Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-30
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The values of a central character are algebraic integers

Statement

Let G be a finite group, let χ be an irreducible complex character of G, and let C be a conjugacy class of G. Then ωχ(C^) is an algebraic integer.

Facts & Assumptions

Given: A finite group G, an irreducible complex character χ of G, and a conjugacy class C of G.

[F1]

The class sum C^ acts on the irreducible representation affording χ as the scalar ωχ(C^) (Class sums act on an irreducible representation by central-character scalars).

[F2]

An element is integral over Z exactly when it lies in a faithful module that is finitely generated over Z (Integrality and finite-module characterizations for one element).

[F3]

Integral elements over a nonzero base ring form a subring (Integral elements over a nonzero base ring form a subring).

Proof

technique · direct
1.1

Let C1,,Cr be the conjugacy classes of G, and let A be the Z-subring of Z(C[G]) generated by the class sums C^1,,C^r. If C^iC^j=knijkC^k, then each coefficient nijk counts pairs (x,y)Ci×Cj with xyCk, so nijkZ. Hence the additive group of A is contained in the free abelian group on the finitely many class sums, so A is finitely generated as a Z-module.

givenalgebra
2.1

The ring A is a faithful A-module over itself, and step 1.1 makes it finitely generated over Z. Therefore [F2] shows that each generator C^i is integral over Z. Then [F3] implies that every element of A is integral over Z.

F2F3step 1.1algebra
3.1

Let V be an irreducible representation affording χ. By [F1], the action of A on V sends each aA to a scalar λχ(a)C, and aλχ(a) is a ring homomorphism because the action of A is by endomorphisms. If m(X)Z[X] is monic with m(C^)=0, then applying λχ gives the monic equation m(ωχ(C^))=0. So ωχ(C^) is integral over Z, that is, an algebraic integer.

F1step 2.1algebra

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Dependency tree · two levels

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Sources