# Understanding divison by monic polynomial in $R[x]$ where $R$ is an arbitrary ring

I read "Algebra: Chapter 0" by P.Aluffi. I encountered a topic where it says you can divide any polynomial in $$R[x]$$($$R$$ is any ring) by a monic polynomial(that is, a polynomial of the form $$x^d + \sum\limits_{i=0}^{d-1} a_i x^{i}$$).

It says if $$g(x)$$ is any polynomial and $$f(x)$$ is a monic polynomial. Then $$\exists \ \ \ q(x), r(x) \in R[x],\: \deg r(x) < \deg f(x): g(x) = f(x)q(x) + r(x).$$

I have two questions regarding this:

1) Why $$f(x)$$ needs to be monic?

2) How can we prove it? The book talk about induction and that if $$\deg g(x) = n > d = deg \ f(x)$$, then $$\forall a \in R$$ we have $$ax^n = ax^{n-d}f(x) + h(x)$$ for some $$h(x): \deg \ h(x) < n$$. It's true, but how do I take it from here?

• $1)$ - How would you divide $X$ by $2X$ in $\mathbb{Z}[X]$? Jun 3, 2016 at 20:21
• @ Aloizio I see, there is no way. Jun 3, 2016 at 20:25
• deg $h(x)$ should be less than $d$ instead of $n$ Oct 9, 2016 at 15:51

The reason $f$ has to be monic is that you can always divide by $1$, but not for other elements. Actually, the leading coefficient of $f$ can be any unit.

The proof is by induction on the degree of $g$; if $g$ is the zero polynomial, there is nothing to prove. If $g$ has degree zero and $f(x)=1$, write $g(x)=f(x)g(x)+0$; if $f$ has degree $>0$, $g(x)=f(x)\cdot 0+g(x)$.

Suppose $g$ has degree $n$ and that the statement holds for polynomials of degree $k<n$. Let's assume $\deg f=m$.

If $n<m$, we are done: $g(x)=f(x)\cdot 0+g(x)$. Otherwise, let $a_n$ be the leading coefficient of $g(x)$ and consider $$g_1(x)=g(x)-a_nx^{n-m}f(x)$$ By construction, $\deg g_1<n$, so $g_1(x)=f(x)q_1(x)+r(x)$, with $\deg r<m$, by the induction hypothesis. Therefore $$g(x)=g_1(x)+a_nx^{n-m}f(x)=f(x)(a_nx^{n-m}+q_1(x))+r(x)$$ and we are finished.

If the leading coefficient of $f(x)$ is a unit $u$, then $u^{-1}f(x)$ is monic, so $$g(x)=u^{-1}f(x)q_0(x)+r(x)$$ by the previous result and now $q(x)=u^{-1}q_0(x)$ solves the problem.

• Good answer! However, I have a question regarding the last part. $u^{-1}f(x)$ is indeed monic, but what if $R$ is not commutative? What if $u^{-1}f(x)q_0(x) \neq f(x)u^{-1}q_0(x)$. Should we use $f(x)u^{-1}$ instead? So do we only need the leading coefficient of $f(x)$ to be left-unit? Jun 5, 2016 at 11:58
• @Jxt921 In all this discussion I assumed the ring to be commutative. If not, that's the solution. Jun 5, 2016 at 12:19
• @atulgangwar Good shot! Oct 9, 2016 at 18:02
• @AkashPatalwanshi Provided the indeterminate is central, that is, it commutes with all elements of the ring. Sep 10, 2021 at 9:02
• @AkashPatalwanshi It just works one way. For instance, the easy polynomial $x^2+1$ has infinitely many roots. Sep 10, 2021 at 12:24

For polynomials over any commutative coefficient ring, the high-school polynomial long division algorithm works to divide with remainder by any monic polynomial, i.e any polynomial $$\rm\:f\:$$ whose leading coefficient $$\rm\:c =1\:$$ (or a unit), since $$\rm\:f\:$$ monic implies that the leading term of $$\rm\:f\:$$ divides all higher degree monomials $$\rm\:x^k,\ k\ge n = deg\ f,\:$$ so the high-school division algorithm works to kill all higher degree terms in the dividend, leaving a remainder of degree $$\rm < n = deg\ f\,$$ (see here for further detail).

The division algorithm generally fails if $$\rm\:f\:$$ is not monic, e.g. $$\rm\: x = 2x\:q + r\:$$ has no solution for $$\rm\:r\in \mathbb Z,\ q\in \mathbb Z[x],\:$$ since evaluating at $$\rm\:x=0\:$$ $$\Rightarrow$$ $$\rm\:r=0,\:$$ evaluating at $$\rm\:x=1\:$$ $$\Rightarrow$$ $$\:2\:|\:1\:$$ in $$\mathbb Z,\,$$ contradiction. Notice that the same proof works in any coefficient ring $$\rm\:R\:$$ in which $$2$$ is not a unit (invertible). Conversely, if $$2$$ is a unit in $$\rm\:R,$$ say $$\rm\:2u = 1\:$$ for $$\rm\:u\in R,\:$$ then division is possible: $$\rm\: x = 2x\cdot u + 0.$$

However, we can generalize the division algorithm to the non-monic case as follows.

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 $$\$$ Hint: use induction on $$\,\deg G.\,$$ See this answer for a full proof.