Let $R$ be a ring. An ideal $I$ of $R$ is called nilpotent if $I^{n} = {0}$ for some positive integer $n$. I want to show that every nilpotent ideal is a nil ideal.
Please help me…. i can't start with the proof.
[Math] Every nilpotent ideal is a nil ideal.
abstract-algebraring-theory
Related Solutions
- 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.
$I$ is nil because every element is in an ideal generated by finitely many nilpotent elements. (Every polynomial in $\mathbb{Z}[x_1,x_2,x_3,...]$ uses only finitely many variables.)
$I$ is not nilpotent because the nilpotency index of $\bar x_n$ is $n+1$ and so there can't be a fixed $N$ such that $I^N=0$.
Best Answer
An ideal $J$ is called nil if every element $t \in J$ is nilpotent; that is, if every $ t \in J$ satisfies $t^m = 0$ for some $m \in \Bbb N$ (and here I take $0 \ne \Bbb N$). Recall the the definition of of $J_1 J_2$ for ideals $J_1$ and $J_2$ is
$J_1 J_2 = \{\sum j_{1i} j_{2i} \mid j_{1i} \in J_1; j_{2i} \in J_2 \}; \tag{1}$
here it is of course understood that that sums occurring in (1) are finite. It is easy to see from this definition that $j^2 \in J^2$ for $j \in J$, and if $j^k \in J^k$ for $k \in \Bbb N$ then clearly $j^{k + 1} = jj^k \in J^{k + 1}$; thus it follows by induction that $j^m \in J^m$ for all $m \in \Bbb N$. Now if $i \in I$ and $I^n = 0$ for some $n \in \Bbb N$, we have $i^n = 0$ as well since as has just been shown $i^n \in I^n$; every $i \in I$ is nilpotent; this shows that $I$ is, in fact, a nil ideal. QED.
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