9 - The Combinatorial Nullstellensatz

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1 9 - The Combinatorial Nullstellensatz Jacques Verstraëte jacques@ucsd.edu Hilbert s nullstellensatz says that if F is an algebraically closed field and f and g 1, g 2,..., g m are polynomials in F[x 1, x 2,..., x n ] where f vanishes on the common zeroes of g 1, g 2,..., g m, then there exists a positive integer k and polynomials h 1, h 2,..., h m such that f k = m g ih i. The combinatorial nullstellensatz, due to Alon (1995) is a refinement of Hilbert s nullstellensatz to the case that the g i are linear polynomials. The practical consequence is as follows, with an elementary proof due to Michalek (2009). Here the degree of f F[x 1, x 2,..., x n ], denoted deg(f) is the maximum of t 1 + t t n such that x t 1 1 x t x tn n with a non-zero coefficient in f. is a monomial appearing Theorem 1 (Combinatorial nullstellensatz). Let F be an arbitrary field and let f F[x 1, x 2,..., x n ]. Suppose x t 1 1 x t x tn n is a monomial of degree deg(f) appearing with a non-zero coefficient in f, and suppose S i F are sets of size at least t i + 1 for i = 1, 2,..., n. Then there exists x S 1 S 2 S n such that f(x) 0. Proof. We proceed by induction on deg(f); the case deg(f) = 1 is easy. If deg(f) > 1, suppose f(s 1 S 2 S n ) = 0 and t 1 > 0, but that the statement of the theorem is true for all polynomials of degree deg(f) 1. For any s S 1, write f = (x 1 s)g + h using the Euclidean division algorithm, and h does not contain the variable x 1. Since f({s} S 1 S n ) = 0, h({s} S 1 S n ) = 0 which means h(s 1 S 2 S n ) = 0. Now deg(g) = deg(f) 1 and a largest degree monomial in g is x t x t x tn n. But g(s 1 \{s} S 2 S n ) = 0, contradicting the induction hypothesis. 1

2 1 Cauchy-Davenport Theorem The combinatorial nullstellensatz has many applications in additive number theory. Recall the sumset of sets A and B in an abelian group is A + B = {a + b : a A, b B}. The following classical theorem was first proved by Cauchy in Theorem 2 (Cauchy-Davenport Theorem). Let A, B Z/pZ. Then A + B min{p, A + B 1}. Proof. If A + B > p, then A (a B) for every a Z/pZ, which implies A+B = Z/pZ. Therefore we assume A + B p. Suppose A+B A + B 2, and let C be a set of size A + B 2 containing A + B. Let F = Z/pZ. Define f F[x, y) by f(x, y) = c C(x + y c). Note deg(f) = C = A + B 2. Since A + B C, f is zero on F F. By the combinatorial nullstellensatz, the coefficient of x A 1 y B 1 in f must be zero. However, that coefficient is ( C A 1) which is not zero in F. Define (A + B) to be {a + b : a A, b B\{a}} in other words it is the sumset minus the diagonal. Then the above proof can be modified to show (A + B) min{p, A + B 3}. We leave this as an exercise. 2 Alon-Füredi Theorem Let Q n denote the n-dimensional cube in R n namely the set of all points in {0, 1} n. A hyperplane covers a point if it contains that point. In other words, if a, x = b defines the hyperplane, then a point P is covered if a, P = b. The following theorem was proved by Alon and Füredi: Theorem 3. Let H 1, H 2,..., H m be hyperplanes covering Q n \{0}. Then m n. Proof. Let a i, x = b i be the equations of the hyperplanes, and define f = m n b j j=1 (1 x i ) m (b j a j, x ). j=1 2

3 Then f R[x 1, x 2,..., x n ] and f(0) = 0 and deg(f) = max{m, n}. Furthermore, if the H i cover Q n \{0}, then is zero on {0, 1} n. If m < n, then the coefficient of x 1 x 2... x n is ( 1) n m j=1 b j, and since then deg(f) = n, this contradicts the combinatorial nullstellensat. Therefore m n. 3 Chevalley-Warning Theorem Here is an application to a theorem in algebraic geometry: Theorem 4 (Chevalley-Warning Theorem). Let f 1, f 2,..., f m F q [x 1, x 2,..., x n ] be homogeneous polynomials such that n > m deg(f i). Then there exists x 0 such that for all i [m], f i (x) = 0. Proof. Consider f = m (1 f q 1 ) i n (1 x q 1 i ). The condition on degrees of the f i implies deg(f) = (q 1)n, and x q 1 1 x q x q 1 n is a largest degree monomial in f. By the nullstellensatz, with S i = F q for i = 1, 2,..., n, there exists x S 1 S 2 S n such that f(x) 0. Note x 0 since f(0) = 0. Now since x 0, the second term in f disappears by Fermat s theorem, and the first is non-zero only if f i (x) = 0 for all i = 1, 2,..., m, again by Fermat s Theorem. 4 Erdős-Ginzburg Ziv Theorem For a finite group Γ, let s (Γ) be the minimum positive integer n such that any sequence of n elements of Γ contains a subsequence of Γ elements whose product is the identity. For instance, if Γ = Z/pZ, then s (Γ) 2p 1, since no sum of elements from the multiset of p 1 ones and p 1 zeroes is zero. It turns out this is tight: Theorem 5 (Erdős-Ginzburg-Ziv). s (Z/pZ) = 2p 1. Proof. Let (a 1, a 2,..., a 2p 1 ) be a multiset of 2p 1 elements of Z/pZ. Consider 3

4 the polynomials f 1 = n a i x p 1 i f 2 = n x p 1 i. Then deg(f 1 ) + deg(f 2 ) = 2p 2, whereas there are 2p 1 variables x i. By the Chevalley-Warning Theorem, there exists x 0 such that f 1 (x) = 0 and f 2 (x) = 0. If S [2p 1] is the set of i for which x i = 0, then by Fermat s Theorem, i S a i = 0 and i S xp 1 i = 0. This means S = 0 modulo p. In general the value of s (Γ) is not not known. The Kemnitz conjecture (1983) states that s (Z n Z n ) = 4n 3, and this was finally proved by Reinher (2007) using the nullstellensatz together with some additional combinatorial ideas: Theorem 6. Given at least 4n 3 integer lattice points in the plane, there exist n of them whose centroid is an integer lattice point. The question of determining s (Z k n) remains open for k 3. Alon and Dubiner (1993) showed s (Z k n) is linear in n as n, but even the asymptotics are not known. 5 The Davenport constant The Davenport constant of a finite group Γ, denoted s(γ), is the minimum n such that any sequence of n elements of Γ has a subsequence whose product is the identity. For instance, if Γ = Z/pZ, then s(γ) p, since no sum of elements from the sequence of p 1 ones is zero. It is not hard to show s(γ) = p in this case. It is in general an open problem to determine s(γ) for a general finite group Γ. Clearly for any finite group Γ, we have s(γ) Γ, since in a sequence (a 1, a 2,..., a k ) of more than Γ elements, two of the rising products j a i for j = 1, 2,..., k must coincide. If we consider a product of cyclic groups Γ = k Z n i, then s(γ) 1 + k (n i 1) by considering vectors with n i 1 copies of 1 in the ith position and zeroes elsewhere. In particular, if Γ = Z n then s(γ) = n since the above two bounds coincide. It is however not true that s(γ) = 1 + k (n i 1) for a Γ product of cyclic groups Z ni 4

5 (the reader can come up with examples). Olson s Theorem generalizes the statement s(z p ) = p to any p-group Γ = Z p n i. The most convenient language for the proof is that of group rings. Recall for an abelian group Γ and a ring R, the group ring R[Γ] comprises the set of all functions f : Γ R with addition defined pointwise and multiplication defined by the convolution f g(u) = f(x)g(y) x+y=u where we use additive notation for Γ. The following was proved by Olson (1969): Theorem 7 (Olson s Theorem). Let Γ be a finite product of cyclic groups Z p n i where p is prime. Then D(Γ) = 1 + k (pn i 1). Proof. Let (a 1, a 2,..., a m ) have no zero sum and let 1 be the identity in R[Γ]. Consider f = (1 a 1 ) (1 a 2 ) (1 a m ) Z p [Γ]. Let δ 0 be the number of even length subsequences of a 1, a 2,..., a m with sum equal to zero, and let δ 1 be the number of odd length subsequences with sum zero. Expanding the above product using inclusion-exclusion and considering f(0) in Z p, we get f(0) 1 = δ 1 δ 0 mod p and therefore if f(0) = 0 we have δ 0 0 or δ 1 0 and so there is a sum of elements that is zero. Now let us show that if m 1 + (p n i 1), then f(0) = 0, by showing more strongly that f = 0. Let x i denote the function with a 1 in position i and zero elsewhere. Then every monomial in the expansion of f in terms of the x i looks like k (1 x i ) d i where d 1 + d d k = m. The lower bound on m shows d i p n i for some i, which means that this monomial is zero since the additive order of any element of Z p n i is at most p n i. In other words, every monomial in the expansion of f vanishes, and so f = 0. A proof independent of group rings was given by Alon, Friedland and Kalai (1984). 5

6 6 Regular subgraphs Erdős and Berge raised the extremal problem for r-regular subgraphs: determine the maximum number of edges ex(n, r reg) in an n-vertex graph which has no r-regular subgraph. Clearly ex(n, 2 reg) = n 1, but for r 3 the problem is much more difficult. Erdős and Berge asked whether every 4-regular graph contains a 3-regular graph, and this was first proved by Tâskinov (1984). Here is a consequence of the nullstellensatz: Theorem 8. Let p be a prime, and let G be an n-vertex graph with more than (p 1)n edges. Then G has a non-empty subgraph in which the degrees are all multiples of p. Proof. Let G = (V, E) and consider f = v V ( 1 ( e v x e ) p 1 ) e E(1 x e ). Here x e {0, 1} for all e E. Then deg(f) = E > (p 1)n and a largest degree monomial is e E x e. The nullstellensatz applies with S e = {0, 1} for e E. This can also be derived from the Chevalley-Warning theorem. corollary: Here is an easy Corollary 9. Let p be a prime, and let G be a graph of average degree more than 2p 2 and maximum degree 2p 1. Then G has a p-regular subgraph. Proof. By the preceding theorem there is a subgraph where the degrees are multiples of p, and since the maximum degree is less than 2p, such a subgraph must be p- regular. This extends to powers of p using Olson s Theorem. 7 Degree-constrained subgraphs of graphs The notion of an f-factor of a graph where f assigns to each vertex a set of integers was introduced and studied by Lovász (1969). 6

7 Definition. Let f : V N ω be a function into subsets of non-negative integers. An f-factor of a graph G = (V, E) is a spanning subgraph H of G such that d H (v) f(v) for all v V. A partial f-factor of G is a non-empty subgraph H of G such that d H (v) f(v) {0} for all v V. Tutte s f-factor theorem gives a necessary and sufficient condition for a graph to have an f-factor when f : V N 1. Lovász [?] showed that a similar condition exists if the set f(v) does not contain a pair of integers differing by more than 2 for all v V, but that the problem becomes computationally hard otherwise. Using the nullstellensatz, it is possible to prove the existence of an f-factor if f(v) > d(v)/2 and f(v) {0, 1, 2,..., d(v)} for every v V (proved by Shirazi and the author (2008)): Theorem 10. Let G = (V, E) be a graph and let f : V N ω satisfy f(v) {0, 1, 2,..., d(v)} for all v V and f(v) > d(v)/2 for all v V. Then G has an f-factor. Proof. Let f(v) c = {0, 1, 2,..., d(v)}\f(v) and let S v be a set of integers of size d(v)/2 in {0, 1, 2,..., d(v)} containing f(v) c. Let (x e : e E) be 0-1 variables over R and define P R[x e : e E] by P (x) = ( ) X e d. v V e v d S v If P (x) 0, then for every v V and d f(v) c, x e d. e v This implies e v x e f(v) for every v V, and then F = {e : x e = 1} is the required subgraph. To show P (x) 0, note first that f(v) c d(v)/2 for all v V. We consider the coefficient of e E x e. Add a vertex to G joined to all vertices of G of odd degree. are eulerian. The result is a graph G all of whose components Therefore G contains a union of euler tours, and therefore some orientation of G assigns to every vertex v at most d (v)/2 edges oriented out from v, where d (v) is the degree of v in G. Removing the additional vertex, we obtain an orientation of G such that each vertex v has d(v)/2 edges oriented out from v. 7

8 The coefficient of e E x e is therefore positive: the edges out of v correspond to a choice of one x e from each of the terms ( ) x e d d S v e v and no x e is picked twice. By the combinatorial nullstellensatz, there is x {0, 1} E such that P (x) 0 in R. Similarly one obtains a condition for partial f-factors: Theorem 11. Let G = (V, E) be a graph and let f : V N ω {0, 1, 2,..., d(v)} for all v V and f(v) c \{0} < E. v V satisfy f(v) Then G has a partial f-factor. Proof. Let S v = f(v) c \{0}. Let x = (x e : e E) be 0-1 variables over R and define P R[x] by P (x) = v V ( d d S v e v x e d ) e E(1 x e ). Certainly P (0) = 0. If P (x) 0 for some x 0, then for all v V, x e f(v) {0} e v and then F = {e G : x e = 1} is a partial f-factor. To apply the nullstellensatz, note that the degree of the first term is at most S v < E v V and therefore ( 1) E +1 e E x e is the unique largest degree monomial. The nullstellensatz gives the result. Definition. Let f : V N ω be a function and p N. An f-factor mod p of a graph G = (V, E) is a spanning subgraph H of G such that d H (v) d mod p for 8

9 some d f(v) for all v V. A partial f-factor mod p of G is a non-empty subgraph H of G such that d H (v) = d mod p for some d f(v) {0} for all v V. The results in the last section say that if p be a prime, G = (V, E) has average degree more than 2p 2, and f(v) = 0 for all v V, then G contains a partial f-factor mod p. An interesting line of work is to determine conditions whereby a graph has an f-factor or a partial f-factor mod d. For instance, we can state Petersen s 2-factor theorem, that every 4-regular graph has a 2-factor, as stating that every 4-regular graph has a 2-factor mod 3. Is there an algebraic proof of this result along the lines of the nullstellensatz? 8 Counting solutions One can say a little bit more about the number of solutions in Chevalley s Theorem: Theorem 12. Let F be a finite field of characteristic p and let P 1, P 2,..., P m be homogeneous polynomials over F. Then the number of solutions to P 1 = 0, P 2 = 0,..., P m = 0 is zero mod p. Proof. The number of solutions is exactly x F n m (1 P i (x) F 1 ). Expanding the product in terms of monomials, each monomial has a term x d i i where d i < F 1, since the sum of the degrees of the P i is less than n. Adding that particular monomial over all x F n we get zero: for k < F 1 it is always true that x k = x F F 1 γ ik = 1 γk( F 1) 1 γ k = 0 where γ is a generator of the multiplicative group of F. This means that the monomial contributes zero mod p to the total number of solutions. This gives a condition for f-factors mod p: 9

10 Theorem 13. Let O k (G) be the number of orientations of a graph G in which every vertex has outdegree exactly k. If G has average degree 2p 2, then either O p 1 (G) 0 mod p, or for every f : V N, G has an f-factor mod p. Proof. As in the last proof, let (x e : e E) be 0-1 variables in Z p and define P (x) = v V ( 1 ( e v ) p 1 ) x e f(v). Then G has an f-factor mod p if P (x) 0 for some x. By the nullstellensatz, this is guaranteed if for each v V we can associate uniquely to v a set of exactly p 1 edges containing v and the total number of ways of doing it is non-zero mod p. This is independent of f, and equal to O p 1 (G). Here is a surprising corollary: Corollary 14. The number of eulerian orientations of a 2p 2-regular bipartite graph when p is a prime is zero mod p. Proof. If f(v) = {0} for all but one vertex v in a bipartite graph, and f(v ) = {1}, then the graph has no f-factor mod p. The last theorem gives the corollary. It appears to be difficult to determine the residue mod p of the number of Eulerian orientations. This comes up a lot in computing coefficients in problems involving the nullstellensatz. One open problem is to determine the complexity of computing the number of Eulerian tours in a 4-regular graph this is an open problem of M. Dyer. 8.1 Counting zero sums Let Γ be a abelian finite group. A non-trivial bound on the number of zero sums in a sequence of length at least s(γ) can be deduced quickly from Olson s Theorem on the number of zero sums. Here we declare the empty sequence to have zero sum. Theorem 15 (Olson). Let Γ be a finite abelian group with D(Γ) = m. Then any sequence of n m elements of Γ contains at least 2 n m+1 subsequences with zero sum. 10

11 Proof. This is clear for n = m by Olson s Theorem. Suppose n > m and let the elements be a 1, a 2,..., a n. Assume a 1 +a 2 + +a s = 0 for some s m. By induction there are at least 2 n m zero sums in the sequence ( a 2, a 3,..., a s, a s+1,..., a n ). When we add such a zero sum to a 1 + a a s, we get a zero sum involving a 1. So we have at least 2 n m zero sums using a 1. We also have 2 n m zero sums avoiding a 1 by induction, so we are done. This allows us for the degree problems to find large subgraphs with prescribed degrees: Corollary 16. Let q be a prime power and let G be a graph with m edges and n vertices. Then G has a subgraph H in which every vertex degree is a multiple of q where In particular, e(h) e(g) as e(g). ( ) e(g) 2 e(g) (q 1)n. e(h) A probabilistic approach due to Alon can be used to show that every graph G = (V, E) has a subgraph in which all vertices have degrees equal to a power of two and 0.01 E edges. Alon asks whether every graph G = (V, E) has a subgraph where the degrees are powers of three and with c E edges for some c > 0, and more generally: Conjecture. For every d there exists c d such that any graph G = (V, E) of average degree at least c d has a partial f-factor for every function f : V (G) N {0}. 9 The extremal problem for regular subgraphs The aim of this section is to estimate the extremal function for the family of k- regular graphs: how many edges can an n-vertex graph have if it has no k-regular subgraph? For k = 2 the answer is n 1 but for k > 2 the situation is expectedly much more complicated. Unlike subgraphs where the degrees are zero mod k, k- regular subgraphs have a much higher extremal function. The following is due to Pyber, Rödl and Szemerédi (1991): 11

12 Theorem 17. Any n-vertex graph with at least 128k 2 n log n edges has a k-regular subgraph. The proof requires a lemma. A bipartite graph G(A, B) is called r-half-regular if A B and every vertex in A has degree exactly r. Lemma 18. Let G be a bipartite graph of minimum degree at least r. Then G contains an r-half regular bipartite graph. Furthermore, any r-half-regular bipartite graph has an r-half-regular subgraph with a perfect matching. Proof. Suppose G = G(A, B) with A B. Take X A with X Γ(X) such that X is a minimum. Note X exists since A B. Now for Y X, we cannot have Γ(Y ) Y otherwise we would take Y instead of X. Also X = Γ(X) otherwise delete a vertex of X. Proof of Theorem 17. It is enough to prove that if p is prime, then an n-vertex graph G with at least 32p 2 n log n edges has a p-regular bipartite subgraph, since if k is any integer, then there is a prime p {k, k + 1,..., 2k} by Bertrand s Postulate, and if G has 128k 2 n log n 32p 2 n log n edges, then G has a p-regular bipartite subgraph and hence a k-regular subgraph. Pass to a bipartite subgraph H of G with minimum degree at least d = 8p 2 log n. By Lemma 18, H has a d-half-regular subgraph H 1 with a perfect matching M 1. Then again by Lemma 18, H 1 M 1 has a d 1-half-regular subgraph H 2 with a perfect matching M 2. Continuing in this way, we obtain matchings M 1, M 2,..., M d with V (M 1 ) V (M 2 ) V (M d ). If for any matching M i we have e(m i M i+1 M i+2p 1 ) > (2p 2) V (M i ) then by Theorem 3 there is a non-empty 0-factor mod p in M i M i+1 M i+2p 1. However, this graph has maximum degree 2p 1 so the 0-factor must be a p- regular subgraph. Therefore the union of those matchings always has size at most (2p 2) V (M i ). In particular we have ( 2p 3 ) d/(2p 1) M d n < 1 2p 2 by the choice of d. This contradiction completes the proof. 12

13 In fact it is possible to show that if G is a graph of maximum degree with average degree at least 256k 2 log, then G has a k-regular subgraph. One can then deduce the following Ramsey-type result here log n is an extremely slowly growing function of n, namely the inverse of the tower function T defined by T (n) = 2 T (n 1) with T (1) = 2. Corollary 19. If a graph G on n vertices has no k-regular subgraph, then G has an independent set of size Ω k (n/ log n). The proof is left as an exercise. We conjecture that the independence number show actually be Ω(n) and that the chromatic number is a constant. The above theorem is complemented by the following construction of Szemerédi, which gives a lower bound of order n log log n for the extremal function for regular subgraphs: Theorem 20. There exists an n by n bipartite graph of average degree 0.1 log log n not containing any k-regular subgraphs for k 3. Sketch Proof. The construction is random. Fix an increasing sequence of integers d 1 d 2 d m n such that d i n for all i and we ensure n/d 1 + n/d n/d m n. Form a bipartite graph with parts A and B where B = n and A = A 1 A 2 A m and A i = n/d i. For each i m, replace each vertex v of A i with a set D v of d i new vertices, and then randomly and uniformly place a perfect matching between {D v : v A i } and B, and then contract for each v A i each set D v to the single vertex v. Then the number of edges in this construction is exactly mn and every vertex of A i has degree d i, and every vertex of B has degree m. We consider the expected number of 3-regular subgraphs. Let f(t) denote the number of 3-regular subgraphs with t vertices in each part. Counting f(t) by the number t i of vertices in A i, the expected value of f(t) is at most ( )( ) n 3t m ( ) ti ( ) di n/di (n 3ti )!(3t i )!. t 3t 1 3t t m 3 t i n! Using the bound ( a b) (ea/b) b, we have f(t) ( 100t n ) t max t Ω(t) m ( di t ) 2ti. t i Here Ω(t) is the set of vectors (t 1, t 2,..., t m ) with sum equal to t and 0 t i min{n/d i, t/3}. Taking logaritheorems and applying the method of Lagrange Multipliers shows that a maximum occurs when for all but at most one value of i, 13

14 t i {0, t/3, n/d i }. There are at most three values of i such that t i = t/3. If j is the largest positive integer such that t/3 n/d j, then the contribution of those t i for which t i = t/3 is maximized when t j = t j 1 = t j 2 = t/3. The contribution to f(t) in that case is at most ( 100t ) t (27dj d j 1 d j 2 ) 2t/3 10 5t( d 2 j 1 d2 ) j 2 2t/3. n d j If d i d 5 i 1 (10m)5 for all i, then this quantity less than (2m) t. Under the same condition, it is possible but technical to show that the contribution of all terms t i = n/d i with i > j is also at most (2m) t, due to the rapid increase of the d i s. Therefore we have f(t) < 2(2m) t for all t, and this shows that the expected number of 3-regular subgraphs is at most Ω(t) 2(2m) t < t=3 ( ) m + t 1 2(2m) t < m 1 t=3 t=3 1 2 t < 1. The number of edges in the graph is mn and we take d 1 = (10m) 5, and d i d 5 i 1 for i 2. So it is enough to take d i = (10m) 10i for i 1, provided that (10m) 10m n. This is satisfied if we take m = 0.1 log log n. 14