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

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Centerless does not imply semisimple

Counterexample

The two-dimensional affine Lie algebra $a = k x \oplus k y$ with $[x, y] = y$ is centerless but solvable. It therefore refutes Centerless implies semisimple.

Facts & Assumptions

Given: The displayed nonabelian Lie algebra over a characteristic-zero field.

[L1]

The derived series defines solvability (Derived series and solvable Lie algebras).

[L2]

A finite-dimensional Lie algebra is semisimple when its solvable radical is zero (Semisimple Lie algebras).

Refutation

technique · counterexample

1.1

For $u = a x + b y$ , $[u, x] = - b y, [u, y] = a y .$ If $u$ is central, both brackets vanish, so $a = b = 0$ . Hence $Z (a) = 0$ .

givenalgebra

1.2

On the other hand, $a^{(1)} = [a, a] = k y$ and $a^{(2)} = [k y, k y] = 0$ . Thus $a$ is solvable by [L1]. As a solvable ideal of itself, its radical is all of $a$ , which is nonzero; by [L2] it is not semisimple.

L1L2algebra

2.1

Step 1.1 satisfies the proposed centerless hypothesis and step 1.2 fails the semisimplicity conclusion. The witness is nonzero, two-dimensional, and completely explicit; no choice is used.

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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, affine algebra and semisimplicity (standard reference, not scraped)