TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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Alon's Combinatorial Nullstellensatz: if $\deg f = \sum_{i} t_{i}$ , the coefficient of $x_{1}^{t_{1}} \dots x_{n}^{t_{n}}$ in $f$ is nonzero, and $∣ S_{i} ∣ > t_{i}$ , then $f (s_{1}, \dots, s_{n}) \neq 0$ for some $s_{i} \in S_{i}$

Statement

Let $F$ be a field, let $f \in F [x_{1}, \dots, x_{n}]$ , and let finite sets $S_{1}, \dots, S_{n} \subseteq F$ . Suppose

$\deg f = \sum_{i} t_{i}$ ;
the coefficient of $x_{1}^{t_{1}} \dots x_{n}^{t_{n}}$ in $f$ is nonzero; and
$∣ S_{i} ∣ > t_{i}$ for every $i$ .

Then there is a point $(s_{1}, \dots, s_{n}) \in S_{1} \times \dots \times S_{n}$ with $f (s_{1}, \dots, s_{n}) \neq 0$ .

Facts & Assumptions

Given: a field $F$ , a polynomial $f \in F [x_{1}, \dots, x_{n}]$ , finite subsets $S_{1}, \dots, S_{n} \subseteq F$ , and exponents $t_{1}, \dots, t_{n}$ satisfying the three hypotheses above.

[L1]

The reduction lemma gives a polynomial $\tilde{f}$ with $\deg_{x_{i}} \tilde{f} < ∣ S_{i} ∣$ , agreeing with $f$ on the whole grid and preserving the top coefficient of $x_{1}^{t_{1}} \dots x_{n}^{t_{n}}$ (Reducing $f$ modulo $g_{i} (x_{i}) = \prod_{s \in S_{i}} (x_{i} - s)$ lowers each $\deg_{x_{i}}$ below $∣ S_{i} ∣$ , preserves the values on the grid, and preserves any top-degree coefficient whose exponents stay below the grid sizes).

[L2]

A polynomial with separate degrees below the grid sizes that vanishes on the whole grid is the zero polynomial (If $\deg_{x_{i}} P < ∣ S_{i} ∣$ for each $i$ and $P$ vanishes on $S_{1} \times \dots \times S_{n}$ , then $P = 0$ ).

Proof

technique · contradiction

1.1assume-contraL1

Suppose, for contradiction, that $f$ vanishes at every point of $S_{1} \times \dots \times S_{n}$ . Apply [L1] to obtain the reduced polynomial $\tilde{f}$ .

2.1L1L2step 1.1

By [L1], the polynomial $\tilde{f}$ still vanishes on the whole grid and satisfies $\deg_{x_{i}} \tilde{f} < ∣ S_{i} ∣$ for every $i$ , so [L2] gives $\tilde{f} = 0$ .

3.1L1step 2.1discharge-contradiction∎

But [L1] also says that the coefficient of $x_{1}^{t_{1}} \dots x_{n}^{t_{n}}$ is the same in $\tilde{f}$ as in $f$ , hence nonzero. That contradicts $\tilde{f} = 0$ . Therefore some grid point satisfies $f (s_{1}, \dots, s_{n}) \neq 0$ .

Remarks

The top-coefficient hypothesis is load-bearing. The companion page carries the false statement obtained by deleting it.

Depends on

Used by

Dependency tree · two levels

22 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

N. Alon, Combinatorial Nullstellensatz, Theorem 1.2 (standard reference, not scraped)

Alon's Combinatorial Nullstellensatz: if deg⁡f=∑itidegf=∑i​ti​, the coefficient of x1t1⋯xntnx1t1​​⋯xntn​​ in ff is nonzero, and ∣Si∣>ti∣Si​∣>ti​, then f(s1,…,sn)≠0f(s1​,…,sn​)=0 for some si∈Sisi​∈Si​