CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

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Finite iterated tensor products represent multilinear maps independently of parenthesization

Statement

Let $R$ be a commutative ring and let $M_{1}, \dots, M_{k}$ be a finite list of $R$ -modules. Any parenthesized tensor product

$T = M_{1} \otimes_{R} \dots \otimes_{R} M_{k}$

represents $R$ -multilinear maps from $M_{1} \times \dots \times M_{k}$ : for every $R$ -module $P$ , composition with $(m_{1}, \dots, m_{k}) \mapsto m_{1} \otimes \dots \otimes m_{k}$ is a bijection from ${Hom}_{R} (T, P)$ to the set of multilinear maps into $P$ . Different parenthesizations are connected by the unique isomorphism preserving pure tensors.

For $k = 0$ , take $T = R$ and identify zero-variable multilinear maps with chosen elements of $P$ . For $k = 1$ , take $T = M_{1}$ .

Facts & Assumptions

Given: A commutative ring $R$ , a finite list of $R$ -modules, and an $R$ -module $P$ .

[L1]

The binary tensor product represents balanced, hence over a commutative ring bilinear, maps (Universal property of the tensor product for balanced maps into abelian groups).

[L2]

Tensor products over a commutative ring have canonical symmetry and associativity isomorphisms preserving elementary tensors (Symmetry and associativity isomorphisms for tensor products over a commutative ring).

[L3]

The regular module is a tensor unit (The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ ).

Proof

technique · induction

1.1baseL3

For $k = 0$ , an $R$ -linear map $R \to P$ is uniquely determined by the image of $1_{R}$ , and every $p \in P$ defines such a map by $r \mapsto r p$ ; this is the required representation of maps from the one-point empty product.

1.2base

For $k = 1$ , the identity $M_{1} \to M_{1}$ represents linear maps from $M_{1}$ by composition.

1.3ihL1

Assume a parenthesized product $T_{k}$ represents $k$ -linear maps. A $(k + 1)$ -linear map is equivalently a bilinear map $T_{k} \times M_{k + 1} \to P$ : first use the induction bijection with the last variable fixed, and then use multilinearity to see that the resulting dependence on the last variable is linear.

2.1step 1.3L1

By [L1], the bilinear maps in step 1.3 correspond uniquely to linear maps $T_{k} \otimes_{R} M_{k + 1} \to P$ , proving the representing property for $k + 1$ .

3.1step 2.1L2

By [L2], any two parenthesizations are joined by composites of elementary associativity isomorphisms preserving pure tensors. Any two such comparison maps agree on every pure tensor, so the representing uniqueness proved in step 2.1 makes them equal.

4.1step 1.1step 1.2step 2.1step 3.1discharge-induction∎

The base cases and induction step establish the representation for every finite $k$ , including the empty and singleton cases, and step 3.1 proves independence of parenthesization.

Depends on

Used by

Dependency tree · two levels

12 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 Project, Section 10.12: Tensor products (standard reference, not scraped)