Ring of Polynomial Forms is Integral Domain

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

Finally we note that if neither $\displaystyle f \left({x}\right) = \sum_{k=0}^n a_k x^k$ nor $\displaystyle g \left({x}\right) = \sum_{k=0}^m b_k x^k$ are the null polynomial, then their leading coefficients $a_n$ and $b_m$ are non-zero.

Hence so is their product, which is then non-null.

It follows that $D \left[{X}\right]$ has no proper zero divisors.

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