Let $f(x), g(x) \in R[x]$ where $R$ is a domain, if the leading coefficient in $f(x)$ is a unit in $R$ then the division algorithm gives a quotient $q(x)$ and a remainder $r(x)$ after dividing $g(x)$ by $f(x)$. Prove that $q(x)$ and $r(x)$ are uniquely determined by $g(x)$ and $f(x)$.

I understand this Rotman exercise as a proof for the division algorithm for $R[x]$ where $R$, is a domain, I suppose it refers to an integer domain. But for the division algorithm for $f(x), g(x) \in R[x]$ where $R$ is a domain we don't use the fact $K$ is a field, just the fact that the leading coefficient in $f(x)$ is a unit in $R$ in the existence part. Im troubled because the hint for this exercise mention as a hint using $\operatorname{Frac}(R)$ so maybe I didn't understand what Im supposed to prove. Any help showing me what I'm supposed to prove and how to do it? Thanks

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  • $\begingroup$ If $c_f$, $c_g$ and $c_q$ are the leading coefficients of $f$, $g$ and $q$ respectively, then $c_g=c_qc_f$, and so it is necessary that $c_f$ divides $c_g$. If we assume that $c_f$ is a unit, this is certainly true. I think the hint suggests to first prove that the division algorithm yields unique $q(x)$ and $r(x)$ in $K[x]$, where $K=\operatorname{Frac}(R)$, and then prove that $q(x),r(x)\in R[x]$. $\endgroup$ – Servaes Mar 5 at 20:16
  • $\begingroup$ Thanks but the problem is everybody is showing me how to prove uniqueness as if R[x] for R field but what is struggling me is that the nobody besides you mention the Hint. And I'm not sure how to use it in the proof @Servaes $\endgroup$ – Cos Mar 5 at 20:40
  • $\begingroup$ I do not see how the hint is relevant in proving uniqueness either. All you need is that $$\deg\left((q-q')f\right)\geq\deg f,$$ which follows from the assumption that $R$ is a domain. $\endgroup$ – Servaes Mar 5 at 20:49
  • $\begingroup$ Which Rotman book and what page? $\endgroup$ – Will Jagy Mar 5 at 20:50
  • $\begingroup$ Advanced Modern Álgebra, page 142 , exercise 3,32 @WillJagy $\endgroup$ – Cos Mar 5 at 20:54

You're supposed to prove that if one can write \begin{cases}g(x)q(x)f(x)+r(x),\quad & r=0\;\text{ or }\;\deg r<\deg f,\\ g(x)q'(x)f(x)+r'(x),\quad & r=0\;\text{ or }\;\deg r'<\deg f \end{cases} in two ways, then $q=q'$ and $r=r'$.

Hint: deduce from these equalities that $$\bigl(q(x)-q'(x)\bigr)f(x)=r'(x)-r(x).$$ Suppose $r\ne r'$and compare the degrees of both sides.

  • $\begingroup$ Thanks! But I use the Hint telling me to use Fracc(R), here? @Bernard $\endgroup$ – Cos Mar 5 at 20:38
  • $\begingroup$ No you make the proof again, taking into account the leading coefficient is a unit in $R$. You also can embed $R$ in its field of fractions $K$, hence $R[X}$ in $K[X]$, and use the uniqueness of Euclidean division for polynomial with coefficients in a field. $\endgroup$ – Bernard Mar 5 at 20:43
  • $\begingroup$ @Cos We don't need to use the fraction Field (for existence) because Rotman already proves the monic case in Cor 3.22, so we can use this for monic $u^{-1}f$. But this is a moot point since the questions assumes existence is given. It concerns only uniqueness. The Hint is mistaken. $\endgroup$ – Bill Dubuque Mar 5 at 20:52

If the exercise given to you is

Let $f(x), g(x) \in R[X]$ where $R$ is a domain, if the leading coefficient in $f(x)$ is a unit in $R$ then the division algorithm gives a quotient $q(x)$ and a remainder $r(x)$ after dividing $g(x)$ by $f(x)$. Prove that $q(x)$ and $r(x)$ are uniquely determined by $g(x)$ and $f(x)$.

then first of all there is a lot of sloppy notation; the symbols $x$ and $X$ are not interchangeable. Also, it seems to be implicit that $\deg r<\deg f$.

Second, it seems to be assumed that the division algorithm in $R[X]$ works, i.e. that it gives $q,r\in R[X]$ such that $g=qf+r$ and $\deg r<\deg f$. The question only asks to prove that these $q$ and $r$ are unique. That is to say, if $q',r'\in R[X]$ are such that $g=q'f+r'$ and $\deg r'<\deg f$, then $q'=q$ and $r'=r$.

To prove uniqueness, let $q,q,r,r'\in R[X]$ with $\deg r<f$ and $\deg r'<f$ be such that $$g=qf+r\qquad\text{ and }\qquad g=q'f+r'.$$ Then subtracting the two from eachother shows that $$(q-q')f=r'-r.$$ Of course $\deg(r'-r)<f$. Because $R$ is a domain, if $q-q'\neq0$ then $\deg\left((q-q')f\right)\geq\deg f$, a contradiction. Hence $q=q'$, from which it immediately follows that $r=r'$.

Note that this proof makes no use of the fraction field, but only of the fact that $R$ is a domain.

  • $\begingroup$ Anyone care to explain the downvote? $\endgroup$ – Servaes Mar 5 at 20:30

Hint If $\, \deg r,\deg R < \deg\,f\,$ and $\,qf+r=Qf+R\,$ then $\,\color{#c00}{(Q−q)f}=\color{#0a0}{r−R}.\,$ If $\,Q\neq q\,$ then, since lead coef of $f$ is a unit, $\,\deg\rm \color{#c00}{LHS} \ge \deg f > \deg {\rm\color{#0a0}{ RHS}}\Rightarrow\!\Leftarrow\,$ Thus $\,Q=q\,$ so $\,r−R=0$

  • $\begingroup$ -1 This is incrediby hard to parse, and does not answer the question "What am I supposed to prove?". $\endgroup$ – Servaes Mar 5 at 20:29
  • $\begingroup$ Thanks but what does LHS and RHS mean? And how I use the exercise hint here? $\endgroup$ – Cos Mar 5 at 20:35
  • $\begingroup$ @Cos The red LHS refers to the matching red Left Hand Side of the equation (RHS is its Right Hand Side $= r-R$. The proof does not require any use of the fraction field. $\endgroup$ – Bill Dubuque Mar 5 at 20:46
  • $\begingroup$ @cos If you're also interested in proving existence (not part of this exercise) then you can find a few ways here. That's what the (misplaced) Hint is aiming at. $\endgroup$ – Bill Dubuque Mar 5 at 21:12

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