A basic question about finding ideals of rings and proving that these are all the ideals I am a student taking a "discrete maths" course. Teacher seems to jump from one subject to another rapidly and this time he covered ring theory, Z/nZ, and polynomial rings.
It is hard for me to understand anything in his class, and so the reports he gives become very hard.
I did my best to find answers using google, but I just couldn't find it.
Specifically he asked us to find all ideals of Z/6Z, and prove that these are in fact all of them.
He also asked us to find all ideals of F[X]/(X^3-1) where F stands for Z/2Z.
I understand the idea behind ideals, like I can see why {0,3} is ideal of Z/6Z, but how do I find ALL the ideals?
And regarding polynomials, is there some kind of a mapping between polynomials and Z/nZ?
Because otherwise I have no idea how to find ideals of polynomials.
 A: Since $\mathbb{Z}/6\mathbb{Z}$ is finite, it is not difficult to try to find all ideals: you've got $\{0\}$ and you've got $\mathbb{Z}/6\mathbb{Z}$. Suppose the ideal contains $a\neq 0$. Then it must also contain $a+a$, $a+a+a$, and so on. Check the possibilities. 
No, there generally is no mapping between $F[X]/(p(x))$ and a $\mathbb{Z}/n\mathbb{Z}$. But first, notice that by doing long division, every polynomial $p(x)$ in $F[X]$ can be written as $p(x) = q(x)(x^3-1) + r(x)$, where $r(x)=0$ or else $\deg(r)\lt 3$. That means that every element of $F[x]/(x^3-1)$ corresponds to one of the "remainders", and there are only $8$ possible remainders (the remainder must be of the form $a+bx+cx^2$, with $a,b,c\in\mathbb{Z}/2\mathbb{Z}$), so again $F[x]/(x^3-1)$ is finite, and you can check the possibilities. Here adding an element to itself is not going to help much (because $p+p=0$ for all $p$) but you can instead consider a given $p(x)$ and all $8$ multiples of it that you get when you multiply by elements of $F[x]$.
Alternatively, the ideals of $R/I$ correspond to ideals of $R$ that contain $I$. So the ideals of $\mathbb{Z}/6\mathbb{Z}$ correspond to ideals of $\mathbb{Z}$ that contain $6\mathbb{Z}$, and ideals of $F[X]/(x^3-1)$ correspond to ideals of $F[x]$ that contain $(x^3-1)$. Notice that $(a)$ contains $(b)$ if and only if $a$ divides $b$. 
A: They're both products of fields: $\mathbb{Z}/6\cong\mathbb{Z}/2\times\mathbb{Z}/3$ and $F[x]/(x^3-1)\cong GF(2)\times GF(2^2)$ (if you've covered this kind of thing) which makes it easy to see the idea structure.
