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Universal property of algebraic differentials
Statement
Let be a ring homomorphism of commutative rings and let with its universal derivation be as in Universal algebraic differentials and A-derivations. For every -module the assignment is a -module isomorphism
natural in : for every -linear the diagram of the two assignment maps commutes, and the isomorphism is additive in and .
Facts & Assumptions
Given: A ring homomorphism of commutative rings, the presented module with universal derivation , and a -module .
Universal algebraic differentials and A-derivations: , where is the free -module on the basis symbols , , and is generated by , and ; ; an -derivation into a -module is a map that is additive, -constant and satisfies the Leibniz rule.
Proof
Every gives : additivity, -constancy and the Leibniz rule for are those of transported by the additive map , and has them by construction. The assignment is additive and -linear: and pointwise. It is injective, because the classes generate as a -module, so forces .
Conversely let . Since is free on the , there is a unique -linear with . Each relator is killed: by additivity of , similarly the Leibniz relator, and by -constancy. Hence factors through , giving a -linear with , that is, .
The two assignments are mutually inverse: the derivation attached to is , and the homomorphism attached to agrees with on every generator , hence equals because those generators span. Each assignment is additive, so the bijection is a -module isomorphism; and for -linear one has , which is exactly naturality in .
Depends on
Used by
- Differentials of a polynomial quotient and the Jacobian cokernel Lemma
- Localization, base change and functoriality of differentials Lemma
- Separable residue and the cotangent sequence of a local algebra Lemma
- Transitivity sequence for differentials Lemma
- Differentials of a separably generated field extension Theorem
Dependency tree · two levels
3 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
- Stacks Algebra 10.131.3 (standard reference, not scraped)
- Vakil §22.2.17, pp.582–583 (standard reference, not scraped)