Ring of Polynomial Forms over Integral Domain is Integral Domain

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

Let $\struct {D \sqbrk X, \oplus, \odot}$ be the ring of polynomial forms over $D$ in the indeterminate $X$.

Then $\struct {D \sqbrk X, \oplus, \odot}$ 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 $\struct {D \sqbrk X, +, \circ}$ is a commutative ring with unity.

Suppose $f, g \in D \sqbrk X$ such that neither $f$ nor $g$ are the null polynomial.

Let $\map \deg f = n$ and $\map \deg g = m$.

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 $D \sqbrk X$.

Thus neither $f$ nor $g$ is a proper zero divisor of $D \sqbrk X$.

This holds for any two arbitrary non-null polynomials elements of $D \sqbrk X$.

Hence $\struct {D \sqbrk X, \oplus, \odot}$ is a commutative ring with unity with no proper zero divisors.

That is, $\struct {D \sqbrk X, \oplus, \odot}$ is an integral domain.