Division by a monic polynomial over a commutative ring

If $d=0$, monicity gives $g=1$, so $q=f,r=0$ works for every $f$. Hence suppose $d>0$; for the zero dividend, $q=r=0$ works.

Let $f\ne0$ have degree $n$ and assume, as the strong induction hypothesis, that division exists for every zero polynomial or nonzero polynomial of degree below $n$. If $n<d$, take $q=0,r=f$.

If instead $n\ge d$, put $a=\operatorname{lc}(f)$; the polynomial $f_1=f-a x^{n-d}g$ has its degree-$n$ coefficient cancelled and is zero or has degree below $n$. The induction hypothesis gives $f_1=q_1g+r$ with $r=0$ or $\deg r<d$, whence $f=(q_1+a x^{n-d})g+r$.

Steps 1.1–2.1 and strong induction [L3] prove existence for every dividend.

If $f=qg+r=q'g+r'$ are two such expressions, then $(q-q')g=r'-r$; if $q-q'\ne0$, the leading coefficient of $(q-q')g$ equals the nonzero leading coefficient of $q-q'$ because $g$ is monic, so [L1] gives degree at least $d$, whereas $r'-r$ is zero or has degree below $d$, a contradiction. Thus $q=q'$ and then $r=r'$, proving uniqueness and completing the induction proof.