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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?

user26857
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tony
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  • It does depend on the ring R. For instance you would require R[x] to be an Euclidean Domain, which would need R to be minimally an integral domain. – Chris C Feb 12 '17 at 05:15
  • And yeah , for any such R , you need the polynomial to be monic, which is not required in a field. – Chirantan Chowdhury Feb 12 '17 at 08:34
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    There do exist rings $R$ (even nice rings like $R = \mathbb R[y]$ and $R = \mathbb Z$) for which $R[x]$ does not have a division algorithm; i.e. is not a Euclidean domain. This can be proved, for example, by finding a non-principal ideal in $R[x]$. (Recall that Euclidean domains are necessarily principal ideal domains.) However, the usual proof for $F[x]$ does provide a division algorithm for dividing by monic polynomials (or, slightly more generally, those with unit leading coefficients) if $R$ is any commutative ring with 1. – Ravi Fernando Feb 14 '17 at 07:36

1 Answers1

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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.

Bill Dubuque
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  • Thanks. Did you get this proof from Nathan Jacobson? – tony Feb 16 '17 at 03:28
  • @Tony No, I wrote the proof from scratch (it is straightforward if you know the idea). – Bill Dubuque Feb 16 '17 at 03:55
  • Mr. Dubuque, where could one find a reference for the non-monic version of the division algorithm? textbooks usually gives the monic version and even many of them only state the result when $A$ is a field. – Xam Jul 09 '17 at 18:35
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    @Xam I don't recall at the moment, but the above comment seems to indicate it may appear in one of Jacobson's Algebra textbooks. In any case, I give a complete proof in the linked answer, so one can always refer to that. – Bill Dubuque Jul 09 '17 at 18:50
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    @Xam, try Basic ALgebra 1 by Nathan Jacobson. You will find the proof in chapter 2.11 , page128 – tony Jul 29 '17 at 11:40
  • @tony I'd already found it in Jacobson's book following the suggestion of Mr. Dubuque. I also found the above result in the book "Polynomials: An Algorithmic Approach" by Mignotte and Stefanescu. In that book it's theorem 1.3.6. – Xam Jul 29 '17 at 16:50
  • cool, will check it out – tony Jul 30 '17 at 10:03
  • Idea of the proof: multiply $F$ by a convenient power of $a$, so you can rewrite the result as a monic polynomial in $ax$. – Matemáticos Chibchas Oct 05 '19 at 19:50
  • @Matemáticos Yes, if you follow my link to proofs you will see that is precisely the way one of the proofs works, and I give a link to the ring theory behind it (AC-method $\leadsto$ Schreier rings and closely related concepts). – Bill Dubuque Jun 22 '21 at 23:02