THE COMBINATORIAL NULLSTELLENSÄTZE REVISITED

We revisit and further explore the celebrated Combinatorial Nullstellensätze of N. Alon in several different directions. Notation and Terminology: Let N be the non-negative integers. For d = (d1, . . . , dn), e = (e1, . . . , en) ∈ N, we write d ≤ e if di ≤ ei for all 1 ≤ i ≤ n. We write d < e if d ≤ e and ∑n i=1 di < ∑n i=1 ei. Our rings are commutative with multiplicative identity. A domain is a ring R in which a, b ∈ R \ {0} =⇒ ab ̸= 0. A ring R is reduced if for all x ∈ R,n ∈ Z, we have x = 0 =⇒ x = 0. We abbreviate the polynomial ring R[t1, . . . , tn] by R[t].


Introduction
1.1.The Combinatorial Nullstellensätze.This note concerns the following celebrated results of N. Alon.
Theorem 1.1.Let F be a field, let X 1 , . . ., X n ⊂ F be nonempty and finite, and Let f ∈ F [t] be a polynomial which vanishes on all the common zeros of φ 1 , . . ., φ n : that is, for all x ∈ F n , if φ 1 (x) = . . .= φ n (x) = 0, then f (x) = 0. Then: a) (Combinatorial Nullstellensatz I, or CNI) There are q 1 , . . ., q n ∈ F [t] such that (2) b) (Supplementary Relations) Let R be the subring of F generated by the coefficients of f and φ 1 , . . ., φ n .Then the q 1 , . . ., q n may be chosen to lie in R[t] and satisfy Theorem (ii) The coefficient of t d1 1 • • • t dn n in f is nonzero.Then, for any subsets X 1 , . . ., X n of F with #X i = d i + 1 for 1 ≤ i ≤ n, there is x = (x 1 , . . ., x n ) ∈ X = ∏ n i=1 X i such that f (x) ̸ = 0. Alon used his Combinatorial Nullstellensätze to derive various old and new results in number theory and combinatorics, starting with Chevalley's Theorem that a homogeneous polynomial of degree d in at least d + 1 variables over a finite field has a nontrivial zero.The use of polynomial methods has burgeoned to a remarkable degree in recent years.We recommend the recent survey [Ta14], which lucidly describes the main techniques but also captures the sense of awe and excitement at the extent to which these very simple ideas have cracked open the field of combinatorial number theory and whose range of future applicability seems almost boundless.
One easily deduces CNII from CNI and the Supplementary Relations, but (apparently) not conversely.For appplications in combinatorics and number theory, CNII seems more useful: [Al99] organizes its applications into seven different sections, and only in the last is CNI applied.Most later works simply refer to Theorem 1.2 as the Combinatorial Nullstellensatz.We find this trend somewhat unfortunate.On the one hand, CNI is stronger and does have some applications in its own right.On the other hand, it is CNI which is really a Nullstellensatz in the sense of algebraic geometry, and we find this geometric connection interesting and suggestive.
Theorem 1.3.(Coefficient Formula) Let F be a field, and let .
In this note we revisit and further explore these theorems, in three different ways.
• In §2 we improve CNI to a full Nullstellensatz for polynomial functions on arbitrary finite subsets X ⊂ F n over a field F (Theorem 2.4).When F = F q and X = F n q , we recover the Finite Field Nullstellensatz of G. Terjanian (Corollary 2.5).
• CNI and CNII hold with F replaced by any domain R. Schauz showed that Theorem 1.3 holds over any ring R so long as X satisfies "Condition (D)": no two distinct elements of any X i differ by a zero-divisor.Moreover, one can view Alon's proof of CNI and CNII as a restricted variable analogue of Chevalley's proof of Chevalley's Theorem, and Schauz's work shows that one can do this over any ring with Condition (D) in hand.We do so in §3: following Chevalley, we establish versions of Theorems 1.1, 1.2 and 1.3 over any ring.It turns out that Condition (D) is necessary and sufficient for Theorem 1.1 to hold.On the other hand, if we clear denominators in (4) we get a formula which is meaningful even in the absence of Condition (D).This Integral Coefficient Formula (Theorem 3.9b)) follows by "the permanence of algebraic identities".We close up this circle of ideas by establishing a Restricted Variable Chevalley-Warning Theorem (Theorem 3.10), a refinement of the Restricted Variable Chevalley Theorem [Sc08], [Br11] which is complementary to the restricted variable version of Warning's Second Theorem [CFS14].
• In §4 we further analyze the evaluation map from polynomials to functions on an arbitrary subset X ⊂ R n for an arbitrary ring.We aim to show that the (perhaps rather arid-looking) formalism of a restricted variable Nullstellensatz leads to interesting open problems in polynomial interpolation over commutative rings.

A Nullstellensatz for Finitely Restricted Polynomial Functions
2.1.Alon's Nullstellensatz versus Hilbert's Nullstellensatz.The prospect of improving Theorem 1.1 as a Nullstellensatz has not been explored, perhaps because the notion of a Nullstellensatz, though seminal in algebra and geometry, is less familiar to researchers in combinatorics.But it was certainly familiar to Alon, who began [Al99] by recalling the following result.
Theorem 2.1.(Hilbert's Nullstellensatz) Let F be an algebraically closed field, let g 1 , . . ., g m ∈ F [t], and let f ∈ F [t] be a polynomial which vanishes on all the common zeros of g 1 , . . ., g m .Then there is k ∈ Z + and q 1 , . . ., q m ∈ F [t] such that Let us compare Theorems 1.1 and 2.1.They differ in the following points: • In Theorem 1.1, F can be any field.In Hilbert's Nullstellensatz, F must be algebraically closed.Really must: if not, there is a nonconstant polynomial g(t 1 ) without roots in F ; taking m = 1, g 1 = g and f = 1, the conclusion fails.
• In CNI, the conclusion is that f itself is a linear combination of the φ i 's with polynomial coefficients, but in Hilbert's Nullstellensatz we must allow taking a power of f .Really must: e.g.take k ∈ Z + , m = 1, g 1 = t k 1 and f = t 1 .• The Supplementary Relations give upper bounds on the degrees of the polynomials q i : they make CNI effective.Hilbert's Nullstellensatz is not effective.Effective versions have been given by Brownawell [Br87], Kollár [Ko88] and others, but their bounds are much more complicated than the ones in Theorem 1.1.
• In Theorem 1.1 the φ i 's are extremely restricted.On the other hand, in Hilbert's Nullstellensatz the g i 's can be any set of polynomials.Thus Theorem 2.1 is a full Nullstellensatz, whereas Theorem 1.1 is a partial Nullstellensatz.
We will promote Theorem 1.1 to a full Nullstellensatz for all finite subsets.

The Restricted Variable Formalism.
For a set Z, let 2 Z be its power set.For a ring R, let I(R) be the set of ideals of R. For a subset J of a ring R, let ⟨J⟩ be the ideal of R generated by J and let rad Let R be a ring, and let X ⊂ R n .For x ∈ X, f ∈ R[t], we put When R is an algebraically closed field and X ⊂ R n is Zariski-closed (c.f.§4.2), this is the usual connection between subsets of an affine variety and its coordinate ring.We will see that the case of R any field and X finite is even better behaved.In §4 we return to the general case and find some new phenomena.
Put V = V R n .We may extend I and V X to maps on power sets as follows: We have X = V X (0), so (5) Proof.a) We immediately reduce to the case m = 2. Since ) is a radical ideal containing both I(X) and I(V (J)) ⊃ J, so it contains rad(J + I(X)).The other inclusions are immediate.
It is well known (see Theorem 3.3) that when F is infinite we have I(F n ) = {0}.This serves to motivate the following restatement of Hilbert's Nullstellensatz.
Theorem 2.3.Let F be an algebraically closed field.For all J ⊂ F [t], Theorem 2.4.(Finitesatz) Let F be a field, and let X ⊂ F n be a finite subset.a) For all ideals J of F [t], we have In particular, if Proof.a) Let F be a field, and let X ⊂ F n be finite.Let so by the Chinese Remainder Theorem [L, Cor.2.2], Let F A be the set of all maps f : A → F , so F A is an F -algebra under pointwise addition and multiplication and Thus ι is an injective F -linear map between F -vector spaces of equal finite dimension, hence it an is an isomorphism of rings.It follows that Identifying I(F [t]/I(X)) with {J ∈ I(F [t]) | J ⊃ I(X)} and restricting V X to ideals containing I(X), we get maps Since I(F [t]/I(X)) and 2 X have the same finite cardinality, it follows that V X and I are mutually inverse bijections.Thus for any ideal J of F [t], using (5) we get and thus there is an induced surjective F -algebra homomorphism ẼX : Corollary 2.5.(Finite Field Nullstellensatz [Te66]) Let F q be a finite field.Then for all ideals J of F q [t], we have Proof.Apply Theorem 2.4 with F = X 1 = . . .= X n = F q .
3. Cartesian Reduction, the Atomic Formula, and the Nullstellensätze when b is monic, the standard division algorithm involves no division of coefficients so works in any ring.Part b) follows by contemplating the division algorithm.
Proposition 3.2.(Cartesian Reduction) Let R be a ring.For then divide the remainder r 1 by φ 2 , then divide the remainder r 2 by φ n , and so forth, getting f = ∑ n i=1 q i φ i + r n .Apply Lemma 3.1b).c) Part b) implies that Ψ is surjective.For the injectivity: let q 1 , . . ., q n ∈ R[t] be such that f = ∑ n i=1 q i φ i ∈ R c .We must show that f = 0.For each i, by dividing q i by φ j for i < j ≤ n and absorbing the quotient into the coefficient q j of φ j , we may assume that deg tj q i < d j for all j > i.It follows inductively that for all 1 ≤ m ≤ n, ∑ m i=1 q i φ i is either 0 or has t i -degree at least d i for some 1 ≤ i ≤ m.Applying this with m = n shows f = 0. d) This follows from part c).
Let R be a ring and X 1 , . . ., X n ⊂ R be finite, nonempty subsets.Put Condition (F) implies Condition (D).Conversely, Condition (D) implies Condition (F) with R replaced by its total fraction ring.A ring is a field (resp.a domain) iff every subset satisfies Condition (F) (resp.Condition (D)).(Schauz) The following are equivalent: (i) X satisfies condition (D).(ii) If f ∈ R X and f (x) = 0 for all x ∈ X, then f = 0. (iii) We have Φ = I(X).b) (Chevalley-Alon-Tarsi) The above conditions hold when R is a domain.Proof.a) (i) =⇒ (ii): By induction on n: suppose n = 1.Write X = {x 1 , . . ., x a1 }, and let f ∈ R[t 1 ] have degree less than a 1 − 1 such that f (x i ) = 0 for all 1 ≤ i ≤ a 1 .By Polynomial Division, we can write f = ( ), and comparing degrees shows f = 0. Suppose n ≥ 2 and that the result holds in n − 1 variables.Write has degree less than a n and vanishes for all a n elements x n ∈ X n , so it is the zero polynomial: f i (x 1 , . . ., x n−1 ) = 0 for all 0 ≤ i ≤ a n .By induction, each f i (t 1 , . . ., t n−1 ) is the zero polynomial and thus f is the zero polynomial.(ii) =⇒ (iii): We have Φ ⊂ I(X).Let f ∈ I(X).Since f − r X (f ) ∈ Φ ⊂ I(X), for all x ∈ X we have r X (f )(x) = f (x) = 0. Then (ii) gives r X (f ) = 0, so f ∈ Φ. (iii) =⇒ (i): We argue by contraposition: suppose X does not satisfy Condition (D).Then for some 1 ≤ i ≤ n, we may write X i = {x 1 , x 2 , . . ., x ai } such that there is 0 is a domain then Condition (D) holds for every X.

Theorem 3.5. (Atomic Formula) Suppose Condition (F). Then for all f ∈ R[t],
we have Let c = (c 1 , . . ., c n ) ∈ N n .We say a polynomial f ∈ F [t] is c-topped if for every e = (e 1 , . . ., e n ) with c < e, the coefficient of An elementary reduction of f consists of identifying a monomial which is divisible by t di+1 i and replacing t di+1 i by ψ i (t i ).Elementary reduction on a d-topped polynomial yields a d-topped polynomial with the same coefficient of t d .We obtain r X (f ) from f by finitely many elementary reductions.
Theorem 3.8.Let R be a ring, let X 1 , . . ., X n ⊂ R be finite nonempty subsets, and define d, X, φ 1 , . . ., φ n , Φ as above.Suppose f ∈ I(X): i.e., f (x) = 0 for all x ∈ X. Then: a) (Combinatorial Nullstellensatz I) The following are equivalent: (i) X satisfies Condition (D).(ii) We have f ∈ Φ: there are q 1 , . . ., q n ∈ R[t] such that f (t) = ∑ n i=1 q i (t)φ i (t).b) (Supplementary Relations) Suppose the equivalent conditions of part a) hold.Let r be the subring of R generated by the coefficients of f and φ 1 , . . ., φ n .We can take q 1 , . . ., Proof.If X satisfies Condition (D), replace R by r and apply Proposition 3.2b) and Theorem 3.3.We get q 1 , . . ., q n ∈ r[t] such that f = ∑ n i=1 q i φ i and deg q i ≤ deg f − deg φ i for all 1 ≤ i ≤ r.If X does not satisfy Condition (D), then by Theorem 3.3, there is a nonzero element f ∈ R X ∩ I(X), and then f / ∈ Φ by Proposition 3.2c).
The right hand side of (9) lies in R if X satisfies Condition (F).b) (Integral Coefficient Formula) In general, we have Proof.a) Replace R by its total fraction ring and apply (8).b) There is a domain R and a surjective ring homomorphism q : R → R. For instance, let R be a polynomial ring over Z in a set of indeterminates {T r } r∈R indexed by the elements of R and let q be the unique homomorphism with q(T r ) = r.
There is a unique extension of q to a ring homomorphism q : R[t 1 , . . ., Applying q to both sides of (11) gives c) This follows from part a).d) This is Lemma 3.7.

The Restricted Variable Chevalley-Warning Theorem.
For a ring R and . Then: a) As elements of F q , we have Proof.a) We define and thus the coefficient of Applying the Coefficient Formula (Theorem 1.3), we get 0 = Parts b) through d) follow from part a) by taking X to be, respectively, F n q , {0, 1} n and (F × q ) n , and computing the φ ′ i (t i ) ′ s.The details are left to the reader.

Further Analysis of the Evaluation Map
4.1.The Finitesatz holds only over a field.
If a ring R is not a field and X ̸ = ∅, the assertion of Theorem 2.4a) remains meaningful with R in place of F , but it is false.Let x ∈ X.Since R[t]/m x ∼ = R is not a field, m x is not maximal.Let J be an ideal with m x J R[t], and let

Towards an Infinitesatz.
We revisit the formalism of § 2.2: let R be a ring and let X ⊂ R n .
For a subset A ⊂ R n we define the Zariski closure A = V (I(A)).Thus A is the set of points at which any polynomial which vanishes at every point of A must also vanish.A subset A is algebraic if A = A and Zariski-dense if A = R n .When R is a domain the algebraic subsets are the closed sets of a topology, the Zariski topology.Over an arbitrary ring this need not hold and some strange things can happen: for instance if R = Z/6Z and n = 1 then {2, 3} = {0, 2, 3, 5}.In fact for any composite positive integer m ̸ = 4, there is a subset A ⊂ Z/mZ which is not algebraic [Si54], [Ch56].Some partial results towards an explicit description of the operator A → A for subsets of Z/mZ have recently been obtained by B. Bonsignore.
If F is an algebraically closed field and X ⊂ F n is algebraic, then using Hilbert's Nullstellensatz, for all ideals J of F [t], ).When X is infinite, we claim the "rad" cannot be removed in general.proof of claim: Suppose rad(J + I(X)) = J + I(X) for all J. Equivalently, every ideal J ⊃ I(X) is a radical ideal.Then for any element x in the quotient ring F [t]/I(X), since (x 2 ) is radical we must have (x) = (x 2 ) = (x) 2 .It follows (e.g.[AM,p. 35,p. 44,p. 90]) that F [t]/I(X) is Noetherian and absolutely flat, hence is Artinian, hence has only finitely many maximal ideals.Since x → m x is an injection from X to the set of maximal ideals of F [t]/I(X), X is finite.
The case of an arbitrary subset over an arbitrary ring R is much more challenging.In fact, even determining whether the evaluation map E X : R[t] → R X is surjective -existence of interpolation polynomials -or injective -uniqueness of interpolation polynomials -becomes nontrivial.In the next section we address these questions, but we are not able to resolve them completely.Lemma 4.2.Let R be a ring, and let X be an infinite set.Then R X is not a countably generated R-module.
an uncountable linearly ordered family of nonempty subsets of Q.Since X is infinite, there is an injection ι : Q → X; then C = {ι(A x )} x∈R is an uncountable linearly ordered family of nonempty subsets of X.
Step 2: For each A ∈ C, let 1 A be the characteristic function of A. Then {1 A } A∈C is an R-linearly independent set: let A 1 , . . ., A n ∈ C and α 1 , . . ., α n ∈ R be such that α 1 1 A1 + . . .+ α n 1 An ≡ 0. We may order the A i 's such that A 1 ⊂ . . .⊂ A n and thus there is Evaluating at x gives α n = 0.In a similar manner we find that α n−1 = . . .= α 1 = 0.
Step 3: Suppose R X is countably generated: thus there is a surjective R-module homomorphism Φ : Step 2, S is uncountable and R-linearly independent, so it spans a free R-module with an uncountable basis which is an R-submodule of Let π i : R n → R be the ith projection map: π i : (x 1 , . . ., x n ) → x i .For a subset X ⊂ R n , we define the Cartesian hull C(X) as ∏ n i=1 π i (X): it is the unique minimal Cartesian subset containing X, and it is finite iff X is.
Proof.a) Since X ⊂ C(X), it suffices to show that E C(X) is surjective, and we have essentially already done this: under Condition (F) we may define r X (f ) = ∑ x∈X f (x)δ X,x (t), and as in § 3.3 we see that E(r X (f )) = f .b) There is 1 ≤ i ≤ n and y i ̸ = y ′ i ∈ Y i such that y 1 − y 2 / ∈ R × , hence a maximal ideal m of R with y 1 −y 2 ∈ m.For all j ̸ = i, choose y j ∈ Y j ; let y = (y 1 , . . ., y n ); and let y ′ be obtained from y by changing the ith coordinate to y ′ i .For any f ∈ F [t], f (y) ≡ f (y ′ ) (mod m), so f (y) − f (y ′ ) ∈ m.Hence the function δ Y,y : Y → R which maps y to 1 and every other element of Y to 0 does not lie in the image of the evaluation map.Thus E Y is not surjective, so E X cannot be surjective.
Thus if X is itself Cartesian, the evaluation map is surjective iff X satisfies Condition (F): this result is due to Schauz.Proposition 4.4 is the mileage one gets from this in the general case.When every Cartesian subset of X satisfies condition (F) but C(X) does not, the question of the existence of interpolation polynomials is left open, to the best of my knowledge even e.g. over Z.
We say that a ring R is (F)-rich (resp.(D)-rich) if for every d ∈ Z + there is a d-element subset of R satisfying Condition (F) (resp.Condition (D)).If ι : R → S is a ring embedding and R is (F)-rich, then S is (F)-rich, hence also (D)-rich.Proposition 4.5.Let R be a ring and X ⊂ R n .Consider the following assertions: (i) E X is injective.(ii) X is infinite and Zariski-dense.a) We always have (i) =⇒ (ii).b) If R is (D)-rich -e.g. if it contains an (F)-rich subring -then (ii) =⇒ (i).c) If R is finite, a domain, or an algebra over an infinite field, then (ii) =⇒ (i).d) If R is an infinite Boolean ring -e.g.R = ∏ ∞ i=1 Z/2Z -and X = R n , then (ii) holds and (i) does not.
Proof.a) By contraposition: suppose first that X is finite.Then F X is a free Fmodule of finite rank #X and F [t] is a free F -module of infinite rank, so E cannot be injective.Now suppose X is not Zariski-dense: then there is y ∈ F n \ X and f ∈ F [t] such that E(f )| X ≡ 0 and E X (f )(y) ̸ = 0, hence 0 ̸ = f ∈ Ker E. b) Let f ∈ Ker E X = I(X), and let d = deg f .Since X is Zariski-dense in F n , f (x) = 0 for all x ∈ F n .Since R is (D)-rich, there is a S ⊂ R of cardinality d + 1 satisfying Condition (D).Put X = ∏ n i=1 S. Then f ∈ R X and f (x) = 0 for all x ∈ X, so f = 0 by Theorem 3.3.c) This is immediate from part b).d) Since R is infinite, R n is infinite and Zariski-dense.Since R is Boolean, the polynomial t 2 1 − t 1 evaluates to zero on every x ∈ R n .