Ring of Polynomial Functions is Commutative Ring with Unity

Theorem
Let $\left({R, +, \circ}\right)$ be a commutative ring with unity.

Let $R \left[{\left\{{X_j: j \in J}\right\}}\right]$ be the ring of polynomial forms over $R$ in the indeterminates $\left\{{X_j: j \in J}\right\}$.

Let $R^J$ be the free module on $J$.

Let $A$ be the set of all polynomial functions $R^J \to R$.

Let $\left({A, +, \circ}\right)$ be the ring of polynomial functions on $R$.

Then $\left({A, +, \circ}\right)$ is a commutative ring with unity.

Proof
First we check that the operations of ring product and ring addition are closed in $A$.

Let $Z$ be the set of all multiindices indexed by $J$.

Let:
 * $\displaystyle f = \sum_{k \mathop \in Z} a_k \mathbf X^k,\ g = \sum_{k \mathop \in Z} b_k \mathbf X^k \in R \left[{\left\{{X_j: j \in J}\right\}}\right]$.

Under the evaluation homomorphism, $f$ and $g$ map to:


 * $\displaystyle A \owns \hat f: \forall x \in R^J: \hat f \left({x}\right) = \sum_{k \mathop \in Z} a_k x^k$


 * $\displaystyle A \owns \hat g: \forall x \in R^J: \hat g \left({x}\right) = \sum_{k \mathop \in Z} b_k x^k$

Then the induced sum of $\hat f$ and $\hat g$ is:

Thus polynomial functions are closed under ring addition.

The induced product of $\hat f$ and $\hat g$ is:

Thus polynomial functions are closed under ring product.

Finally, we invoke Structure Induced by Ring Operations is Ring, which shows that $\left({A, +, \circ}\right)$ is a commutative ring with unity.