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For example $4x^4 + x + 1$ divided by $3x + 1$ is $\frac{4x^3}{3} - \frac{4x^2}{9} + \frac{4x}{27} + \frac{23}{81}$ remainder $\frac{58}{81}$.

Now I want to do the same division mod $9$, but I can't because the denominators are not coprime to it.

Is there a way to get around this somehow or is the quotient/remainder not defined mod $9$?

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The polynomial division algorithm only works in general for polynomials with coefficients in a field. It will work for divisors whose leading coefficient is a unit for coefficients over an arbitrary commutative ring. $3$ is not a unit in the ring $\mathbb{Z}/9\mathbb{Z}$ of integers modulo $9$, so you can't expect the division algorithm to work for a divisor with leading coefficient $3$ when the coefficients are in $\mathbb{Z}/9\mathbb{Z}$. The quotient and remainder you are looking for in your example don't exist: there are no polynomials $q$ and $r$ over $\mathbb{Z}/9\mathbb{Z}$, such that $4x^4 + x + 1 = q\times(3x+1) + r$ and $\deg(r) = 0$.

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  • $\begingroup$ Yes. "unit" is the standard jargon for an invertible element in a ring. $\endgroup$ – Rob Arthan Apr 8 '15 at 15:48
  • $\begingroup$ Yes. "unit" means invertible element in this context. $\endgroup$ – Rob Arthan Apr 8 '15 at 15:55
  • $\begingroup$ @Bill Dubuque: the algorithm you suggest meets a different specification from the standard algorithm, so I was focussing on that avoid confusion and to find out what the OP actually wanted the algorithm to achieve. The algorithm you propose is not very useful if the leading coefficient $a$ satisfies $a^2 = 0$ as in the example in the question. $\endgroup$ – Rob Arthan Apr 8 '15 at 16:31
  • $\begingroup$ I moved my comment to an answer so the OP will get pinged. $\endgroup$ – Bill Dubuque Apr 8 '15 at 16:57
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It is easy to prove that the division is impossible if $\,3\,$ doesn't divide the lead coef of $\,f.\,$

For $\ f \equiv (3x\! +\! 1)q + r\pmod{\color{#c00}9},\ r\in \Bbb Z\,$ $\Rightarrow$ $\ f = (3x\!+\!1)q + r + \color{#c00}{9g}\ $ for $\,q,g\in \Bbb Z[x]\,$ therefore $\, {\rm mod}\ 3\!:\ f\equiv q+r\,$ has reduced degree $(\le \deg q),\,$ so the lead coef of $\,f\,$ is $\,\equiv 0$

The same idea works more generally to disprove certain nonmonic divisions.

Remark $ $ The division algorithm works for nonmonic divisors too if one scales the dividend by a (sufficiently large) power of the lead coeff of the divisor, see the nonmonic division algorithm. In your example this amounts to scaling the division by $3^4$ to make everything integral. Whether or not this proves useful depends on the context.

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