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Give an example of a commutative ring $R$ and $f(x), g(x) R[x]$ with $f$ monic such that the remainder after dividing $g$ by $f$ is not unique; that is, there are $q,q',r,r' \in R[x]$ with $qf + r = g = q' f + r'$ and $\deg (r)$ and $\deg (r')$ are both strictly less than $\deg(f)$.

Okay. I have been thinking about this problem for a rather long time ---I've been foiling for hours (perhaps an exaggeration). I have tried several pairs of polynomials when the base ring is $\Bbb{Z}_4$, $\Bbb{Z}_6$, and $\Bbb{Z}_8$. My strategy was to solve for $q$ when $r=0$; and then find a $q'$ given some nonzero $r'$. I set up equations with five variables so that I could find a nonzero remainder with $\deg (r) < \deg (f)$, but I could find no values for these five variables. I could use a hint on finding a ring $R$ and polynomials $f(x)$ and $g(x)$. But please don't just hand them to me; I would like to figure out this problem on my own as much as possible.

Note: The polynomial division algorithm has just been introduced, so the solution should involve relatively elementary ideas.

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    $\begingroup$ Perhaps I miss some key point, but I don't see how this could be possible. If $f$ is monic, then for $p\in R[x]$ with $p\neq 0$ we must have $\deg(pf)≥\deg(f)$ (or if you like $\deg(pf)=\deg(p)+\deg(f)$), even if the ring contains zero divisors. Thus if $qf+r=q'f+r'$ we have $(q-q')f=r'-r$, so $q=q'$ because otherwise $\deg((q-q')f)≥\deg(f)>\deg(r)≥\deg(r'-r)$. But then we also have $r=r'$. Where is my mistake? $\endgroup$ – Redundant Aunt Oct 3 '17 at 16:31
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In $\mathbb{Z}_8$, let $f(x)=2x+3$ and $g(x)=4x$. Then

\begin{align} g &= 4x(f) + 0\\ g &= 2(f) + 2 \end{align}

EDIT: Just noticed you wanted $f$ monic. Then I don't believe this is possible. Because for any such $q(x)$, the leading coefficient is fixed. You can show every following coefficient is fixed as well, by noting that you know what the first few coefficients of $fq$ should look like (reading off from $g$). This holds all the way down to the constant term of $q(x)$, and so $q$ is unique.

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    $\begingroup$ Yeah, it's impossible if $f$ is really meant to be monic. $\endgroup$ – darij grinberg Oct 3 '17 at 18:19
  • $\begingroup$ @darijgrinberg I checked my book and it does indeed say that f is monic; perhaps it is a misprint. $\endgroup$ – user193319 Oct 3 '17 at 18:23

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