LemmaStatement: Literature-sourcedProof: Literature-sourcedverified 2026-09-26 (gpt-6-sol)

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.

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The character field is the stabilizer fixed field

Statement

Let $F \subseteq E \subseteq C$ with $E / F$ finite Galois, let $G$ be finite, and let $U$ be an absolutely irreducible $E$ -representation with character $χ$ . Then $E^{Stab (U)} = F (χ) .$

Facts & Assumptions

Given: $F \subseteq E \subseteq C$ , $G$ , $U$ , and $χ$ as in the statement.

[L1]

$χ_{^{σ} U} = σ \circ χ$ , and $Stab (U)$ consists of conjugates equivalent to $U$ (Galois conjugates of a representation).

[L2]

Finite-dimensional complex representations of a finite group are equivalent exactly when their characters agree (Finite-dimensional complex representations of a finite group are determined up to isomorphism by their characters).

[L3]

Intertwiner spaces commute with extension of scalars (Base change for intertwiner spaces).

[L4]

For a finite Galois extension, intermediate fields are fixed fields of their fixing subgroups (The fundamental theorem of finite Galois theory).

Proof

technique · direct

1.1L1L2L3algebra

For $σ \in Gal (E / F)$ , equivalent $E$ -representations have equal characters. Conversely, if $σ (χ (g)) = χ (g)$ for every $g \in G$ , [L1] and [L2] make the complex scalar extensions of $^{σ} U$ and $U$ isomorphic. By [L3] with the extension $C / E$ , a nonzero complex intertwiner comes from a nonzero $E$ -intertwiner. Both $E$ -representations are irreducible, so this intertwiner is an isomorphism. Thus $σ \in Stab (U) ⟺ σ (χ (g)) = χ (g)$ for every $g \in G$ .

2.1L1step 1.1

The right side says exactly that $σ$ fixes the field generated by $F$ and all character values, namely $F (χ)$ .

3.1L4step 2.1∎

Thus $Stab (U) = Gal (E / F (χ))$ ; [L4] gives $E^{Stab (U)} = F (χ)$ .

Depends on

Used by

The Schur index of an irreducible character Definition
Character formula over a nonsplitting field Theorem
The Schur index equals the division-algebra index Theorem

Dependency tree · two levels

23 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 2.5.5 (standard reference, not scraped)
Weizhe Zheng, Lectures on Algebra, Proposition 4.6.14 (standard reference, not scraped)