TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-16

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Hom-tensor adjunction: ${Hom}_{R} (M \otimes_{R} N, P) ≅ {Hom}_{R} (M, {Hom}_{R} (N, P))$

Statement

Let $R$ be a commutative ring and let $M, N, P$ be $R$ -modules. There is a natural $R$ -module isomorphism

${Hom}_{R} (M \otimes_{R} N, P) ≅ {Hom}_{R} (M, {Hom}_{R} (N, P)) .$

It sends $F$ to the map $m \mapsto [n \mapsto F (m \otimes n)]$ and sends $u : M \to {Hom}_{R} (N, P)$ to the homomorphism determined by $m \otimes n \mapsto u (m) (n)$ .

Facts & Assumptions

Given: A commutative ring $R$ and $R$ -modules $M, N, P$ .

[L1]

The internal Hom is an $R$ -module under $(r f) (n) = r f (n)$ (The $R$ -module ${Hom}_{R} (M, N)$ over a commutative ring).

[L2]

Bilinear maps from $M \times N$ into $P$ correspond uniquely to homomorphisms $M \otimes_{R} N \to P$ (Universal property of the tensor product for balanced maps into abelian groups).

[L3]

Scalar multiplication on the tensor product satisfies $r (m \otimes n) = (r m) \otimes n = m \otimes (r n)$ (Over a commutative ring, $M \otimes_{R} N$ is an $R$ -module with $r (m \otimes n) = (r m) \otimes n = m \otimes (r n)$ ).

Proof

technique · direct

1.1givenL1L3algebra

Given $F : M \otimes_{R} N \to P$ , define $cur (F) (m) (n) = F (m \otimes n)$ . For fixed $m$ this is $R$ -linear in $n$ by [L3], and the dependence on $m$ is $R$ -linear by the same formula, so $cur (F) : M \to {Hom}_{R} (N, P)$ is an $R$ -module homomorphism.

1.2givenL1L2

Given $u : M \to {Hom}_{R} (N, P)$ , the pairing $(m, n) \mapsto u (m) (n)$ is bilinear by $R$ -linearity of $u$ and each $u (m)$ ; [L2] therefore induces a unique $uncur (u) : M \otimes_{R} N \to P$ .

2.1step 1.1step 1.2L2

For every $F, m, n$ , $uncur (cur (F)) (m \otimes n) = F (m \otimes n)$ , so uniqueness in [L2] makes $uncur cur$ the identity.

2.2step 1.1step 1.2

For every $u, m, n$ , $cur (uncur (u)) (m) (n) = u (m) (n)$ , so the two Hom-valued maps are equal and $cur uncur$ is the identity.

2.3step 1.1step 1.2L1algebra

Both assignments are $R$ -linear pointwise, and precomposition or postcomposition with homomorphisms commutes with evaluation; hence the bijection is an $R$ -module isomorphism natural in all three variables, contravariantly in the Hom source variables and covariantly in $P$ .

3.1step 2.1step 2.2step 2.3∎

Steps 2.1 through 2.3 prove the natural Hom-tensor adjunction.

Depends on

Used by

Dependency tree · two levels

11 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)
M. Barr, Acyclic Models, Chapter 2 (standard reference, not scraped)

Hom-tensor adjunction: Hom⁡R(M⊗RN,P)≅Hom⁡R(M,Hom⁡R(N,P))HomR​(M⊗R​N,P)≅HomR​(M,HomR​(N,P))