2
$\begingroup$

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

enter image description here

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Anyone care to explain the downvote? $\endgroup$ – Servaes Mar 5 at 20:30
0
$\begingroup$

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 $\,r−R=0$

$\endgroup$
  • $\begingroup$ -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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.