LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck 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.

Coextension of scalars ${Hom}_{R} (S, M)$ carries its canonical left $S$ -module structure

Statement

Let $φ : R \to S$ be a unital ring homomorphism and let $M$ be a left $R$ -module. Regard $S$ as a left $R$ -module by $r \cdot t = φ (r) t$ . Then ${Hom}_{R} (S, M)$ is a left $S$ -module under

$(s \cdot h) (t) : = h (t s) .$

For an $R$ -linear map $u : M \to M^{'}$ , postcomposition $h \mapsto u \circ h$ is $S$ -linear, so this construction is functorial in $M$ .

Facts & Assumptions

Given: A unital ring homomorphism $φ : R \to S$ , a left $R$ -module $M$ , elements $r \in R$ , $s, s^{'} \in S$ , and $h, h^{'} \in {Hom}_{R} (S, M)$ .

[F1]

A ring homomorphism preserves addition, multiplication, zero, and one (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ).

[F2]

A left module action satisfies distributivity, $(r s) m = r (s m)$ , and $1_{R} m = m$ (Unital left and right modules over a ring; unqualified module means left module).

[F3]

An $R$ -linear map satisfies $h (x + y) = h (x) + h (y)$ and $h (r x) = r h (x)$ (Module homomorphism and isomorphism, kernel, image and cokernel).

[F4]

The functions from one set to another form a set (The set $B^{A}$ of all functions $A \to B$ ).

Proof

technique · direct

1.1F4

The set ${Hom}_{R} (S, M)$ is a subset of the function set $M^{S}$ , which exists by [F4].

1.2F1F3algebra

For $r \in R$ and $t \in S$ , $(s \cdot h) (φ (r) t) = h (φ (r) t s) = r h (t s) = r (s \cdot h) (t)$ , and additivity is similar, so $s \cdot h$ is $R$ -linear.

1.3F1F2algebra

Pointwise, $((s s^{'}) \cdot h) (t) = h (t s s^{'}) = (s \cdot (s^{'} \cdot h)) (t)$ and $(1_{S} \cdot h) (t) = h (t)$ , so associativity and the unit law hold.

1.4F3algebra

Pointwise additivity of $h$ gives $((s + s^{'}) \cdot h) (t) = (s \cdot h + s^{'} \cdot h) (t)$ , and linearity in $h$ gives $(s \cdot (h + h^{'})) (t) = (s \cdot h + s \cdot h^{'}) (t)$ .

2.1step 1.1step 1.2step 1.3step 1.4F2

Steps 1.1 through 1.4 verify all left $S$ -module axioms in [F2].

3.1F3step 1.2∎

If $u : M \to M^{'}$ is $R$ -linear, then $u \circ h$ is $R$ -linear and $u ((s \cdot h) (t)) = u (h (t s)) = (s \cdot (u \circ h)) (t)$ , so postcomposition is $S$ -linear. Identities and composites are preserved by associativity of function composition.

Depends on

Used by

Coextension of scalars is right adjoint to restriction of scalars Theorem

Dependency tree · two levels

13 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

Emily Riehl, Category Theory in Context, 2nd ed., Example 4.1.10 (standard reference, not scraped)

Coextension of scalars Hom⁡R(S,M)HomR​(S,M) carries its canonical left SS-module structure