# One-sided nilpotent ideal not in the Jacobson radical?

Problem XVII.5a of Lang's Algebra, revised 3rd edition, is:

Suppose $N$ is a two-sided nilpotent ideal of a ring $R$. Show that $N$ is contained in the Jacobson radical $J: = \{ \cap\, I: I \text{ a maximal left ideal of } R \}$.

I put my solution below the fold. My question is: can't we generalize a bit more? It seems that all we need is that $N$ is a nil ideal; further, I don't see why $N$ can't be merely a one-sided ideal. I assume there's some error in my thinking here...

Solution: Take $y \in N$, and show that $1-xy$ has a left inverse for all $x\in R$ (this is an equivalent characterization of the Jacobson radical, see here). The way to construct the left inverse is to note that $xy \in N$, so $\exists k$ s.t. $(xy)^k= 0$, so $(1 + xy + \dotsb + (xy)^{k-1})(1-xy)=1$.