Prove that there exists unique polynomials $r, q$ such that $g=fq+r$, and $r$ has a degree less than $f$.

Question

I have been trying to understand a proof of the following theorem for a while now.

For polynomials $g,f$, there exists unique polynomials $q, r$ such that $g=fq+r$ and the degree $D(r)$ of $r$ is strictly less than the degree $D(f)$ of $f$. The author first uses induction to prove the existence of $q$ and $r$. Everything good for the moment, but then he gives the following proof of uniqueness

"Suppose that $fq_1+r_1=fq_2+r_2$, and $D(r_1), D(r_2) < D(f)$, then $$f(q_1-q_2)=r_2-r_1$$

By proposition $2.2$, ..."

Proposition $2.2$ says that if $f, g$ are polynomials over $\mathbb {C}$ , then $$D(f+g)≤max(D(f), D(g))$$

$$D(fg)=D(f)+D(g)$$

"...the polynomial on the left has a higher polynomial than that on the right, unless both are zero. Since $f≠0$ we must have $q_1=q_2$ and $r_1=r_2$, so $q$ and $r$ are unique."

---

I understand proposition $2.2$, I just can't see why this proves that $D(f(q_1-q_2)>D(r_2-r_1)$, nor why even if this is true, then $q_1$ has to equal $q_2$ and $r_1$ has to equal $r_2$.

---

Any thoughts would be really appreciated.

Answer 1

Suppose that $q_1\neq q_2$, $D(f(q_1-q_2))=D(f)+D(q_1-q_2)\geq D(f)>D(r_1-r_2)$ contradiction.

Answer 2

If $q_1\neq q_2$, then $D(q_1-q_2)\geq 0$ and $D(f(q_1-q_2))\geq D(f)$.

On the other hand $D(r_1),D(r_2)<D(f)$ so $D(r_1-r_2)<D(f)$.