# Help proving polynomials division algorithm in $R[x]$ where $R$ is a domain.

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

• 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 $q(x),r(x)\in R[x]$. – Servaes Mar 5 at 20:16
• 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 – Cos Mar 5 at 20:40
• 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. – Servaes Mar 5 at 20:49
• Which Rotman book and what page? – Will Jagy Mar 5 at 20:50
• Advanced Modern Álgebra, page 142 , exercise 3,32 @WillJagy – 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.

• Thanks! But I use the Hint telling me to use Fracc(R), here? @Bernard – Cos Mar 5 at 20:38
• 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. – Bernard Mar 5 at 20:43
• @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. – 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 $$\deg r and $$\deg r' 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 $$\deg(r'-r). Because $$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.

• Anyone care to explain the downvote? – 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 $$\,r−R=0$$

• -1 This is incrediby hard to parse, and does not answer the question "What am I supposed to prove?". – Servaes Mar 5 at 20:29
• Thanks but what does LHS and RHS mean? And how I use the exercise hint here? – Cos Mar 5 at 20:35
• @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. – Bill Dubuque Mar 5 at 20:46
• @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. – Bill Dubuque Mar 5 at 21:12