Ring of Polynomial Forms is not necessarily Isomorphic to Ring of Polynomial Functions

Theorem

Let $D$ be an integral domain.

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

Let $\map P D$ be the ring of polynomial functions over $D$.

Then it is not necessarily the case that $D \sqbrk X$ is isomorphic with $\map P D$.

Proof

Proof by Counterexample:

Consider the integral domain $\struct {\Z_2, +, \times}$.

From Ring of Integers Modulo Prime is Integral Domain, it is seen that $\struct {\Z_2, +, \times}$ is indeed an integral domain.

Consider the ring of polynomial forms $\Z_2 \sqbrk X$.

This is an infinite ring, as it can be seen that $S \subseteq \Z_2 \sqbrk X$ where:

$S = \set {1, X, X^2, X^3, \dotsc}$

But the ring of polynomial functions $\map P D$ is finite, as:

$\map P D \subseteq \Z_2^{Z_2}$

where $\Z_2^{Z_2}$ is the set of all mappings from $\Z_2$ to $\Z_2$, and has $4$ elements.

$\blacksquare$

Also see

• Epimorphism from Polynomial Forms to Polynomial Functions, where it is shown that there exists an epimorphism from $D \sqbrk X$ to $\map P D$.

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 64$ Polynomial rings over an integral domain