Multiplication of Polynomials is Associative

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Theorem

Multiplication of polynomials is associative.


Proof

Let $\struct {R, +, \circ}$ be a commutative ring with unity with zero $0_R$.

To improve readability of the expressions used, we will write the ring product $\circ$ in multiplicative notation.


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:

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

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


Then it follows from Polynomials Closed under Ring Product that:

$\ds f \circ \paren {g \circ h} = \sum_{k \mathop \in Z} m_k \mathbf X^k$

for some $m_k \in R$, and:

$\ds \paren {f \circ g} \circ h = \sum_{k \mathop \in Z} n_k \mathbf X^k$

for some $n_k \in R$.

To establish associativity of $\circ$ we compute $m_k$ and $n_k$, and check that they are equal.


We have:

\(\ds m_k\) \(=\) \(\ds \sum_{r + s \mathop = k} a_r \sum_{p + q \mathop = s} b_p c_q\) Definition of Multiplication of Polynomial Forms
\(\ds \) \(=\) \(\ds \sum_{r + s \mathop = k} \, \sum_{p + q \mathop = s} a_r b_p c_q\) Distributive property of finite sums
\(\ds \) \(=\) \(\ds \sum_{p + q + r \mathop = k} a_r b_p c_q\) Properties of finite sums



Similarly we compute:

\(\ds n_k\) \(=\) \(\ds \sum_{r + s \mathop = k} \left({ \sum_{p + q \mathop = r} a_p b_q }\right) c_s\) Definition of Multiplication of Polynomial Forms
\(\ds \) \(=\) \(\ds \sum_{r + s \mathop = k} \, \sum_{p + q \mathop = r} a_p b_q c_s\) Distributive property of finite sums
\(\ds \) \(=\) \(\ds \sum_{p + q + s \mathop = k} a_p b_q c_s\) Properties of finite sums



Since $p$, $q$, $r$ and $s$ are all dummy variables, it follows that $m_k = n_k$ for all $k \in Z$.

Therefore, $f \circ \paren {g \circ h} = \paren {f \circ g} \circ h$ for all polynomials $f$, $g$ and $h$.


Hence multiplication of polynomials is associative.

$\blacksquare$