False statementConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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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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Not every Lie subalgebra is an ideal

Statement

Every Lie subalgebra is an ideal.

Facts & Assumptions

Given: The asserted implication from Lie subalgebra to ideal.

[L1]

A subalgebra is closed under its internal brackets, whereas an ideal must be closed under brackets with every ambient element (Lie subalgebras, ideals, and center).

Refutation

technique · direct counterexample

1.1

Over any field, inside the commutator Lie algebra of $2 \times 2$ matrices take $h = (\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array})$ and $e = (\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array})$ . Their span $g = k h \oplus k e$ is bracket-closed and satisfies $[h, e] = e$ .

constructalgebra

2.1

The line $k h$ is a Lie subalgebra because $[h, h] = 0$ , but it is not an ideal because $[e, h] = - e \notin k h$ . This violates the asserted implication in every characteristic.

step 1.1L1∎

Depends on

Lie subalgebras, ideals, and center

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 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, MIT 18.745 notes, Lie subalgebras and ideals in §8.3, printed pp. 50–51 (standard reference, not scraped)