Unique Representation in Polynomial Forms
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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.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 64.2$ Polynomial rings over an integral domain