组合零点定理的简单证明

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The following theorem is due to Alon [1]. Theorem 1 (Combinatorial Nullstellensatz [1]). Let F be an arbitrary field, and let P (x1 , . . . , xn ) be a polynomial in F[x1 , . . . , xn ]. Suppose the degree deg( P ) of P is in=1 ki , where each ki is a nonnegative integer, and suppose the coefficient of k k kn in P is nonzero. Then for any subsets A1 , . . . , An of F satisfying | Ai | ≥ x11 x22 · · · xn ki + 1 for i = 1, 2, . . . , n, there are a1 ∈ A1 , . . . , an ∈ An so that P (a1 , . . . , an ) = 0.
ACKNOWLEDGMENT. The assistance of Ronald van Luijk is gratefully acknowledged. Mathematisch Instituut, Universiteit Leiden, Postbus 9512, 2300 RA Leiden, The Netherlands
x ∈F5
( X − x ).
ˆ 2 are polynomials of degree 6 with the same leading coefficient, one has Since h 2 and h actually v ∈ F5 . Comparing the leading coefficients and the values at r , one discovers ˆ , so that v is nonzero. h = ±h ˆ )/2 and G = (h + h ˆ )/2. Then one has deg F = 3 and F · G = Define F = (h − h (v/4) · x ∈F5 ( X − x ), and consequently deg G = 2 and ( F , G ) ∈ S . Since one also has F + G = h , it remains to verify F = f and G = g . The zeroes of F are those ˆ (x ), so they coincide with the elements of Y . In particular, x ∈ F5 for which h (x ) = h Y has size 3, consisting of r and two elements adding up to 2r . Since the leading coefficient of F equals the leading coefficient s of h , one concludes F = f . Similarly, the ˆ (x ), which are the elements of Z . zeroes of G are those x ∈ F5 for which h (x ) = −h Comparing coefficients of X 2 in F + G = h , one finds that the leading coefficient of G equals the element t computed by the algorithm, so that G = g . This completes the proof. The argument given provides an independent proof of the surjectivity of the map S → T and, combined with the injectivity proof, of the equality #T = # S = 160.
using the usual long division algorithm for polynomials. Formally equation (∗) is an identity in the ring of polynomials in one variable x1 with coefficients in the ring F[x2 , . . . , xn ]. Since the degree of R in the variable x1 is strictly less than deg(x1 − a ), the polynomial R does not contain x1 at all. By the assumption on P it follows that k −1 k kn Q must have a nonvanishing monomial of the form x1 1 x2 2 · · · xn and deg( Q ) = n k − 1 = deg ( P ) − 1. i =1 i Take any x ∈ {a } × A2 × · · · × An and substitute into the equation (*). Since P (x ) = 0 it follows that R (x ) = 0, too. But R does not contain x1 , so R vanishes also on ( A1 − {a }) × A2 × · · · × An . Now substitute any x ∈ ( A1 − {a }) × A2 × · · · × An into (*). Since x1 − a is nonzero, it follows that Q (x ) = 0. So Q vanishes on ( A1 − {a }) × A2 × · · · × An , which contradicts the inductive assumption. The theorem is also true when F is any integral domain. If we omit the integrality assumption and take any commutative ring with unity as F, then the proof works if for any two elements a , b ∈ Ai (i = 1, . . . , n ), the difference a − b is not a zero divisor in F. Theorem 1 has many useful applications in combinatorics, graph theory, and additive number theory (see [1]). As an illustration we apply it in the proof of the Erd˝ osHeilbronn conjecture. It was first proved by da Silva and Hamidoune, and then by Alon, Nathanson, and Ruzsa in a more general setting. Theorem 2. Let p be a prime and let A and B be two nonempty subsets of Z p . Let C = {x ∈ Z p : x = a + b for some a ∈ A , b ∈ B , a = b}. Then |C | ≥ min( p , | A| + | B | − 3). Proof. Only the case p > 2 is interesting. If min( p , | A| + | B | − 3) = p , then for any g ∈ Z p the sets A and g − B have at least two different elements in common. Let a be one that is different from g /2. Then g = a + b for some b ∈ B different from a . This proves that g ∈ C , so C = Z p . Suppose now that | A| + | B | − 3 < p and the theorem does not hold. In that case there exists a set D such that C ⊆ D and | D | = | A| + | B | − 4. We define two polynomials: P (x , y ) =