PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)verified 2026-10-03 (gpt-6.1-sol)

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A proper geometrically connected group variety is commutative

Statement

Assume the Axiom of Choice. Every abelian variety over a field is commutative. More generally, if $A$ is an abelian variety and $H$ a connected separated finite-type $k$ -group scheme, every group homomorphism $u : A \to H$ has central image, scheme theoretically.

Facts & Assumptions

[F1]

Abelian varieties are proper geometrically integral group varieties, with a rational identity. (Abelian varieties over a field)

[F2]

A morphism from $X \times Y$ with $X$ proper geometrically integral and rationally pointed, $Y$ connected, and separated target, constant on a rational fibre, factors through $Y$ , under AC. (Rigidity for a proper geometrically integral factor)

Proof

Given: AC, an abelian variety $A$ , a connected finite-type group scheme $H$ , and a homomorphism $u : A \to H$ .

1.1F1F2givenalgebra

Form the morphism $c : A \times_{k} H \to H$ given on scheme-valued points by $c (a, h) = u (a) h u (a)^{- 1} h^{- 1}$ . Group multiplication and inverse show that it is a scheme morphism. On $A \times {e_{H}}$ it is identically $e_{H}$ . Apply [F2] with $X = A$ , $Y = H$ , and $Z = H$ ; algebraic group schemes here are separated. Therefore $c (a, h) = c (e_{A}, h) = e_{H}$ as an identity of scheme morphisms. Equivalently, conjugation of $u (a)$ by every scheme-valued point of $H$ fixes it. This is precisely the statement that $u$ factors through the scheme-theoretic centre.

2.1step 1.1given∎

Take $H = A$ and $u = {id}_{A}$ in step 1.1. The equality $a h a^{- 1} h^{- 1} = e$ then gives $a h = h a$ on every $k$ -scheme of points and hence as a morphism identity. Thus the group law of $A$ is commutative. AC is used through [F2].

A proper geometrically connected group variety is commutative

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources