LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-09

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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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Invertibility of a positive natural scalar in a field

Statement

Let $k$ be a field and $n > 0$ an integer. Then $n \cdot 1_{k}$ is invertible if and only if $char (k) ∤ n$ . Divisibility is in $Z$ ; in particular, $0$ divides no positive integer.

Facts & Assumptions

Given: A field $k$ and an integer $n > 0$ .

[F1]

In a field $0 \neq 1$ , distributivity holds and every nonzero scalar has an inverse (Field).

[F2]

For any integer $n$ , $n \cdot 1_{k} = 0$ if and only if $char (k) ∣ n$ (The characteristic of a ring is the additive order of $1_{R}$ , with $0$ recording infinite order; $n \cdot 1_{R} = 0$ holds exactly when $char (R) ∣ n$ ; and in an integral domain every nonzero element has the same additive order as $1_{R}$ ).

Proof

technique · direct

1.1

Put $s = n \cdot 1_{k}$ . For every $t \in k$ , distributivity gives $0 t = (0 + 0) t = 0 t + 0 t$ , so cancellation gives $0 t = 0$ . Since $0 \neq 1$ , the scalar zero is not invertible. If $s$ is invertible, it is therefore nonzero, and the equivalence in F2 gives $char (k) ∤ n$ .

F1F2

1.2

Conversely, if $char (k) ∤ n$ , F2 gives $s \neq 0$ , and the field inverse axiom gives $s^{- 1} \in k$ with $s s^{- 1} = s^{- 1} s = 1$ .

F1F2

2.1

By integer divisibility, $0 ∣ n$ would mean $n = 0 q = 0$ for some integer $q$ , impossible for $n > 0$ . Thus characteristic zero is included. At $n = 1$ , the scalar is $1_{k}$ with inverse $1_{k}$ . Together with the two implications this proves the claim.

step 1.1step 1.2F2F1∎

Sources

Milne, Fields and Galois Theory, pp. 8–9, characteristic cases 1–2. The normalization motivating this interface occurs in Etingof et al., Theorem 4.1.1, pp. 61–62.

Depends on

Used by

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Dependency tree · two levels

15 results within two dependency steps 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

Etingof et al., Introduction to Representation Theory (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, v5.10 (2022) (standard reference, not scraped)