# 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.

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This of course won't always work, but in your two cases it does! First note that $\mathbb{Z}/6\mathbb{Z}$ and $F[X]/(X^3-1)$ both only have finitely many elements. So you could list all subsets, and then just check if they are ideals. This would prove that you've found them all. (You could also try to streamline this a little by saying things like, if $\{n\}\subset \mathbb{Z}/6\mathbb{Z}$ and $n\neq 0$, then $2n\notin \{n\}$, so it isn't an ideal to take care of one element sets all at once). –  Matt Feb 21 '11 at 18:22
For small rings such as Z/6Z, you can simply use brute force to find them all: take the ideals generated by each element, by pairs of elements, etc. –  Harry Stern Feb 21 '11 at 18:30

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 $b$.

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Does it mean that F[X]/(x^3-1) = F[X]/(X^3) = F[X]/(X^3-X^2-X-1) Because all of them must be a set of all polynomials of degree less than 3? –  Sunny88 Feb 21 '11 at 18:40
@Sunny88: No! There are many different rings that have $8$ elements; the multiplication rules are different. –  Arturo Magidin Feb 21 '11 at 19:07
Thanks! I think I solved: nontrivial deals of Z/6Z are {0,3}, {0,2,4}. And nontrivial ideals of F[X]/(x^3-1) are {x+1, 0, x^2+1, 1, x^2+x}, {x^2+x+1, x^2, 1, 0}. –  Sunny88 Feb 22 '11 at 7:36

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.

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