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?
Best Answer
What you call "side classes" I'm used to hearing called cosets. Anyway, in the first example, we have the ideal $I=2R+xR$. Every element in the quotient ring (it is also called a "factor ring") is of the form $a+I$ with $a\in R$. If $a$ and $b$ differ by an element of $I$, then $a+I$ and $b+I$ represent the same coset. We may say that $a=a_0+a_1x+\cdots$, in which case this is $a_0+(a_1+\cdots)x+I$ and thus equal to $a_0+I$, because $a$ and $a_0$ differ by a multiple of $x$, and multiples of $x$ are in $I$. We no longer have any need to consider $a\in R$ anymore, but rather just $a\in\mathbb{Z}$. Furthermore, if $a$ and $b$ are integers that differ by a multiple of two, then $a+I$ and $b+I$ are the same coset because $2\in I$.
Finally, every integer is either a multiple of two different from $0$ (the integer is even) or from $1$ (the integer is odd), so all cosets are of one of the forms $\{0+I,1+I\}$. Since $0\ne1$ in the ring and $1\not\in I$, we know these represent distinct cosets.
In the case of $\mathbb{F}_2[x]/(x^2)$, if we have a polynomial $p=a_0+a_1x+a_2x^2+\cdots$, then this polynomial differs from $a_0+a_1x$ by a multiple of $x^2$ (namely $(a_2+\cdots)x^2)$, so $p+I$ and $a_0+a_1x+I$ refer to the same coset. The only values $a_0$ and $a_1$ can take are $0$ or $1$ so we're looking at the only four possible representatives being $0+0x,0+1x,1+0x,1+1x$. None of these or their differences are in the ideal $I$ (except $0$), so they are all distinct, and hence exhaust all ideals in this quotient ring.