Polynomials Contain Multiplicative Identity

From ProofWiki
Jump to navigation Jump to search

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.

$\blacksquare$