Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

how can I divide for example $\frac{x^2+1}{2x+1}$ in $\frac{\mathbb{Z}_3[x]}{x^3+1}$? It's like normal polynomial dividing but here I got in first step $\frac{x}{2} (\frac{x^2}{2x}=\frac{x}{2}$).What is the modulo value of this statement? I need it for solve a linear equation solve, it's part of the Euclidan algorithm..but here I am stuck and I don't know what to do..Right answer is 2x+2 with remainder 2..But I don't know WHY:D

Please help..And please be patience for my poor English and absence of right formatting, today I've an exam so I am a bit hurry:D

share|cite|improve this question
As you're calculating modulo $3$, you have $2 = -1$. – j.p. Jan 3 '11 at 11:11
Oh!Excellent..So if I am dividing with larger factor(multiple of x), I flip it like this..and I got -x equals 2x..So thanks a lot!! – simekadam Jan 3 '11 at 11:20
Be a bit careful. In $Z_3, 2=\frac{1}{2}$, but in other fields it does not. In $Z_{13}$, for example, $7=\frac{1}{2}$, so you don't just "flip it". – Ross Millikan Jan 3 '11 at 11:22
I don't mean to flip the fraction ($\frac{1}{2}$ to $\frac{2}{1}$), but to flip or revert the modulo value to a better one usable for dividing..So in your example it will be like $7=-6 Z_{13}$..Another way.I am not sure if I can divide two "undivisible" numbers…Wait, maybe I am dummy..I need an inverse for every divide:DSorry Sorry..It's to confusing for me these congruence modulo things.. – simekadam Jan 3 '11 at 11:32
$Z_p$ is a field for prime $p$, so every non-zero element will have an inverse. If $p$ is not a prime, you need the element to be coprime to $p$ – Ross Millikan Jan 3 '11 at 13:16

HINT $\rm\ \ 3 \equiv 0\ \Rightarrow\ 1 + 2\ x \:\equiv\: 1-x\:.\ $ Now $\rm\ x^3\:\equiv\:-1\ \Rightarrow\ x^2+1\ \equiv\ x^2 -x^3\ \equiv\ x^2\ (1-x)\ $

Therefore $\rm\ (x^2+1)/(1-x)\ \equiv\ \ldots$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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