Equivalence of Definitions of Polynomial Function on Subset of Ring

Theorem
Let $R$ be a commutative ring with unity.

Let $S \subset R$ be a subset.

Outline of proof
This follows by interpreting both definitions as $f$ being a polynomial in the element $\iota$.

1 implies 2
Let $\map P X \in R \sqbrk X$ be the polynomial:
 * $P = \displaystyle \sum_{k \mathop = 0}^n a_k \cdot X^k$

where $\sum$ denotes indexed summation.

We show that $\map P \iota = f$.

2 implies 1
Let $P \in R \sqbrk X$ be a polynomial and $f = \map P \iota \in R^S$.

By Polynomial is Linear Combination of Mononomials, there exist:
 * $n \in \N$
 * $a_0, \ldots, a_n \in R$

such that:
 * $P = \ds \sum_{k \mathop = 0}^n a_k \cdot X^k$

where $\sum$ denotes indexed summation.

Let $\operatorname{ev}_{\iota}$ denote the evaluation homomorphism at $\iota$.

Then $\map {\operatorname{ev}_{\iota} } P = f$.

We have: