Timeline for Is every finitely generated idempotent ring singly generated as a two-sided ideal?
Current License: CC BY-SA 3.0
20 events
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| Dec 10, 2011 at 19:55 | history | edited | Andreas Thom | CC BY-SA 3.0 |
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| Dec 3, 2011 at 9:30 | comment | added | Andreas Thom | Benjamin, I do not know. That is precisely the problem. | |
| Dec 3, 2011 at 0:39 | comment | added | Benjamin Steinberg | @andreas, a priori how does bounding the number of generators as a left ideal control the number of generators as 2-sided ideal? | |
| Dec 2, 2011 at 13:39 | comment | added | Andreas Thom | Benjamin, the augmentation ideal of a group ring, where the group is perfect, finitely generated and has positive first $\ell^2$-Betti number. With homological invariants, such as $\ell^2$-Betti numbers, one can bound from below the number of generators as a left ideal. | |
| Dec 2, 2011 at 12:33 | comment | added | Benjamin Steinberg | @Andreas, do you know an example of an idempotent ring which is finitely generated as an ideal but not principal? | |
| Dec 2, 2011 at 5:58 | history | edited | Andreas Thom | CC BY-SA 3.0 |
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| Dec 1, 2011 at 20:29 | comment | added | Benjamin Steinberg | @Mark, this was the problem my first few attempts ran into. But I couldn't prove it in general. | |
| Dec 1, 2011 at 20:10 | comment | added | user6976 | @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 $K$. | |
| Dec 1, 2011 at 19:51 | comment | added | Benjamin Steinberg | 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. | |
| Dec 1, 2011 at 19:21 | comment | added | user6976 | @Andreas: Perhaps if Ozawa's proof is not too long, you can post it here? | |
| Dec 1, 2011 at 19:12 | answer | added | kassabov | timeline score: 0 | |
| Dec 1, 2011 at 16:34 | comment | added | Andreas Thom | 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. | |
| Dec 1, 2011 at 16:21 | comment | added | user6976 | @Andreas: Never mind. The problem is nice and well written. Normal people still call a ring a ring. | |
| Dec 1, 2011 at 15:48 | comment | added | Todd Trimble | 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.) | |
| Dec 1, 2011 at 14:38 | comment | added | Andreas Thom | Sorry, this was not intended. I changed it. | |
| Dec 1, 2011 at 14:38 | history | edited | Andreas Thom | CC BY-SA 3.0 |
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| Dec 1, 2011 at 14:18 | comment | added | Todd Trimble | I find the first two sentences subjective and argumentative. | |
| Dec 1, 2011 at 13:03 | answer | added | user6976 | timeline score: 4 | |
| Dec 1, 2011 at 11:11 | history | edited | Andreas Thom | CC BY-SA 3.0 |
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| Dec 1, 2011 at 11:05 | history | asked | Andreas Thom | CC BY-SA 3.0 |