Timeline for An example of a rank one projective R-Module that is not invertible
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 5, 2011 at 17:40 | comment | added | Andrew Parker | @Hailong: The rough definition for invertible that I was considering was essentially just local freeness: $M$ where $M \otimes R_P \cong R_P$ for all primes $P$. @Tom Goodville gave an interesting example that fit my 'definition' for a non-finitely generated invertible module, but was not projective. I also see what 'fails' for that example, in that there is no element $r \in R$ such that $rM \subset R$, unlike the finitely generated case. | |
| Jul 5, 2011 at 17:17 | vote | accept | Andrew Parker | ||
| Jul 3, 2011 at 21:54 | comment | added | Karl Schwede | Georges, yes, I misread the question. Sorry about that. | |
| Jul 3, 2011 at 20:43 | answer | added | YCor | timeline score: 10 | |
| Jul 3, 2011 at 18:41 | comment | added | Hailong Dao | @S\andor: you are right, I meant to say, if $M$ is not f.g. (as the OP wanted), then it is trivially not invertible. So I am not sure what the question asked. Thank you for the correction. | |
| Jul 3, 2011 at 17:32 | comment | added | Sándor Kovács | @Hailong: the actual question does not ask $M$ to be invertible, in fact it asks that it is not that, so it seems that that could not force anything. But it does ask that $M$ be of rank $1$ and perhaps this is what you have in mind. So my question is the same as Martin's: @Andrew: How do you define rank? | |
| Jul 3, 2011 at 15:51 | comment | added | Hailong Dao | Andrew, what definition of "invertible" you are using? The ones I know would force $M$ to be finitely generated automatically, so the answer is trivially no. | |
| Jul 3, 2011 at 8:23 | comment | added | Georges Elencwajg | Dear Karl, I am not able to extract an answer to Andrew's question from your link. Could you please state a precise statement there that answers his question? | |
| Jul 2, 2011 at 19:40 | comment | added | Martin Brandenburg | How do you define the rank in this case? Dimension of the residue fields? Of course your alluded generalized Picard group will be no group. | |
| Jul 2, 2011 at 18:47 | history | asked | Andrew Parker | CC BY-SA 3.0 |