Timeline for Are f.g. projective modules free over total quotient ring of a reduced non-noetherian commutative ring
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 19, 2013 at 9:58 | comment | added | manoj | Thanks Laurent. I did not realize that $A$ is local. Further, $A=C$, since any element of $B-(x_1,...)$ has non-zero constant and is not a zerodivisor in $B$, hence a unit in $C$. | |
| Feb 16, 2013 at 18:11 | comment | added | Neil Epstein | Well, I guess one could always look at $B=R/I$ and then let $C$ be the total ring of quotients of $B$. Does $C=A$? I kinda think so, but if not, then there's no reason to think $C$ is local, and it could theoretically be a useful example in this context. However, in general it can be hard to find non-free f.g. projective modules (e.g. Serre's conjecture=Quillen-Suslin theorem). | |
| Feb 16, 2013 at 15:19 | comment | added | user26857 | manoj's example comes from that of QiL given here: math.stackexchange.com/questions/294384/… | |
| Feb 16, 2013 at 15:07 | comment | added | Laurent Moret-Bailly | This ring is local, so projective=free. | |
| Feb 16, 2013 at 6:36 | history | edited | manoj | CC BY-SA 3.0 |
added 24 characters in body
|
| Feb 16, 2013 at 6:31 | history | answered | manoj | CC BY-SA 3.0 |