Unique Representation in Polynomial Forms

Theorem
Let $f$ be a polynomial in the indeterminates $\{X_j: j \in J\}$ 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 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:


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

has only finitely many non-zero terms. Moreover it is everywhere equal to $f$, and is the unique such sum.

Corollary
Dropping the zero terms from the sum we can write the polynomial $f$ as


 * $f=a_{k_1}\mathbf X^{k_1}+\cdots+a_{k_r}\mathbf X^{k_r}$

for some $a_{k_i}\in R$, $i=1,\ldots,r$.

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


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

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


 * $\tilde f(\mathbf X^m)=b_m\neq a_m = f(\mathbf X^m)$

therefore $\hat f$ above is the only such representation.