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

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

Therefore, polynomial multiplication is commutative.