Timeline for What are the units in the ring of Laurent polynomials?
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 12, 2010 at 10:38 | vote | accept | Seb | ||
| Jul 28, 2010 at 7:25 | |||||
| Jul 8, 2010 at 22:48 | comment | added | Jose Brox | Thanks for the reference to that article, it may be useful for me! (btw, I think you should accept your own answer!) | |
| Jul 8, 2010 at 15:36 | comment | added | Seb | Um, thank you very much for the explanation! I have to admit I did not understand how to interpret your answer before. | |
| Jul 8, 2010 at 15:23 | comment | added | Boyarsky | That's exactly a concrete description of my comment/answer even in the general case (no connectedness or reducedness hypotheses): those idempotents correspond to a finite decomposition of $R$ into factor rings, and your unit in the $j$th factor ring of $R[X,1/X]$ is the product of the $j$th component of $r$ times $X^{i_j}$ times something which is 1 mod nilpotents. I think my proof in terms of fibering over Spec($R$) is elegant enough. :) | |
| Jul 8, 2010 at 15:04 | history | answered | Seb | CC BY-SA 2.5 |