Powers in finite fields

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1 Powers in finite fiels Péter Sziklai Department of Computer Science an Information Theory Technical University Buapest Magyar Tuósok krt, Buapest, H 1117 Hungary sziklai@csbmehu an Department of Computer Science Eötvös University, Buapest Pázmány P s 1/c, Buapest, H 1117 Hungary sziklai@cseltehu Abstract: There are lots of results on the ranom-like behaviour of square elements in finite fiels For example, they can be use in combinatorial constructions an algorithms, as their properties somehow imitate a ranom istribution In this paper we investigate the more general question concerning the behaviour of -th powers in finite fiels (where is a fixe value) Surprisingly, they are istribute in a way which is ranom-like an very regular at the same time Keywors: finite fiels, powers, ranomness 1 Introuction There are lots of results on the ranom-like behaviour of square elements in finite fiels For example, they can be use in combinatorial constructions (like Paley-graphs an esigns) an algorithms, as their properties somehow imitate a ranom istribution In this paper we investigate the more general question concerning the behaviour of -th powers in finite fiels (where is a fixe value) Surprisingly, they are istribute in a way which is ranom-like an very regular at the same time By regularity we mean the following Consier the affine space AG(n,q) of imension n over the finite (Galois) fiel GF(q) of orer q (where q is a prime power); it is isomorphic to GF(q n ) (meant as vector spaces over GF(q)) Sometimes we call this representation of AG(n, q) as the big fiel representation After ientifying AG(n,q) an GF(q n ) this way, one can speak about the geometrical properties of any subset S GF(q n ) In particular, if S is the set {x : x GF(q n ) } for some, one can ask how the (affine) subspaces of GF(q n ), so for example lines grant Research was partially supporte by OTKA Grants F0077, D817, FKFP 006/001 an Magyary Zoltán 1

2 or hyperplanes of GF(q n ) intersect S As we can show, this multiplicative subgroup S an its cosets meet these subspaces in a very regular way, the intersection sizes follow a few patterns only In this paper we go into the etails for the special case of n = only, most of the proofs work in general after some easy moifications Throughout the paper θ k enotes q k + q k q + 1 = qk+1 1 q 1 Distribution of -th powers 1 Motivation Bruen an Fisher prove in [8], as an illustration of the Jamison metho, that in the big fiel representation AG(,q) GF(q ) every line contains at least one square an one non-square element of GF(q ), provie q is an o prime-power (See Hirschfel-Szőnyi [11] as well) One can think that in general, if (q n 1), then the -th powers an their cosets in the multiplicative group GF(q n ) are istribute quite evenly on the lines of the corresponing AG(n,q) A very similar conjecture can be that if (q n +q n 1 ++q +1), then the points in cosets of a subcycle of inex in the Singer cycle of PG(n,q) are istribute quite evenly on the lines This question is iscusse in Brouwer [7] Biliotti an Korchmáros investigate the very similar problem of transitive blocking sets in [] There is a very extensive literature on other cyclic configurations as well, we omit the list for the sake of brevity Note that if gc(,q n 1) = 1 (eg if q n ) then there is nothing to o as every element of GF(q n ) is a -th power We remark that by -th powers we sometimes mean GF(q n ) () an sometimes GF(q n ) (), but it will be always clear, which one We may use the following Result 1 (Folklore) (i) In the big fiel representation AG(n,q) GF(q n ), for any two hyperplanes H 1,H not through the origin there is a unique a GF(q n ) such that ah 1 = H (ii) Also for any two hyperplanes H 1,H through the origin there is a subset A H1,H = agf(q) of GF(q n ) such that αh 1 = H for any α A H1,H It means that all the hyperplanes not containing the origin (or containing it) behave similarly Proof: (i) The number of hyperplanes not through the origin is θ n θ n 1 1 = q n 1 = GF(q n ) This transformation group shoul be transitive hence regular on these hyperplanes (ii) The number of hyperplanes through the origin is θ n 1, the stabilizer of any hyperplane containing the origin is isomorphic to GF(q) GF(q n ) In this section we go into etails with the easiest case, with the plane, ie n =, where q = p a prime, an the case =, as = is very easy The higher imensional cases can be treate in a similar way Remark Note that in the planar case, if q + 1 then GF(q) is a set of -th powers, so the points of any line through the origin fall in one coset of the -th powers, an any other line, with slope in a coset S, contains q+1 1 points from S an q+1 points from each of the other cosets

3 -th powers in GF(q ) Let (q 1) One can easily see that any the line through the origin consists of q 1 points from each of the cosets (plus the origin, as the q-th point) Using Result 1 about the similar behaviour of the lines not through the origin, we have that there is a fixe -tuple, such that all these lines have (a 1,a,,a ) points from the cosets (the orer of the a i -s vary from line to line) We want to estimate these numbers Counting the points on a line an the orere (L,P 1,P ) triples, where L is a line containing P 1 P an the points P 1 an P are in the same coset, we have a i = q; i or (q 1) i a i (a i 1) + (q + 1) q 1 q 1 i a i = q + ( 1)q i α i = ( 1)q+1 = q 1 q 1, Let a i = q 1 + α i, then i α i = 1; It shows that α i 1 1 q (1 1 ) q, which is a bit better boun than Weil s In particular if =,q = p a prime: in this case we can calculate the intersection numbers explicitly Let α = α 1,β = α,γ = α, then α + β + γ = 1; α + β + γ = p + 1 Expressing α from the first equation an substituting to the secon one, we gain an equation in β with iscriminant 4p 1 γ +γ Since α,β an γ are integers, we have 4p 1 γ +γ = δ for some integer δ Hence 4p = δ + (γ 1) Consier Q[ ], it is a unique factorization omain, its integers are Z[ ] Here p = ( 1 (γ 1) + 1 δ )( 1 (γ 1) 1 δ ) = N( 1 (γ 1) + 1 δ ) This factorization is unique up to multiplication with units (as both factors have norm p an it is prime) The units are ( 1 + i)k, k = 0,1,,5 As we know that there exist some solutions, it follows that there are exactly six ones Since α an β have to satisfy the same equation, an (α 1)+(β 1)+(γ 1) = 0, we have that the solutions corresponing to α,β an γ form a regular triangle of sie p in the complex plane We have to choose between the two triangles such that the real parts shoul result in integral α,β an γ So we get an estimate for the iscrepancy of the istribution: p 1 < α,β,γ < p + 1 It means that we have p p < a,b,c < p+ p Note that for p = k(k + 1) + 1 we have {a,b,c} = {k, p 1 = k(k + 1),(k + 1) }, so in this case we have a set of type {k, p 1 = k(k + 1),(k + 1) }

4 -th power ifferences 1 Introuction In [1] van Lint an MacWilliams conjecture (an prove for q = p) that the only q-subset X of GF(q ), with the properties 0,1 X an x y is a square for all x,y X, is the set GF(q), as a subset of GF(q ) Such a set X, consiere as a set of points in AG(,q) GF(q ), by the big fiel representation, etermine at most q+1 irections So for q a prime the conjecture is a consequence of a theorem of Réei, cf [16] p 7 Satz 4 an [1], stating that a set of q points in AG(,q) etermines at least p+ irections, unless it consists of the points of an affine line (q is a prime here) This case was also prove in an elementary way by Lovász an Schrijver [14] Then in [4] Aart Blokhuis prove the conjecture for arbitrary o q Another version of the proof can be foun in Bruen Fisher [8] In [17] we prove the similar theorem for higher ( -th ) powers (instea of squares) Theorem If (q + 1), then in GF(q ) the only q-subsets with the property that the ifference of any two elements is always a -th power, are certain lines of GF(q ), consiere as the affine plane AG(,q) Note that if gc(,q 1) = 1, then every element is a -th power in GF(q ) If this is not the case an (q + 1), then GF(q) is not such a set The Paley graph P q is efine as follows: take V (P q ) = GF(q) as the vertex set, an the pair (x,y) is an ege in E(P q ) iff y x is a square element of GF(q) If 1 GF(q) is a non-square (so when q (mo 4)), then it is a irecte graph; on the other han, when 1 GF(q) is a square (so when q 1 (mo 4)), then it can be consiere as an unirecte graph as in this case x y is a square iff y x is a square These graphs have several nice properties, eg they are strongly regular, vertex-transitive, etc The natural generalization is the following: let P q, be the irecte graph on the vertices V = GF(q) with eges E = {(i,j) : i j GF(q) () } If 1 GF(q) (), then P q, can be consiere as an unirecte graph One may ask how similar Paley an generalize Paley graphs behave For example, these graphs are not strongly regular any more if >, see [1] They are very similar to ranom graphs; more precisely, they are ( 1,q/4 ) jumble (if q = r + 1), see Thomason [1] This property can be prove using Weil s estimate It is interesting to examine the largest cliques of Paley graphs This is the point where they eviate from ranom graphs In a ranom graph G1 (p) the expecte size of the largest clique is (1 + o(1))(log p)/log But the largest clique of P p is clog p log log log p for infinitely many primes p by a result of Graham an Ringrose [9] Assuming the Generalize Riemann Hypothesis we woul have clog p log log p infinitely often (Montgomery [15]) For prime-power q, in P q the largest clique can be substantially bigger than than log q An easy counting argument shows that the maximum possible size is q, an it is reache for every o square q, which was the case characterize by Blokhuis As one can see easily, the following theorem, which is the reformulation of Theorem, generalizes this situation for P q, Theorem 4 In the generalize Paley graph P q, the largest possible clique is of size q, an the clique of size q containing 0,1 GF(q ) consists of the vertices in GF(q) 4

5 A last remark is, that the eep result of Blokhuis, Ball, Brower, Storme, Szőnyi [6] may be use as an alternative way to prove the theorem above This alternative way oes not seem to be trivial or straightforwar however Instea of the quite complicate proof of Theorem 4 (or Theorem ) we prove an extenability result here In finite geometry results of this type are very important They, combine with estimates about the size of a given structure, can be formulate saying that in a class containing extremal ones of a given (sub-)structure (maximal arcs, minimal blocking sets, maximal sets with -th power ifferences, etc) there are certain gaps in the range of the possible sizes As an example we cite a theorem of Segre Bose prove that an arc can have at most q +1 or q + points accoring as q is o or even Then Segre prove that if the size of an arc is close to it (it is larger than q q/4 + 7/4 if q is o, an larger than q q + 1 if q is even), then it is containe in an arc of maximum carinality (For improvements on these bouns see eg Voloch []) Here we have a very similar situation as the following statement shows Theorem 5 Let be a given integer, 1 <, (q + 1); an S GF(q ) such that 0,1 S an S = q ε, where ε < (1 1 ) q Suppose that S has the property that the ifference of any two elements in it is always a -th power in GF(q ) Then S GF(q) GF(q ) Before proving this theorem, first we sketch the proof of a generalization of the main result of Szőnyi [19] Let D be a set of irections in AG(,q) A set U AG(,q) is calle a D-set if U etermines precisely the irections belonging to D Theorem 6 Let U be a D-set of AG(,q) consisting of q n points, where n α q an D < (q + 1)(1 α), 1/ α 1 Then U is incomplete, ie it can be extene to a D-set Y with Y = q We will sketch the proof of it only, after the following key lemma, which is a variant of a lemma in Szőnyi s paper: Lemma 7 Let C n ( n q+1 ) be a curve of orer n efine over GF(q), an enote by N the number of its points in PG(,q) Choose a constant 1 α Moreover, suppose that C n oes not contain a linear component efine over GF(q) an n α q Then N n(q + 1)α Proof of Lemma 7: Suppose first that C n is absolutely irreucible Then Weil s theorem ([], [10]) gives N q (n 1)(n ) q n(q + 1)α If C n is not absolutely irreucible, then it can be written as C n = D i1 D is, where D ij is an absolutely irreucible component of orer i j, so s j=1 i j = n If D ij can not be efine over GF(q), then it has at most N ij (i j ) i j (q + 1)α points in PG(,q) (see [10], Lemma 1011) If D ij is efine over GF(q), then the first part of the proof shows that N ij i j (q + 1)α Hence s s N N ij i j (q + 1)α = n(q + 1)α j=1 j=1 The proof of Theorem 6 goes as follows: we associate a polynomial H(X,Y ) to U such that it reflects its geometrical properties Then another polynomial (curve) f(x, Y ) is efine (in an 5

6 algebraic way), which in some sense completes H(X,Y ) Lemma 7 implies that f has some linear components; finally it is shown that such a linear component X + ay b means that U can be extene by the point (a,b) st U {(a,b)} etermines the same irections as U i Now we are reay for the Proof of Theorem 5: In our case the number of irections etermine by S is at most D q+1 (so α = 1 1 ) Now Theorem 6 gives that if S q (1 1 ) q, then S can be extene to a D-set of size q But, as we have seen it, the fact that the ifference of two elements in GF(q ) is a -th power, epens only on the irection they etermine in AG(,q) using the big fiel representation 4 A lemma on ranom-like behaviour In this section we prove a lemma, being interesting in itself, which is a common generalization of a result of Szőnyi an another by Babai, Gál an Wigerson In fact this lemma is a consequence of the character sum version of Weil s estimate In orer to formulate it, we nee a Definition 8 Let f 1 (x),,f m (x) GF(q)[x] be given polynomials We say that their system is power inepenent, if no partial prouct f s 1 i 1 f s i f s j i j (1 j m; 1 i 1 < i < < i j m; 1 s 1,s,,s j 1) can be written as a constant multiple of a -th power of a polynomial Equivalently, one may say that if any prouct f s 1 i 1 f s i f s j i j is a constant multiple of a -th power of a polynomial, then this prouct is trivial, ie for all the exponents s i,i = 1,,j Now Lemma 9 Let f 1 (x),,f m (x) GF(q)[x] be a set of power inepenent polynomials, where (q 1) If m m 1 eg(f i ) < q + q( m 1) ( 1) q + 1, then there is an x 0 GF(q) such that every f i (x 0 ) is a -th power in GF(q) for every i = 1,,m More precisely, if we enote the number of these x 0 -s by N, then 1 N q m ( 1 1 q + ) m eg(f i ) q(1 1 m) q m ( 1) For the sake of simplicity one can say that if m eg(f i ) < then N q m q m eg(f i ), if m 4 Note that this lemma implies that, uner some natural conitions, one can solve a system of equations χ (f i (x)) = δ i (i = 1,,m), where the δ i -s are -th complex roots of unity, an χ is the multiplicative character of orer So the -th power behaviour can be prescribe if the polynomials are inepenent 6

7 It can be interprete as being a -th power is like a ranom event of probability 1 Some wors about the conition (q 1): if an q 1 are co-primes, then every element is a -th power in GF(q) If gc(,q 1) = 1 an we write = 1 an gc(,q 1) =gc( 1, ) = 1, then the lemma can be applie with 1, as -th an 1 -th powers are the same in this case We remark that Szőnyi [0] prove this lemma for = L Babai, A Gál an Wigerson [1] have a similar result for linear polynomials We nee Result 10 (character sum version of Weil s estimate, [1], Thm 541) Let f(x) be a polynomial over GF(q) an r the number of istinct roots of f in its splitting fiel If χ e is a multiplicative character (of orer e) of GF(q) an f(x) cg(x) e, then χ e (f(x)) (r 1) q x GF(q) Proof of Lemma 9 : First note that we use the efinition χ(x) = χ (x) = x q 1 Let {ε 0 = 1,ε 1,ε,,ε 1 } be the set of -th complex roots of unity Define the following expression: H = m (χ(f i (x)) ε 1 )(χ(f i (x)) ε )(χ(f i (x)) ε 1 ) = x GF(q) m (χ(f i (x)) 1 + χ(f i (x)) + + χ(f i (x)) + 1) x GF(q) As N enotes the number of solutions, H is roughly m N (because the prouct (χ(f i (x)) ε 1 )(χ(f i (x)) ε )(χ(f i (x)) ε 1 ) is zero if χ(f i (x)) 1 or 0; it is ±1 if f i (x) = 0 an it is as big as iff χ(f i (x)) = 1) An error term comes from the zeros of the polynomials: m H m N m 1 eg(f i ) Let s examine H: H = q + m x GF(q) j=1 1 i 1 <<i j m 1 s 1,,s j 1 χ(f i1 (x) s 1 f i (x) s f ij (x) s j ) The secon term (which is a real integer in fact, but it is not important for us now) has absolute value less than H q m j=1 1 i 1 <<i j m 1 s 1,,s j 1 j ( eg(f ik )s k 1) q k=1 by Weil But this is equal to m q j=1 1 i 1 <<i j m j 1 eg(f ik )( 1) j 1 l k=1 l=1 7 q m j=1 ( ) m ( 1) j = j

8 Now, using the assumption we have ( 1) m ( ) m 1 q ( 1) j 1 m eg(f i ) q( m 1) = j 1 j=1 ( 1) m 1 q ( 1) m 1 q m eg(f i ) q( m 1) m m 1 eg(f i ) < q + q( m 1), q + 1 ( 1) m eg(f i ) q( m 1) + m 1 m eg(f i ) < q, so N > 0 an the existence of x 0 is prove For the inequality we can ivie the left han sie by m to get N q m ( 1 1 m q + ) eg(f i ) q(1 1 m) References [1] L Babai, A Gál an A Wigerson, Superpolynomial lower bouns for monotone span programs, Combinatorica (1999) 19, [] RD Baker, GL Ebert, J Hemmeter an A Wolar, Maximal cliques in the Paley graph of square orer, J of Stat Plan an Inf (1996) 56, [] M Biliotti an G Korchmáros, Transitive blocking sets in cyclic projective planes, Mit Math Sem Giessen (1991) 01 (Proceeings of the first international conference on blocking sets, e A Beautelspacher, F Eugeni, F Mazzocca), 5-8 [4] A Blokhuis, On subsets of GF(q ) with square ifferences, Inag Math (1984) 46, 69-7 [5] A Blokhuis, Blocking sets in Desarguesian planes, Combinatorics: Paul Erős is Eighty, Vol, János Bolyai Mathematical Society, Buapest (199), [6] A Blokhuis, S Ball, A Brouwer, L Storme an T Szőnyi, On the number of slopes etermine by a function on a finite fiel, J of Combin Theory, Ser A (1999) 86, [7] A Brouwer, A series of separable esigns with application to pairwise orthogonal Latin squares, Europ J Combinatorics (1980) 1, 9-41 [8] AA Bruen an JC Fisher, The Jamison metho in Galois geometries, Designs, Coes an Cryptography (1991) 1,

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