Multiplication of Polynomials is Commutative

Theorem

Multiplication of polynomials is commutative.

Proof

Let $(R, +, \circ)$ be a commutative ring with unity.

Let $\left\{{X_j: j \in J}\right\}$ be a set of indeterminates.

Let $Z$ be the set of all multiindices indexed by $\left\{{X_j: j \in J}\right\}$.

Let:

$\displaystyle f = \sum_{k \in Z} a_k \mathbf X^k$

$\displaystyle g = \sum_{k \in Z} b_k \mathbf X^k$

be arbitrary polynomials in the indeterminates $\left\{{X_j: j \in J}\right\}$ over $R$.

Then

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle f \circ g\)	\(=\)	\(\displaystyle \sum_{k \in Z} \left({ \sum_{p + q = k} a_p b_q }\right) \mathbf X^k\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	applying the definition of polynomial multiplication
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{k \in Z} \left({ \sum_{p + q = k} a_q b_p }\right) \mathbf X^k\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	because addition of multiindices is commutative, and $p$ and $q$ are dummy variables
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle g \circ f\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	applying the definition of polynomial multiplication

Therefore, $f \circ g = g \circ f$ for all polynomials $f$ and $g$.

Therefore, polynomial multiplication is commutative.

$\blacksquare$