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 $f\mathrm{\ne }0$ 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 $a=\mathrm{lc}(f)$; 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 $f_1=q_{1}g+r$ with $r=0$ or $\mathrm{deg}r<d$, 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 $f=qg+r=q^{\mathrm{\prime }}g+r^{\mathrm{\prime }}$ are two such expressions, then $(q-q^{\mathrm{\prime }})g=r^{\mathrm{\prime }}-r$; if $q-q^{\mathrm{\prime }}\mathrm{\ne }0$, the leading coefficient of $(q-q^{\mathrm{\prime }})g$ equals the nonzero leading coefficient of $q-q^{\mathrm{\prime }}$ because $g$ is monic, so [L1] gives degree at least $d$, whereas $r^{\mathrm{\prime }}-r$ is zero or has degree below $d$, a contradiction. Thus $q=q^{\mathrm{\prime }}$ and then $r=r^{\mathrm{\prime }}$, proving uniqueness and completing the induction proof.