I am doing some Noncommutative Algebra and I am studying the notion of the nilradical of a (not necessarily) commutative ring. More precisely:
Let $R$ be a ring with $1_R$. The nilradical $N(R)\subseteq R$ of $R$ is the sum of all nilpotent ideals of $R$. That is,
$$\boxed{N(R):=\sum\{ I\trianglelefteq R:\ I \text{ is a nilpotent ideal of } R\}\subseteq R}.$$
So, I have some questions:
- I searched some books and on the internet but I didn't find this definition. All the sources said that it is the set consisting of all nilpotent elements of $R$. Is there a reason for using this? And is there a kind of equivalence between them?
- Does this definition mean that $N(R)$ is equal to $I_1+I_2+\dotsb+I_n+\dotsb \trianglelefteq R$, where $I_1,I_2,\dots,I_n,\dots\trianglelefteq R$ are all (possibly infinite) nilpotent two-sided ideals of $R$?
- If the above are correct, we realise that $N(R)$ is a two-sided ideal of $R$. Now, in my notes,
$$N(R) \subseteq \sum\{ I \underset{\ell}{\trianglelefteq} R:\ I \text{ is a nilpotent left ideal of } R\}.$$
But I can not see why this happens.
Thanks.
Best Answer
The upper nilradical $Nil^\ast(R)$ is defined to be the sum of nil ideals.
The lower nilradical $Nil_\ast(R)$ is defined to be the intersection of prime ideals of the ring.
In the context of Artinian rings, nil radicals are nilpotent, so the sum of all nilpotent ideals is again a nilpotent ideal, and this is sometimes referred to as the Wedderburn radical. But I'm not sure if anyone uses the sum of all nilpotent ideals outside this context.
We have $N(R)\subseteq Nil_\ast(R)\subseteq Nil^\ast(R)$, and I don't think the first two are always equal.
No, it is not the set of nilpotent elements. Take $M_2(F)$ for any field $F$, and $N(R)$ is the zero ideal, yet there are nilpotent elements.
Yes.
Every nilpotent ideal is a nilpotent left ideal. So the set of nilpotent ideals is a subset of the set of nilpotent left ideals. Therefore the sum of the former is contained in the sum of the latter. There is nothing hard here.