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
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 21, 2013 at 12:32 | vote | accept | manoj | ||
| Feb 21, 2013 at 12:26 | vote | accept | manoj | ||
| Feb 21, 2013 at 12:32 | |||||
| Feb 21, 2013 at 11:47 | answer | added | user26857 | timeline score: 2 | |
| Feb 18, 2013 at 6:39 | comment | added | Fred Rohrer | Dear @Qing, thank you for your explanation. | |
| Feb 17, 2013 at 17:55 | comment | added | Qing Liu | Fred: one can descend the module to a f.g. module $M_n$ over some $k[x_1,..,x_n]$. As $k[x_1,..,x_n,...]=k[x_1,...,x_n][x_{n+1},...]$ is faithfully flat over $ k[x_1,..,x_n]$, $M_n$ is flat hence free. So the answer is yes (any number of indeterminates). | |
| Feb 16, 2013 at 15:17 | comment | added | user26857 | This question is also posted on MSE: math.stackexchange.com/questions/296109/… | |
| Feb 16, 2013 at 8:18 | comment | added | Fred Rohrer | Is it known that projective modules of finite type over polynomial algebras in countably many indeterminates over fields are free? | |
| Feb 16, 2013 at 6:31 | answer | added | manoj | timeline score: 0 | |
| Feb 15, 2013 at 14:46 | comment | added | user30379 | Fred's comment is appropriate, as Neil notes. More specifically, an inclusion between rings doesn't necessarily extend to a ring homomorphism between total quotient rings, since a nonzero element of a subring that isn't a zero-divisor there may be a zero-divisor in a bigger ring. For example, the inclusion $k[x] \subset k[x,y]/(xy)$ doesn't extend to the total quotient rings as a ring homomorphism (because the composite map $k[x] \hookrightarrow k[x,y]/(xy) \twoheadrightarrow k[y]$ kills $x$ and thus doesn't extend to a map of rings $k(x) \rightarrow k(y)$). | |
| Feb 15, 2013 at 14:40 | comment | added | Neil Epstein | Sure, but that still doesn't work. That is, if $A$ is non-Noetherian and its own total quotient ring, it does not follow that a finitely generated subring is also its own total quotient ring. | |
| Feb 15, 2013 at 13:35 | comment | added | Martin Brandenburg | The total quotient ring of a reduced noetherian ring is a finite product of fields, therefore with trivial class group. | |
| Feb 15, 2013 at 13:07 | comment | added | Fred Rohrer | @Martin: Why does Tom's comment answer the question? | |
| Feb 15, 2013 at 13:05 | comment | added | Martin Brandenburg | Why are still comment boxes used for (often excellent) answers? | |
| Feb 15, 2013 at 12:54 | comment | added | Tom Goodwillie | If you had a counterexample for a given ring $A$, you could describe the module and demonstrate its projectiveness using only finitely many elements of $A$. This would lead to a counterexample for a finitely generated, therefore Noetherian, subring. | |
| Feb 15, 2013 at 11:52 | history | asked | manoj | CC BY-SA 3.0 |