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:
- $\ds f = \sum_{k \mathop \in Z} a_k \mathbf X^k$
- $\ds g = \sum_{k \mathop \in Z} b_k \mathbf X^k$
be arbitrary polynomials in the indeterminates $\set {X_j: j \in J}$ over $R$.
Then:
\(\ds f \circ g\) | \(=\) | \(\ds \sum_{k \mathop \in Z} \paren {\sum_{p \mathop + q \mathop = k} a_p b_q} \mathbf X^k\) | Definition of Polynomial Multiplication | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop \in Z} \paren {\sum_{p \mathop + q \mathop = k} a_q b_p} \mathbf X^k\) | Addition of multiindices is commutative, and $p$ and $q$ are bound variables | |||||||||||
\(\ds \) | \(=\) | \(\ds g \circ f\) | Definition of Polynomial Multiplication |
Therefore, $f \circ g = g \circ f$ for all polynomials $f$ and $g$.
Therefore, polynomial multiplication is commutative.
$\blacksquare$