Unique Representation in Polynomial Forms

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

Let $$\left({D, +, \circ}\right)$$ be an integral domain such that $$D$$ is a subring of $$R$$.

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

Let $$D \left[{X}\right]$$ be the ring of polynomial forms in $$X$$ over $$D$$.

Then each non-zero member of $$D \left[{X}\right]$$ can be expressed in just one way in the form $$f \in D \left[{X}\right]: f = \sum_{k=0}^n {a_k \circ X^k}$$.

Proof
Suppose $$f \in D \left[{X}\right] - \left\{{0_R}\right\}$$ 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.