often More generall, for each d dividing e, the trace function defined for α F q e b Tr d (α = α + α qd + + α qe d maps onto the subfield F q d uniforml in the sense that it maps onto each element of the subfield F q d equall often These subfield ideas can be used to construct sets of mutuall orthogonal frequenc squares (MOFS; see [7] A connection to maximal curves is given in [3] From now on, let V f (q e ; q d = {f(c F q d : c F q e} denote the subfield value set of f that lies in the subfield F q d as c ranges over the extension field F q e Further let V f (q e ; q d denote the cardinalit of V f (q e ; q d, that is, the number of distinct elements in the image of f that lie in F q d as c ranges over the extension field F q e As a special case we note that V f (q e ; q e denotes the usual value set {f(c: c F q e} of a polnomial f over the field F q e Further let N f (q e ; q d denote the number of images f(c (counting multiplicities of f : F q e F q e that lie in the subfield F q d, as c ranges over the elements of the extension field F q e We clearl have V f (q e ; q d N f (q e ; q d, and of course N f (q e ; q e = q e for an polnomial f over the field F q e In this paper, we consider subfield value sets for several classes of polnomials, namel linearized polnomials, power polnomials and Dickson polnomials

3 Subfield Value Sets 3 The structure of the paper is as follows Section gives the value set in the subfield for linearized polnomials and includes an extension of the König-Rados Theorem which gives the distribution of zeroes of a polnomial ling within a subfield In Section 3 we give the value set in the subfield for power polnomials Section 4 presents the main result of this paper on the value set of Dickson polnomials within subfields We conclude in Section 5 with an open problem on the general case of the subfield value set of a Dickson polnomial König-Rados and linearized polnomials 1 König-Rados theorem for subfields Let n > 0, let f F q [x] and consider the equation f(x = 0 The distinct roots of f can be found as the roots of gcd(f, x q x, which have multiplicit 1 Thus, the number of distinct solutions of f(x = 0 is equal to the degree of gcd(f, x q x It is trivial to determine if f(0 = 0 and so we consider onl the solutions to gcd(f, x q 1 1 Furthermore, since α q 1 = 1 for all α F q, we ma assume, without loss of generalit, that n q when we consider the number of nonzero solutions of f(x = 0 The König-Rados Theorem expresses the number of nonzero roots of a polnomial in terms of the rank of a coefficient matrix Theorem 1 [6, Theorem 61] Let q be a power of a prime, let q f(x = a s x s F q [x] s=0 and denote b C the left circulant matrix a 0 a 1 a q 3 a q a 1 a a q a 0 C = a q a 0 a q 4 a q 3 The number of nonzero solutions of the equation f(x = 0 in F q is equal to q 1 rk(c, where rk(c is the rank of the matrix C We further extend the König-Rados Theorem to determine the number of roots of the polnomials occurring within a subfield

4 4 W-S Chou et al Theorem Let q be a power of a prime, and let e, d be positive integers with d dividing e Let f(x = q e s=0 a s x s F q e[x], and denote b C and B d the matrices C = a 0 a 1 a q e 3 a q e a 1 a a q e a 0 a q e a 0 a q e 4 a q e 3, B d = c 1 c c q d 1 c 1 c c q d 1 c qe 1 c qe c qe q d 1, where c 1, c,, c q d 1 are the distinct elements of F q d Then, the number of non-zero solutions of the equation f(x = 0 in F q d is equal to q d 1 rk(cb d Proof Let N d be the number of solutions of f(x = 0 occurring within F q d Let c 1, c,, c q d 1 be ordered so that f(c i 0 for 1 i q d 1 N d and c q d N d, c q d N d +1,, c q d 1 F Let the columns of B q d d be ordered in this wa Since the elements of F which are solutions of f(x = 0 appear q d in the final columns of B d, the final N d columns of CB d are equal to 0 and the rank of CB d is at most q d 1 N d Now consider the submatrix E of CB d E = f(c 1 f(c f(c q d 1 N d c 1 1 f(c 1 c 1 f(c c 1 q d 1 N d f(c q d 1 N d c 1 f(c 1 c f(c c q d 1 N d f(c q d 1 N d c (qd N d 1 f(c 1 c (qd N d f(c c (qd N d q d 1 N d f(c q d 1 N d The matrix E is non-singular since det(e = f(c 1 f(c f(c q d 1 N d det(e, where E is Vandermonde with defining row (c 1 1, c 1,, c 1 q d 1 N d Thus, rk(cb d = q d 1 N d We comment that if e = d, then B d = B Since B d has full rank, rk(cb d = rk(c, and Theorem reduces to Theorem 1

5 Subfield Value Sets 5 Linearized polnomials Let F q e be the finite field with q e elements A linearized polnomial over F q e is a polnomial of the form L(x = α i x qi F q e[x] An affine polnomial over F q e is given b A(x = L(x + α, where L(x is a linearized polnomial over F q e and α F q e Note that ever linearized polnomial L over F q e is indeed linear over F q We can consider L as a linear operator F q e F q e, when F q e is seen as a vector space over F q For the remainder of this paper we use the notation F q e both to denote the finite field of degree e over F q and to denote the vector space F e q over F q In addition, we do not make the distinction between a linearized polnomial L F q e[x] and the linear operator F e q F e q It is well known when linearized polnomials define permutations over finite fields, see [6, Theorem 79] We use a technique similar to an alternate discussion, given in [6, Page 36], to determine the value set of a linearized polnomial Theorem 3 Let q be a power of a prime, and let e be a positive integer Denote b F q e the finite field with q e elements and let L(x = e 1 s=0 α ix qi be a linearized polnomial over F q e Denote b M the e e matrix α 0 α q e 1 α qe 1 1 α 1 α q 0 α qe 1 α e 1 α q e α qe 1 0 Then, L is a permutation polnomial if and onl if det(m 0, where det(m denotes the determinant of the matrix M Furthermore, the value set of L, denoted V L, satisfies V L = q rk(m Proof The statement of the theorem is proven in [6, Page 36], except for the final line For the final assertion, we fix a basis {β 0, β 1,, β e 1 } of F q e over F q and let γ i = L(β i, i = 0, 1,, e 1 For 0 i, j e 1 we have γ qj i = s=0 α qj s β qs+j i,

6 6 W-S Chou et al and taking subscripts (mod e, we have γ qj i = s=0 α qj s j βqs i We therefore have a matrix equation relating the conjugates of the γ i, β i and α s j of the following form γ 0 γ q 0 γ qe 1 0 γ 1 γ q 1 γ qe 1 1 γ e 1 γ q e 1 γ qe1 e 1 β 0 β q 0 β qe 1 0 β 1 β q 1 β qe 1 1 = β e 1 β q e 1 β qe 1 e 1 α 0 α q e 1 α qe 1 1 α 1 α q 0 α qe 1 α e 1 α q e α qe 1 0 Labelling the corresponding matrices Γ, B and M, respectivel, b [6, Corollar 38], the matrix B is non-singular and thus the rank of Γ is equal to the rank of M Since the value set of the linearized polnomial L is equal to the image set of the linear operator, we have V L = q rk(m Corollar 4 Let L F q e[x] be a linearized polnomial with value set of cardinalit q rk(m, as given in Theorem 3 Ever image is repeated q e rk(m times Furthermore, N L (q e ; q d = V L (q e ; q d q e rk(m, where N L (q e ; q d denotes the total number of images of L in F q d, including repetitions Proof Since L defines a linear operator F q e F q e, we have, b the first isomorphism theorem, F q e/ ker(l = V L Since dim(ker(l = e rk(m, the claim follows Suppose L F q e[x] is a linearized polnomial and let A(x = L(x + α, for some α F q e Consider the subfield value set of A, V A (q e ; q d, for an d dividing e We have triviall that V A (q e ; q e = V L (q e ; q e If α F q d, then V A (q e ; q d = V L (q e ; q d Example 5 Let L(x = Tr d (x Then L is a linearized polnomial and L maps F q e onto F q d Let α F q e with α F q d and let A(x = L(x + α Then V A (q e ; q d = If α V L (q e ; q e, that is, if α is an image of L, then for all β F q e, A(β = L(β + α = L(β + γ, where α = L(γ Thus, running over all β F q e, we

7 Subfield Value Sets 7 have that V L (q e ; q d = V A (q e ; q d for all d dividing e If α is not an image of L, then the subfield value set of A depends on the additive cosets of the subfield value set of L It can be easil verified with a computer algebra program, such as SAGE or Maple, that the cardinalities of subfield value sets of affine polnomials most often var from the subfield value sets of their corresponding linearized polnomials Lemma 6 Let q be a power of a prime, and let e be a positive integer Let F q e be the finite field with q e elements and let L be a linearized polnomial over F q e defined b L(x = e 1 a ix qi Then { N L (q e ; q d ( = β : a qd e d+i a i β qi = 0} and V L (q e ; q d = N L (q e ; q d /q e rk(m, where M is the matrix given in Theorem 3 Proof Let L(x = a i x qi and suppose that L(α lies in F q d That is, Rearranging, we find L(α qd = a qd i α qi+d = L(α = a i α qi a qd i α qi+d a i α qi = (a qd e d+i a iα qi = 0, where the subscripts are taken (mod e Thus L(α lies in the subfield F q d of F q e if and onl if α is a root of the polnomial (1 b(x = (a qd e d+i a ix qi The final expression for V L (q e ; q d is given b Corollar 4 Counting the number of zeroes of the polnomial b in Equation (1 can be done b the König-Rados theorem, see Theorem 1

8 8 W-S Chou et al Theorem 7 Let L be a linearized polnomial over F q e given b L(x = q e 1 a ix i, that is a j = 0 for j 1, q, q,, q e 1 Let C be the left-circulant matrix of size q e 1 with defining row 0 b 0 0 } {{ 0 } b 1 } 0 {{ 0 } b b e 0 } {{ 0 } b e 1 0 } {{ 0 }, q times q q 1 times q e 1 q e 1 times q e q e 1 times where b 0 are the coefficients of b in Equation (1 Then, where M is given b Corollar 4 V L (q e ; q d = qe rk(c q e rk(m, Proof Theorem 1 gives the number of non-zero roots of b is q e 1 rk(c Since 0 is a root of b, the claim follows 3 Power polnomials We now consider the subfield value set V x n(q e ; q d of the polnomial f(x = x n Power polnomials are a special case of Dickson polnomial D n (x, a with a = 0, as we will see in the next section It is well known and eas to see that V x n(q e ; q e = qe 1 (n, q e We first show the number of preimages of the subfield value set of a power polnomial Theorem 31 The number of preimages of the power polnomial x n given b N x n(q e ; q d = (n(q d 1, q e is Proof Recall that if α F q e, then α F q d if and onl if α qd = α For c F q e, if (cn qd = c n, we have c n(qd 1 = 1 The number of solutions of this equation for c F q e, is given b (n(qd 1, q e 1, and the result follows Since the multiplicative group F q e is cclic, we have in F q e V x n(q e ; q d = N x n(qe ; q d (n, q e = (n(qd 1, q e (n, q e 1 We note that if (n, q e 1 = 1 so that x n is a permutation polnomial on F q e, then V x n(q e ; q d = N x n(q e ; q d = q d since x n must map F q d onto itself In fact, if (n, q d 1 = 1, then x n is a permutation polnomial on F q d and so V x n(q e ; q d = q d

9 Subfield Value Sets 9 4 Dickson polnomials For a F q e, the Dickson polnomial D n (x, a of degree n and parameter a is defined b D n (x, a = n/ n n i ( n i i ( a i x n i Dickson polnomials have been studied extensivel since the pla ver important roles in the theor of permutation polnomials over finite fields, and in the Schur conjecture; see [5] In [4], the use of Dickson polnomials in crptographic sstems, particularl over finite fields, is generalized b considering Dickson polnomials over Galois rings Dickson polnomials have man properties which are closel related to properties of the power polnomial x n = D n (x, 0, see [5] For example, for a F q, D n (x, a induces a permutation on the field F q if and onl if (n, q 1 = 1 Moreover, from [1] we have V Dn(x,a(q; q = q 1 (n, q 1 + q + 1 (n, q α, where α can be explicitl stated and is usuall 0 In [3], if n is odd and n divides q 1, it was shown that N Dn(x,1(q ; q = (q(n + 1 (n 1/, and N Dq 1 (x,1(q ; q = (q + 1/ Let a F q e If c F q e, then we can write c = + a/ for some F q e, and we obtain a functional equation for Dickson polnomials, D n (c, a = n + a n / n Thus, in order to have the image D n (c, a in the subfield F q d, we must have (41 ( n + an n q d = n + an n If a n F q d, Equation (41 becomes, after simplification, (4 ( n(qd 1 1( n(qd +1 a n = 0; that is, either n(qd 1 = 1 or n(qd +1 = a n The following lemma is essential but has an elementar proof which is omitted Lemma 41 For a F q e, let C a be the set C a = { + a/ : F q e qe +1 = a} Then, C a = F q e or We consider onl the case a n F q d, for otherwise, when a n F q d, Equation (41 does not seem to lead to a convenient factorization as in Equation (4; see Section 5 We derive V Dn(x,a(q e ; q d in detail for q odd and note that the derivation for q even is similar and therefore omitted

10 10 W-S Chou et al Let q be odd In the following lemma, η q l is the quadratic character on F q l, so that η q l(a = 1 if a F is a non-zero square and η q l q l(a = 1 if a F is not a square Moreover a is a square root in F q l q l of a F For q l an number m, let r m be the non-negative integer satisfing rm m, that is, r m is the highest non-negative power of dividing m Lemma 4 Let F q d be a subfield of F q e with q odd If a n F q d, then c V Dn(x,a(q e ; q d if and onl if c = n + a n / n, where satisfies at least one of the following requirements: I (qe 1,n(q d 1 = 1, II a for η q e(a = 1 and η q d(a n = 1, 1 ( a (qe +1,n(q d 1 = 1, ( a (qe 1,n(q d +1 = 1, 3 ( a (qe +1,n(q d +1 = 1, b for η q e(a = 1 and η q d(a n = 1, 1 ( a (qe +1,n(q d 1 = 1 and r q e +1 < r n(q d 1, ( a (qe 1,n(q d +1 = 1 and r q e 1 < r n(q d +1, 3 ( a (qe +1,n(q d +1 = 1 and r q e +1 < r n(q d +1, c for η q e(a = 1 and η q d(a n = 1, 1 ( a (qe +1,n(q d 1 = 1 and r n(q d 1 < r q e +1, ( a (qe 1,n(q d +1 = 1 and r n(q d +1 < r q e 1, 3 ( a (qe +1,n(q d +1 = 1 and r n(q d +1 < r q e +1, d for η q e(a = 1 and η q d(a n = 1, 1 ( a (qe +1,n(q d 1 = 1 and r q e +1 = r n(q d 1, ( a (qe 1,n(q d +1 = 1 and r q e 1 = r n(q d +1, 3 ( a (qe +1,n(q d +1 = 1 and r q e +1 = r n(q d +1 Proof B Lemma 41, {D n (c, a : c F q e} = { n + (a/ n : F q e or qe +1 = a} Since a n F q d, n + (a/ n V Dn(x,a(q e ; q d if and onl if qe 1 = 1 or qe +1 = a and n(qd 1 = 1 or n(qd +1 = a n b Equation (4 If qe 1 = 1 and n(qd 1 = 1, then (qe 1,n(q d 1 = 1 and Case I holds In Case II, we prove onl (b1 All other cases can be proved in similar was Suppose η q d(a n = 1 and η q e(a = 1 Then ( a n(qd 1 = 1 and ( a qe 1 = 1 The last equalit is equivalent to ( a qe +1 = a

11 Subfield Value Sets 11 (Case IIb1 Suppose qe +1 = a and n(qd 1 = 1 These two equations ( q e +1 ( n(q d 1 are equivalent to a = 1 and a = 1, respectivel Thus, ( ((q e +1,n(q d 1 ( (q e +1,n(q d 1 a = 1 but a = 1, and so rq e +1 < r n(q 1 d We can now evaluate N Dn(x,a(q e ; q d Theorem 43 Let q be odd and let a F q e with a n F q d For integers m and k, let δ m<k = 1, if m < k, and δ m<k = 0, if m k Also, let δ m=k = 1, if m = k, and δ m=k = 0, if m k a If η q e(a = 1 and η q d(a n = 1, then N Dn(x,a(q e ; q d = (qe 1, n(q d 1 + (q e 1, n(q d + 1 (q e 1, n + (qe + 1, n(q d 1 + (q e + 1, n(q d + 1 (q e + 1, n b If η q e(a = 1 and η q d(a n = 1, then N Dn(x,a(q e ; q d (q e 1, n(q d 1 + δ rq e 1 <r n(q d +1 (q e 1, n(q d + 1 = (1 δ rn<rq e 1 (q e 1, n + δ rq e +1 <r n(q d 1 (q e + 1, n(q d 1 + δ rq e +1 <r n(q d +1 (q e + 1, n(q d + 1 (1 δ rn<rq e +1 (q e + 1, n + c If η q e(a = 1 and η q d(a n = 1, then N Dn(x,a(q e ; q d (q e 1, n ( q d 1 + δ rn(q d +1 <r q e 1 (q e 1, n(q d + 1 = δ rn(q d 1 <r q e +1 (q e + 1, n(q d 1 + δ rn(q d +1 <r q e +1 (q e + 1, n(q d d If η q e(a = 1 and η q d(a n = 1, then N Dn(x,a(q e ; q d (q e 1, n(q d 1 + δ rq e 1 =r n(q d +1 (q e 1, n(q d + 1 = δ rq e +1 =r n(q d 1 (q e + 1, n(q d 1 + δ rq e +1 =r n(q d +1 (q e + 1, n(q d + 1 +

12 1 W-S Chou et al Proof We prove this theorem according to the cases in Lemma 4 We onl prove Case b and comment that the proof of this case is a tpical example of the proofs of the remaining cases (Case b η q e(a = 1 and η q d(a n = 1 Let } E 1 = { F q e : (qe 1,n(q d 1 = 1, { ( } (qe +1,n(qd 1 E = F q e : a = 1 and r q e +1 < r n(q d 1, E 3 = E 4 = { { F q e : F q e : ( } (qe 1,n(qd +1 a = 1 and r q e 1 < r n(q d 1, and ( } (qe +1,n(qd +1 a = 1 and r q e +1 < r n(q d +1 The definition of E 1 comes from Case I of Lemma 4 We note that E 1 = (q e 1, n(q d 1, E = (q e + 1, n(q d 1, E 3 = (q e 1, n(q d + 1, and E 4 = (q e + 1, n(q d + 1 For E, can be written as = u a with u ((qe +1,n(q d 1 = 1 and u qe +1 = 1 This implies that r (q e +1 divides the order of u Moreover, if E 3, then r (q e 1 divides the order of u Since either q e 1 0 (mod 4 or q e (mod 4, 8 divides the order of u However u (qe +1 = 1 = u (qe 1 would impl u 4 = 1, a contradiction So, E E3 = Similar arguments show that E 1 E = E 1 E4 = E 3 E4 = Let = u a Then E E4 if and onl if u ((qe +1,n(q d 1 = 1, u qe +1 = 1 and u ((qe +1,n(q d +1 = 1 These are equivalent to u ((qe +1,n = 1 and u qe +1 = 1 So, if r q e +1 > r n, then E E4 = 0, while if r q e +1 r n, then E E4 = (q e +1, n B similar arguments, we have that E 1 E3 = 0 if r q e 1 > r n, and E 1 E3 = (q e 1, n if r q e 1 r n Combining all of the results above together, we have, b the inclusionexclusion principle, N Dn(x,a(q e ; q d = E 1 E E3 E4 (q e 1, n(q d 1 + δ rq e 1 <r n(q d 1 (q e 1, n(q d + 1 = (1 δ rn<rq e 1 (q e 1, n + δ rq e +1 <r n(q d 1 (q e + 1, n(q d 1 + δ rq e +1 <r n(q d +1 (q e + 1, n(q d + 1 (1 δ rn<rq e +1 (q e + 1, n +

13 Subfield Value Sets 13 This completes the proof The result of Theorem 43, Case a is a generalization of the results in [3] stated before Indeed, if n is an odd divisor of q 1 (and so n properl divides q 1, then (q 1, n(q 1 = (q 1(q + 1, n = q 1, (q 1, n(q + 1 = n(q + 1, (q 1, n = n, and (q + 1, n(q 1 = = (q + 1, n(q + 1 = (q +1, n So, N Dn(x,1(q ; q = (q(n+1 (n 1/ Moreover, in the case n = q 1, we obtain N Dq 1 (x,1(q ; q = (q + 1/ using similar arguments We are now read to compute V Dn(x,a(q e ; q d with a n F q d Theorem 44 Let q be odd and let a F q with e an F q d For integers m and k, let δ m<k = 1, if m < k, and δ m<k = 0, if m k Also, let δ m=k = 1, if m = k, and δ m=k = 0, if m k Suppose that r (q e 1 a If η q e(a = 1 and η q d(a n = 1, then V Dn(x,a(q e ; q d = (q e 1, n(q d 1 + (q e 1, n(q d + 1 (q e 1, n + (qe + 1, n(q d 1 + (q e + 1, n(q d + 1 (q e + 1, n b If η q e(a = 1 and η q d(a n = 1, then V Dn(x,a(q e ; q d = 3 + ( 1n+1 (q e 1, n(q d 1 + δ rq e 1 <r n(q d +1 (q e 1, n(q d + 1 δ r 1<rn + (q e 1, n δ rq e +1 <r n(q d 1 (q e + 1, n(q d 1 + δ rq e +1 <r n(q d +1 (q e + 1, n(q d (q e + 1, n c If η q e(a = 1 and η q d(a n = 1, then V Dn(x,a(q e ; q d = (q e 1, n(q d 1 + δ rn(q d +1 <r q e 1 (q e 1, n(q d + 1 (q e 1, n δ rn(q d 1 <r q e +1 (q e + 1, n(q d 1 + δ rn(q d +1 <r q e +1 (q e + 1, n(q d (q e + 1, n d If η q e(a = 1 and η q d(a n = 1, then V Dn(x,a(q e ; q d = (q e 1, n(q d 1 + δ rq e 1 =r n(q d +1 (q e 1, n(q d + 1 (q e 1, n δ rq e +1 =r n(q d 1 (q e + 1, n(q d 1 + δ rq e +1 =r n(q d +1 (q e + 1, n(q d (q e + 1, n

14 14 W-S Chou et al Proof We prove onl Case a The proofs of the remaining cases are similar Let } E 1 = { F q e : (qe 1,n(q d 1 = 1, { ( } (qe +1,n(qd 1 E = F q e : a = 1, E 3 = E 4 = { { F q e : F q e : ( } (qe 1,n(qd +1 a = 1, and ( } (qe +1,n(qd +1 a = 1 Similar to the proof of Theorem 43, we have = u a E 1 E3 if and onl if the order of u divides (q e 1, n and = u a E E4 if and onl if the order of u divides (q e + 1, n In both situations, we have n = (a/ n From [1, Lemma 7], for x 0 = 0 + a/ 0, 0 (E 1 E3 (E E4 if and onl if D n (x 0, a = ±a n/ Ever element x 0 = 0 + a/ 0 with 0 (E 1 E3 \ (E 1 E3 satisfies η q e(x 0 4a = 1 and D n (x 0, a a n/ From [1, Theorem 9], the total number I 1 of images D n (x 0, a with x 0 = 0 + a/ 0 for all 0 (E 1 E3 \ (E 1 E3 is I 1 = (qe 1, n(q d 1 + (q e 1, n(q d + 1 (q e 1, n (q e 1, n Similarl, the total number I of images D n (x 0, a with x 0 = 0 + a/ 0 for all 0 (E E4 \ (E E4 is I = (qe + 1, n(q d 1 + (q e + 1, n(q d + 1 (q e + 1, n (q e + 1, n We have seen that E 1 E3 = (q e 1, n, E E4 = (q e + 1, n, and (E 1 E3 (E E4 = {± a} Let t, r satisf t n and r (q e 1 1, respectivel If 1 t r, then either (q e 1, n = (q e 1, n and (q e +1, n = (q e +1, n or (q e 1, n = (q e 1, n and (q e +1, n = (q e + 1, n Furthermore, (q e 1, n = (q e 1, n and (q e + 1, n = (q e + 1, n if t = 0, while (q e 1, n = (q e 1, n and (q e + 1, n = (q e + 1, n if t r 1 For x 0 = 0 + a/ 0 F q e, if 0 E 1 E3, then η q e(x 0 4a = 1, while if ± a 0 E E4, then η q e(x 0 4a = 1 Moreover, ever element 0 E 1 E3 can be written as 0 = u a with u (qe 1,n = 1

15 Subfield Value Sets 15 So, if x 0 = 0 + a/ 0 F q e with 0 E 1 E3, then D n (x 0, a = 0 n + a n /0 n = u (qe 1,n ( a n Hence, if (q e 1, n = (q e 1, n, then for all elements 0 E 1 E3, D n ( 0 + a/ 0, a = ( a n If (q e 1, n = (q e 1, n, then for half of elements 0 E 1 E3, D n ( 0 + a/ 0, a = ( a n and for all other elements 0 E 1 E3, D n ( 0 + a/ 0, a = ( a n Similarl, if (q e + 1, n = (q e + 1, n then for all elements 0 E E4, D n ( 0 + a/ 0, a = ( a n, and if (q e + 1, n = (q e + 1, n then for half of elements 0 E E4, D n ( 0 +a/ 0, a = ( a n and for all other elements 0 E E4, D n ( 0 + a/ 0, a = ( a n Combining all of these results together, we have, from [1, Theorem 9], that the total number I 3 of images D n ( 0 + a/ 0, a with 0 (E 1 E3 (E E4 equals either 1 if t n and t r 1, or otherwise We have for the value set V Dn(x,a(q e ; q d = I 1 +I +I 3 We now compute V Dn(x,a(q e ; q d according to the value of t Case 1: t = 0 (n is odd We have (q e 1, n = (q e 1, n, (q e +1, n = (q e + 1, n and I 3 = From the result above, we have V Dn(x,a(q e ; q d =+ (qe 1, n(q d 1 + (q e 1, n(q d + 1 (q e 1, n (q e 1, n + (qe + 1, n(q d 1 + (q e + 1, n(q d + 1 (q e + 1, n (q e + 1, n = (qe 1, n(q d 1 + (q e 1, n(q d + 1 (q e 1, n + (qe + 1, n(q d 1 + (q e + 1, n(q d + 1 (q e + 1, n Case : 1 t r We have either (q e 1, n = (q e 1, n and (q e + 1, n = (q e + 1, n, or (q e 1, n = (q e 1, n and (q e + 1, n = (q e + 1, n In this case, I 3 = and Thus, I 1 + I = (qe 1, n(q d 1 + (q e 1, n(q d + 1 (q e 1, n + (qe + 1, n(q d 1 + (q e + 1, n(q d + 1 (q e + 1, n V Dn(x,a(q e ; q d = (qe 1, n(q d 1 + (q e 1, n(q d + 1 (q e 1, n 3 + (qe + 1, n(q d 1 + (q e + 1, n(q d + 1 (q e + 1, n 1

16 16 W-S Chou et al Case 3: t r 1 We have (q e 1, n = (q e 1, n, (q e +1, n = (q e +1, n and I 3 = 1 Therefore, the value set satisfies V Dn(x,a(q e ; q d = (qe 1, n(q d 1 + (q e 1, n(q d + 1 (q e 1, n This completes the proof + (qe + 1, n(q d 1 + (q e + 1, n(q d + 1 (q e + 1, n 1 The following corollar is the most important special case, which appears as [1, Theorem 10] Corollar 45 Let q be odd and let a F q Suppose that r (q 1 Then we have V Dn(x,a(q; q = q 1 (n, q 1 + q + 1 (n, q α, where 1 if e 1 n and η q (a = 1, 1 α = if t n with 1 t r, 0 otherwise Proof This is a special case of Theorem 44 with q e = q d = q Determining the value of α depends on the parit of n and the value of η q (a; the details are omitted We now state the results of N Dn(x,a(q e ;q d and V Dn(x,a(q e ; q d for q even The derivation of these results is similar, but slightl simpler, to those of q odd and is therefore omitted Theorem 46 Let q be even and let a F q e with an F q d Then N Dn(x,a(q e ; q d = (qe 1, n(q d 1 + (q e 1, n(q d + 1 (q e 1, n + (qe + 1, n(q d 1 + (q e + 1, n(q d + 1 (q e + 1, n Theorem 47 Let q be even and let a F q with e an F q d Then V Dn(x,a(q e ; q d = (qe 1, n(q d 1 + (q e 1, n(q d + 1 (q e 1, n + (qe + 1, n(q d 1 + (q e + 1, n(q d + 1 (q e + 1, n Corollar 48 Let q be even and let a F q Then we have V Dn(x,a(q; q = q 1 (n, q 1 + q + 1 (n, q + 1 1

17 Subfield Value Sets 17 5 Open problem We conclude with an open problem In all of our work concerning Dickson polnomials, we have assumed that the parameter a has the propert that a n F q d The reason for this assumption is that in this case, Equation (41 leads to the simple factorization in Equation (4 For this equation, we are able to calculate the number of solutions to each factor as well as the number of solutions which simultaneousl satisf both factors For a n F q d, we do not know how to find V Dn(x,a(q e ; q d in general However, we know that if gcd(n, q e 1 = 1, then V Dn(x,a(q e ; q = q = N Dn(x,a(q e ; q because D n (x, a is a permutation polnomial over F q e The following is an example in the other extreme case, namel V Dn(x,a(q e ; q = 0 = N Dn(x,a(q e ; q Proposition 51 Let q 7 (mod 8 be a prime power and let a F q a primitive element of F q Then V D(q 1(q +1 (x,a(q ; q = 0 be Proof Since a F q is a primitive element, a (q 1(q +1 F q At first, we consider D (q 1(q +1 ( + a, a with F q In this case, q 1 = 1 and so we have D (q 1(q +1 ( + a, a = (q 1(q +1 + a(q 1(q +1 = (q 1(q +1 (q 1 + a(q 1 (q 1 Hence, D (q 1(q +1 ( + a, a F q if and onl if (q 1 + a(q 1 (q 1 = q ( (q 1 + a(q 1 = (q q q + a(q (q 1 = (q 1 + a (q 1 (q 1 = 1 a (q 1 (q q ( (q 1 + a(q 1 (q 1 This implies that D (q 1(q +1 ( + a, a F q if and onl if either a (q 1 = 1 or 4(q 1 = a (q 1 We have that a (q 1 = 1 cannot hold because a is primitive in F q Also, 4(q 1 = a (q 1 cannot hold because a is primitive in F q and a (q 1 = a (q 1(+(q+1/ with the fact that + q+1 (mod 4 from q 7 (mod 8 So, what we have shown is that D (q 1(q +1 ( + a, a F q in this case Finall, we consider q +1 = a In this case, D (q 1(q +1 ( + a, a = (q 1(q +1 + a(q 1(q +1 = (q 1(q +1 aq 1, is trivall not in F q Combining this with the result above, we have V D(q 1(q +1 (x,a(q ; q = 0 In general, a F q e is such that a n F q d, then Equation (41 seems to not ield a simple factorization like that occurring in Equation (4 In such

18 18 W-S Chou et al a setting, how does one proceed to calculate the cardinalit of the subfield value set V Dn(x,a(q e ; q d? References [1] W-S Chou, J Gomez-Calderon, and GL Mullen, Value sets of Dickson polnomials over finite fields, Journal of Number Theor Vol 30 (1988, [] P Das and GL Mullen, Value sets of polnomials over finite fields, Finite Fields with Applications in Coding Theor, Crptograph and Related Areas, edited b GL Mullen, H Stichtenoth, and H Tapia- Recillas, Springer (00, [3] A Garcia and H Stichtenoth, On Chebshev polnomials and maximal curves, Acta Arithmetica, Vol 90 (1999, [4] J Gomez-Calderon and GL Mullen, Galois rings and algebraic crptograph, Acta Arithmetica, Vol 59 (1991, [5] R Lidl, GL Mullen, and G Turnwald, Dickson Polnomials, Longman Scientific and Technical, Essex, United Kingdom (1993 [6] R Lidl and H Niederreiter, Finite Fields, Encclopedia of Mathematics and its Applications, Vol 0 (nd ed, Cambridge Universit Press, Cambridge (1997 [7] SJ Suchower, Polnomial representation of complete sets of frequenc hperrectangles with prime power dimensions, Journal of Combinatorial Theor, Series A, Vol 6 (1993, 46-65