Ring of Polynomial Forms over Integral Domain is Integral Domain

Theorem
Let $\left({D, +, \circ}\right)$ be an integral domain whose zero is $0_D$.

Let $\left({D \left[{X}\right], \oplus, \odot}\right)$ be the ring of polynomial forms over $D$ in the indeterminate $X$.

Then $\left({D \left[{X}\right], \oplus, \odot}\right)$ is an integral domain.

Proof
By definition an integral domain is a commutative ring with unity.

From Ring of Polynomial Forms is Commutative Ring with Unity it follows that $\left({D \left[{X}\right], +, \circ}\right)$ is a commutative ring with unity.

Suppose $f, g \in D \left[{X}\right]$ such that neither $f$ nor $g$ are the null polynomial.

Let $\deg \left({f}\right) = n$ and $\deg \left({g}\right) = m$.

From Degree of Product of Polynomials over Integral Domain the degree of $f \odot g$ is $n + m$.

Thus by definition $f \odot g$ is not the null polynomial of $D \left[{X}\right]$.

Thus neither $f$ nor $g$ is a proper zero divisor of $D \left[{X}\right]$.

This holds for any two arbitrary non-null polynomials elements of $D \left[{X}\right]$.

Hence $\left({D \left[{X}\right], \oplus, \odot}\right)$ is a commutative ring with unity with no proper zero divisors.

That is, $\left({D \left[{X}\right], \oplus, \odot}\right)$ is an integral domain.