Timeline for Is every finitely generated idempotent ring singly generated as a two-sided ideal?
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5 events
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| Dec 2, 2011 at 1:23 | comment | added | user6976 | 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.). | |
| Dec 1, 2011 at 22:05 | comment | added | user6976 | @Martin: I do not think structure can help here at all. Some manipulation with words. That is why I asked Agata. | |
| Dec 1, 2011 at 21:56 | comment | added | kassabov | @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. | |
| Dec 1, 2011 at 19:17 | comment | added | user6976 | @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? | |
| Dec 1, 2011 at 19:12 | history | answered | kassabov | CC BY-SA 3.0 |