Polynomial Addition is Commutative

Theorem

Addition of polynomials is commutative.

Proof

Let $(R, +, \circ)$ be a commutative ring with unity with zero $0_R$.

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle f + g\)	\(=\)	\(\displaystyle \sum_{k \in Z} \left({ a_k + b_k }\right) \mathbf X^k\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	applying the definition of polynomial addition
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{k \in Z} \left({ b_k + a_k }\right) \mathbf X^k\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	because addition in $R$ is commutative
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle g + f\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	applying the definition of polynomial addition

Therefore, $f + g = g + f$ for all polynomials $f$ and $g$.

Therefore, polynomial addition is commutative.

$\blacksquare$