Equivalence of Definitions of Polynomial Ring in One Variable

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Theorem

Let $R$ be a commutative ring with unity.


The following definitions of polynomial ring are equivalent in the following sense:

For every two constructions, there exists an $R$-isomorphism which sends indeterminates to indeterminates.


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Definition 1: As a Ring of Sequences

Let $R^{\left({\N}\right)}$ be the ring of sequences of finite support over $R$.

Let $\iota : R \to R^{\left({\N}\right)}$ be the mapping defined as:

$\iota \left({r}\right) = \left \langle {r, 0, 0, \ldots}\right \rangle$.

Let $X$ be the sequence $\left \langle {0, 1, 0, \ldots}\right \rangle$.


This article, or a section of it, needs explaining.
In particular: Please clarify the role of $X$. This does not look like a ring to me. What is the multiplication? --Wandynsky (talk) 17:17, 30 July 2021 (UTC)
What is not clear? $R^{\left({\N}\right)}$ is a ring. --Usagiop (talk) 19:16, 28 September 2022 (UTC)
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The polynomial ring over $R$ is the ordered triple $\left({R^{\left({\N}\right)}, \iota, X}\right)$.


Definition 2: As a Monoid Ring on the Natural Numbers

Let $\N$ denote the additive monoid of natural numbers.

Let $R \sqbrk \N$ be the monoid ring of $\N$ over $R$.


The polynomial ring over $R$ is the ordered triple $\struct {R \sqbrk \N, \iota, X}$ where:

$X \in R \sqbrk \N$ is the standard basis element associated to $1 \in \N$
$\iota : R \to R \sqbrk \N$ is the canonical mapping.


Outline of Proof

We show that they all satisfy the same universal property.


Proof

Use Polynomial Ring of Sequences Satisfies Universal Property


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Also see

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