Multiplication of Polynomials is Associative

Theorem
Multiplication of polynomials is associative.

Proof
Let $(R, +, \circ)$ 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 $\left\{{X_j: j \in J}\right\}$ be a set of indeterminates.

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

Let:


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


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


 * $\displaystyle h = \sum_{k \mathop \in Z} c_k \mathbf X^k$

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

Then it follows from Polynomials Closed under Ring Product that:


 * $\displaystyle f \circ \left({ g \circ h }\right) = \sum_{k \mathop \in Z} m_k \mathbf X^k$

for some $m_k \in R$, and:


 * $\displaystyle \left({ f \circ g }\right) \circ h = \sum_{k \mathop \in Z} n_k \mathbf X^k$

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:

Similarly we compute:

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

Therefore, $f \circ \left({ g \circ h }\right) = \left({ f \circ g }\right) \circ h$ for all polynomials $f$, $g$ and $h$.

Hence multiplication of polynomials is associative.