Existence of Ring of Polynomial Forms in Transcendental over Integral Domain

Theorem
Let $\struct {R, +, \circ}$ be a commutative ring with unity.

Let $\struct {D, +, \circ}$ be an integral subdomain of $R$ whose zero is $0_D$.

Let $X \in R$ be transcendental over $D$

Then the ring of polynomials $D \sqbrk X$ in $X$ over $D$ exists.

Proof
Suppose that $D \sqbrk X$ exists.

Let $\ds \map P X = \sum_{k \mathop = 0}^n a_k X^k$, where $a_n \ne 0_D$, be an arbitrary element of $D \sqbrk X$.

Then $\map P X$ corresponds to, and is completely described by, the ordered tuple of coefficients $\tuple {a_0, a_1, \dotsc, a_n, 0_D, 0_D, 0_D, \dotsc}$.

Consider the set $S$ of infinite sequences of elements of $D$ which are eventually $0_D$.

That is, whose elements are of the form $\tuple {b_0, b_1, \dotsc, b_n, 0_D, 0_D, 0_D, \dotsc}$ where $b_0, \ldots, b_n \in D$.

Consider the polynomial ring over $S$ by defining the operations:

From Polynomial Ring of Sequences is Ring we have that $\struct {S, +, \circ}$ is a ring.