TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Coextension of scalars is right adjoint to restriction of scalars

Statement

For a unital ring homomorphism $φ : R \to S$ , restriction of scalars

$Res : S - M o d \to R - M o d$

is left adjoint to coextension of scalars

$Coind (M) : = {Hom}_{R} (S, M) .$

Naturally in an $S$ -module $N$ and an $R$ -module $M$ ,

${Hom}_{S} (N, {Hom}_{R} (S, M)) ≅ {Hom}_{R} (Res N, M) .$

Facts & Assumptions

Given: The ring homomorphism $φ : R \to S$ , a left $S$ -module $N$ , and a left $R$ -module $M$ .

[L1]

On ${Hom}_{R} (S, M)$ the canonical left $S$ -action is $(s \cdot h) (t) = h (t s)$ (Coextension of scalars ${Hom}_{R} (S, M)$ carries its canonical left $S$ -module structure).

[F1]

Left modules and module homomorphisms form locally small categories (Left modules over a fixed ring and module homomorphisms form the large locally small category $R - M o d$ ).

[L2]

A natural family of hom-set bijections determines an adjunction (Under local smallness, transposition gives the natural hom-set bijection, and conversely).

Proof

technique · direct

1.1

Given an $S$ -linear map $α : N \to {Hom}_{R} (S, M)$ , define $E (α) : Res N \to M$ by $E (α) (n) = α (n) (1_{S})$ . It is $R$ -linear because $α (φ (r) n) = φ (r) \cdot α (n)$ and [L1] gives $(φ (r) \cdot α (n)) (1) = α (n) (φ (r)) = r α (n) (1)$ .

L1algebra

1.2

Given an $R$ -linear map $u : Res N \to M$ , define $C (u) (n) (s) : = u (s n)$ . For $r \in R$ , $C (u) (n) (φ (r) s) = u (φ (r) s n) = r u (s n)$ , so $C (u) (n)$ is $R$ -linear.

constructalgebra

2.1

For $a, s \in S$ , $C (u) (a n) (s) = u (s a n) = C (u) (n) (s a) = (a \cdot C (u) (n)) (s)$ by [L1], so $C (u)$ is $S$ -linear.

step 1.2L1

2.2

Evaluation gives $E (C (u)) (n) = C (u) (n) (1) = u (n)$ . Conversely, $S$ -linearity of $α$ gives $C (E (α)) (n) (s) = E (α) (s n) = α (s n) (1) = (s \cdot α (n)) (1) = α (n) (s)$ .

step 1.1step 1.2L1

3.1

Thus $E$ and $C$ are inverse bijections. Precomposition in $N$ and postcomposition in $M$ commute with the two displayed formulas, so the bijections are natural.

step 2.1step 2.2F1

4.1

By [L2], these natural bijections define $Res ⊣ Coind$ . No tensor-product or extension-of-scalars construction is used.

step 3.1L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 30 results over 10 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

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