Existence of Ring of Polynomial Forms in Transcendental over Integral Domain

Theorem
Let $\left({R, +, \circ}\right)$ be a commutative ring with unity.

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

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

Then the ring of polynomials $D \left[{x}\right]$ in $X$ over $D$ exists.

Proof
Suppose that $D \left[{X}\right]$ exists.

Let $\displaystyle P \left({X}\right) = \sum_{k \mathop = 0}^n a_k X^k$, where $a_n \ne 0_D$, be a typical element of $D \left[{x}\right]$.

Then $\displaystyle P \left({X}\right)$ corresponds to, and is completely described by, the sequence of coefficients $\left({a_0, a_1, \ldots, a_n, 0_D, 0_D, 0_D, \ldots}\right)$.

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

That is, whose elements are of the form $\left({b_0, b_1, \ldots, b_n, 0_D, 0_D, 0_D, \ldots}\right)$ 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 $\left({S, +, \circ}\right)$ is a ring.