Null Polynomial is Additive Identity

Theorem

The set of polynomials has an additive identity.

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:

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

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

Let :

$\displaystyle N = \sum_{k\in Z} 0_R \mathbf X^k$

be the null polynomial.

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle f + N\)	\(=\)	\(\displaystyle \sum_{k\in Z} \left({ a_k + 0_R }\right) \mathbf X^k\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	applying the definition of polynomial addition
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{k\in Z} a_k \mathbf X^k\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle f\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	applying the definition of polynomial addition

Therefore, $N + f = f$ for all polynomials $f$.

Therefore, $N$ is an additive identity for the set of polynomials.

$\blacksquare$