TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-23 (gpt-6-sol)

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The Schur index equals the division-algebra index

Statement

Assume the Axiom of Choice. 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: AC, $χ$ , $K$ , $V$ , 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

1.1L1L2L4given

Under the stated AC premise, the attached $V$ and $m_{K} (χ)$ exist. 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 $E \otimes_{K} D_{V} ≅ M_{m_{K} (χ)} (E)$ .

2.1step 1.1algebra

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, $Z (D_{V}) = K$ .

3.1L3step 1.1step 2.1∎

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

Depends on

Used by

The faithful quaternion character has Schur index two Example

Dependency tree · two levels

25 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)