# Unique Representation in Polynomial Forms/General Result

## Theorem

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 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.

### 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:

 $\displaystyle \map {\hat f} {\mathbf X^m}$ $=$ $\displaystyle \paren {\sum_{k \mathop \in Z} a_k \mathbf X^k} \paren {\mathbf X^m}$ $\displaystyle$ $=$ $\displaystyle \paren {a_m \mathbf X^m} \paren {\mathbf X^m} + \sum_{k \mathop \ne m \mathop \in Z} \paren {a_k \mathbf X^k} \paren {\mathbf X^m}$ $\displaystyle$ $=$ $\displaystyle a_m$

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:

$\map {\tilde f} {\mathbf X^m} = b_m \ne a_m = \map f {\mathbf X^m}$

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

$\blacksquare$