Set of Polynomials over Integral Domain is Subring

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

Let $\left({D, +, \circ}\right)$ be an integral domain such that $D$ is a subring of $R$.

Then $\forall x \in R$, the set of polynomials in $x$ over $D$ is a subring of $R$.

This subring is denoted $D \left[{x}\right]$.

$D \left[{x}\right]$ contains $D$ as a subring and $x$ as an element.

$D \left[{x}\right]$ is the smallest subring of $R$ with these properties.

Proof
By application of the Subring Test:

As $D$ is an integral domain, it has a unity $1_D$ and so $x = 1_D x$.

Hence $x \in D \left[{x}\right]$ and so $D \left[{x}\right] \ne \varnothing$.

Let $p, q \in D \left[{x}\right]$.

Then let:
 * $\displaystyle p = \sum_{k \mathop = 0}^m a_k \circ x^k, q = \sum_{k \mathop = 0}^n b_k \circ x^k$

Thus:
 * $\displaystyle -q = -\sum_{k \mathop = 0}^n b_k \circ x^k = \sum_{k \mathop = 0}^n \left({-b_k}\right) \circ x^k$ and so $q \in D \left[{x}\right]$

Thus as Polynomials Closed under Addition‎, it follows that:
 * $p + \left({-q}\right) \in D \left[{x}\right]$

Finally, from Polynomials Closed under Ring Product, we have that $p \circ q \in D \left[{x}\right]$.

All the criteria of the Subring Test are satisfied.

Hence the result.