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LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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In a field, the additive multiple n⋅1F is the canonical natural ι(n): the additive power of the group-power definition and the canonical natural are the same function, both being the unique one given by the recursion ι(0)=0F, ι(σ(n))=ι(n)+1F

Statement

Let F be a field (Field), which is a ring by Every field is a commutative ring with 1≠0; it is an integral domain, and it is a commutative division ring. Two functions N→F are in play:

These are the same function: ι(n)=n⋅1F for every n∈N. In particular the notation n⋅1F used by The canonical natural ι(n)=n⋅1F of a field and the notation n⋅1F used by Powers gn: natural exponents in a monoid and integer exponents in a group, with g0=e denote the same element of F, and no second notion is in play.

Facts & Assumptions

Given: A field F with 0F and 1F, the map ι:N→F of The canonical natural ι(n)=n⋅1F of a field, and the additive natural powers of Powers gn: natural exponents in a monoid and integer exponents in a group, with g0=e in the group (F,+,0F).

[L2]

ι(0)=0F and ι(n+1)=ι(n)+1F for every n∈N (The canonical natural ι(n)=n⋅1F of a field).

[L3]

The additive natural powers satisfy 0⋅a=0F and σ(n)⋅a=n⋅a+a for every n∈N and every a∈F (Powers gn: natural exponents in a monoid and integer exponents in a group, with g0=e).

[L4]

On N: m+0=m and m+σ(n)=σ(m+n), so n+1=σ(n) (Addition of natural numbers, The natural numbers N (von Neumann)).

[L5]

The recursion theorem: for a set A, an element a∈A and a function u:A→A there is exactly one g:N→A with g(0)=a and g(σ(n))=u(g(n)) (The recursion theorem).

Proof

technique · direct
1.1L1L5

Let u:F→F be the function u(t)=t+1F, which is a function from F to F because addition is a binary operation on F. By [L5] applied with A=F, a=0F and this u, there is exactly one function g:N→F satisfying g(0)=0F and g(σ(n))=g(n)+1F for every n∈N.

1.2L2L4

The map ι satisfies those two equations: ι(0)=0F by [L2], and ι(σ(n))=ι(n+1)=ι(n)+1F by [L2] together with n+1=σ(n).

1.3L3

The map n↦n⋅1F satisfies them too: 0⋅1F=0F and σ(n)⋅1F=n⋅1F+1F, both by [L3] with a=1F.

2.1step 1.1step 1.2step 1.3L5∎

By the uniqueness clause of step 1.1, the two functions of steps 1.2 and 1.3 are equal, so ι(n)=n⋅1F for every n∈N.

Remarks

Depends on

Dependency tree · 2 levels

Sources