Polynomials Closed under Ring Product

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 closed under the operation $\circ$.

Proof
Let $f, g$ be polynomials in $x$ over $D$ such that $f \ne 0_R, g \ne 0_R$.

We can express them as $\displaystyle f = \sum_{k=0}^n a_k \circ x^k, g = \sum_{k=0}^m b_k \circ x^k$ where:
 * 1) $a_k, b_k \in D$ for all $k$;
 * $m, n$ are some non-negative integers.

Multiplying termwise exploiting the distributivity of $\circ$ over $+$, we find $f \circ g$ is also a polynomial in $x$ over $D$ whose leading coefficient is $c_{m + n} = a_n \circ b_m$.

This proof can be made rigorous by induction.