Polynomials Contain Multiplicative Identity

Theorem
The set of polynomials has a multiplicative identity.

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

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:


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

where $b_k = 0_R$ if $k \neq 0$ 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.