Set systems with no union of cardinality 0 modulom

Let q be a prime power. It is shown that for any hypergraph ~,~ = {F~,..., Fdtq_~)+~ } whose maximal degree is d, there exists Z ¢ ~o c ~, such that IUF~oFI =-0 (rood q). For integers d, m __ 1 let fe(m) denote the minimal t such that for any hypergraph -~ = {Fz . . . . . Ft} whose maximal degree is d, there exists ~ ¢ o~ o c Y, such that I~F~ ~oFI -= 0 (mod m). Here we determine fd(m) when m is a pr ime power, and remark on the general case . Example. Let Aij 1 < i <_ m 1, 1 <_ j < d, be pairwise disjoint sets, each of cardinality m, and let {v~ . . . . . v,,_a } be disjoint from all the Aii's. Now ~" = {Aq U {vi}: 1 < i < m 1,1 _< j _< d} satisfies ] ~ l = d ( m 1 ) b u t ] U e ~ o t ~ 0 (mod m) for any ~Z~ # ~o c ~ . Hence fa(m) _> d(m 1) + 1. T h e o r e m 1. I f q is a prime power than fd(q) = d(q 1) + 1. Proof. Let ~ = {F 1 . . . . . Ft}, t = d(q 1) + 1, be a hypergraph of degree _ d(q 1) + 1 and h ( x l , . . . , x t ) ~ Z [ x l . . . . . xt] satisfies h(O)= O, and degh <_ d, then there exists an 0 ~ ~ e {0, 1} t such that h(e) =0 (mod q). 98 N. Alon, D. Kleitman, R. Lipton, R. Meshulam, M. Rabin, J. Spencer Proof. Suppose h(~) ~ 0 (mod q) f o r all 0 ¢ e • {0, 1} t, and let u(x) = I]~-_2~ (h(x) i). Denote by s the smallest power of p that does not divide ( q 1)[, i.e., s = p . m a x { p ' ; p q ( q 1)[}. The proof of the following simple fact is omitted: Lemma 1. For every integer a, 1-[~2_~ (a i) = 0 (mod s) i f f a ~ 0 (mod q). [] By Lemma 1 u(e) = 0 (rood s) for all 0 ¢ e • {0, 1} t, and u(0) ~ 0 (mod s). Let g(x) denote the multilinear polynomial obtained from u(x) by replacing each monomial xi~ ~ . . . x l f j, el . . . . . ej > 1, by xh . . . xij. The following Lemma can be easily proved by induction on t: Lemma 2 [1]. I f g (x 1 . . . . . x ,) is a mult i l inear polynomial in Z [x l , . . ., x t] and g(e) =0 (mod s) f o r all e • {0, 1}', then g ( x l , . . . , x , ) = 0 (mod s) [] Now g(x) = g(x) u(O). [I~=1 (1 xi) satisfies the assumptions of Lemma 2, hence g(x) = u(0)" M~=I (1 x i ) (mod s), and so deg g _> t. But deg ~ _< deg u = (deg h) q-1 <_ d(q 1) < t, a contradiction. [] Returning to the proof of Theorem 1, we note that deg p <_ d and p(0) = 0. Hence by Theorem 2 p(e) =0 (rood q) for some 0 ¢ e • {0, 1} t. Now by Inclusion Exclusion p(e) = q ~){~,=1} F~[, and so IU{i:,,=I}F~[ = 0 (mod q). [] Following [2] let g~(m) denote the minimal t such that for any h • Z [ x ~ . . . . , x , ] which satisfies h(0)= 0, and deg h < d, there exists an 0 ¢ ~ • {0, 1}' such that h(e) = 0 (rood m). The proof of Theorem 1 shows that fd(m) < (q 1)d vertices and with e edges, each of size at most d, contains an induced sub-hypergraph on less than n vertices whose number of edges is congruent to e modulo q. []