TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-07-31

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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.

For every prime $p$ , the two operations on $Z / p$ make it a field

Statement

For every prime $p$ , the operations of addition and multiplication on $Z / p$ make it a field (Field).

Facts & Assumptions

Given: A prime integer $p$ .

[L1]

Modular addition makes $Z / p$ an abelian group, modular multiplication makes it a commutative monoid, and both distributive laws hold (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

[L2]

A prime satisfies $p > 1$ , and if $p ∤ a$ then $\gcd (a, p) = 1$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ , For a prime $p$ and any integer $a$ , $\gcd (p, a)$ is $p$ when $p ∣ a$ and $1$ otherwise; so $p ∤ a$ makes $p$ and $a$ coprime).

[L3]

For positive modulus $p$ , $[a]_{p}$ is invertible exactly when $\gcd (a, p) = 1$ (For $n \geq 1$ , $[a]_{n}$ is a unit if and only if $\gcd (a, n) = 1$ ).

[L4]

Equality $[a]_{p} = [b]_{p}$ is equivalent to $a \equiv b (m o d p)$ , which means $p ∣ (a - b)$ (The congruence class $[a]_{n}$ and the quotient set $Z / n$ , Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ ).

[L5]

A field is a commutative ring with $0 \neq 1$ in which every nonzero element is multiplicatively invertible (Field).

Proof

technique · direct

1.1L1L5

By [L1], all additive, multiplicative, identity, inverse, commutativity, associativity and distributive field axioms already hold except possibly $[0]_{p} \neq [1]_{p}$ and invertibility of nonzero classes.

1.2L2L4

Since $p > 1$ , it does not divide $1$ , so [L4] gives $[0]_{p} \neq [1]_{p}$ .

1.3L2L3L4

If $[a]_{p} \neq [0]_{p}$ , then $p ∤ a$ by [L4]. Hence $\gcd (a, p) = 1$ by [L2], and [L3] makes $[a]_{p}$ invertible.

2.1step 1.1step 1.2step 1.3L5∎

Steps 1.1 through 1.3 verify every clause of [L5], so $Z / p$ is a field.

For every prime $p$ , the two operations on $Z / p$ make it a field

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

For every prime pp, the two operations on Z/pZ/p make it a field

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

For every prime $p$ , the two operations on $Z / p$ make it a field