Monic polynomial division over a ring ($g=qf+r,\mathrm{deg}r<\mathrm{deg}f$)

If $f(T)$ is monic of degree $k$ over a commutative ring, every polynomial has a unique expression $g=qf+r$ with $\mathrm{deg}r<k$. Cancel the highest remaining term successively; the leading coefficient one needs no inversion. Uniqueness follows because multiplication by a monic polynomial raises the degree of every nonzero polynomial by $k$, even when the ring has zero divisors. The same argument applies to an even central variable and central coefficients in a graded commutative algebra.

For any commutative ring $A$ and a monic polynomial $F$ of positive degree $D$, monic polynomial division over a ring gives unique representatives of degree less than $D$. Thus the classes of $1,T,\dots ,T^{D-1}$ form a basis of a module over $A$. This is a finite free algebra, hence flat, and the map $A\to A[T]/(F)$ is injective. The monic hypothesis makes leading-degree arguments valid even when $A$ has zero divisors.

For monic $p\in \mathbb{Z}[X]$ irreducible in $\mathbb{Q}[X]$, monic division gives an injection $\mathbb{Z}[X]/(p)\hookrightarrow \mathbb{Q}[X]/(p)$: an integer polynomial in the rational ideal has zero integer remainder. The target is a field, so the source is an integral domain. For positive degree $d$, unique remainders make the source a free abelian group on $1,X,\dots ,X^{d-1}$. The nonzero integer $2$ is not a unit, since multiplying an integer remainder by two cannot give remainder one. Thus this integral domain is not a field.