Division algorithm for polynomials in R[x], where R is a commutative ring with unity.

Is there a division algorithm for polynomials in R[x], where R is a commutative ring with unity?

All the algebra books I read mention division algorithm for polynomials in F[x], where F is a field. Since the leading coefficient of a non zero polynomial in R[x] is not necessarily invertible, I find it difficult to proceed the same way we do for polynomials in F[x]. Can I get some help?

Solution 1:

For polynomials over any commutative coefficient ring, the (high-school) polynomial long division algorithm shows how to divide with remainder by any monic polynomial, i.e any polynomial $\rm\:f\:$ whose leading coefficient $\rm\:a=1\:$ (or a unit, i.e. $\,\rm a\mid 1),\,$ since this implies the leading monomial $\rm\,ax^n\,$ of $\rm\:f\:$ divides all higher degree monomials $\rm\:x^k,\,$ so the division algorithm works to kill all higher degree terms in (successive) dividends, leaving a remainder of degree $\rm < n = deg\ f.$

But this generally fails if $\rm\:f\:$ is $\rm\color{#c00}{not\ monic}$, e.g. $\rm\: x = ax\:q + r,\ \color{#c00}{a\nmid 1}\,$ has no solution since evaluating at $\rm\:x=0\:$ $\Rightarrow$ $\rm\:r=0,\:$ then eval at $\rm\:x=1\:$ $\Rightarrow$ $\rm\,1 = aq\,\Rightarrow\,\color{#c00}{a\mid 1\Rightarrow\!\Leftarrow}\,$ Conversely, if $\rm\,a\,$ is a unit, then $\rm\,ab = 1$ for some $\rm\,b\,$ so the division is possible: $\rm\: x = ax\cdot b + 0.$

However, it is possible to generalize division with remainder to non-monic polynomials by scaling the dividend by a sufficiently large power $\,\color{#0a0}{a^i}$ of the leading coefficient of the divisor, i.e.

Theorem (nonmonic Polynomial Division Algorithm) $\ $ Let $\,0\neq F,G\in A[x]\,$ be polynomials over a commutative ring $A,$ with $\,a\,$ = lead coef of $\,F,\,$ and $\, i \ge \max\{0,\,1+\deg G-\deg F\}.\,$ Then
$\qquad\qquad \phantom{1^{1^{1^{1}}}}\color{#0a0}{a^{i}} G\, =\, Q F + R\ $ $\ {\rm for\ some}\ \ Q,R\in A[x],\ \deg R < \deg F$

Proof $\ $ See here for a few proofs.