Defintion 1 An ideal I of R is an additive subgroup so that $a\in R , s\in I \Rightarrow as,sa \in I$
The ring R/I is called the quotient ring.
Example 1 : $R=\mathbb{Z}[x], I=nR (n\in \mathbb{Z})$. Then R/I "=" (\mathbb{Z}/n\mathbb{Z})[x]. Or $I=xR$. What are the $A+I (A\in \mathbb{Z}[x])?$
A=a+bx+... , so they are equal to the $a+I (a\in \mathbb{Z})$ because if $A=a+bx+...$, then $A-a = bx+... \in I$. So R/I "=" \mathbb{Z} or : $I=2R+xR = \{ 2A+xB; A,B \in R\}$. So we can say that there are at most $2$ different side classes $0+I, 1+I$ and they can not be equal because otherwise $1\in I$, so $1=2A+xB; A,B\in \mathbb{Z}[x]$. It is true that $R/I$ is identical to $\mathbb{F}_{2}=\{0,1\}$
Example 2: R=$\mathbb{F}_{2}[x]$, I=xR, R/I = \{0+I, 1+I\} "=" \mathbb{F}_{2}. Or $I=x^{2}R, A+I = (a+bx+...)+I = (a+bx)+I (a,b \in \mathbb{F}_{2})$. So there are at most 4 side classes : $0+I,1+I,x+I,x+1+I$ and they are all different from eachother.
This is an outtake from my script. I do not understand how in example 1 it is concluded that there are at most 2 different side classes and 4 side classes with $(\mathbb{F}_{2}/x^{2}\mathbb{F}_{2})[x]$ in the example 2. How does one find the side classes? Are they elements of the ring? It seems to be easy to find them for simple rings (the elements of $\mathbb{F}_{5}$ are the equivalence classes 0,1,2,3,4) , but how to find them (the elements) for quotient rings?