# Polynomials Closed under Ring Product

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## Theorem

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

Let $\ds f = \sum_{k \mathop \in Z} a_k \mathbf X^k$, $\ds g = \sum_{k \in Z} b_k \mathbf X^k$ be polynomials in the indeterminates $\set {X_j: j \in J}$ over $R$, where $Z$ is the set of all multiindices indexed by $\set {X_j: j \in J}$.

Define the product

- $\ds f \otimes g = \sum_{k \in Z} c_k \mathbf X^k$

where

- $\ds c_k = \sum_{\substack {p + q = k \\ p, q \in Z} } a_p b_q$

Then $f \otimes g$ is a polynomial.

## Proof

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It is immediate that $f \otimes g$ is a map from the free commutative monoid to $R$, so we need only prove that $f \otimes g$ is nonzero on finitely many $\mathbf X^k$, $k \in Z$.

Suppose that for some $k \in Z$:

- $\ds \sum_{\substack {p + q = k \\ p, q \mathop \in Z}} a_p b_q \ne 0$

Therefore if $c_k \ne 0$ there exist $p, q \in Z$ such that $p + q = k$ and $a_p$, $b_q$ are both nonzero.

Since $f$ and $g$ are polynomials, the set $\set {p + q: a_p \ne 0, b_q \ne 0}$ is finite.

Hence the result.

$\blacksquare$

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