Ring of Polynomial Forms is Integral Domain

Let $$\left({R, +, \circ}\right)$$ be a commutative ring whose zero is $$0_R$$ and whose unity is $$1_R$$.

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

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

Let $$D \left[{X}\right]$$ be the ring of polynomial forms in $$X$$ over $$D$$.

Then the $$D \left[{X}\right]$$ is an integral domain.

Proof
Follows directly from Polynomials Closed under Ring Product and the fact that $$D \left[{X}\right]$$ is a commutative ring with unity.