Polynomial division in $\mathbb{Z}[x]$.

Question

How is polynomial division defined in $\mathbb{Z}[x]$? For example, how to divide $x$ by $2$ or any other two polynomials in $\mathbb{Z}[x]$?

Answer 1

Given two polynomials $f=\sum_{k=0}^na_kx^k$ and $g=\sum_{k_0}^mb_kx^k$ in $\mathbb{Z}[x]$, the principal problem when we try to divide $f$ by $g$ is that if $g$ is not monic then maybe there are problems when we divide coefficients by the leading coefficient and maybe the algorithm of division maybe go outside $\mathbb{Z}[x]$. In order to prevent that, a modification is made by multiplying $f$ by $b_m^n$ and in this way there is not problem of going outside $\mathbb{Z}$ when dividing by $g$ by the usual algorithm. In this way we obtain uniquely determined $q,r\in\mathbb{Z}[x]$ with $\deg r<m$ and $$b_m^nf=qg+r$$ This method can be extended to arbitrary rings different from $\mathbb{Z}$, although you can see that has the problem of multiplying by the leading coefficient of $g$ in order to don't go outside the ring you are working in.

Answer 2

Polynomial division of $f$ by $g$ consists of finding quotient $q$ and remainder $r$ with $f=qg+r$ and $\deg r<\deg g$. However, if $g$ is not monic, this may not be possible (while in $\mathbb Q[X]$ you have $X=2\cdot \frac X2+0$).

Answer 3

If you are talking about Euclidean division, it is pretty much the same as in $\mathbb Z$: If $P$ can be written $P(X)=Q(X)S(X)$ with $Q,S\in\mathbb{Z}[X]$, then $P/Q=S$. In general, the euclidean division of $P$ by $Q$ comes from writing $P(X)=Q(X)S(X)+R(X)$ with $\mathrm{deg}(R) < \mathrm{deg}(Q)$ and $Q,S,R\in\mathbb{Z}[X]$ when it is possible (see comment below).

For instance, $(X^4-1)/(X-1) = (X+1)(X^2+1)$.