TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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Over a finite-dimensional algebra, a module annihilated modulo its radical is zero, and generators lift from the head

Statement

Let $A$ be a finite-dimensional algebra with Jacobson radical $J$ , and let $M$ be a finite-dimensional left $A$ -module.

If $M = J M$ , then $M = 0$ .
If elements $m_{1}, \dots, m_{r} \in M$ span $M / J M$ , then they generate $M$ .

Facts & Assumptions

Given: A finite-dimensional algebra $A$ with Jacobson radical $J$ and a finite-dimensional left $A$ -module $M$ .

[L1]

The module radical equals $J M$ (For a finite-dimensional algebra, the module radical is exactly the action of the Jacobson radical).

[L2]

The algebra radical $J$ is nilpotent (For a finite-dimensional algebra, the Jacobson radical is nilpotent and the quotient by it is semisimple).

Proof

technique · direct

1.1

If $M = J M$ , then iterating the equality gives $M = J^{n} M$ for every $n \geq 1$ . Choose $n$ with $J^{n} = 0$ from [L2]. Then $M = J^{n} M = 0$ .

L2givenalgebra

2.1

Let $N : = A m_{1} + \dots + A m_{r}$ . By hypothesis, the images of the $m_{i}$ span $M / J M$ , so $M = N + J M$ . Hence $M / N = J (M / N)$ . Applying part 1 to the module $M / N$ and using [L1], we obtain $M / N = 0$ . Therefore $M = N$ , so $m_{1}, \dots, m_{r}$ generate $M$ .

L1step 1.1givenalgebra

3.1

Steps 1.1 and 2.1 are exactly the two asserted conclusions.

step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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

Peter Webb, A Course in Finite Group Representation Theory (23 Feb 2016 draft) (standard reference, not scraped)