Unique Representation in Polynomial Forms/General Result

Theorem
Let $f$ be a polynomial 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 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 that the sum has infinitely many non-zero terms. Then infinitely many $a_k$ are non-zero, which contradicts the definition of a polynomial. Therefore the sum consists of finitely many non-zero terms.

Let $\mathbf X^m \in M$ be arbitrary. Then

So $\hat f = f$.

Finally suppose that:


 * $\displaystyle \tilde f = \sum_{k \mathop \in Z} b_k \mathbf X^k$

is another such representation with $b_m \ne a_m$ for some $m \in Z$.

Then:


 * $\tilde f \left({\mathbf X^m}\right) = b_m \ne a_m = f \left({\mathbf X^m}\right)$

Therefore $\hat f$ as defined above is the only such representation.