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

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Trace functionals of split simple modules are independent

Statement

For $A$ and $S_{i}$ as in the separation lemma, the functions $τ_{i} : A \to k$ , $τ_{i} (a) = tr (a ∣ S_{i})$ , are linearly independent. If $A = k G$ for a finite group, their restrictions to $G$ are linearly independent.

Facts & Assumptions

Given: The finite-dimensional algebra and distinct split simple modules in the Statement.

[F1]

The action map onto the product of the endomorphism algebras is surjective (A finite-dimensional algebra separates its split simple modules).

Proof

1.1

In a basis of the nonzero $S_{i}$ , the matrix $E_{11}$ has trace $1$ , including when the characteristic divides $\dim S_{i}$ . By surjectivity choose $a_{i}$ acting as $E_{11}$ on $S_{i}$ and as zero on every other $S_{j}$ . Hence $τ_{j} (a_{i}) = δ_{i j}$ .

F1given

2.1

If $\sum_{j} c_{j} τ_{j} = 0$ , evaluation at $a_{i}$ gives $c_{i} = 0$ for every $i$ . For $k G$ , a relation vanishing on $G$ vanishes on $a = \sum_{g} a_{g} g$ since $\sum_{j} c_{j} τ_{j} (a) = \sum_{g} a_{g} \sum_{j} c_{j} τ_{j} (g) = 0$ . It is therefore the zero relation by the first assertion. An empty family has only the empty relation.

step 1.1algebra∎

Depends on

A finite-dimensional algebra separates its split simple modules

Used by

Irreducible Brauer characters are independent on p-regular elements Theorem

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

Yanqi Lake Lectures on Algebra I, Theorems 11.1.5 and 11.2.2, pp.128 and 130 (standard reference, not scraped)