ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)

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.

Eilenberg-Watts recovers extension of scalars

Example

Let $f : R \to S$ be a unital homomorphism of commutative rings and let $_{S} S_{R}$ be the $(S, R)$ -bimodule with left action by multiplication and right action $s \cdot r = s f (r)$ of Restriction of scalars and extension of scalars $S \otimes_{R} M$ along a ring homomorphism $R \to S$ . Then the tensor functor $T_{S} = S \otimes_{R} -$ is exactly the extension of scalars along $f$ ; it is additive and cocontinuous; and its Eilenberg-Watts kernel is $T_{S} (R) = S \otimes_{R} R ≅ S$ with the right action of $F (A)$ is a $(B, A)$ -bimodule for every additive functor $F$ equal to the displayed action $s \cdot r = s f (r)$ . Thus Eilenberg-Watts theorem for arbitrary unital rings recovers extension of scalars with kernel $_{S} S_{R}$ . The same bimodule and computation apply to an arbitrary unital ring homomorphism, but the published definition of extension of scalars is stated for commutative rings. No choice is used.

Facts & Assumptions

Given: A unital homomorphism $f : R \to S$ of commutative rings, the $(S, R)$ -bimodule $_{S} S_{R}$ with $s \cdot r = s f (r)$ , and the tensor functor $T_{S} = S \otimes_{R} - : R - M o d \to S - M o d$ .

[F1]

Extension of scalars along $f$ is the functor $S \otimes_{R} -$ , with $S$ the $(S, R)$ -bimodule whose left action is multiplication and whose right action is $s \cdot r = s f (r)$ ; the outer action $s^{'} (s \otimes m) = (s^{'} s) \otimes m$ makes $S \otimes_{R} M$ an $S$ -module, and extension sends $u$ to $1_{S} \otimes u$ (Restriction of scalars and extension of scalars $S \otimes_{R} M$ along a ring homomorphism $R \to S$ ).

[F2]

Extension of scalars is left adjoint to restriction of scalars (Extension of scalars is left adjoint to restriction of scalars), and a left adjoint preserves every colimit that exists (Left adjoints preserve every colimit that exists).

[F3]

Every tensor functor $T_{N} = N \otimes_{R} -$ , for $N$ an $(S, R)$ -bimodule, is additive, right exact, coproduct-preserving and therefore cocontinuous (Eilenberg-Watts theorem for arbitrary unital rings).

[F4]

The tensor-unit map $ρ_{S} : S \otimes_{R} R \to S$ , $ρ_{S} (s \otimes x) = s x$ , is an isomorphism with inverse $s \mapsto s \otimes 1$ (The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ ).

[F5]

For an additive functor $F$ and the regular module $R$ , the formula $m a = F (r_{a}) (m)$ with $r_{a} (x) = x a$ turns $F (R)$ into an $(S, R)$ -bimodule ( $F (A)$ is a $(B, A)$ -bimodule for every additive functor $F$ ).

Verification

Given: The data of the Example.

1.1F1F2F3

The functor $T_{S} = S \otimes_{R} -$ is by definition the extension of scalars along $f$ by [F1]. It is additive and cocontinuous: it is additive by [F3], and it is cocontinuous because extension of scalars is left adjoint to restriction of scalars by [F2] and a left adjoint preserves every colimit that exists by [F2] (equivalently, additive and coproduct-preserving with right exactness gives cocontinuity directly by [F3]).

2.1F1F4F5step 1.1

Kernel: apply the evaluation lemma [F5] to $F = T_{S}$ and the regular module $R$ . For $a \in R$ one has $T_{S} (r_{a}) = 1_{S} \otimes r_{a}$ , so the reconstructed right action on $T_{S} (R) = S \otimes_{R} R$ is $(s \otimes x) a = s \otimes (x a)$ ; under the unit isomorphism $ρ_{S}$ of [F4] this corresponds to $ρ_{S} (s \otimes (x a)) = s f (x a) = s f (x) f (a)$ , since $f$ is a unital ring homomorphism, and the last term is $ρ_{S} (s \otimes x) \cdot a$ for the displayed right action $s^{'} \cdot a = s^{'} f (a)$ of [F1]. Hence the evaluation right action on the kernel $S \otimes_{R} R ≅ S$ is exactly $s \cdot a = s f (a)$ , the published action of $_{S} S_{R}$ .

3.1F3step 1.1step 2.1

By step 1.1 the extension-of-scalars functor is the tensor functor $T_{S}$ with kernel the $(S, R)$ -bimodule $_{S} S_{R}$ , as computed in step 2.1, and by [F3] it is additive and right exact as the theorem requires; hence the Eilenberg-Watts classification reproduces extension of scalars together with its kernel bimodule.

4.1F1step 3.1∎

The same bimodule $_{S} S_{R}$ and the same unit-isomorphism computation make sense for an arbitrary unital ring homomorphism, but the published definition of extension of scalars is stated only for commutative rings, so only the commutative case is claimed here; that caveat is recorded and not used. No element of an auxiliary set is chosen, so no choice is involved.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

23 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

M. Kamensky, Non-Commutative Algebra (BGU course notes, Spring 2017), §5.1, Theorem 5.1.43, Proposition 5.1.40, Lemma 5.1.46, Corollaries 5.1.48-5.1.49 (standard reference, not scraped)
A. Nyman and S. P. Smith, A Generalization of Watts's Theorem: Right Exact Functors on Module Categories, arXiv:0806.0832, Theorem 1.1-1.2, Propositions 3.2-3.3, Lemma 3.4 (standard reference, not scraped)