Multiplication of Polynomials is Commutative

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Theorem

Multiplication of polynomials is commutative.


Proof

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

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

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

Let:

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

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

Then:

\(\displaystyle f \circ g\) \(=\) \(\displaystyle \sum_{k \mathop \in Z} \paren {\sum_{p \mathop + q \mathop = k} a_p b_q} \mathbf X^k\) Definition of Polynomial Multiplication
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop \in Z} \paren {\sum_{p \mathop + q \mathop = k} a_q b_p} \mathbf X^k\) Addition of multiindices is commutative, and $p$ and $q$ are dummy variables
\(\displaystyle \) \(=\) \(\displaystyle g \circ f\) Definition of Polynomial Multiplication

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

Therefore, polynomial multiplication is commutative.

$\blacksquare$