Let $R$ be a commutative unital ring $f(x),g(x) \in R[x]$, let $b$ be the leading coefficient of $g(x)$, and let $l:=\max \{0,\deg f -\deg g+1\}$. Then is it true that $\exists g(x),r(x) \in R[x]$ such that $\deg r(x) < \deg g(x)$ and $b^lf(x)=q(x)g(x)+r(x)$ ?

I know a division algorithm where the leading coefficient of $g(x)$ is a unit, but the division algorithm here seems to be different. Please help. Thanks in advance.


Below are a few methods to prove it.

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
$\qquad\qquad \phantom{1^{1^{1^{1^{1^{1}}}}}}a^{i} G\, =\, Q F + R\ \ {\rm for\ some}\ \ Q,R\in A[x],\ \deg R < \deg F$

Proof $\,\ $ If $\ \deg G < \deg F\,$ then let $\, Q = 0 ,\ R = a^i G.\, $ Else we induct on $\,\deg G.\,$ Let $\, k = \deg F,\,$ so $\,\deg G = k+j\,$ for $\, j \geq 0.\,$ Splitting $\,G,F\,$ into $ $ lead $\color{#c00}+$ rest $ $ terms:

$\begin{array}{lrl} \ G = b x^{k+j\!}\color{#c00} + G',\ \deg G'<k\!+\!j\!\!\!\!\!\! &\Rightarrow\quad aG\!\!\!\! &=\,abx^{k+j}+aG'\\ \ F = a x^k\color{#c00}+F',\quad \deg F' \!< k & bx^jF\!\!\!\! &=\, abx^{k+j}+bx^jF'\\ \qquad\ a,b\neq 0\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! & aG-bx^jF\!\!\!\! &=\, aG'-bx^jF'\ \ {\rm has\ \ deg} < k\!+\!j \end{array}$

$\begin{array}{lrl}{\rm Therefore,\ \ by\ \ induction} &a^j(aG-bx^jF) =&\!\!\! Q F + R\ \ {\rm for}\ \ Q,R\in A[x], \ \deg R < k\\ &\Rightarrow\quad a^{j+1} G\, =&\!\!\! \bar QF + R\ \ {\rm for}\ \ \bar Q = Q\!+b(ax)^j\end{array}$

Remark $\ $ Alternatively, if localizations are known, we can divide by the monic $\,a^{-1} F\in A[a^{-1}][x]\,$ then pullback the result to $\,A[x].$

Or, as in the AC-method, by scaling $\,F\,$ by $\,a^{k-1}\,$ for $\,k = {\rm deg} F,\,$ we can rewrite $\,F\,$ as a monic polynomial in $\,X = ax,\,$ and similarly we can scale $\,G\,$ by $\,a^i\,$ to make it a polynomial in $\,X.\,$ Then we divide $\,G(X)\,$ by the monic $\,F(X),\,$ and finally replace $\,X\,$ by $\,ax$ in the result.

  • $\begingroup$ The "nonmonic Polynomial Division Algorithm" is also proved here. $\endgroup$ – user26857 Aug 24 '16 at 15:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy