Timeline for Freeness of modules along ring homomorphisms
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 23, 2011 at 17:50 | vote | accept | Hailong Dao | ||
| Mar 23, 2011 at 17:50 | answer | added | Hailong Dao | timeline score: 1 | |
| Mar 9, 2011 at 20:40 | comment | added | Hailong Dao | Dear Laurent, thank you for the hint. In hindsight this one probably should be asked on math.stackexchange.com! | |
| Mar 9, 2011 at 18:20 | comment | added | Hailong Dao | @Andrew: you are right. My comment about Pic(R) was a response to Laurent's suggestion. | |
| Mar 9, 2011 at 17:43 | comment | added | Andrew Parker | Indeed. I misread, but you still have a counterexample for $R=S^2$ and $S= R \otimes \mathbb{C}$ with your module being the tangent bundle on $S^2$, no? | |
| Mar 9, 2011 at 17:29 | comment | added | Hailong Dao | @Andrew, this is $S^1$. | |
| Mar 9, 2011 at 17:08 | comment | added | Andrew Parker | So, I guess $Pic(R)$ doesn't need to be non-trivial, then? (Since all line bundles over $S^2$ are free...) | |
| Mar 9, 2011 at 14:41 | comment | added | Laurent Moret-Bailly | Nice and simple! | |
| Mar 9, 2011 at 14:19 | comment | added | Hailong Dao | Actually, $R=\mathbb R[x,y]/(x^2+y^2-1)$, $S=R\otimes \mathbb C$ seems to work. | |
| Mar 9, 2011 at 14:02 | comment | added | Hailong Dao | Dear Laurent, that sounds promising! So we want $S$ to be $R$-free, $Pic(S)$ is trivial but $Pic(R)$ is not? | |
| Mar 9, 2011 at 13:47 | comment | added | Laurent Moret-Bailly | Your assumption on $S$ is satisfied in particular if $S$ is a free $R$-module. I wouldn't be surprised if you could find a counterexample in this case (e.g. with $R$ a ring of algebraic integers, and $M$ invertible?) | |
| Mar 9, 2011 at 13:32 | history | edited | Hailong Dao | CC BY-SA 2.5 |
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| Mar 9, 2011 at 13:27 | history | asked | Hailong Dao | CC BY-SA 2.5 |