Partitions of nonzero elements of a finite field into pairs
aa r X i v : . [ m a t h . C O ] M a r PARTITIONS OF NONZERO ELEMENTS OF A FINITE FIELD INTOPAIRS
R.N. KARASEV AND F.V. PETROV
Abstract.
In this paper we prove that the nonzero elements of a finite field with oddcharacteristic can be partitioned into pairs with prescribed difference (maybe, with somealternatives) in each pair. The algebraic and topological approaches to such problems areconsidered. We also give some generalizations of these results to packing translates in afinite or infinite field, and give a short proof of a particular case of the Eliahou–Kervaire–Plaigne theorem about sum-sets. Introduction
In this paper we prove several theorems on combinatorics of finite fields. Denote F p thefinite field of size p , where p is a prime. As it was shown in [20], if p is an odd prime, thepartitioning F p ∗ into pairs with strictly prescribed differences is possible. Let us denote[ n ] = { , , . . . , n } and give the formal statement. Theorem 1.
Let p be an odd prime, m = p − . Suppose we are given m elements d , d , . . . , d m ∈ F p ∗ . Then there exist pairwise distinct x , . . . , x m , y , . . . , y m ∈ F p ∗ suchthat for every i = 1 , . . . , m we have y i − x i = d i . In this paper we present new proofs of this theorem, using algebraic and topologicaltechniques. We also prove some generalizations of Theorem 1, for example the followingresult on packing translates in a field.
Definition 1.
Let F be some field. For X ⊆ F and t ∈ F denote X + t = { x + t : x ∈ X } , and for X ⊆ F and Y ⊆ F denote X ± Y = { x ± y : x ∈ X, y ∈ Y } . Mathematics Subject Classification.
Key words and phrases.
Finite fields, Combinatorial Nullstellensatz, the Borsuk–Ulam theorem.The research of R.N. Karasev is supported by the Dynasty Foundation, the President’s of RussianFederation grant MK-113.2010.1, the Russian Foundation for Basic Research grants 10-01-00096 and 10-01-00139, the Federal Program “Scientific and scientific-pedagogical staff of innovative Russia” 2009–2013.The research of F.V. Petrov is supported by the Russian Foundation for Basic Research grant 08-01-00379.
Theorem 2.
Suppose F is a field, m and d are positive integers such that ( md )!( d !) m = 0 in F .Let X , . . . , X m and T , . . . , T m be subsets of F such that ∀ i < j | X i − X j | ≤ d, ∀ i | T i | ≥ ( m − d + 1 . Then there exists a system of representatives t i ∈ T i such that the sets X + t , . . . , X m + t m are pairwise disjoint. In particular, if F = F p , T i = F p , and md < p , then we can translate the sets X , . . . , X m ⊂ F p so that they become pairwise disjoint, provided | X j − X i | ≤ d forall i < j . Theorem 1 is a particular case of Theorem 2 with F = F p , d = 2 , X i = { , d i } ,and T i = F \ { , − d i } .Let us return to partitions into pairs of other finite Abelian groups. For finite fields ofsize p k (we treat them as F p -vector spaces) the differences cannot be prescribed strictly.The simple counterexample is when the difference is the vector d = (1 , , , . . . ,
0) for allpairs, then every line { v + td } t ∈ F p can have at most ⌊ p ⌋ pairs and the partition is impossible.The same obstruction arises if we try to generalize Theorem 1 for the rings Z / ( n ) for oddcomposite n . Here we may require all d i to be d = n/p , where p is prime divisor of n . It isclear that in every set { v + dt } t ∈ F p one element would be not paired. Further conjecturesfor partitions of Z / ( n ) are discussed in Section 6.Now return to the positive results. In the case of the finite field of size p k (treated as F p -vector space here), it is sufficient to give some alternatives for each pair, which is donein the following theorem. Theorem 3.
Let p be an odd prime, and let V be the F p -vector space of dimension k .Denote V ∗ = V \ { } and put m = | V ∗ | / p k − . Suppose we are given m linear basesof the vector space V ( v , . . . , v k ) , ( v , . . . , v k ) , . . . , ( v m , . . . , v mk ) Then there exist pairwise distinct x , . . . , x m , y , . . . , y m ∈ V ∗ and a map g : [ m ] → [ k ] suchthat for every i = 1 , . . . , m we have y i − x i = v ig ( i ) . The paper is organized as follows. We discuss different proofs of Theorem 1 in Section 2.In Section 3 we give a new algebraic proof of a lemma on discriminant-like polynomials,known as the Dyson conjecture [8], which is used in the proofs of Theorems 1 and 2. InSection 4 we prove Theorem 2. The topological proofs for Theorems 1 and 3 are given inSection 5. In Sections 6 and 7 we discuss some conjectures and similar results for cyclicgroups.Generally, the algebraic methods in combinatorics are known to be very useful andpowerful, see [11, 14, 1] for examples of their application. The topological methods incombinatorics and discrete geometry also proved to be very useful, starting from the lower
ARTITIONS OF NONZERO ELEMENTS OF A FINITE FIELD INTO PAIRS 3 bounds for the chromatic number of the Kneser graphs in [17], other examples of topologicalmethods can be found in [18, 24].The authors thank Noga Alon, Doron Zeilberger, and Michal Adamaszek for usefuldiscussion and comments. We thank D¨om¨ot¨or P´alv¨olgyi for drawing our attention to theknown proofs of Theorem 1; and we thank the unknown referee for numerous remarks,corrections, and references.2.
The algebraic proofs of Theorem 1
First, we sketch a simplified version of the proof in [20]. It is in the spirit of theCombinatorial Nullstellensatz [1], see also [2] for a similar proof of a theorem on distinctpairwise sums in F p .For a polynomial f ( x , x , . . . , x m ) ∈ F p [ x , x , . . . , x m ] denote Z f = X ( c ,...,c m ) ∈ F pm f ( c , c , . . . , c m ) . Recall the following lemma, often used to study Diophantine equations over finite fields.
Lemma 1. If deg f < m ( p − , then R f = 0 .Proof. This is well-known for m = 1 (recall the proof: it suffices to consider f ( x ) = x k ,1 ≤ k ≤ p −
2. There exists g ∈ F p ∗ , which is not a root of polynomial x k −
1, forsuch g we have S := P x ∈ F p f ( x ) = P x ∈ F p f ( gx ) = g k S , hence S = 0). In the generalcase, note that each monomial of degree less then m ( p −
1) has degree less then p − (cid:3) We interpret our problem as follows. We need to find elements c , c , . . . , c m from F p such that elements c i and c i + d i are all distinct and nonzero. That is, it suffices to provethat the following polynomial f takes non-zero values f ( x , . . . , x m ) == x . . . x m ( x + d ) . . . ( x m + d m ) Y i Let us discuss another approach to this theorem via the Combinatorial Nullstellensatz,similar to the technique of [2], this approach to Theorem 1 was also discovered indepen-dently in [16].Recall the Combinatorial Nullstellensatz. Theorem 4. Suppose a polynomial f ( x , x , . . . , x n ) over field F has degree at most c + c + · · · + c n , where c i are non-negative integers, and denote by C the coefficient at x c . . . x c n n in f (maybe, C = 0 ). Let A , A , . . . , A n be arbitrary subsets of F such that | A i | = c i + 1 for any i . Denote also ϕ i ( x ) = Q α ∈ A i ( x − α ) . Then (1) C = X α i ∈ A i f ( α , . . . , α n ) ϕ ′ ( α ) . . . ϕ ′ n ( α n ) In particular, if C = 0 , then there exists a system of representatives α i ∈ A i such that f ( α , α , . . . , α n ) = 0 .Proof. For n = 1, (1) just follows from the Lagrange interpolation formula, which gives therepresentation f ( x ) = X α ∈ A f ( α ) ϕ ( x ) ϕ ′ ( α )( x − α ) . By induction on n (1) also holds for any monomial of degree at most c i in each x i . Next,by linearity of both parts of (1) it suffices to prove (1) for h := f − Cx c . . . x c n n . Eachmonomial of h has degree less then c i for at least one index i . If we fix all values α j for j = i , then summation over α i ∈ A i gives 0, as follows again from the one-dimensionalcase. (cid:3) Remark. The formula (1) after multiplying by the common denominator holds also forcommutative rings with unity. Indeed, for fixed c i ’s both parts are some polynomials withintegral coefficients in x i ’s, elements of A i ’s and coefficients of f . Since (1) holds over R ,these polynomials should be identically equal.Let us use this theorem for the polynomial f = Y i 3, and A i = F p ∗ \ {− d i } . The only thing left to check isthat the coefficient of Q x m − i in f does not vanish. It equals the coefficient of Q x m − i in Q i The Combinatorial Nullstellensatz is often used for getting information on values ofpolynomials from the knowledge of their coefficients. But (1) allows to use it in otherdirection, as we show by deriving the Dyson conjecture. Theorem 5. Let a i , ≤ i ≤ n be positive integers. Denote by C the free term in Y ≤ i = j ≤ n (1 − x i /x j ) a i . In other words, with a = P a i , C equals the coefficient of Q x a − a i i in (2) f ( x , . . . , x n ) := Y ≤ i In notations of Theorem 4, we have c i = a − a i . The idea is to add terms of lowerdegree to f , it does not change the coefficient C , but may significantly change the RHSof (1). Also, we are free to choose A i . Let’s try to change f to ˜ f and choose A i so that˜ f takes unique non-zero value on Q A i . Put A i = { , , . . . , a − a i } . So, if x i ∈ A i , thenthe segment ∆ i := [ x i , x i + a i − 1] lies inside [0 , a − . . . ] segments of integers.Now we change f . Replace ( x j − x i ) a i + a j in formula (2) for f by C i,j ( x , . . . , x n ) := a j Y s = − a i +1 ( x j − x i + s ) . Non-vanishing of C i,j means that the segments ∆ i , ∆ j are disjoint and ∆ i may not be thesegment following ∆ j (that is, min ∆ i = max ∆ j + 1). All this together may happen onlyif ∆ , ∆ , . . . , ∆ n are consecutive segments [0 , a − a , a + a − . . . , [ a − a n , a − x i = β i := a + · · · + a i − .So, C equals C = Q ≤ i Choose some sets M ij ⊇ X j − X i of size exactly 2 d and consider the polynomial f ( x , . . . , x m ) = Y ≤ i The statement of Theorem 2 holds also if we replace the inequality | X i − X j | ≤ d by the following condition: the polynomial Y i Now we go to the topological proofs, as usual they use a certain generalization of theBorsuk–Ulam theorem.The general examples of using the Borsuk–Ulam theorem in combinatorics can be foundin [18, 24]. In particular, the topological proof of Theorem 1 uses the ideas in [22], wherethe lower bound on the number of Tverberg partitions is proved. The proof of Theorem 3uses the technique from [13], where the number of Tverberg partitions was estimated forthe case when the number of parts is a prime power.Let us state the generalized Borsuk–Ulam theorem that we need (see [21] for example). Lemma 2. Let G = ( F p ) k be the additive group. Let X and Y be G -CW-complexes withfixed point free action of G . Let X be n -connected and Y be n -dimensional. Then therecannot exist a continuous map f : X → Y , commuting with the action of G . ARTITIONS OF NONZERO ELEMENTS OF A FINITE FIELD INTO PAIRS 7 We are going to use this lemma in the case, when X and Y are simplicial complexes,the action of G and the map f are simplicial. In this case the spaces are indeed G -CW-complexes. Such a point of view allows to state everything purely in combinatorial terms,without appealing to topological spaces.Let us prove Theorem 1. Consider the following simplicial complex ( ∗ means join) K = V ∗ S ∗ · · · ∗ S m , where V is a discrete set equal to F p , S i is a one-dimensional subcomplex of V ∗ V , withedges of type ( x, x ) and ( x, x + d i ). Clearly K is a join of a discrete set and m circles(equivalently, a join of V with p − p − p − L , having the same vertices as V , and all subsets of ≤ p − f : K → L is defined naturally on vertices. Note theimportant thing: this map would be simplicial if the required permutation does not exist.Indeed, if a simplex in K is mapped to a non-simplex in L , then it is mapped onto theentire set V . Hence, it is p − v ∗ [ a , b ] ∗ · · · ∗ [ a m , b m ] ∈ V ∗ S ∗ · · · ∗ S m , and the set { v, a , b , . . . , a m , b m } equals V . Shifting by − v we have v = 0, and frombijectivity b i = a i + d i (the case a i = b i is impossible). Hence we may assume the contrary: f is simplicial.Note that the map f is F p -equivariant. Here we identify V = F p and consider the actionof V on itself by shifts and on S i by shifting both coordinates by the same value. Thusarises a free action on K (by shifting all the coordinates by the same value), and a freeaction on L by shifts. Besides, f maps a p − p − V = ( F p ) k and action of V on itself by shifts.Let K = V ∗ S ∗ · · · ∗ S m , where the complex S i has vertices V ∗ V and edges ( x, x ), ( x, x + v ij ) for all possible x ∈ V, j = 1 , . . . , k . It is essential that S i is connected iff { v ij } kj =1 linearly span V , whichis required in the theorem. In this case K is also p k − p k − L on vertices V is defined the same way, its simplices are all subsets of sizeat most p k − 1, hence it is p k − V on itself by shifts, and on S i by shifting both coordinates bythe same value. Thus arises a free action on K (by shifting all the coordinates by the samevalue). The action by shifts on L is only fixed point free (not free ) this time. In this caseLemma 2 is essentially needed, while for Theorem 1 we only need its simple particular casefor free actions (the Dold theorem, see [18] for example). The rest of the proof is the same,applying Lemma 2 for the group of shifts V = ( F p ) k . R.N. KARASEV AND F.V. PETROV Conjectures on partitions of Z / ( n )We conjecture the following generalizations of Theorem 1 for rings Z / ( n ). We denote by Z / ( n ) ∗ the invertible (coprime with n ) elements of Z / ( n ). Conjecture 1. Let n = 2 m + 1 be a positive integer. Suppose we are given m elements d , d , . . . , d m ∈ Z / ( n ) ∗ . Then there exists a partition of Z / ( n ) \ { } into pairs with differ-ences d , d , . . . , d m . The following conjecture was proposed (and verified for n = 24) by Michal Adamaszek(private communication). Conjecture 2. Let n = 2 m be a positive integer. Suppose we are given m elements d , d , . . . , d m ∈ Z / ( n ) ∗ . Then there exists a partition of Z / ( n ) into pairs with differences d , d , . . . , d m . Let us discuss the possible approach to these conjectures using the Combinatorial Null-stellensatz, based on ideas from [6]. Let us embed Z / ( n ) into C as the n -th roots of unity(denote these roots by C n ). Denote w = cos 2 πn + i sin 2 πn . The numbers d i are transformed into w i = w d i . Consider the polynomial F ( x , . . . , x n ) = Y ≤ i In the general case the authors cannot prove that this coefficient is nonzero, but in case n = p is a prime we obtain an alternative proof of Theorem 1 as follows. Let us divide (3)by Q ni =1 (1 − w i ) to obtain(4) X π ( w π + · · · + w m − − π ) · · · · · ( w π m m + · · · + w m − − π m m ) , and apply the following lemma. Lemma 3. Let f be a polynomial with integer coefficients, p be a prime, w be the p -th rootof unity. If f ( w ) = 0 , then f (1) is divisible by p .Proof. The minimal polynomial of w is g = 1 + w + ... + w p − . Hence f = hg , where h isa polynomial with integer coefficients, and therefore f (1) = h (1) g (1) = h (1) p . (cid:3) Now we note that after replacing all w i (by definition they are w d i ) in (4) by 1, we obtain X π (2 m − π − · · · · · (2 m − π m − , Every summand equals (2 m − m ! summands, hence the total valueequals m !(2 m − p .7. Some remarks on the sum-sets The technique of the previous section allows to give a short proof of a particular case ofthe Cauchy–Davenport type theorem from [3, 10] (see also [5, 7, 9, 1, 15]). The Cauchy–Davenport type theorems estimate the cardinality of A + B = { a + b : a ∈ A, b ∈ B } , where A and B are finite subsets of an Abelian group. The technique we are going touse was already used in [15] in application to the sum-sets problem (and the restrictedsum-sets problem). Definition 2. Define β p ( r, s ) to be smallest integer n such that p | (cid:0) nk (cid:1) for all k in therange n − r < k < s. Theorem 6. Let A, B ⊂ Z / ( p α ) , where p is prime. Then | A + B | ≥ β p ( | A | , | B | ) . Proof. The reasoning is essentially the same as in [15, Theorem 3, Lemma 5]. Consider Z / ( p α ) as the multiplicative group of solutions of the equation z p α − C . Put | A | = r and | B | = s . By considering − B instead of B , we pass to studying the set C = { a/b : a ∈ A, b ∈ B } and prove that | C | ≥ β p ( r, s ). Assume the contrary, | C | = n < β p ( r, s ), then the polynomial f ( x, y ) = Y c ∈ C ( x − cy ) is zero on A × B . By the definition of β p ( r, s ) there exists n − r < k < s such that (cid:18) nk (cid:19) = 0 mod p. Consider the coefficient in f ( x, y ) at the monomial x k y n − k , which is d = ( − n − k σ n − k ( c , . . . , c n ) , where σ is the elementary symmetric function, C = { c , . . . , c n } . By the CombinatorialNullstellensatz (Theorem 4) this coefficient d should be zero.Let us write every c i as a power of z , the primitive p α -th root of unity. Then d becomesa polynomial of z with integer coefficients, denote it by d ( z ). Note that d ( z ) = 0, andtherefore d ( z ) is divisible by the minimal polynomial of zr ( z ) = z p α − z p α − − p − X i =0 z ip α − . Substituting into the equality for polynomials with integer coefficients d ( z ) = r ( z ) q ( z )the value z = 1, we obtain( − n − k (cid:18) nk (cid:19) = ( − n − k σ b (1 , . . . , 1) = r (1) q (1) ≡ p, which is a contradiction with (cid:0) nk (cid:1) = 0 mod p . (cid:3) Remark. The Combinatorial Nullstellensatz can also prove this theorem (with the sameestimate β p ( · , · )) for groups of type ( Z p ) k . These groups are additive groups of fields andthe proof is even simpler compared to the above reasoning, the relevant coefficients of Q c ∈ C ( x + y − c ) are already equal to (cid:0) nk (cid:1) . A much stronger result is proved in [3, 10]:Theorem 6 holds for any finite Abelian p -group. References [1] N. Alon. Combinatorial Nullstellensatz. // Combin. Probab. 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E-mail address : r n [email protected] Roman Karasev, Dept. of Mathematics, Moscow Institute of Physics and Technology,Institutskiy per. 9, Dolgoprudny, Russia 141700 E-mail address : [email protected]@gmail.com