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

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Absolute irreducibility via the endomorphism division algebra

Statement

Assume the Axiom of Choice. Let $F$ have characteristic $0$ , $G$ be finite, and $V$ be an irreducible finite-dimensional $F$ -representation. Then $V$ is absolutely irreducible if and only if $D_{V} = F$ (via scalar endomorphisms).

Facts & Assumptions

Given: $F$ , $G$ , and $V$ as in the statement, and the Axiom of Choice.

[A1]

The stated Axiom of Choice is the direct premise used to obtain an algebraic closure (The Axiom of Choice).

[L1]

Base change gives $E \otimes_{F} D_{V} ≅ {End}_{G} (E \otimes_{F} V)$ for every field extension $E / F$ (Base change for intertwiner spaces).

[L2]

Assuming Choice, $F$ has an algebraic closure $\overline{F}$ (Assuming Choice, every field has an algebraic closure).

[L3]

Over an algebraically closed field, every endomorphism of an irreducible representation is scalar (Over an algebraically closed field, every endomorphism of an irreducible representation is scalar).

[L4]

Since the characteristic is zero, every finite-dimensional representation of the finite group $G$ over any extension field of $F$ is completely reducible (If $char k ∤ ∣ G ∣$ , every finite-dimensional representation of $G$ is completely reducible).

Proof

technique · direct

1.1A1L1L2L3algebra

If $V$ is absolutely irreducible, then $\overline{F} \otimes_{F} V$ is irreducible for an algebraic closure supplied by [L2] under [A1]. By [L3] its endomorphism algebra is $\overline{F}$ , and [L1] gives $\overline{F} \otimes_{F} D_{V} ≅ \overline{F}$ . Since $D_{V} \subseteq {End}_{F} (V)$ is finite-dimensional, comparison of $\overline{F}$ -dimensions gives $\dim_{F} D_{V} = 1$ . The scalar copy of $F$ in $D_{V}$ is therefore all of $D_{V}$ .

1.2L1L4algebra

Conversely assume $D_{V} = F$ , and let $E / F$ be any field extension. By [L1], ${End}_{G} (E \otimes_{F} V) ≅ E \otimes_{F} D_{V} ≅ E$ , so every $G$ -endomorphism is scalar. By [L4], $E \otimes_{F} V$ is completely reducible.

2.1step 1.2algebra∎

If $E \otimes_{F} V$ were reducible, complete reducibility would split it into two nonzero $G$ -subrepresentations. Projection onto one would be a nonzero, nonidentity idempotent $G$ -endomorphism and therefore could not be scalar over the field $E$ , contradicting step 1.2. Thus $E \otimes_{F} V$ is irreducible for every extension $E / F$ , which is absolute irreducibility.

Depends on

Used by

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Dependency tree · two levels

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Sources

Gabor Wiese, Galois Representations, Theorem 2.3.11 (standard reference, not scraped)
Weizhe Zheng, Lectures on Algebra, Theorem 4.2.3 (standard reference, not scraped)