LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-25

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 field has no zero divisors: $a b = 0 \Rightarrow a = 0$ or $b = 0$

Statement

In any field $F$ (Field), if $a, b \in F$ satisfy $a b = 0$ , then $a = 0$ or $b = 0$ .

Facts & Assumptions

Given: A field $F$ and elements $a, b \in F$ with $a b = 0$ .

[L1]

Field axioms (Field): if $a \neq 0$ then $a$ has a multiplicative inverse $a^{- 1}$ with $a^{- 1} a = 1$ ; multiplication is associative; and $1$ is the multiplicative identity, so $1 \cdot x = x$ .

[L2]

For every $x \in F$ , $x \cdot 0 = 0$ (Multiplication by zero: $0 \cdot a = 0$ , using commutativity).

Proof

technique · direct

1.1suffices: a ne 0 implies b = 0

If $a = 0$ the disjunction $a = 0$ or $b = 0$ already holds, so it suffices to assume $a \neq 0$ and derive $b = 0$ .

1.2L1

Assuming $a \neq 0$ , the inverse $a^{- 1}$ exists with $a^{- 1} a = 1$ .

1.3L1

Since $1$ is the multiplicative identity, $b = 1 \cdot b$ .

1.4L1

By associativity, $(a^{- 1} a) \cdot b = a^{- 1} (a b)$ .

1.5given

We are given $a b = 0$ .

1.6L2

$a^{- 1} \cdot 0 = 0$ .

2.1step 1.2step 1.3step 1.4

Substituting $1 = a^{- 1} a$ into $b = 1 \cdot b$ and using associativity gives $b = (a^{- 1} a) b = a^{- 1} (a b)$ .

2.2step 1.5

Since $a b = 0$ , we have $a^{- 1} (a b) = a^{- 1} \cdot 0$ .

3.1step 2.1step 2.2step 1.6

Combining, $b = a^{- 1} (a b) = a^{- 1} \cdot 0 = 0$ .

4.1step 1.1step 3.1∎

Hence $a \neq 0$ forces $b = 0$ , so in every case $a = 0$ or $b = 0$ .

Depends on

Used by

Dependency tree · one level

2 results within one dependency step 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

W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 1 (standard reference, not scraped)
M. Spivak, Calculus, 4th ed., Ch. 1 (standard reference, not scraped)
Tom Denton, Introduction to Algebraic Structures: The Problem of Division (standard reference, not scraped)

A field has no zero divisors: ab=0⇒a=0ab=0⇒a=0 or b=0b=0