Polynomials Contain Multiplicative Identity

Theorem
The set of polynomials has a multiplicative identity.

Proof
Let $\struct {R, +, \circ}$ be a commutative ring with unity with multiplicative identity $1_R$ and additive identity $0_R$.

Let $\set {X_j: j \in J}$ be a set of indeterminates.

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

Let:


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

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

Let:


 * $\ds N = 1_R \mathbf X^0 = \sum_{k \mathop \in Z} b_k \mathbf X^k$

where $b_k = 0_R$ if $k \ne 0$ and $b_0 = 1_R$.

Then:

Therefore:
 * $f \circ N = f$

for all polynomials $f$.

Therefore $N$ is a multiplicative identity for the set of polynomials.