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So, let $A=\mathbb{Z}/2\mathbb{Z}[X]$ and $ I=\{(X^2+X+1)P | P \in A \}$ I succeeded showing that this is the ideal, but now i have to find $A/I$, show that it have only 4 classes and find to what class corresponds elements like $X^3+X^2$ and $X^2+1$. I have some problems understanding the quotient groups. Mainly I considered looking at the wiki's page on quotients but there was sets and I arrived to the question.

It's very easy intuitevely follow that quotient is just mod n. But I dont understand the idea how to arrive to such result from having that $G/H=\{ xH | x \in G \}$.

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For rings, $G/H=\{H+x|x\in G\}$. –  Alex Becker Apr 21 '12 at 1:49
Could you please elaborate what sort of problems you have understanding quotient structures. It's difficult to answer that part of your query without knowing more precisely what the problems are. –  Bill Dubuque Apr 21 '12 at 2:08
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1 Answer

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Hint $\ $ Use the Division Algorithm to write $\rm\:f(x) = (x^2+x+1)\:q(x) + r(x),\ deg\ r \le 1.\:$ Thus, mod $\rm I = (x^2+x+1),\:$ every $\rm\:f(x)\in \mathbb F_2[x]\:$ is equivalent to a linear polynomial (degree $\le 1).\:$ Further, distinct linear polynomials are inequivalent mod I, else their difference would be a linear polynomial $\ne 0$ divisible by $\rm\:x^2 + x + 1.$ Thus the linear polynomials $\rm\: a\:x + b,\ a,b\in \mathbb F_2,\:$ are a complete system of representatives of the equivalence classes (cosets) of the quotient ring$\rm\:A/I.$

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Thus, mod I = (x^2+x+1), every $\rm\:f(x)\in \mathbb F_2[x]\:$ is equivalent to a linear polynomial (i.e. degree $\le 1)\:$ Cant understand this line.. Why? –  user974514 Apr 21 '12 at 2:07
Modulo $\rm\:(g),\ \ g\:\equiv\: 0\ \Rightarrow\ g\:q + r\:\equiv\: 0\cdot q + r\:\equiv\: r.\:$ Above $\rm\: g = x^2+x+1.\:$ Such polynomial congruence arithmetic is analogous to congruence arithmetic of integers mod $\rm\:m\:$. Both cases employ the Euclidean Algorithm to reduce elements to their "smaller" normal-form representatives. –  Bill Dubuque Apr 21 '12 at 2:18
As far as i can get $A$ is any polynomial of degree $n$. $I$ is the same polynomials but multiplied by $(X^2+X+1)$? Zero class will be equal to class of $X^2+X+1$ so we have left $1$, $X$, $X+1$ ? –  user974514 Apr 21 '12 at 2:18
$A = \mathbb F_2[x]$ is the ring of all polynomials with coefficients in the field of two elements. $I$ is the prinicpal ideal $(g) = g A$ of all multiples of $g = x^2+x+1$. Yes, those four classes form a complete system of reps of $A/I$, by the reasoning in the above answer. –  Bill Dubuque Apr 21 '12 at 2:29
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