Upper Nilradical of a Ring

Question

If we define the upper nilradical of a ring as the sum of all nil ideals of the ring, how could we deduce from just this definition that this is a nil ideal? Thanks!

Answer 1

If your nil ideals are two-sided proceed like this:

Let $I,J$ be two-sided nil ideals of a Ring $R$. We want to show that $I+J$ is also nil. To do this consider $R/J$ and $(I+J)/J$.

You can see rather easily that $(I+J)/J$ is a two-sided nil ideal of $R/J$: Let $[x] \in (I+J)/J$, then there exists a representative $a$ of $[x]$ that lies in $I$. Since then $a^n=0$ for some $n$, you get $[x]^n=[a]^n=[a^n]=0$ and $[x]$ is nilpotent. That $(I+J)/J$ is a two-sided ideal is a standard proof.

Let $x \in (I+J)$, since $(I+J)/J$ is nil, you have that $[x^n]=[x]^n=0$ for some $n$, but this implies $x^n \in J$. Since $J$ is nil you then have $(x^n)^m=0$ for some $m$, but this is the same as $x^{n+m}=0$ and $x$ is nilpotent.

So the sum of two (two-sided) nil ideals is again a (two-sided) nil ideal. Since elements of $\oplus_{\alpha \in A} I_\alpha$ are finite sums of the elements of $I_\alpha$ this statement shows inductively that the elements of any sum of two-sided nil ideals are nilpotent. So any sum of two-sided nil ideals is again a two-sided nil ideal.

If the nil ideals are not necessarily two-sided, then it is an open problem whether or not the sum of two ideals is again nil, see https://en.wikipedia.org/wiki/Köthe_conjecture

The proof I have written down is basically Lemma 1 that can be found here but with more details.