Some extensions of Alon s Nullstellensatz

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1 Some extensions of Alon s Nllstellensatz arxiv: v2 [math.co] 15 Ag 2011 Géza Kós Compter and Atomation Research Institte, Hngarian Acad. Sci; Dept. of Analysis, Eötvös Loránd Univ., Bdapest kosgeza@szta.h Tamás Mészáros Dept. of Mathematics, Central Eropean University Meszaros Tamas@ce bdapest.ed Janary 18, 2013 Lajos Rónyai Compter and Atomation Research Institte, Hngarian Acad. Sci. Dept. of Algebra, Bdapest Univ. of Technology and Economics lajos@ilab.szta.h This paper is dedicated to professors Kálmán Győry, Attila Pethő, János Pintz, and András Sárközy, on the occassion of their rond birthdays. Abstract Alon s combinatorial Nllstellensatz and in particlar the reslting nonvanishing criterion is one of the most powerfl algebraic tools in combinatorics, with many important applications. The nonvanishing theorem has been extended in two directions. The first and the third named athors proved a version allowing mltiple points. Micha lek established a variant which is valid over arbitrary commtative rings, not merely over sbrings of fields. In this paper we give new proofs of the latter two reslts and provide a common generalization of them. As an application, we prove extensions of the theorem of Alon and Füredi on hyperplane coverings of discrete cbes. 1 Introdction Alon s combinatorial Nllstellensatz Theorem 1.1 from [2] and the reslting nonvanishing criterion Theorem 1.2 from [2] is one of the most powerfl algebraic tools in combinatorics. It has several beatifl and strong applications, see [7], [9], [10], [11], [12], [15], [17] for some recent examples. Let F be a field, S 1,S 2,...,S n be finite nonempty sbsets of F. Let F[x] = F[x 1,...,x n ] stand for the ring of polynomials over F in variables x 1,...,x n. Alon s theorem is a specialized, precise version of the Hilbertsche Nllstellensatz for the ideal IS of all polynomial fnctions vanishing on the set S = S 1 S 2 S n F n, and for the basis f 1,f 2,...,f n of IS, where f i = f i x i = s S i x i s F[x] 0 Mathematics Sbject Classification MSC2010: 05-XX, 05E40, 12D10. Key words and phrases: Combinatorial Nllstellensatz, polynomial method, interpolation, mltiset, zero divisor, mltiple point, covering by hyperplanes. Research spported in part by OTKA grants NK 72845, K77476, and K

2 2 Géza Kós, Tamás Mészáros, and Lajos Rónyai for i = 1,...,n. From this a simple and widely applicable nonvanishing criterion Theorem 1.2 in [2] has been dedced. It provides a sfficient condition for a polynomial f F[x] for not vanishing everywhere on S. Herewith we consider extensions of the latter reslt in two directions. Before formlating these, we need first some notation and definitions. Let N denote the set of nonnegative integers, and let n be a fixed positive integer. Throghot the paper R will denote a commtative ring with 1, as sal, and F stands for a field. Vectors of length n are denoted by boldface letters, for example s = s 1,...,s n R n stands for points in the space R n. For vectors a,b N n, the relation a b etc. means that the relation holds at every component. We se the same notation for constant vectors. e.g. 0 = 0,0,...,0 or 1 = 1,1,...,1. For w N n, we write x w for the monomial x w x wn n R[x 1,...,x n ]. If s R n, then x s w stands for the polynomial x 1 s 1 w 1...x n s n wn. as Itiswellknownthatforanarbitrarys R n wecanexpressapolynomialfx R[x 1,...,x n ] fx = N n f sx s, 1 where the coefficients f s R are niqely determined by f, and s. In particlar we have f 0 s = fs for all s R n. Observe that if n degf, then f = f s does not depend on s. Sppose now that S 1,S 2,...,S n are nonempty finite sbsets of R, and assme frther that we have a positive integer mltiplicity m i s attached to every element s S i. This way we can view the pair S i,m i as a mltiset which contains the element s S i precisely m i s times. We shall consider the sm d i = ds i := s S i m i s as the size of the mltiset S i,m i. We pt S = S 1 S 2 S n. For an element s = s 1,...,s n S we set the mltiplicity vector ms as m 1 s 1,...,m n s n, and write ms = m 1 s 1 + +m n s n. We formlate first a version of Alon s powerfl nonvanishing theorem Theorem 1.2 in [2] for mltiple points over fields. From this one can obtain Alon s reslt by setting m i s = 1 identically. Theorem 1. Let F be a field, f = fx 1,...,x n F[x 1,...,x n ] be a polynomial of degree n t i, where each t i is a nonnegative integer. Assme, that the coefficient in f of the monomial x t 1 1 x t 2 2 x tn n is nonzero. Sppose frther that S 1,m 1,S 2,m 2,...,S n,m n are mltisets of F sch that for the size d i of S i,m i we have d i > t i i = 1,...,n. Then there exists a point s = s 1,...,s n S 1 S n and an exponent vector = 1,..., n with i < m i s i for each i, sch that f s 0. Theorem 1 was first proved in [13]. In another direction, over an arbitrary commtative ring R Micha lek proved the following extension of the nonvanishing theorem: Theorem 2. Let R be a commtative ring, and let fx be a polynomial in R[x 1,...,x n ]. Sppose the degree degg of f is n t i, where t i is a nonnegative integer, and sppose the coefficient of n xt i i in f is nonzero. Sppose frther that S 1,...,S n are sbsets of R with S i > t i, and with the property that if s s S i, then s s is a nit in R. Then there exists a vector s S = S 1 S n, sch that

3 Some extensions of Alon s Nllstellensatz 3 fs 0. In the case R = F, when we work over a field, the above reslt specializes to Alon s nonvanishing theorem; note that if R is a field, then s s is always a nit, whenever s and s are different. We give two new proofs of Theorem 2. The first one will be on the basis of an identity obtained from an interpolation argment. The second proof follows closely the original line of reasoning by Alon. In fact, for a sbset S of R n let s denote the set of polynomials from R[x 1,...,x n ] vanishing at all s S by IS. It is easy to see that IS is an ideal of R[x 1,...,x n ]. Now Theorem 2 will be a simple conseqence of the following reslt, which can be considered as an extension of Alon s Nllstellensatz. Theorem 3. Let R be a commtative ring, and let fx be a polynomial from R[x 1,...,x n ]. Let S 1,...,S n be nonempty finite sbsets of R with the property that if s s S i, then s s is a nit in R, S = S 1 S n and define g i x i = s S i x i s. Then for every polynomial fx R[x 1,...,x n ] there are polynomials h 1,...,h n,r R[x 1,...,x n ] sch that degh i degf S i for all i, the degree of r is less then S i in every x i, for which n fx = rx+ h i xg i x i. Moreover, f IS if and only if r is identically zero, hence g 1 x 1,...,g n x n is a basis of IS. We have the following common generalization of Theorems 1 and 2: Theorem 4. Let R be a ring, f = fx 1,...,x n R[x 1,...,x n ] be a polynomial of degree n t i, where each t i is a nonnegative integer. Assme, that the coefficient in f of the monomial x t 1 1 x t 2 2 x tn n is nonzero. Sppose frther that S 1,m 1,S 2,m 2,...,S n,m n are mltisets of R sch that for the size d i of S i,m i we have d i > t i i = 1,...,n, and for each i any nonzero element s s from S S is a nit in R. Then there exists a point s = s 1,...,s n S 1 S n and an exponent vector = 1,..., n with i < m i s i for each i, sch that f s 0. In the next section we prove Theorem and 4, which implies Theorems 1, and 2. The proof will be based on an argment extending nivariate Hermite interpolation to or setting. In Section 3 we give an alternative proof for Theorem 2, wich will follow from Theorem 3. This line of reasoning is an adaptation of Alon s original proofs from [2]. In Section 4 we give two applications. These will be extensions of a theorem of Alon and Füredi on almost covering a discrete hypercbe by hyperplanes. In Section 5 some conclding remarks are given. 2 Proof of Theorem 4 Lemma 5. Let S,m be a nonempty mltiset in a commtative ring R sch that all nonzero elements s s in S S are nits, and let gx = s Sx s ms. For all polynomials fx R[x], the following statements are eqivalent: a f s = 0 for every s S and 0 < ms; b the polynomial x s ms divides fx for every s S; c gx divides fx.

4 4 Géza Kós, Tamás Mészáros, and Lajos Rónyai Proof. The relations a b and c b are trivial. To prove a,b c, apply an indction on S. The initial case S = 1 is trivial. Let n 2 and assme the statement of the Lemma for S = n 1. Choose an s 0 S arbitrarily. By b, there is a polynomial f R[x] sch that fx = x s 0 ms0 f x. Let g x = x s ms. We have to prove that g divides f. s S\{s 0 } First we show that fs = 0 for every s S \ {s 0 } and 0 < ms. Sppose the contrary, and take a pair s, for which f s 0 and is minimal, i.e. f 0 s = f 1 s =... = s = 0. Then f 1 f s = x s 0 ms0 f x s = v=0 x s 0 ms 0 v s f vs = s s 0 ms f s. Since s s 0 is a nit in R and f s 0, this contradicts f s = 0. So we have f s = 0 for every s S\{s 0} and 0 < ms. By the indction hypothesis, f is divisible by g. Lemma 6. Let S,m be a nonempty mltiset in a commtative ring R sch that all nonzero elements s s in S S are nits, and let s 0 S and 0 0 < ms 0. Then there exists a polynomial h s 0, 0 x R[x] with degh s 0, 0 < ds sch that for every s S and 0 < ms we have h s 0, 0 s = Proof. For every v = 0,1,...,ms 0 1 let f v x = x s 0 v { 1 if s = s 0 and = 0 ; 0 otherwise. s S\{s 0 } ms x s. s 0 s These axiliary polynomials have the following obvios properties: For every s S \{s 0 } and every v, the polynomial f v x is divisible by x s ms, so f v s = 0 for all 0 < ms. For every 0 < v, since f v x is divisible by x s 0 v, we have f v s 0 = 0. For every v we have f v v s 0 = 1. Now we can constrct h s 0, 0 s as a linear combination of the axiliary polynomials, indctively: if h s 0, 0 +1,...,h s 0,ms 0 1 are already defined then let h s 0, 0 x = f 0 x 0 <<ms 0 f 0 s 0 h s0, x. Lemma 7 Hermite interpolation. Let S,m be a mltiset in a commtative ring R sch that all nonzero elements s s in S S are nits. For each s S and 0 < ms, let y s, be an arbitrary element in R. Then

5 Some extensions of Alon s Nllstellensatz 5 a there exists a niqe polynomial fs R[x], with degf < ds, satisfying f s = y s, for every pair s,; b this polynomial can be constrcted as Proof. Let f = s S f s = r S <ms v<mr f = s S <ms y s, h s,. y s, h s,. For every s S and < ms we have y r,v h r,v s = r S v<mr so the polynomial f satisfies the reqested property. { } 1 if r = s and v = y 0 otherwise r,v = y s, Fortheniqeness, sppose that thereexists anotherpolynomial f with thesameproperty. Then, for all s S and < ms we have f f s = f s f s = 0. By Lemma 5, this implies that f f is divisible by s Sx s ms. Since the degree of the latter polynomial is ds and its leading coefficient is 1, this contradicts degf f < ds. Lemma 8. Let S,m be a mltiset in a commtative ring R sch that all nonzero elements s s in S S are nits, and let t = ds 1. Then there exist elements αs, R for all s S and 0 < ms with the following property: for every l 0, s S 0 <ms Here the symbol means ndetermined. l 0 if l < t; αs, s l = 1 if l = t; if l > t. Proof. Take the polynomials h s, x provided by Lemma 6, and let αs, be the coefficient of x t in the polynomial h s, x. For every s S we have x l = s+x s l = =0 l s l x s, and therefore x l s = l s l for every 0. For > l we have l = 0 and the negative exponent in s l does not matter. Now take an arbitrary l < ds, and apply Lemma 7 to the vales y s, = l s l. From these vales, Lemma 7 reconstrcts the polynomial x l : l s l h s, x = x l. s S <ms

6 6 Géza Kós, Tamás Mészáros, and Lajos Rónyai Comparing the coefficients of x t, we get s S <ms l αs, s l = { 0 if l < t; 1 if l = t. Proof of Theorem 4. Withot loss of generality, we may assme that t i = ds i 1 for every i = 1,2,...,n. Expand f as fx = c k x k. For every s S, from k fx = k k n c k s+x s = c k = n c k k k i s k i i i i s k i i i x s x s = we get f s = k c k n i s k i i i. For each i, by Lemma 8, there exist elements α i s, R for all s S i and 0 < m i s sch that l 0 if l < t i ; α i s, s l = 1 if l = t i ; s S i 0 <m i s if l > t i. Now let αs, = n α i s i, i and consider the following expression: s = s S <msαs,f s S = c k... k s 1 S 1 1 <m 1 s 1 = n c k α i s i, i k si Si i <m i s i ms n n α i s i, i c k i n s n S n n<m ns n s k i i i k α i s i, i = k i i s k i i i = s k i i i = n 0 if k i < t i c k 1 if k i = t i. if k i > t i Since degf = t t n, the last prodct is zero except for k = t when every factor is 1. Therefore, we have αs,f s = c t. s S <ms On the left-hand side, there stands a linear combination of the vales f s. Since c t 0 on the right-hand side, there mst be at least one nonzero among these vales.

7 Some extensions of Alon s Nllstellensatz 7 3 An alternative proof for Theorem 2 Here we intend to give a proof of Theorem 2 which follows closely the original line of reasoning from Alon [2]. Proof of Theorem 3. We denote by V the R modle of all fnctions from S to R. V is a free R modle, n rank R V = S = d i, where S i = degg i = d i. In fact, for s S we denote by f s the S R fnction tang vale 1 at s and 0 everywhere else in S. Then the set F = {f s s S} is a free generating set of V over R, and F = n d i = S. Next we observe, that every f s can be written as a polynomial from R[x 1,...,x n ], sing interpolation. For s = s 1,...,s n S we have f s x = n α S i,α s i x i αs i α 1 Note that since s i α S i, the element s i α is a nit in R, hence the definition of f s x makes sense. Consider the following set of monomials M = {x w ; w i d i 1, i = 1,...,n}. Set G = {g 1 x 1,...,g n x n }. An arbitrary polynomial f from R[x 1,...,x n ] can be redced with G. This means that an occrrence of the monomial x d i i is replaced by g i x i x d i i as long as it is possible. Note that this redction does not change f as a fnction on S. Clearly any f can be redced into a R-linear combination of monomials from M. In particlar, the elements of F are redced this way into a collection of S polynomials which are independent over R. Using also that M = S, we infer that M, as a set of fnctions from S to R, is also linearly independent over R. Now consider an arbitrary polynomial f from R[x 1,...,x n ], and redce f with G as mch as possible. Denote the reslting redced polynomial by r, which is a R-linear combination of monomials from M. The fact that f redces to r means that there are polynomials h 1,...,h n R[x 1,...,x n ] sch that degh i degf d i for all i, the degree of r is less than d i in every x i, and n fx = rx+ h i xg i x i. Becase of the linear independence of the monomial fnctions from M, we have that f IS if and only if r is the all zero linear combination. This concldes the proof.. We remark that the proof actally gives that G is a Gröbner basis of IS with respect to an arbitrary term order on R[x 1,...,x n ]. For an introdction to Gröbner bases the reader is referred to [1]. From Theorem 3 the original argment of Alon gives Theorem 2 qite simply. Below we reprodce Alon s proof for the convenience of the reader.

8 8 Géza Kós, Tamás Mészáros, and Lajos Rónyai An alternative proof of Theorem 2. Clearly we may assme that S i = t i +1 for all i. Sppose that the reslt is false, i.e. f IS, and define g i x i = s S i x i s. By Theorem 3 there are polynomials h 1,...,h n R[x 1,...,x n ] sch that degh j n t i degg j for all j, for which f = n h i g i. Here the degree of h i g i is at most degf, and if there are any monomials of degree degf in it, then they are divisible by x t i+1 i. It follows that the coefficient of n xt i i on the right hand side is zero. However by or assmption the coefficient of n xt i i on the left hand side is nonzero, and this contradiction completes the proof. 4 Two applications We can extend a reslt of Alon and Füredi [4] on the covering of a discrete cbe by hyperplanes in the following way. The original reslt is the special case when every mltiplicity is 1. Theorem 9. Let S 1,m 1,...,S n,m n be finite mltisets from the field F. Sppose that 0 S i, with m i 0 = 1 for every i, and H 1,...,H k are hyperplanes in F n sch that every point s S \{0} is covered by at least ms n+1 hyperplanes and the point 0 is not covered by any of the hyperplanes. Then k ds 1 +ds 2 + +ds n n. Proof. Let l j x be a linear polynomial defining the hyperplane H j, set fx = k l j x, and t i = ds i 1. Let and Px = n s S i \{0} x i s m is Fx = Px P0 f0 fx. Note that we have f0 0, becase the hyperplanes do not cover 0. If the statement is false, then the degree of F is t 1 +t 2 + +t n and the coefficient of x t 1 1 x tn n is 1. Theorem 1 applies with S 1,m 1,...,S n,m n and t 1,...,t n : there exists a vector s S, and an exponent vector < ms sch that F s 0. We observe that s cannot be 0, becase F0 = 0. Ths s mst have at least one nonzero coordinate, implying that P s = 0. Moreover, as s is a nonzero vector, fx mst vanish at s at least ms n + 1 times, implying that f s = 0 expand the prodct at s; for every term x s v obtained there will be an index j sch that v j m j s j. These facts imply that F s = 0, a contradiction. This finishes the proof. Next, as an application of Theorem 2, we present a generalization of Theorem 6.3. from [2] to the Boolean cbe over a commtative ring R. By a hyperplane H in R n we nderstand the set of zeros of a polynomial of the form a 1 x 1 + +a n x n b := a,x b, where a i,b R. j=1

9 Some extensions of Alon s Nllstellensatz 9 Theorem 10. Let R be a commtative ring, and let H 1,...,H m be hyperplanes in R n sch that H 1,...,H m cover all the vertices of the nit cbe {0,1} n R n, with the exception of 0. Let a i,x b i be the polynomial defining H i. If m b i 0, then m n. Proof. The proof is essentially the same as the one in [2]. Assme that the assertion is false: m < n, and consider the polynomial m n Px = 1 n+m+1 b j x i 1 j=1 m [a i,x b i ]. Thedegreeofthispolynomialisclearlyn,andthecoefficientof n x i inp is 1 n+m+1 m j=1 b j, which is nonzero by or assmption. By applying Theorem 2 to S i = {0,1}, t i = 1, we obtain a point s {0,1} n for which Ps 0. This point is not the all zero vector, as P vanishes on 0. Bt otherwise a i,s b i = 0 for some i as s is covered by an H i, implying that P does vanish on s, a contradiction. Alon and Füredi have obtained the preceding statement for the case R = F, sing the original nonvanishing argment. If we pt R = Z n for some sqare free integer n N, then an application of the original nonvanishing theorem for a sitable prime factor of n proves the statement. However, if n has sqare factors, then Theorem 10 appears to give a new reslt. 5 Conclding remarks Or interest in developing a version of the nonvanishing theorem for mltisets has grown ot of an attempt to prove Snevily s conjectre [16] in a particlar case. Let G be a finite grop of odd order and sppose that a 1,...,a k G are pairwise distinct and b 1,...,b k G are pairwise distinct. Snevily s Conjectre states that there is a permtation π of the indices 1,2,...,k for which a 1 b π1,a 2 b π2,...,a k b πk are pairwise distinct. The conjectre has been proved for cyclic grops of prime order by Alon [3], for cyclic grops by Dasgpta, Károlyi, Serra, and Szegedy [8] and recently for all commtative grops by Arsovs [5]. Note that Alon s proof, which is based on the Combinatorial Nllstellensatz, allows one of the seqences a 1,...,a k and b 1,...,b k to contain repeated elements when k < G. Or approach was the following. Let N be a normal sbgrop of G. We look for the permtation to be applied to the factor grop G/N first. Of corse, in the factor grop, some of the cosets a 1 N,...,a n N andb 1 N,...,b n N may coincide toeach other; it is even possible that a 1 N,...,a n N all coincide and b 1 N,...,b n N all coincide. So we mst allow a i Nb πi N = a j Nb πj N in some cases. The idea is to allow a i Nb πi N = a j Nb πj N only in those cases, when a i N = a j N and b πi N = b πj N holds. Sppose that we fond sch a permtation π. If some classes a i1 Nb πi1 N,a i2 Nb πi2 N,...,a il Nb πil N coincide, then the elements a i1,...,a il are all in the same coset cn, and similarly b πi1,...,b πil are in the same coset Nd. Then we cold applysnevily sconjectreindctivelytotheseqencesc 1 a i1,...,c 1 a il andb πi1 d 1,...,b πil d 1 which all lie in N. By the indction hypothesis, the vales πj 1,...,πj l can be permted in sch a way, that c 1 a i1 b πi1 d 1,...,c 1 a il b πil d 1 are pairwise distinct. Hence the elements a i1 b πi1,...,a il b πil are all in the coset cdn = cnd, and they will be pairwise distinct. If we choose N to be a maximal normal sbgrop of G, then the order G/N = p will be a prime G is solvable, since G is odd. In this case we can se the additive grop of the prime field F p and re-formlate the existence of the desired permtation.

10 10 Géza Kós, Tamás Mészáros, and Lajos Rónyai Conjectre 11. Sppose that k p and a 1,...,a k F p and b 1,...,b k F p are arbitrary elements. Then there is a permtation π of the indices 1,2,...,k with the following property: a i = a j and b πi = b πj holds whenever a i +b πi = a j +b πj. This conjectre wold imply Snevily s conjectre for all cases when k is not greater than the smallest prime divisor of G. To se the polynomial method to prove Conjectre 11 and similar statements, it appears that one is reqired to handle mltiple vales in the seqences a 1,...,a k and b 1,...,b k. The reqirements in Theorems 2 and 4 abot the invertibility in R of the differences s s can be somewhat relaxed. The theorems still hold if we assme, instead of invertibility, that the mltiplicative monoid D R generated by the differences s s and f t1,...,t n does not contain 0. In this case the map from R to its ring of fractions D 1 R lifts or proofs from D 1 R to R the reader is referred to [6] for basic facts on rings of fractions. Actally, it sffices to assme that a certain specific prodct from D is not zero. References [1] W. W. Adams, P. Lostana, An introdction to Gröbner bases, American Mathematical Society, [2] N. Alon, Combinatorial Nllstellensatz, Combinatorics, Probability and Compting , [3] N. Alon, Additive Latin transversals, Israel J. Math , [4] N. Alon, Z. Füredi, Covering the cbe by affine hyperplanes, Eropean J. Combinatorics , [5] B. Arsovs, A proof of Snevily s conjectre, Israel Jornal of Math , [6] M. F. Atiyah, I. G. Macdonald, Introdction to commtative algebra, Addison-Wesley, [7] M. Cámara, A. Lladó, J.Moragas, On a conjectre of Graham and Häggkvist with the polynomial method, Eropean Jornal of Combinatorics , [8] S. Dasgpta, Gy. Károlyi, O. Serra, B. Szegedy, Transversals of additive Latin sqares, Israel J. Math , [9] B. Felszeghy, On the solvability of some special eqations over finite fields, Pblicationes Mathematicae Debrecen , [10] B. Green, T. Tao, The distribtion of polynomials over finite fields, with applications to the Gowers norms, Contribtions to Discrete Mathematics , [11] Gy. Károlyi, An inverse theorem for the restricted set addition in abelian grops, Jornal of Algebra ,

11 Some extensions of Alon s Nllstellensatz 11 [12] Gy. Károlyi, Restricted set addition: the exceptional case of the Erdős-Heilbronn conjectre, Jornal of Combinatorial Theory, Ser. A , [13] G. Kós, L. Rónyai, Alon s Nllstellensatz for mltisets, to appear in Combinatorica. [14] M. Micha lek, A short proof of Combinatorial Nllstellensatz, Amer. Math. Monthly , [15] H. Pan, Z-W. Sn, A new extension of the Erdős-Heilbronn conjectre, Jornal of Combinatorial Theory, Ser. A , [16] H. Snevily, Unsolved Problems: The Cayley Addition Table of Z n, Amer. Math. Monthly , [17] Z-W. Sn, On vale sets of polynomials over a field, Finite Fields and Applications ,