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 $(1 + xy + \dotsb + (xy)^{k-1})(1-xy)=1$.

That is totally correct. 
If you require further validation, then check out Lam's First course in noncommutative rings pg 53, lemma 4.11 which has exactly the generalization you describe, with the same proof.
