If $f, d$ are polynomials over a field $F$ and $d$ is different from $0$ then there exist polynomials $q, r$ in $F[x]$ such that (i) $f = dq + r$. (ii) either $r=0$ or $\deg (r)\lt \deg (d)$.
Hoffman’s Proof: If $f$ is $0$ or $\deg (f)\lt \deg (d)$ we may take $q=0$ and $r=f$. In case $f\neq 0$ and $\deg (f) \geq \deg (d)$, the preceding lemma shows we may choose a polynomial $g$ such that $f-dg=0$ or $\deg (f - dg) \lt \deg (f)$. If $f - dg \neq 0$ and $\deg (f- dg) \geq \deg (d)$ we choose a polynomial $h$ such that $(f-dg)- dh=0$ or $\deg [f-d(g+h)]\lt \deg(f-dg)$. Continuing this process as long as necessary, we ultimately obtain polynomials $q,r$ such that $r=0$ or $\deg (r)\lt deg (d)$,and $f=dq+r$.
Que: I don’t understand following step: If $f - dg \neq 0$ and $\deg (f- dg) \geq \deg (d)$ we choose a polynomial $h$ such that $(f-dg)- dh=0$ or $\deg [f-d(g+h)]\lt \deg(f-dg)$. Continuing this process as long as necessary, we ultimately obtain polynomials $q,r$ such that $r=0$ or $\deg (r)\lt \deg (d)$,and $f=dq+r$.
In this video (time stamp 13:32) professor uses induction explicitly. Inductive hypothesis is suppose theorem holds for polynomial with degree $\lt n$. Que: I think, it is not the “usual” inductive hypothesis, i.e suppose $P(n-1)$ is true, for some $n$. I know there are different variation of mathematical induction. How to rigioursly proof present case of induction is correct?
Edit: I found this amazing lecture on division algorithm for polynomial. The induction I was looking for is strong induction.
