In this post, a *ring* is understood to be what one usually calls a ring, not assuming that it has a unit. Some people call such objects *rng*.

Question:Let R be a finitely generated (non-unital and associative) ring, such that $R=R^2$, i.e. the multiplication map $R \otimes R \to R$ is surjective (every element is a sum of products of other elements). Is it possible that every element of $R$ is contained in a proper two-sided ideal of $R$? Or, must it be the case that $R$ is singly generated as a two-sided ideal in itself?

Note, if $Z \subset R$, then the ideal generated by $Z$ is the span of $Z \cup RZ \cup ZR \cup RZR$, which in the case of idempotent rings is equal to the span of $RZR$.

More generally, one can ask:

Question:For a fixed natural number $k$, can it happen that every set of $k$ elements of $R$ generates a proper ideal of $R$?

So far, I do not know of any example where the ring $R$ is not generated by a single element as a two-sided ideal in itself. I first thought that it must be easy to find counterexamples, but I learned from Narutaka Ozawa that the free non-unital ring on a finite number of idempotents is singly generated as a two-sided ideal in itself. He also showed that no finite ring can give an interesting example. The commutative case is also well-known; Kaplansky showed that every finitely generated commutative idempotent ring must have a unit.

Update: Some partial results about this question and a relation to the Wiegold problem in group theory can be found in http://arxiv.org/abs/1112.1802