Example of nil and commutator ideal I have a problem with understanding a few definition. I can't find it anywhere.
Does the set of all nilpotent elements is called nil? And is it always ideal? Could you give me some example of nil? Is nil ideal diffrent from nilradical?
Could you also give me an example of commutator ideal? 
Could you recommend any books which elaborate on this subcject?
 A: *

*the set of nilpotent elements of a commutative ring form an ideal

*the set of nilpotent elements of a ring in general don't have to form an ideal (consider nilpotent elements in a full matrix ring, for example.)

*an ideal is called nil if each of its elements are nilpotent.

*an ideal $I$ is called nilpotent if there exists $n$ such that $I^n=\{0\}$. This is much stronger than being nil because it says in particular the product of any $n$ elements of $I$ is zero, not just the same element $n$ times. But of course, nilpotent ideals are nil.

*the radical of an ideal $I$ is the intersection of all prime ideals containing $I$. In particular, you can talk about the radical of the ideal $\{0\}$. You can find out why the radical of $\{0\}$ is the set of all nilpotent elements at this recent question.

*I'm not 100% sure what definition of "commutator ideal" you have in mind, but I'm guessing it's like this. Usually you define the commutator of $a,b\in R$ to be $ab-ba$, and maybe denote it with something like $[a,b]$. I think the commutator ideal of a ring would therefore be the ideal $\langle\{[a,b]\mid a,b\in R\}\rangle$. This ideal would be zero exactly when $R$ is commutative.

*By extension, you could restrict what commutators go into the ideal. So, for example, you could take two ideals $A,B$ of $R$ and look at $[A,B]=\{[a,b]\mid a\in A,b\in B\}$ and that would also probably be a commutator ideal. Again you can see that this ideal would be zero exactly when $A$ and $B$ centralize each other. With this definition you can also easily see that $[A,B]\subseteq A\cap B$.
Examples
Let $T=\{\begin{bmatrix}a&b\\0&c\end{bmatrix}\mid a,b,c\in \Bbb R\}$. The ideal $I=\{\begin{bmatrix}0&b\\0&0\end{bmatrix}\mid b\in \Bbb R\}$ is a nil (in fact, nilpotent) ideal. The ideals $J=\{\begin{bmatrix}0&b\\0&c\end{bmatrix}\mid a,b,c\in \Bbb R\}$ and $J=\{\begin{bmatrix}a&b\\0&0\end{bmatrix}\mid a,b,c\in \Bbb R\}$ are not nilpotent. Finally, $[J,K]=I$.
It also just happens that the nilradical (=intersection of prime ideals) is the set of all nilpotent elements for this particular noncommutative ring, but as mentioned before, if you take all the $2\times 2$ matrices you will for sure have nilpotent elements, and yet there are only two ideals: the zero ideal and the whole ring.
A: Let $R$ be your ring.
For an ideal $I$, define the radical of $I$ as $\operatorname{rad}\left(I\right):=\left\{x \in R\mid \exists n \in \Bbb N^*, x^n\in x\right\}$.

If your ring is commutative, then $\operatorname{rad}\left(I\right)$ is an ideal.
Proof:


*

*$\operatorname{rad}\left(I\right)$ is not empty since $I$ is not empty and $\forall x \in I, x^1\in I$ so $x \in \operatorname{rad}\left(I\right)$, that is $I\subseteq \operatorname{rad}\left(I\right)$

*Let $x,y\in \operatorname{rad}\left(I\right)$. You have $p\in \Bbb N^*$ so that $x^p\in I$ and $p\in \Bbb N^*$ so that $y^q\in I$. Since the ring is commutative, $(x+y)^{2p+2q}=\sum\limits_{k=0}^{2p+2q}{2p+2q\choose k}x^ky^{2p+2q-k}=\sum\limits_{k=0}^{p+q}{2p+2q\choose k}x^ky^{2p+2q-k}+\sum\limits_{k=p+q+1}^{2p+2q}{2p+2q\choose k}x^ky^{2p+2q-k}\\=y^q\sum\limits_{k=0}^{p+q}{2p+2q\choose k}x^ky^{2p+q-k}+x^p\sum\limits_{k=p+q+1}^{2p+2q}{2p+2q\choose k}x^{k-p}y^{2p+2q-k}\in I$ So $x+y \in \operatorname{rad}\left(I\right)$

*Let $x\in \operatorname{rad}\left(I\right)$ and $y\in R$. You have $n\in\Bbb N^*$ so that $x^n\in I$. Since the ring is commutative, $(xy)^n=x^ny^n\in I$ so $xy \in \operatorname{rad}\left(I\right)$ (and $yx\in\operatorname{rad}\left(I\right)$ since our ring is commutative)

The set of nilpotent elements of a ring is exactly $\operatorname{rad}\left(\{0\}\right)$, hence the name nilradical.
And nil ideal is simply an ideal whose elements are nilpotent which means that a nil radical is a radical contained in $\operatorname{rad}\left(\{0\}\right)$. In the case where $R$ is commutative, since $\operatorname{rad}\left(\{0\}\right)$ is an ideal and it is contained in itself, it is a nil radical. Another example would be $\{0\}$.
