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 $$k$$;
 * 2) $$m, n$$ 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.