Unique Representation in Polynomial Forms

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Theorem

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

Let $\left({D, +, \circ}\right)$ be an integral subdomain of $R$.

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

Let $D \left[{X}\right]$ 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:

$\displaystyle f \in D \left[{X}\right]: f = \sum_{k \mathop = 0}^n {a_k \circ X^k}$


General Result

Let $f$ be a polynomial form in the indeterminates $\left\{{X_j: j \in J}\right\}$ 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 mononomials.


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


Then the sum representation:

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

has only finitely many non-zero terms.

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


Proof

Suppose $f \in D \left[{X}\right] \setminus \left\{{0_R}\right\}$ 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 $D \left[{X}\right]$ equal to zero.

As $f$ is transcendental, the result follows.

$\blacksquare$


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