Polynomials Closed under Ring Product

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Theorem

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

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

Define the product

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

where

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


Then $f \otimes g$ is a polynomial.


Proof

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 $k \in Z$:

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

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 $\left\{{p + q: a_p \ne 0, \ b_q \ne 0}\right\}$ is finite, and we are done.

$\blacksquare$