Sum of nil right ideals

Question

Is an arbitrary sum of nil (nilpotent) right ideals a nil (nilpotent) right ideal?

If $I=\sum I_i$ is a sum of nil ideals then each element $x$ of $I$ is a finite sum $x=x_1+...+x_n$ of elements $x_k\in I_k$ a power ${x_k}^{p_k}$of each of which is zero. But, since each power of $x$ is a finite sum constituting of summands of the form $∏{x_i}^{t_i} $( of course, in this product the $x_i$'s may be repeated with different powers), I guess that one could choose the power of $x$ so "big" that it be annihilated.

Is my guess true? How?

Thanks!

Answer 1

Is the arbitrary sum of nilpotent right ideals nilpotent?

The counterexample writes itself: Take $R=F[x_1,x_2,\ldots]/(\{x_n^n\mid n\in \Bbb N\})$ for a field $F$. While the individual ideals $(x_i)$ are nilpotent, their sum is merely a nil ideal.

Is the arbitrary sum of nil right ideals nil?

This is not even known for finite sums. The statement "the sum of any two nil right ideals is a nil right ideal" is one of the equivalent formulations of the Koethe conjecture which is still unsolved.