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.

$\blacksquare$

Also see

• Ring of Polynomial Forms over Integral Domain is Integral Domain

Sources

• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $1$: Integral Domains: $\S 5$. Further Examples of Integral Domains: Example $5$
• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 64.3$ Polynomial rings over an integral domain