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

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Tensor products commute with arbitrary direct sums

Statement

Let $R$ be a commutative ring, let $(M_{i})_{i \in I}$ be any family of $R$ -modules, and let $N$ be an $R$ -module. The homomorphism induced by the coordinate inclusions is a natural $R$ -module isomorphism

$Φ : ⨁_{i \in I} (M_{i} \otimes_{R} N) ⟶ (⨁_{i \in I} M_{i}) \otimes_{R} N,$

$Φ (ȷ_{i} (m \otimes n)) = ȷ_{i} (m) \otimes n .$

The same holds with the direct sum in the second variable: there is a natural $R$ -module isomorphism

$Φ^{'} : ⨁_{i \in I} (N \otimes_{R} M_{i}) ⟶ N \otimes_{R} (⨁_{i \in I} M_{i}), Φ^{'} (ȷ_{i} (n \otimes m)) = n \otimes ȷ_{i} (m) .$

The assertion includes $I = \emptyset$ , when both sides are zero.

Facts & Assumptions

Given: A commutative ring $R$ , a family $(M_{i})_{i \in I}$ of $R$ -modules, and an $R$ -module $N$ .

[L1]

Balanced pairings induce unique homomorphisms from tensor products (Universal property of the tensor product for balanced maps into abelian groups).

[L2]

Tensor products over a commutative ring carry their canonical $R$ -module structure (Over a commutative ring, $M \otimes_{R} N$ is an $R$ -module with $r (m \otimes n) = (r m) \otimes n = m \otimes (r n)$ ).

[L3]

The direct sum consists of finite-support tuples, with coordinate inclusions $ȷ_{i}$ , and the empty direct sum is zero (The direct sum of an indexed family of modules).

[L4]

Every family of homomorphisms $f_{i} : M_{i} \to P$ extends uniquely to a homomorphism $⨁_{i} M_{i} \to P$ whose value is the finite sum of the coordinate values (Universal property of a direct sum of modules).

[L5]

Over a commutative ring there is a natural isomorphism $σ_{M, N} : M \otimes_{R} N \to N \otimes_{R} M$ with $σ_{M, N} (m \otimes n) = n \otimes m$ , and $σ_{N, M} σ_{M, N}$ is the identity (Symmetry and associativity isomorphisms for tensor products over a commutative ring).

Proof

technique · direct

1.1givenL1L2L4

For each $i$ , the pairing $(m, n) \mapsto ȷ_{i} (m) \otimes n$ is bilinear, so [L1] gives $ϕ_{i} : M_{i} \otimes_{R} N \to (⨁_{i} M_{i}) \otimes_{R} N$ . By [L4], the family $(ϕ_{i})$ induces the displayed map $Φ$ .

1.2L2L3algebra

Define $b : (⨁_{i} M_{i}) \times N \to ⨁_{i} (M_{i} \otimes_{R} N)$ by $b ((m_{i}), n) = (m_{i} \otimes n)_{i}$ . This tuple has finite support by [L3], and coordinatewise calculation shows that $b$ is bilinear.

2.1step 1.2L1

By [L1], the pairing in step 1.2 induces $Ψ : (⨁_{i} M_{i}) \otimes_{R} N \to ⨁_{i} (M_{i} \otimes_{R} N)$ with $Ψ ((m_{i}) \otimes n) = (m_{i} \otimes n)_{i}$ .

3.1step 1.1step 2.1L1L4

The composite $Ψ Φ$ fixes every coordinate generator $ȷ_{i} (m \otimes n)$ , so it is the identity by [L4]; the composite $Φ Ψ$ fixes every elementary tensor $(m_{i}) \otimes n$ , so it is the identity by [L1].

4.1step 3.1L1L4L3

Thus $Φ$ and $Ψ$ are inverse isomorphisms. They are natural because maps $u_{i} : M_{i} \to M_{i}^{'}$ and $v : N \to N^{'}$ make the two composites send each coordinate generator $ȷ_{i} (m \otimes n)$ to $ȷ_{i} (u_{i} (m) \otimes v (n))$ . When $I = \emptyset$ , [L3] makes both sides zero and the same construction yields the unique isomorphism.

5.1step 4.1L3L4L5∎

For the second variable, put $Φ^{'} : = σ_{⨁_{i} M_{i}, N} \circ Φ \circ (⨁_{i} σ_{N, M_{i}})$ , where the middle direct sum of the symmetries is the map induced by [L4] from the family $ȷ_{i} \circ σ_{N, M_{i}}$ . Each $σ$ is an isomorphism by [L5] and $Φ$ is one by step 4.1, so $Φ^{'}$ is an isomorphism; it is natural as a composite of natural isomorphisms. Tracing a coordinate generator gives $Φ^{'} (ȷ_{i} (n \otimes m)) = σ (Φ (ȷ_{i} (m \otimes n))) = σ (ȷ_{i} (m) \otimes n) = n \otimes ȷ_{i} (m)$ , which is the displayed formula. At $I = \emptyset$ both sides are again zero.

Tensor products commute with arbitrary direct sums

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources