LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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.

Presentations and localization under base extension

Statement

Let $A \to C$ be a unital ring map. For any set of variables $(t_{i})$ and any ideal $I \subseteq A [t_{i}]$ , $(A [t_{i}] / I) \otimes_{A} C ≅ C [t_{i}] / I C [t_{i}] .$ Here the extended ideal is generated by the coefficient images of all elements of $I$ . For a multiplicative subset $M \subseteq A$ , $(M^{- 1} A) \otimes_{A} C ≅ {\overline{M}}^{- 1} C .$ These are ring isomorphisms; no flatness, finite-generation or nonzero-ring hypothesis is required.

Facts & Assumptions

Given: The objects, hypotheses and conventions in the statement above.

[F1]

Let $A, B, C$ be commutative $R$ -algebras. For every pair of $R$ -algebra homomorphisms $f : A \to C$ and $g : B \to C$ , there is a unique $R$ -algebra homomorphism $h : A \otimes_{R} B ⟶ C$ such that $h (a \otimes 1) = f (a)$ and $h (1 \otimes b) = g (b)$ . It is given by $h (a \otimes b) = f (a) g (b) .$ Thus $A \otimes_{R} B$ , with its two canonical maps, is the coproduct of $A$ and $B$ among commutative $R$ -algebras. (Universal mapping property of the tensor product of commutative algebras)

[F2]

Let $R$ be a commutative ring, $I ⊴ R$ an ideal, and $M$ an $R$ -module. There is a natural $R$ -module isomorphism $M \otimes_{R} (R / I) ≅ M / I M, m \otimes (r + I) ⟼ r m + I M .$ Both sides also carry the induced $R / I$ -module structure, and the isomorphism is $R / I$ -linear. For $I = 0$ it is the tensor-unit isomorphism, while for $I = R$ both sides are zero. ( $M \otimes_{R} R / I ≅ M / I M$ naturally)

[F3]

Let $R$ be a commutative ring, let $S \subseteq R$ be multiplicative, and let $M$ be a left $R$ -module. The map $Φ : (S^{- 1} R) \otimes_{R} M ⟶ S^{- 1} M, Φ ((a / s) \otimes m) = a m / s,$ is an isomorphism of $S^{- 1} R$ -modules. Its inverse is $Ψ : S^{- 1} M ⟶ (S^{- 1} R) \otimes_{R} M, Ψ (m / s) = (1 / s) \otimes m .$ (Localisation of modules is extension of scalars)

Proof

1.1givenF1

A ring map from $C [t_{i}]$ to a $C$ -algebra $R$ is exactly a choice of elements of $R$ for the variables. Equivalently it is an $A$ -algebra map $A [t_{i}] \to R$ together with the fixed map $C \to R$ . F1 therefore identifies $A [t_{i}] \otimes_{A} C$ with $C [t_{i}]$ , fixing coefficients and variables; empty variable sets are included.

1.2F3algebra

The module isomorphism in F3, after swapping tensor factors, sends $(a / m) \otimes c$ to $a c / \overline{m}$ and has inverse $c / \overline{m} \mapsto (1 / m) \otimes c$ . These formulas preserve multiplication and 1. They also apply when $0 \in \overline{M}$ , in which case both rings are zero, and when $M = {1}$ , in which case both are $C$ .

2.1F1F2step 1.1

A map out of the quotient must kill each element of $I$ , which under the preceding identification means killing $I C [t_{i}]$ . This proves the quotient formula by the same universal property. The module quotient map in F2 is consistent with it: multiplication of pure tensors is sent to the product of their images, so it is a ring map. For $I = 0$ this is the polynomial formula and for $I = (1)$ it is the zero ring.

3.1step 1.1step 2.1algebra∎

For instance, base change of $k [x, y, t] / (y^{2} - x^{3} - t x)$ along $k [t] \to k [u, v]$ , $t \mapsto u v$ , gives $k [x, y, u, v] / (y^{2} - x^{3} - u v x)$ ; along $t \mapsto 3$ it gives $k [x, y] / (y^{2} - x^{3} - 3 x)$ . Finitely many generators and relations remain finite. The substitution follows from the displayed maps and holds in every characteristic.

Depends on

Dependency tree · 2 levels

Sources

Vakil 10.2.A, B, F