A duality based proof of the Combinatorial Nullstellensatz Omran Kouba Department of Mathematics Higher Institute for Applied Sciences and Technology P.O. Box 31983, Damascus, Sy ria omran kouba@hiast.edu.sy Submitted: Dec 21, 2008; Accepted: Mar 5, 2009; Published: Mar 13, 2009 Mathematics Subject Classifications: 05A99, 15A03 Abstract In this note we present a proof of the combinatorial nullstellensatz using simple arguments from linear algebra. The combinatorial nullstellensatz [1] is an elegant tool which has many applications in combinatorial number theory, graph theory and combinatorics (see [1] and [2]). In this note we present a proof of this result using simple arguments from linear algebra. In Theorem 1, we recall the statement of the combinatorial nullstellensatz: Theorem 1. Let P be a polynomial in m variables X 1 , X 2 , . . . , X m over an arbitrary field K. Suppose that the coefficient of the monomial X n 1 1 X n 2 2 · · · X n m m in P i s nonzero, and that the total degree of P is m j=1 n j . Then, if S 1 , S 2 , . . . , S m are subsets of K such that card (S j ) > n j (for 1 ≤ j ≤ n,) there is some (t 1 , t 2 , . . . , t m ) in S 1 × S 2 × · · · × S m so that P (t 1 , t 2 , . . . , t m ) = 0. Our proof is based upon a simple lemma concerning linear forms on the vector space K[T ] of polynomials in one variable T over an arbitrary field K. In the dual space (K[T ]) ∗ , we consider the dual basis (ϕ m ) m≥0 of the canonical basis (T m ) m≥0 of K[T ], this means that ϕ m (P ) is the coefficient of T m in P, in other words ϕ i (T j ) = δ ij where δ ij is the Kronecker symbol. We also denote by K n [T ] the subspace of K[T ] formed of polynomials of degree at most n. With the above nota tion we have the following lemma : Lemma 2. Let S be a subset of K such that card (S) = m + 1. Then there is a family (λ S t ) t∈S of elements in K such that ∀ P ∈ K m [T ], ϕ m (P ) = t∈S λ S t P (t). the electronic journal of combinatorics 16 (2009), #N9 1 Proof. Consider, for t ∈ S, the linear form µ t : K m [T ] −→ K, µ t (P ) = P (t). The family (µ t ) t∈S constitutes a basis of the dual space (K m [T ]) ∗ . (To see this, note that if (ℓ t ) t∈S denotes the basis of K m [T ] formed by the Lagrange intepolation polynomials : ℓ t (T ) = s∈S\{t} T −s t−s , then µ u (ℓ v ) = δ uv . This proves that (µ t ) t∈S is the dual basis of (ℓ t ) t∈S .) Now, t he linear form P → ϕ m (P ) defines an element from (K m [T ]) ∗ and, con- sequently, it has a unique expression as a linear combination of the elements of the basis (µ t ) t∈S . This proves the existence of a familly of scalars (λ S t ) t∈S , such that ϕ m (P ) = t∈S λ S t µ t (P ) for any polynomial P in K m [T ], and achieves the proof of Lemma 2. Before proceeding with the proof of Theorem 1, let us recall that the total degree of a polynomial P from K[X 1 , . . . , X m ] is the largest value of d 1 + d 2 + · · · + d m taken over all monomials X d 1 1 X d 2 2 · · · X d m m with nonzero coefficients in P . Proof of Theorem 1. For each j in {1, . . . , m}, we may assume that card (S j ) = n j + 1 (by discarding the extra elements if necessary,) then, using Lemma 2, we find a familly of scalars (λ S j t ) t∈S j such that ∀ P ∈ K n j [T ], ϕ n j (P ) = t∈S j λ S j t P (t). (1) Then, we consider the linear form Φ on K[X 1 , . . . , X m ] defined by : Φ(Q) = (t 1 , ,t m )∈S 1 ×···×S m λ S 1 t 1 λ S 2 t 2 · · · λ S m t m Q(t 1 , t 2 , . . . , t m ). Clearly, we have Φ(X d 1 1 X d 2 2 · · · X d m m ) = t 1 ∈S 1 t 2 ∈S 2 · · · t m ∈S m λ S 1 t 1 λ S 2 t 2 · · · λ S m t m t d 1 1 t d 2 2 . . . t d m m = m j=1 t∈S j λ S j t t d j . So we have the following two properties: i. If there is some k in {1, . . . , m} such that d k < n k , then by ( 1) we have t∈S k λ S k t t d k = ϕ n k (T d k ) = 0, and therefore, Φ(X d 1 1 X d 2 2 · · · X d m m ) = 0. the electronic journal of combinatorics 16 (2009), #N9 2 ii. On the other hand, Φ(X n 1 1 X n 2 2 · · · X n m m ) = m j=1 ϕ n j (T n j ) = 1. Let us suppose that P = (d 1 ,d 2 , ,d m )∈D b d 1 ,d 2 , ,d m X d 1 1 X d 2 2 · · · X d m m , where we collected in D the multi-indexes (d 1 , d 2 , . . . , d m ) satisfying b d 1 ,d 2 , ,d m = 0. Now, if (d 1 , d 2 , . . . , d m ) is an element from D which is different from (n 1 , n 2 , . . . , n m ), then there is some k in {1, . . . , m} such that d k < n k because deg(P ) = m j=1 n j . Therefore, by (i.), if (d 1 , d 2 , . . . , d m ) is an element from D which is different from (n 1 , n 2 , . . . , n m ), then Φ(X d 1 1 X d 2 2 · · · X d m m ) = 0, and if we use (ii.) we conclude that Φ(P ) = (d 1 ,d 2 , ,d m )∈D b d 1 ,d 2 , ,d m Φ(X d 1 1 X d 2 2 · · · X d m m ) = b n 1 ,n 2 , ,n m = 0. Finally, the conclusion of the theorem follows since (t 1 , ,t m )∈S 1 ×···×S m λ S 1 t 1 λ S 2 t 2 · · · λ S m t m P (t 1 , t 2 , . . . , t m ) = Φ(P ) = 0. This ends the proof of Theorem 1. References [1] Alon, N., Combinatorial Nullstellensatz. Recent trends in combinatorics. (M´atra- h´aza, 1995). Combin. Pro bab. Comput. 8 (1999), 7–29. [2] Shirazi, H. and Verstra ¨ ete, J., A note on polynomials and f-factors of graphs. Electronic J. of Combinatorics, 15 (2008), #N22. the electronic journal of combinatorics 16 (2009), #N9 3 . and achieves the proof of Lemma 2. Before proceeding with the proof of Theorem 1, let us recall that the total degree of a polynomial P from K[X 1 , . . . , X m ] is the largest value of d 1 + d 2 +. A duality based proof of the Combinatorial Nullstellensatz Omran Kouba Department of Mathematics Higher Institute for Applied Sciences and Technology P.O. Box 31983, Damascus, Sy ria omran. 2009 Mathematics Subject Classifications: 05A99, 15A03 Abstract In this note we present a proof of the combinatorial nullstellensatz using simple arguments from linear algebra. The combinatorial nullstellensatz