Unique Representation in Polynomial Forms

Theorem
Let $$(R,M,f)$$ 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,


 * is everywhere equal to $f$,


 * 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.