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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

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Thanks, and very sorry to nitpick again, but what one "usually" calls a ring is still subject to question. For many people nowadays, it is usual to suppose that an identity is included when one says "ring". (This is just FYI, not at all a call for another edit.) – Todd Trimble Dec 1 '11 at 15:48
@Andreas: Never mind. The problem is nice and well written. Normal people still call a ring a ring. – Mark Sapir Dec 1 '11 at 16:21
Todd, I intended to make clear that I will not assume that a unit is included.… Anyhow, I think the interesting part of the question starts with the third sentence. – Andreas Thom Dec 1 '11 at 16:34
Is it possible that the algebra of a finitely generated semigroup $S$ satisfying $S^2=S$ could work (or can you prove that such an example never works? I can prove that inverse semigroups don't work. – Benjamin Steinberg Dec 1 '11 at 19:51
@Ben: I tried. It is not easy, and may be not possible. In fact, take one representative $a_i$ of each maximal $J$-class of $S$. Then $\sum a_i$ seem to generate $KS$ for every field $K$. – Mark Sapir Dec 1 '11 at 20:10

Consider the ring generated by $a,b,c,d,e$ subject to the relation $a=bc+de$ and all its cyclic shifts: $b=cd+ea$, and 3 more. It is "idempotent" obviously. Can it be killed by one relation? I will ask Agata Smoktunowicz. She should be able to figure it out quickly.

Update Agata responded saying that the problem, while interesting, is too difficult. She did try using Groebner-Shirshov bases but without success. She did manage to prove the statement for semigroup algebras using an argument similar to Ozawa's (as Ben Steinberg asked here). If $S$ is a semigroup, $S^2=S$, then $KS$ is generated as an ideal by one element.

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Thanks. I asked Agata already, and she could not come up with an example. – Andreas Thom Dec 1 '11 at 13:45
Anyway, your example looks very interesting. – Andreas Thom Dec 1 '11 at 13:52
If Agata cannot do it, nobody can. I have sent her the example. – Mark Sapir Dec 1 '11 at 15:29
Mark, thanks. Does K have to be a field or does it work over the integral semigroup ring? I assume the latter. – Benjamin Steinberg Dec 2 '11 at 18:59
George Bergman has shown me a proof which works over every unital commutative ring; even commutativity is not necessary I think. – Andreas Thom Dec 7 '11 at 12:43

What about the ring generated by $a,b,c$ subject to the relations $a^2=a$, $b^2=b$ and $c^2=c$? If this ring is generated as an ideal by a single element $r$ one can show that $r= \pm a \pm b \pm c + h$ where $h$ is of order at least $2$ (in the ideal generated by $ab, ba, ac, \dots$). I can not inagine how this can happen.

If this works then one can add extra generators which is likely to answer also the second question.

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@Martin: Andreas wrote in his question that for rings generated by idempotents the statement is true. By the way, do you know how to deal with my example? – Mark Sapir Dec 1 '11 at 19:17
@Mark: Andreas has a very cute proof that this will not work. Since for finite rings everything is OK, one need to build some elaborate obstruction -- the only thing which comes to my mind is something build out of ideal class groups, but I do not see how to encode this into a ring.... Your example "lacks" any structure and I have no idea how even to start. – kassabov Dec 1 '11 at 21:56
@Martin: I do not think structure can help here at all. Some manipulation with words. That is why I asked Agata. – Mark Sapir Dec 1 '11 at 22:05
To elaborate on my last comment: to study a f.g. ring $R$ one usually takes its Jacobson radical $J$ and study the semi-simple part $R/J$ where the density theorem gives structure. But for semi-simple rings, I think, the statement is true, so the most interesting case is when $R=J$. I think, in particular, that a nil-ring can be an example. If $R=J$, there is no structure theory as far as I know, and the only way to treat such rings is by studying generators and relations (Groebner bases, etc.). – Mark Sapir Dec 2 '11 at 1:23

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