## Is every finitely generated idempotent ring singly generated as a two-sided ideal?

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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 $RZR$.

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

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@Andreas: Never mind. The problem is nice and well written. Normal people still call a ring a ring. – Mark Sapir Dec 1 2011 at 16:21

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Todd, I intended to make clear that I will not assume that a unit is included. en.wikipedia.org/wiki/… Anyhow, I think the interesting part of the question starts with the third sentence. – Andreas Thom Dec 1 2011 at 16:34

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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 2011 at 19:51

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@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 2011 at 20:10

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@Mark, this was the problem my first few attempts ran into. But I couldn't prove it in general. – Benjamin Steinberg Dec 1 2011 at 20:29
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## 2 Answers

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 2011 at 13:45
Anyway, your example looks very interesting. – Andreas Thom Dec 1 2011 at 13:52

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If Agata cannot do it, nobody can. I have sent her the example. – Mark Sapir Dec 1 2011 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 2011 at 18:59

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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 2011 at 12:43
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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 2011 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 2011 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 2011 at 22:05

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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 2011 at 1:23
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