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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?

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The examples are clear enough to me, and perhaps I can say something about them, but the stuff in between I find very difficult to parse. Also, what you call "side classes" are usually called "cosets". – Dylan Moreland Feb 2 '12 at 20:44
The notation $(\mathbb{F}_2/x^2\mathbb{F}_2)[x]$ is meaningless. You mean $\mathbb{F}_2[x]/x^2\mathbb{F}_2[x]$, or $\mathbb{F}_2[x]/(x^2)$. – anon Feb 2 '12 at 20:49
up vote 2 down vote accepted

"Example 1" seems to be three examples: (i) $R/I$ with $R=\mathbb{Z}[x]$, $I=nR$; (ii) $R/I$ with $I=xR$; and (iii) $R/I$ with $I=2R+xR$.

In the third case, $I$ consists of all polynomials with integer coefficients that have even constant term: all such polynomials can be written as a multiple of $2$ plus a multiple of $x$; and conversely, if a polynomial is a multiple of $2$ plus a multiple of $x$, then its constant term must be even.

When are $a_0 + a_1x+\cdots+a_nx^n$ and $b_0+b_1x+\cdots+b_mx^m$ in the same lateral class modulo this $I$? If and only if $$(a_0+\cdots + a_nx^n) - (b_0 + \cdots + b_mx^m) = (a_0-b_0) + (a_1-b_1)x + \cdots \in I.$$ In order for the difference to be in $I$ what do we need? We need $a_0-b_0$ to be even; so the two polynomials are in the same coset if $a_0$ and $b_0$ have the same parity. So there are at most two lateral classes: one for polynomials with even constant term, one for polynomials with odd constant term.

(Conversely, polynomials with constant terms of opposite parity are not in the same lateral class, so there are in fact exactly two lateral classes).

In Example 2, you have $R=F_2[x]$ and $I=x^2R$. A polynomial is in $I$ if and only if it is a multiple of $x^2$. If $a_0+a_1x+\cdots + a_nx^n$ and $b_0+b_1x+\cdots+b_mx^m$ have $a_0=b_0$ and $a_1=b_1$, then their difference is a multiple of $x^2$; since there are two possible choices for $a_0$ and two possible choices for $a_1$, there are at most four possible lateral classes: one for each choice combination. Because every polynomial will have constant and linear terms equal to $0$, both equal to $1$, constant term $0$ and linear term $1$, or constant term $1$ and linear term $0$. So there are at most 4 possibilities that yield different lateral classes.

(Conversely, those four possibilities are pairwise distinct, since the difference of, say, a poynomial with constant term 1 and linear term 0, and one of linear and constant terms equal to $1$, would not be a multiple of $x^2$).

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Great explanation, thanks Arturo Magidin. – VVV Feb 2 '12 at 21:13

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.

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Thank you, anon. – VVV Feb 2 '12 at 21:12

Two great answers here already. Here are two extra thoughts.

If I have a ring $A$ and two ideals $\mathfrak a$ and $\mathfrak b$, then I can factor out by $\mathfrak a + \mathfrak b$ in steps. Indeed, the image of $\mathfrak b$ in the factor ring $A/\mathfrak a$ is the ideal $(\mathfrak b + \mathfrak a)/\mathfrak a$, and we have an isomorphism $$(A/\mathfrak a)/(\mathfrak b + \mathfrak a) \approx A/(\mathfrak a + \mathfrak b).$$ So at the end of your first example we can set $A = \mathbf Z[x]$, $\mathfrak a = 2A$, and $\mathfrak b = xA$ (see what happens when you switch these). Under the isomorphism $A/\mathfrak a \to \mathbf F_2[x]$, the ideals $\mathfrak b$ and $(\mathfrak b + \mathfrak a)/\mathfrak a$ correspond to the ideal $x\mathbf F_2[x]$. Modding out by that ideal in $\mathbf F_2[x]$ yields $\mathbf F_2$.

It's easiest to understand $A/\mathfrak a$ if you can write down a surjective homomorphism $A \to B$ having $\mathfrak a$ as its kernel, so that $A/\mathfrak a$ is isomorphic to something that you might understand better. At the beginning of your first example, you have a map $\mathbf Z[x] \to (\mathbf Z/n\mathbf Z)[x]$ obtained by reducing the coefficients of polynomials.

On the other hand, it isn't always clear whether you can find a helpful $B$. A good psychological device is to think of $A/\mathfrak a$ as being just like $A$ but for the fact that the elements of $\mathfrak a$ are equal to zero. So in your last example, an element of $\mathbf F_2[x]$ looks like $$ a_0 + a_1x + a_2x^2 + \cdots + a_nx^n $$ where the $a_i$ are $\bar0$ or $\bar1$. But you've declared that $x^2$ and its multiples are zero, so in the quotient it's the same to write $$ a_0 + a_1x. $$ As your notes say, this tells us that there are at most $4$ cosets, and it isn't so hard to check that the four cosets you can write down in this way are all distinct.

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