Enumerating permutation polynomials over finite fields by degree II

Finite Fields and Their Applications12(2006)26–37/locate/ffa Enumerating permutation polynomials overfinitefields by degree IISergei Konyagin a,Francesco Pappalardi b,∗a Department of Mechanics and Mathematics,Moscow State University,Vorobjovy Gory,119992Moscow,Russiab Dipartimento di Matematica,UniversitàDegli Studi Roma Tre,Largo S.L.Murialdo,1,I-00146Roma,ItalyReceived21July2003;revised31August2004Communicated by Gerhard TurnwaldAvailable online5February2005AbstractThis note is a continuation of a paper by the same authors that appeared in2002in the same journal.First we extend the method of the previous paper proving an asymptotic formula for the number of permutations for which the associated permutation polynomial has d coefficients in specifiedfixed positions equal to0.This also applies to the function N q,d that counts the number of permutations for which the associated permutation polynomial has degree<q−d−1. Next we adopt a more precise approach to show that the asymptotic formula N q,d∼q!/q d holds for d q and =0.03983.©2005Elsevier Inc.All rights reserved.Keywords:Permutation polynomials;Finitefields;Exponential sumsCorresponding author.Fax:+390654888080.E-mail addresses:konyagin@ok.ru(S.Konyagin),pappa@mat.uniroma3.it(F.Pappalardi).1071-5797/$-see front matter©2005Elsevier Inc.All rights reserved.doi:10.1016/j.ffa.2004.12.006S.Konyagin,F.Pappalardi/Finite Fields and Their Applications12(2006)26–3727 1.IntroductionLet F q be afinitefield with q>2elements.For any permutation of the elements of F q the permutation polynomialf (x)=c∈F q (c)1−(x−c)q−1∈F q[x](1)has the property that f (a)= (a)for every a∈F q.From the definition,it follows that for every ,the degree*(f ) q−2.In[2]the authors proved that if S(F q)denotes the group of permutations on F q, then#∈S(F q)|*(f )<q−2−(q−1)!2eq q.A similar result has also been proved by Das[1].Here we consider a more general function.Fix d integers k1,k2,...,k d with the property that0<k1<···<k d<q−1and defineN q(k1,...,k d)=#∈S(F q)|∀i=1,...,d,the k i-th coefficient of f is0,where by k-th coefficient of a polynomial f we mean its coefficient of x k. We will extend the method of[2]proving the following:Theorem1.Nq(k1,...,k d)−q!q d<1+1eq((q−k1−1)q)q/2.The above result also applies toN q,d=#∈S(F q)|*(f )<q−d−1sinceN q,d=N q(q−d−1,...,q−3,q−2). We also haveCorollary1.If d qlog q12log log q−log log log q,then N q,d∼q!/q d(q→∞).28S.Konyagin,F.Pappalardi/Finite Fields and Their Applications12(2006)26–37In the case when d is larger with respect to q,the above statement for N q,d can be improved.In Section3we will prove:Theorem2.Suppose =(e−2)/3e=0.08808···,and d< q.ThenN q,d−q!q d<2d dq2+q−dqd2dq−d(q−d)/2.Therefore we haveCorollary2.The asymptotic formula N q,d∼q!/q d holds for q→∞,d 0q and 0=0.03983is a suitable constant.2.The method of[2]—Proof of Theorem1The coefficient of x i in f (x)in(1)equals(−1)q−iq−1ic∈F qc q−i−1 (c)for i>0.Observe thatq−1i=(−1)i for i=1,...,q−1(the equality is consideredin F q;see[3,Exercise7.1]).Hence the j th coefficient of f (x)is0if and only ifc∈F qc q−j−1 (c)=0.We will follow the proof in[2]that uses exponential sums defining auxiliary functions for each S⊆F q:n S(k1,...,k d)=#⎧⎨⎩ff:F q−→S andc∈F qc q−k i−1f(c)=0,for i=1,...,d⎫⎬⎭.By inclusion–exclusion,it is easy to check thatN q(k1,...,k d)=S⊆F q(−1)q−|S|n S(k1,...,k d).(2)S.Konyagin,F.Pappalardi/Finite Fields and Their Applications12(2006)26–3729 Now we need to evaluate n S(k1,...,k d).If e p(u)=e2 iu p and Tr( )∈F p denotes the trace of ∈F q,thenn S(k1,...,k d)=1q d(a1,...,a d)∈F d qf:F q−→Se p⎛⎝c∈F qTrf(c)di=1a i c q−k i−1⎞⎠=1q(a1,...,a d)∈F d qc∈F qt∈Se pTrtdi=1a i c q−k i−1=|S|qq d+R S,(3)where|R S| q d−1q dmax(a1,...,a d)∈F d q\{0}c∈F qt∈Se pTrtdi=1a i c q−k i−1.Furthermore,since the geometric mean is always bounded by the arithmetic mean,c∈F qt∈Se pTrtdi=1a i c q−k i−1⎛⎝1qc∈F qt∈Se pTrtdi=1a i c q−k i−12⎞⎠q/2⎛⎝1qu∈F q(q−k1−1)t∈Se p(Tr(tu))2⎞⎠q/2.By the identityu∈F qt∈Se p(Tr(tu))2=q|S|(4)we getc∈F qt∈Se pTrtdi=1a i c q−k i−1((q−k1−1)|S|)q/2.30S.Konyagin,F.Pappalardi/Finite Fields and Their Applications12(2006)26–37 Using inclusion–exclusion for counting the mappings F q→F q we see thatS⊆F q(−1)q−|S||S|q=q!.(5)Now let us plug the estimate for|R S|in Eq.(3)and then in Eq.(2).By(5)we obtainNq(k1,...,k d)−q!q d=N q(k1,...,k d)−S⊆F q(−1)q−|S|q d|S|qqd−1q dS⊆F q((q−k1−1)|S|)q/2.Next,using the inequality1+x e x(6) we getNq(k1,...,k d)−q!q d<(q−k1−1)q/2qu=0u q/2qu(q−k1−1)q/2qu=0quq exp−q−uqq/2 =((q−k1−1)q)q/2qu=0qu1eq−u=((q−k1−1)q)q/21+1eq.This completes the proof.Proof of Corollary1.From Theorem1since N q,d=N q(q−d−1,q−d,...,q−2), we haveN q,d−q!q d<2q(dq)q/2.S.Konyagin,F.Pappalardi/Finite Fields and Their Applications12(2006)26–3731 By the Stirling formulalim q→∞q!qeq√2 q=1,(7)we obtain that the error term is dominated by the main term if(dq)q/22q=o1q dqeq√q(q→∞).The above is satisfied ifd q =o1q(q→∞).(8)Replace d=u·q log log qlog qand obtain that(8)holds foru 12−log log log qlog log qand q large enough.3.Proof of Theorem2The key ingredient is to use the special properties of N q,d to estimate(3)more accurately.Assume that d is a positive integer and d<q.Let F q(d)denote the vector space of polynomials with degree up to d with constant term equal to0.If P(x)∈F q(d),then define(P)=minT⊂F q,|T|=d|P(T)|.In view of identity(5)and from(2)and(3)we can write:N q,d−q!q d=N q,d−S⊆F q(−1)q−|S||S|qq d=1qS⊆F q(−1)q−|S|P∈F q(d)\{0}c∈F qt∈Se p(Tr(tP(c)))32S.Konyagin,F .Pappalardi /Finite Fields and Their Applications 12(2006)26–37=1qdS ⊆F q(−1)q −|S |dP ∈F q (d)\{0} (P )=c ∈F qt ∈Se p (Tr (tP (c))).(9)In order to estimate the above we will need some lemmas.Lemma 1.Given an integer ∈N ,let H be the set of polynomials P ∈F q (d)such that (P )= .Then|H | dq qd.Proof.Let us fix one of the q dsubsets of F q with d elements and denote it with T .It is sufficient to show that the number of polynomials P ∈F q (d)with |P (T )|= is at most d q.There are qchoices for the set U =P (T ).Take an arbitrary a ∈F q \T and observe that the polynomials of degree d correspond to functions T ∪{a }→F q .Therefore,the number of polynomials P of degree d such that P (T )⊂U is equal to q d and the number of polynomials P of degree d such that |P (T )|= is atmost q d q.Finally,observe that |P (T )|does not change if we add a constant to any polynomial P .Thus the number of polynomials P ∈F q (d)such that |P (T )|= is at most d qas required.Lemma 2.Assume that d q/3and S ⊆F q .If P ∈F q (d)is such that (P ) 2,thenc ∈F q t ∈Se p (Tr (tP (c))) q 2 (q +d)/2 d −1q q −d (q −d)/2while if (P )=1,P =0,thenc ∈F q t ∈Se p (Tr (tP (c))) q 2 (q +d)/2 dq q −d (q −d)/2Proof.Assume (P ) and writeP (F q )={u 1,u 2,...}S.Konyagin,F.Pappalardi/Finite Fields and Their Applications12(2006)26–3733 where the order is chosen in such a way that#P−1(u1) #P−1(u2) ···.Note that since (P) ,#P−1(u1)+···+#P−1(u −1)<d.(10) Therefore for j= −1, ,...we have#P−1(u j)<dprovided that 2.By(10),there is a set T⊂F q such that|T|=d andT⊃{t:P(t)∈{u1,...,u −1}}.Hence,for each c∈T we have that P(c)∈{u ,u +1,...}.Therefore there are at most d/( −1)elements c of F q with P(c )=P(c).If =1then we use the observation that there are at most d elements c with this property.Denoting =max( ,2),we see that there are at most d/( −1)elements c of F q with P(c )=P(c).Now,bounding from the above the geometric mean with the arithmetic mean,we deducec∈F qt∈Se p(Tr(tP(c)))=c∈Tt∈Se p(Tr(tP(c)))×c∈F q\Tt∈Se p(Tr(tP(c)))|S|d⎛⎝1q−dc∈F q\Tt∈Se p(Tr(tP(c)))2⎞⎠(q−d)/2|S|d⎛⎝1q−dd−1u∈F qt∈Se p(Tr(tu))2⎞⎠(q−d)/2=|S|d1q−dd−1q|S|(q−d)/2,where we used once again(4).This concludes the proof of the lemma in the case |S| q/2.34S.Konyagin,F.Pappalardi/Finite Fields and Their Applications12(2006)26–37If|S|>q/2then,by the identityt∈F qe p(Tr(tP(c)))=0for P(c)=0,we can reduce the product for S to the product for F q\S.Also, the identity combined with the supposition d<q shows that the product is zero for |S|=q.Thus,we can consider q/2<|S|<q.If is the number of the zeros of the polynomial P in F q then we havec∈F qt∈Se p(Tr(tP(c)))=|S|q−|S|c∈F qt∈F q\Se p(Tr(tP(c))).Therefore,c∈F qt∈Se p(Tr(tP(c)))|S|q−|S|d(q−|S|)d1q−dd−1q(q−|S|)(q−d)/2=(|S|(q−|S|))d(q−|S|)(q−3d)/21q−dd−1q(q−d)/2q24dq2(q−3d)/2 1q−dd−1q(q−d)/2,as required.We have taken into account that(q−3d)/2 0by our supposition.The proof of the lemma is complete.Lemma3.If d<e−23eq then we have the following estimate:d =21−1(q−d)/2·qd 2d−1(d−1)q2.Proof.The lemma is trivial for d=1.Consider d>1and define the functionf( )=1−1(q−d)/2·qd.Let d=uq,u<e−23e.Taking into account(6),we havef( +1)/f( )=−1(q−d)/2+1dq+1q−1S.Konyagin,F.Pappalardi/Finite Fields and Their Applications12(2006)26–3735<−1(1−3u)q/2q<exp−(1−3u)q2q.By our supposition,(1−3u)/2>1/e.Denoting v=q/ we havef( +1)/f( )<ve−v/e.But,by(6),ve−v/e 1for any v.Therefore,f( +1)/f( )<1.Thus,the function f( )is decreasing for2 d.Hence,d =21−1(q−d)/2·qd=d=2f( ) (d−1)f(2).This implies the statement of Lemma3. We are in the condition to prove Theorem2.Proof of Theorem2.In view of Lemma2(recall that d<e−23e q)and Lemma1,theabsolute value of the right-hand side of(9)isS⊆F q1q dqdq2(q+d)/2 dqq−d(q−d)/2⎛⎝q+d=21−1(q−d)/2·qd⎞⎠.By Lemma3,we obtain that the above is<2qq dqdq2(q+d)/2 dqq−d(q−d)/22d dq2=2d dq2+q−d qd2dq−d(q−d)/2.(11)This concludes the proof by using(9).4.Range of uniformity(proof of Corollary2)In this last section we want to establish how large d can be in order for the asymptotic formula N q,d∼q!/q d to hold.36S.Konyagin,F .Pappalardi /Finite Fields and Their Applications 12(2006)26–37Since 0< ,we can use Theorem 2.Substitute d = q ( 0)in (11)and note that in order to have an asymptotic formula it is enough to verify that2 q q 3+q qq 2 1− q(1− )/2=o(q !)(q →∞).(12)From the Stirling formula (7),qq > 1 (1− )1− q ,we obtain that (12)is satisfied uniformly over 0if we havesup ∈(0, 0]g( )<−1,(13)whereg( )=log 2 −log (1− )1− +log ⎛⎝ 2 1−1− 2⎞⎠.It is easy to see that the functionslog 2 ,−log (1− )1− ,log ⎛⎝ 21− 1− 2⎞⎠are increasing on (0,1/3].Hence,g is also increasing,and (13)follows from the inequality g( 0)<−1.This completes the proof of the corollary.In a future paper we are planning to adopt the method of Theorem 2to prove an asymptotic formula for N q (k 1,...,k d )with arbitrary k 1,...,k d where d is fixed or grows slowly as q →∞.It would be of interest to enumerate permutation polynomials with degree approx-imately √q .Unfortunately our approach cannot reach this range since we know that for q odd there are no permutation polynomials of degree (q −1)/2(see[3,Corollary 7.5]).AcknowledgmentsThe authors are grateful to the referee for a careful reading of the paper.Due to his (or her)remarks,a series of shortcomings have been corrected.Also,we would like toS.Konyagin,F.Pappalardi/Finite Fields and Their Applications12(2006)26–3737 thank Igor Shparlinski and Mike Zieve for useful comments on thefirst draft of this paper.The project was performed during the visit of thefirst author to the UniversitàRoma Tre in April2003supported in part by G.N.S.A.G.A.from Istituto Nazionale di Alta Matematica.References[1]P.Das,The number of permutation polynomials of a given degree over afinitefield,Finite FieldsAppl.8(2002)478–490.[2]S.Konyagin,F.Pappalardi,Enumerating permutation polynomials overfinitefields by degree,FiniteFields Appl.8(2002)548–553.[3]R.Lidl,H.Niederreiter,Finite Fields,second ed.,Cambridge University Press,Cambridge,1997.。