Ring of Polynomial Functions is Commutative Ring with Unity

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Theorem

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

Let $R \sqbrk {\set {X_j: j \in J} }$ be the ring of polynomial forms over $R$ in the indeterminates $\set {X_j: j \in J}$.

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

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

Let $\struct {A, +, \circ}$ be the ring of polynomial functions on $R$.


Then $\struct {A, +, \circ}$ 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:

$\ds f = \sum_{k \mathop \in Z} a_k \mathbf X^k, \ g = \sum_{k \mathop \in Z} b_k \mathbf X^k \in R \sqbrk {\set {X_j: j \in J} }$.

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

$\ds A \owns \hat f: \forall x \in R^J: \map {\hat f} x = \sum_{k \mathop \in Z} a_k x^k$
$\ds A \owns \hat g: \forall x \in R^J: \map {\hat g} x = \sum_{k \mathop \in Z} b_k x^k$


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

\(\ds \map {\hat f} x + \map {\hat g} x\) \(=\) \(\ds \sum_{k \mathop \in Z} a_k x^k + \sum_{k \mathop \in Z} b_k x^k\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop \in Z} \paren {a_k + b_k} x^k\)
\(\ds \) \(=\) \(\ds \map {\widehat {f + g} } x\) Definition of Addition of Polynomial Forms

Thus polynomial functions are closed under ring addition.


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

\(\ds \map {\hat f} x \circ \map {\hat g} x\) \(=\) \(\ds \paren {\sum_{k \mathop \in Z} a_k x^k} \circ \paren {\sum_{k \mathop \in Z} a_k x^k}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop \in Z} \paren {\sum_{p + q \mathop = k} a_p b_q} \mathbf X^k\)
\(\ds \) \(=\) \(\ds \map {\widehat {f \circ g} } x\) Definition of Multiplication of Polynomial Forms

Thus polynomial functions are closed under ring product.

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

$\blacksquare$

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