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Tensor products commute with arbitrary direct sums
Statement
Let be a commutative ring, let be any family of -modules, and let be an -module. The homomorphism induced by the coordinate inclusions is a natural -module isomorphism
The same holds with the direct sum in the second variable: there is a natural -module isomorphism
The assertion includes , when both sides are zero.
Facts & Assumptions
Given: A commutative ring , a family of -modules, and an -module .
Balanced pairings induce unique homomorphisms from tensor products (Universal property of the tensor product for balanced maps into abelian groups).
Tensor products over a commutative ring carry their canonical -module structure (Over a commutative ring, is an -module with ).
The direct sum consists of finite-support tuples, with coordinate inclusions , and the empty direct sum is zero (The direct sum of an indexed family of modules).
Every family of homomorphisms extends uniquely to a homomorphism whose value is the finite sum of the coordinate values (Universal property of a direct sum of modules).
Over a commutative ring there is a natural isomorphism with , and is the identity (Symmetry and associativity isomorphisms for tensor products over a commutative ring).
Proof
For each , the pairing is bilinear, so [L1] gives . By [L4], the family induces the displayed map .
Define by . This tuple has finite support by [L3], and coordinatewise calculation shows that is bilinear.
By [L1], the pairing in step 1.2 induces with .
The composite fixes every coordinate generator , so it is the identity by [L4]; the composite fixes every elementary tensor , so it is the identity by [L1].
Thus and are inverse isomorphisms. They are natural because maps and make the two composites send each coordinate generator to . When , [L3] makes both sides zero and the same construction yields the unique isomorphism.
For the second variable, put , where the middle direct sum of the symmetries is the map induced by [L4] from the family . Each is an isomorphism by [L5] and is one by step 4.1, so is an isomorphism; it is natural as a composite of natural isomorphisms. Tracing a coordinate generator gives , which is the displayed formula. At both sides are again zero.
Depends on
- Universal property of the tensor product for balanced maps into abelian groups
- Over a commutative ring, $M\otimes_RN$ is an $R$-module with $r(m\otimes n)=(rm)\otimes n=m\otimes(rn)$
- The direct sum of an indexed family of modules
- Universal property of a direct sum of modules
- Symmetry and associativity isomorphisms for tensor products over a commutative ring
Used by
- For flat M, one has IM∩ JM=(I∩ J)M Corollary
- Under the stated choice boundary, free modules are projective and hence flat Corollary
- Every projective module over a commutative ring is flat Theorem
- Flatness is equivalent to preserving injections and to the ideal and finitely generated ideal tests Theorem
- The elementary tensors of two bases form the product basis of the tensor product Theorem
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 32 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.
Sources
- Stacks Project, Section 10.12: Tensor products (standard reference, not scraped)
- C. Dennis, Week 4 on tensor products and flatness (standard reference, not scraped)