Proof Of Uniqueness of Division Theorem for Polynomials Trying to prove the uniqueness part of the division theorem for polynomials. Let $F$ be one of $\mathbb{Q}$, $\mathbb{R}$ or $\mathbb{C}$. Namely prove the following lemma.
Let $g(X), p(X), q(X), r(X), s(X)\in F[X]$ with $g(X)\neq 0$.
Suppose that $q(X)g(X) + r(X) = p(X)g(X) + s(X)$, and that $r(X) = 0$ or $deg(r(X)) < deg(g(X))$, and $s(X) = 0$ or $deg_s(X) < deg_g(X)$. Then $p(X) = q(X)$ and $r(X) = s(X)$.
So i have turned it into 
$$(q(X) − p(X))g(X) = s(X) − r(X)$$
And just not sure where to go now to be honest considering degrees.

Suppose $q(X)\not=p(X)$ so that $q(X)-p(X)$ is at least a constant of degre zero. Then $(q(X)-p(X))g(X)$ is at least of degree the degree of $g(X)$, as the degree the product of non-zero polynomials is the sum of their respective degree. (One says that the degree function defined on non zero poolynomials is a valuation.)
Now as $g(X)\not=0(X)$, we know that $s(X)-r(X)$ is not zero, and as it has degree strictly smaller than the degree of $g(X)$, the equation
$$(q(X)-p(X))g(X) = s(X)-r(X)$$
yields a contradiction : $s(X)-r(X)$ of degree strictly smaller than the degree of $g(X)$ should be of degree at least the degree of $g(X)$. So that $q(X)=p(X)$, and that $r=s$ as $g\not=0$.
No need at all the go out from the ring $F[X]$ (in F(X) for instance) for the proof.
