ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 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.

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$S \otimes_{R} R [x] ≅ S [x]$ as $S$ -algebras

Example

Let $f : R \to S$ be a homomorphism of commutative rings. There is an isomorphism of $S$ -algebras

$S \otimes_{R} R [x] ≅ S [x]$

given on elementary tensors by

$s \otimes \sum_{i} r_{i} x^{i} ⟼ \sum_{i} s f (r_{i}) x^{i} .$

Facts & Assumptions

Given: A homomorphism $f : R \to S$ of commutative rings.

[L1]

Restriction along $f$ makes $S$ an $R$ -module, and $S \otimes_{R} -$ is extension of scalars (Restriction of scalars and extension of scalars $S \otimes_{R} M$ along a ring homomorphism $R \to S$ ).

[L2]

Polynomial rings consist of finitely supported coefficient families, with multiplication given by finite convolution (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution).

[L3]

A balanced map from a product of modules induces a unique homomorphism from their tensor product (Universal property of the tensor product for balanced maps into abelian groups).

[L4]

The tensor product of two $R$ -algebras has multiplication $(a \otimes b) (a^{'} \otimes b^{'}) = a a^{'} \otimes b b^{'}$ and its canonical $R$ -algebra structure (The tensor product of $R$ -algebras has multiplication $(a \otimes b) (a^{'} \otimes b^{'}) = a a^{'} \otimes b b^{'}$ ).

Verification

technique · direct

1.1

The displayed coefficient formula is additive in both variables and satisfies $F (s f (r) \otimes p) = F (s \otimes r p)$ , so it is $R$ -balanced. It therefore induces an additive map $F : S \otimes_{R} R [x] \to S [x]$ by [L3].

givenL1L2L3

1.2

Define $G : S [x] \to S \otimes_{R} R [x]$ by $G (\sum_{i} s_{i} x^{i}) = \sum_{i} s_{i} \otimes x^{i}$ . The sum is finite by [L2]. Coefficientwise addition and convolution multiplication show that $G$ is an $S$ -algebra homomorphism, using $(s \otimes x^{i}) (t \otimes x^{j}) = s t \otimes x^{i + j}$ from [L4].

L2L4algebra

2.1

The map $F$ is $S$ -linear, sends $1 \otimes 1$ to $1$ , and, using [L2] and [L4], satisfies $F ((s \otimes p) (t \otimes q)) = F (s t \otimes p q) = F (s \otimes p) F (t \otimes q)$ . Hence it is an $S$ -algebra homomorphism.

step 1.1L2L4

2.2

For every polynomial $\sum_{i} s_{i} x^{i}$ , one has $F (G (\sum_{i} s_{i} x^{i})) = \sum_{i} s_{i} x^{i}$ . For an elementary tensor, balance gives $G (F (s \otimes \sum_{i} r_{i} x^{i})) = \sum_{i} s f (r_{i}) \otimes x^{i} = \sum_{i} s \otimes r_{i} x^{i} = s \otimes \sum_{i} r_{i} x^{i}$ .

step 1.1step 1.2L1

3.1

The two maps $G F$ and the identity induce the same balanced pairing by step 2.2, so uniqueness in [L3] makes them equal; step 2.2 already gives $F G = 1$ coefficientwise. Thus $F$ and $G$ are inverse $S$ -algebra homomorphisms.

step 2.1step 1.2step 2.2L3∎

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: 41 results over 16 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

Wenqi Li, Commutative Algebra, Lecture 9 (standard reference, not scraped)

S⊗RR[x]≅S[x]S⊗R​R[x]≅S[x] as SS-algebras