TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07

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.

The Schur index equals the division-algebra index

Statement

Let $χ$ be an irreducible complex character of a finite group $G$ , put $K = Q (χ)$ , and let $V$ be the irreducible $K$ -representation used in The Schur index of an irreducible character. If $D_{V} = {End}_{G} (V)$ , then $Z (D_{V}) = K$ and $m_{K} (χ) = ind (D_{V}) .$

Facts & Assumptions

Given: $χ$ , $K$ , $V$ , and $D_{V}$ as in the statement, and a finite Galois splitting field $E / K$ inside $C$ .

[L1]

Base change identifies $E \otimes_{K} D_{V}$ with ${End}_{G} (E \otimes_{K} V)$ (Base change for intertwiner spaces).

[L2]

The scalar-extension decomposition is $E \otimes_{K} V ≅ U^{\oplus m_{K} (χ)}$ for an absolutely irreducible $U$ with character $χ$ (Scalar extension of an irreducible finite-group representation, The character field is the stabilizer fixed field).

[L3]

The index of a central division algebra is the square root of its central dimension (Index of a central division algebra).

[L4]

Since $E$ splits $G$ , ${End}_{G} (U) = E$ (A splitting field for a finite group: every irreducible representation has scalar endomorphism ring).

Proof

technique · direct

1.1

By [L2] and [L4], ${End}_{G} (E \otimes_{K} V) ≅ {End}_{G} (U^{\oplus m_{K} (χ)}) ≅ M_{m_{K} (χ)} (E)$ . Thus [L1] gives $E \otimes_{K} D_{V} ≅ M_{m_{K} (χ)} (E)$ .

L1L2L4

2.1

If $z \in Z (D_{V})$ , its image in the matrix algebra of step 1.1 commutes with every matrix, so it is $λ I$ for some $λ \in E$ . Because $1 \otimes z$ is fixed by $Gal (E / K)$ , so is $λ$ ; hence $λ \in K$ . Since $K$ already acts by scalar endomorphisms, $Z (D_{V}) = K$ .

step 1.1algebra

3.1

Taking $E$ -dimensions in step 1.1 gives $\dim_{K} D_{V} = m_{K} (χ)^{2}$ . With step 2.1, [L3] therefore gives $ind (D_{V}) = m_{K} (χ)$ .

L3step 1.1step 2.1∎

Depends on

Used by

Dependency tree · two levels

22 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

Gabor Wiese, Galois Representations, Corollary 2.5.7 and Remark 2.5.15 (standard reference, not scraped)
Weizhe Zheng, Lectures on Algebra, Proposition 4.6.14 (standard reference, not scraped)