Prove that the sum of all the nil right ideals of $R$ is a nil 2-sided ideal.
I need help showing that that the sum of all nil right ideals is is a nil ideal. I've shown it is an ideal already:
$proof:$
First note that if $I$ is a right nil ideal of $R$ and $r \in R$ then $rI$ is also a right nil ideal. This is because of $(xr)^n=0$ then $(rx)^{n+1}=0$
Let $S$ be the sum of all nil right ideals of R and let $y \in S$.
Then $y = y_1 + y_2 + ... +y_k$ where $y_i \in I_i$ where $I_i$ is a right ideal of $R$.
Then $ry = r(y_1 + y_2 + ... +y_k)$ and $ry_i \in rI_i$.
And $S$ is obviously a right ideal.
Thus $S$ is an ideal of $R$. Showing that it is nil is proving to be a little harder, maybe I'll need a generalized binomial formula? Can somebody help walk me through this?
Thanks!
Note: This answer only shows that the sum of nil ideals is nil in a commutative ring.
As pointed out by rschwieb, that question is open for non-commutative rings. I'll leave this here in case anyone finds the argument helpful in proving the commutative case.
We can avoid having to use the multinomial theorem, which would be messy. Elements $s\in S$ are of the form $s=x_1+\cdots+x_n$ for some $n\geq1$ and $x_i\in I_i$, where $I_i$ is a nil ideal. Then we can use induction on $n$ to show that each $s$ is nilpotent.
The result is obvious for $n=1$. If we have $s=x_1+\cdots+x_n$, then $r=x_1+\cdots+x_{n-1}$ is nilpotent by the inductive hypothesis, say $r^a=0$ for some $a\geq1$. We know that $x_n^b=0$ for some $b\geq1$. Then by the usual binomial theorem we have $$s^{a+b}=(r+x_n)^{a+b}=\sum_{k=0}^{a+b}\binom{a+b}{k}r^{a+b-k}x_n^k$$ From this, we can see in the sum that as soon as we have fewer than $a$ lots of $r$ we will have at least $b$ lots of $x_n$, and so the sum will be $0$. Then $s^{a+b}=0$, so $s$ is nilpotent and we are done by induction.