Proof Of Uniqueness of Division Theorem for Polynomials

Question

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.

Answer 1

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.

Answer 2

Let $f$ be such that $$f(x)=q(x)g(x)+r(x)=p(x)g(x)+s(x)$$ so you have ($g(x)\ne 0$) $$\frac{f(x)}{g(x)}=q(x)+\frac{r(x)}{g(x)}=p(x)+\frac{s(x)}{g(x)}$$

Because of the degrees the nul or rational (non-polynomial) functions must be equal, i.e. $$q(x)+\frac{r(x)}{g(x)}=p(x)+\frac{s(x)}{g(x)}\Rightarrow\frac{r(x)}{g(x)}=\frac{s(x)}{g(x)}\Rightarrow r(x)=s(x)$$ and this implies the equality $$q(x)=p(x)$$