# Unique Representation in Polynomial Forms It has been suggested that this page or section be merged into Monomials form Basis of Polynomial Ring. (Discuss)

## Theorem

Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.

Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.

Let $X \in R$ be transcendental over $D$.

Let $D \sqbrk X$ be the ring of polynomials in $X$ over $D$.

Then each non-zero member of $D \left[{X}\right]$ can be expressed in just one way in the form:

$\ds f \in D \sqbrk X: f = \sum_{k \mathop = 0}^n {a_k \circ X^k}$

### General Result

Let $f$ be a polynomial form in the indeterminates $\set {X_j: j \in J}$ such that $f: \mathbf X^k \mapsto a_k$.

For $r \in \R$, $\mathbf X^k \in M$, let $r \mathbf X^k$ denote the polynomial form that takes the value $r$ on $\mathbf X^k$ and zero on all other monomials.

Let $Z$ denote the set of all multiindices indexed by $J$.

Then the sum representation:

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

has only finitely many non-zero terms.

Moreover it is everywhere equal to $f$, and is the unique such sum.

## Proof

Suppose $f \in D \sqbrk X \setminus \set {0_R}$ has more than one way of being expressed in the above form.

Then you would be able to subtract one from the other and get a polynomial in $D \sqbrk X$ equal to zero.

As $f$ is transcendental, the result follows.

$\blacksquare$