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LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passverified 2026-09-23 (gpt-6-sol)
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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 is independent of the splitting field

Statement

Assume the Axiom of Choice. For an irreducible character χ of a finite group, the common multiplicity in the scalar-extension orbit used in The Schur index of an irreducible character is unchanged when the finite Galois splitting field is replaced by a larger finite Galois splitting field.

Facts & Assumptions

Given: AC, K=Q(χ), the irreducible K-module V attached to χ by the definition, and finite Galois splitting fields E⊆L over K.

[L1]

Scalar extension is associative: L⊗KV≅L⊗E(E⊗KV) (Change of rings: N⊗RM≅N⊗S(S⊗RM)).

[L2]

Over either splitting field, an irreducible K-module extends as one Galois orbit with a common multiplicity (Scalar extension of an irreducible finite-group representation).

Proof

technique · direct
1.1L2given

Under AC, the attached V exists by The Schur index of an irreducible character. Write E⊗KV as m times its orbit of pairwise inequivalent absolutely irreducible constituents, using [L2].

2.1L1step 1.1

Each constituent remains irreducible after extension from E to the splitting field L, and distinct constituents remain distinct; therefore [L1] writes L⊗KV with the same coefficient m.

3.1L2step 2.1∎

Applying [L2] directly over L identifies its common multiplicity with that coefficient. Hence both choices give m, proving independence.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

21 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