TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A nonzero polynomial of degree $n$ over an integral domain has at most $n$ distinct roots

Statement

Let $D$ be an integral domain. A nonzero polynomial $f\in D[x]$ of degree $n$ has at most $n$ distinct roots in $D$ .

Facts & Assumptions

Given: An integral domain $D$ and a nonzero polynomial $f\in D[x]$ of degree $n$ .

[L1]

If $a$ is a root of $f$ , then $f=(x-a)q$ for some polynomial $q$ (Factor theorem over a commutative ring).

[L2]

Degrees add when nonzero polynomials over a domain are multiplied (Over an integral domain, degrees add under multiplication of nonzero polynomials).

[L3]

In an integral domain, a product is zero only if one factor is zero (Zero divisor, and integral domain: a commutative ring with $1 \ne 0$ and no zero divisors).

[L4]

If a property holds at $0$ and passes from $n$ to $n+1$ , it holds for every natural number (The principle of mathematical induction).

Proof

technique · induction

1.1

If $n=0$ , then $f$ is a nonzero constant and has no root, proving the base case.

basegiven

2.1

For $n>0$ , if $f$ has no root the claim is immediate; otherwise choose a root $a$ , use [L1] to write $f=(x-a)q$ , and use [L2] to obtain $\deg q=n-1$ .

step 1.1ihL1L2choose

3.1

If $b\ne a$ is another root, then $0=f(b)=(b-a)q(b)$ , and [L3] gives $q(b)=0$ because $b-a\ne0$ ; the induction hypothesis bounds the roots other than $a$ by $n-1$ , so $f$ has at most $n$ roots, and [L4] completes the induction.

step 2.1ihL3L4discharge-induction∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 37 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Thomas W. Judson, Abstract Algebra: Theory and Applications, Corollary 17.9 (standard reference, not scraped)
Neil Donaldson, Math 120B Notes, Theorem 23.14 (standard reference, not scraped)

A nonzero polynomial of degree nnn over an integral domain has at most nnn distinct roots