Ring of Polynomial Forms is Integral Domain

Theorem
Let $\struct {R, +, \circ}$ be a commutative ring with unity.

Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.

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

Let $D \sqbrk X$ be the ring of polynomials in $X$ over $D$.

Then $D \sqbrk X$ is an integral domain.

Proof
By Ring of Polynomial Forms is Commutative Ring with Unity we know that $D \sqbrk X$ is a commutative ring with unity.

Let neither $\displaystyle \map f X = \sum_{k \mathop = 0}^n a_k x^k$ nor $\displaystyle \map g X = \sum_{k \mathop = 0}^m b_k X^k$ be the null polynomial.

Then their leading coefficients $a_n$ and $b_m$ are non-zero.

Therefore, as $D$ is an integral domain and $a_n, b_m \in D$, so is their product $a_n b_m$.

By the definition of polynomial multiplication, it follows that $f g$ is not the null polynomial.

It follows that $D \sqbrk X$ has no proper zero divisors.

Hence $D \sqbrk X$ is an integral domain.

Also see

 * Ring of Polynomial Forms over Integral Domain is Integral Domain