CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17

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A polynomial ring on a finite ordered family agrees canonically with the iterated polynomial-ring construction

Statement

For a commutative ring $R$ and a finite ordered family $(x_{0}, \dots, x_{n - 1})$ , the arbitrary-family construction $R [x_{i} : i < n]$ is canonically isomorphic as an $R$ -algebra to the recursively iterated polynomial ring $R [x_{0}] \dots [x_{n - 1}]$ . The isomorphism fixes $R$ and sends each formal indeterminate to the corresponding iterated indeterminate. For $n = 0$ , both sides are $R$ .

Facts & Assumptions

Given: A commutative ring $R$ and a natural number $n$ indexing an ordered family of indeterminates.

[L1]

A homomorphism out of the family polynomial ring is uniquely determined by its restriction to $R$ and the images of all indeterminates (Universal property of a polynomial ring on an arbitrary family of indeterminates).

[L2]

A homomorphism from $A [x]$ is uniquely determined by a homomorphism from $A$ and the image of $x$ (Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism).

[L3]

The finite multivariate polynomial ring is defined recursively, with the empty iteration equal to $R$ and the successor obtained by adjoining one indeterminate (Polynomial rings in finitely many commuting indeterminates by iteration).

Proof

technique · induction

1.1baseL1L3

For $n = 0$ , [L3] makes the iterated construction $R$ , while the empty-family clause of [L1] makes the family construction canonically $R$ .

1.2ih

Assume the canonical isomorphism has been constructed for a family of length $n$ and fixes $R$ and its indeterminates.

2.1step 1.2L1L2L3

For length $n + 1$ , [L2] extends the induction isomorphism after choosing the image of the new variable, while [L1] gives a homomorphism in the reverse direction fixing $R$ and all $n + 1$ variables.

3.1step 2.1L1L2discharge-induction∎

Both composites fix $R$ and every indeterminate, so uniqueness in [L1] and [L2] makes them identity homomorphisms; the construction therefore holds for every $n$ , including the empty family.

Depends on

Used by

Dependency tree · two levels

10 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

U. Thiel, Commutative Algebra, Section 1.4 (standard reference, not scraped)