TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-12

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The Schwartz-Zippel lemma

Statement

Let $F$ be a field, let $f \in F [x_{1}, \dots, x_{n}]$ be a nonzero polynomial of total degree at most $d$ , and let $S \subseteq F$ be a nonempty finite set. If $a = (a_{1}, \dots, a_{n})$ is chosen uniformly from $S^{n}$ , then

$P (f (a) = 0) \leq \frac{d}{∣ S ∣} .$

Facts & Assumptions

Given: a field $F$ , a nonzero polynomial $f \in F [x_{1}, \dots, x_{n}]$ of total degree at most $d$ , and a nonempty finite set $S \subseteq F$ .

[L1]

Every multivariate polynomial has a finite monomial expansion; in particular one may view it as a univariate polynomial in one chosen variable with coefficients in the remaining variables, and the total degree of a coefficient slice drops by its exponent in that variable (Monomials, coefficients, degree in each variable and total degree in $F [x_{1}, \dots, x_{n}]$ , Polynomial rings in finitely many commuting indeterminates by iteration).

[L2]

A nonzero univariate polynomial of degree $m$ over a domain has at most $m$ roots (A nonzero polynomial of degree $n$ over an integral domain has at most $n$ distinct roots).

Proof

technique · induction

1.1base

If $n = 0$ , then $f$ is a nonzero constant polynomial and $S^{0}$ has the single element $()$ , so $f (a) \neq 0$ always. Hence the vanishing probability is $0$ .

1.2L2base

If $n = 1$ , the statement is exactly [L2]: at most $d$ points of $S$ are roots of $f$ , so the vanishing probability is at most $d / ∣ S ∣$ .

1.3L1ih

Assume the claim for $n - 1$ variables with $n \geq 2$ . Write $f (x_{1}, \dots, x_{n}) = \sum_{j = 0}^{m} f_{j} (x_{2}, \dots, x_{n}) x_{1}^{j}$ , where $m = \deg_{x_{1}} f$ and $f_{m} \neq 0$ . By [L1], the coefficient polynomial $f_{m}$ has total degree at most $d - m$ .

2.1L2step 1.3choose

Choose $(a_{2}, \dots, a_{n}) \in S^{n - 1}$ uniformly. By the induction hypothesis, the probability that $f_{m} (a_{2}, \dots, a_{n}) = 0$ is at most $(d - m) / ∣ S ∣$ . On the complementary event, the specialized univariate polynomial $g (y) : = f (y, a_{2}, \dots, a_{n})$ is nonzero of degree $m$ , so [L2] says that for a uniform $a_{1} \in S$ the conditional probability $g (a_{1}) = 0$ is at most $m / ∣ S ∣$ .

3.1step 1.1step 1.2step 2.1discharge-inductionalgebra∎

Split according to the two cases in step 2.1. The total vanishing probability is at most $(d - m) / ∣ S ∣ + m / ∣ S ∣ = d / ∣ S ∣$ . This proves the claim for every $n \geq 2$ , and together with steps 1.1 and 1.2 completes the induction.

Depends on

Used by

Dependency tree · two levels

15 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

Sanjeev Arora and Boaz Barak, Computational Complexity: A Modern Approach (standard reference, not scraped)
Eric Blais, Models of Computation, 14. Randomized Computation (standard reference, not scraped)