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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-sol)audited 2026-09-27
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Universal property of algebraic differentials

Statement

Let A→B be a ring homomorphism of commutative rings and let ΩB/A with its universal derivation d be as in Universal algebraic differentials and A-derivations. For every B-module M the assignment g↦g∘d is a B-module isomorphism

Hom⁡B(ΩB/A,M)  ≅  Der⁡A(B,M),

natural in M: for every B-linear h ⁣:M→N the diagram of the two assignment maps commutes, and the isomorphism is additive in g and D.

Facts & Assumptions

Given: A ring homomorphism A→B of commutative rings, the presented module ΩB/A=F/R with universal derivation d, and a B-module M.

[F1]

Universal algebraic differentials and A-derivations: ΩB/A=F/R, where F is the free B-module on the basis symbols [b], b∈B, and R is generated by [b+b′]−[b]−[b′], [bb′]−b[b′]−b′[b] and [φ(a)]; db=[b]+R; an A-derivation into a B-module M is a map D that is additive, A-constant and satisfies the Leibniz rule.

Proof

1.1

Every g∈Hom⁡B(ΩB/A,M) gives g∘d∈Der⁡A(B,M): additivity, A-constancy and the Leibniz rule for g∘d are those of d transported by the additive map g, and d has them by construction. The assignment is additive and B-linear: (g1+g2)∘d=g1∘d+g2∘d and (bg)∘d=b (g∘d) pointwise. It is injective, because the classes db generate ΩB/A=F/R as a B-module, so g∘d=0 forces g=0.

F1given
2.1

Conversely let D∈Der⁡A(B,M). Since F is free on the [b], there is a unique B-linear D~ ⁣:F→M with D~([b])=D(b). Each relator is killed: D~([b+b′]−[b]−[b′])=D(b+b′)−D(b)−D(b′)=0 by additivity of D, similarly the Leibniz relator, and [φ(a)]↦D(φ(a))=0 by A-constancy. Hence D~ factors through F/R=ΩB/A, giving a B-linear gD ⁣:ΩB/A→M with gD(db)=D(b), that is, gD∘d=D.

F1step 1.1algebra
3.1

The two assignments are mutually inverse: the derivation attached to gD is gD∘d=D, and the homomorphism attached to g∘d agrees with g on every generator db, hence equals g because those generators span. Each assignment is additive, so the bijection is a B-module isomorphism; and for B-linear h ⁣:M→N one has h∘(g∘d)=(h∘g)∘d, which is exactly naturality in M.

step 1.1step 2.1algebra∎

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