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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
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