Polynomials Contain Multiplicative Identity
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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:
\(\ds f \circ N\) | \(=\) | \(\ds \sum_{k \mathop \in Z} \paren {\sum_{p \mathop + q \mathop = k} a_p b_q} \mathbf X^k\) | Definition of Polynomial Multiplication | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop \in Z} \paren {\sum_{p \mathop + 0 \mathop = k} a_p 1_R} \mathbf X^k\) | by the definition of $b_k$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop \in Z} a_p \mathbf X^k\) | because $1_R$ is a multiplicative identity | |||||||||||
\(\ds \) | \(=\) | \(\ds f\) |
Therefore:
- $f \circ N = f$
for all polynomials $f$.
Therefore $N$ is a multiplicative identity for the set of polynomials.
$\blacksquare$