LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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.

Derived-series terms are characteristic ideals

Statement

For every Lie algebra $g$ and every $r \geq 0$ , the derived-series term $g^{(r)}$ is a characteristic ideal: every Lie-algebra automorphism $f$ of $g$ satisfies $f (g^{(r)}) = g^{(r)}$ .

Facts & Assumptions

Given: A Lie algebra $g$ , an automorphism $f : g \to g$ , and an integer $r \geq 0$ .

[L1]

The derived series starts at $g$ and replaces each term $I$ by $[I, I]$ ; every term is an ideal (Derived series and solvable Lie algebras).

[L2]

A Lie-algebra homomorphism is linear and satisfies $f ([x, y]) = [f (x), f (y)]$ (Homomorphisms of possibly infinite-dimensional Lie algebras).

Proof

technique · direct

1.1

For every subspace $A \subseteq g$ , linearity and bracket preservation give $f ([A, A]) = [f (A), f (A)]$ : each spanning bracket maps to a spanning bracket, and every bracket on the right has a preimage because $f$ is surjective.

givenL2algebra

2.1

Starting with $f (g) = g$ , suppose $f (g^{(j)}) = g^{(j)}$ . Then [L1] and step 1.1 give $f (g^{(j + 1)}) = [g^{(j)}, g^{(j)}] = g^{(j + 1)}$ . Finite induction proves the equality at the given index $r$ ; [L1] already supplies ideality.

L1step 1.1given∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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

Milne, Lie Algebras, §3, Generalities (standard reference, not scraped)