Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-07-31
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  • 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.
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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).

[L3]

For positive modulus p, [a]p is invertible exactly when gcd⁡(a,p)=1 (For n≥1, [a]n is a unit if and only if gcd⁡(a,n)=1).

[L4]

Equality [a]p=[b]p is equivalent to a≡b(modp), which means p∣(a−b) (The congruence class [a]n and the quotient set Z/n, Congruence modulo an integer: a≡b(modn) when n∣(a−b), including the moduli 0 and 1).

[L5]

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

Proof

technique · direct
1.1

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

L1L5
1.2

Since p>1, it does not divide 1, so [L4] gives [0]p≠[1]p.

L2L4
1.3

If [a]p≠[0]p, then p∤a by [L4]. Hence gcd⁡(a,p)=1 by [L2], and [L3] makes [a]p invertible.

L2L3L4
2.1

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

step 1.1step 1.2step 1.3L5∎

Depends on

Used by

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