Timeline for locally isomorphic modules
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Feb 16, 2018 at 8:50 | comment | added | Watson | Exercise 4.13 in Eisenbud's Commutative Algebra with a View Towards Algebraic Geometry states that over a semi-local ring, we can test two finitely-presented modules to be isomorphic by testing them to be isomorphic locally (without any global homomorphism given!). | |
| Aug 10, 2013 at 16:30 | history | reopened |
Steven Landsburg Andrey Rekalo Karl Schwede Ramiro de la Vega Yemon Choi |
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| Aug 10, 2013 at 13:09 | review | Reopen votes | |||
| Aug 10, 2013 at 16:33 | |||||
| Jan 15, 2013 at 21:46 | comment | added | user30035 | This question is I think very poor. A special case of it is "are all projective modules free?" and the answer even to that special case is "go to any commutative algebra class, or read any book on commutative algebra, and get to the point where projective modules are defined, and then read the next two lines, where it is pointed out that there exist projective modules that aren't free". I learnt this in an undergraduate manifolds class. Just because there are 10 people out there that don't know this, doesn't make it a good question...does it?? | |
| Jan 15, 2013 at 16:29 | comment | added | André Henriques | The question got 11 up votes. The accepted answer to 10 up votes. Clearly there's both people who find the question and the answer interesting. What is the benefit of closing such a question? Voting to reopen. | |
| Dec 28, 2012 at 15:44 | comment | added | Fernando Muro | @András: may I ask why? | |
| Dec 27, 2012 at 21:45 | comment | added | András Bátkai | Please, those who closed this should comment on the meta-thread. | |
| Dec 27, 2012 at 21:32 | vote | accept | CommunityBot | ||
| Dec 27, 2012 at 21:16 | comment | added | Will Sawin | I think the simplest explicit example is ring $k[x,y]/(x^2-y^3-1)$. The ideal $(x-1,y)$ is not principal, so as a module it is not isomorphic to the trivial module. But at every point other than $x=1,y=0$, that ideal is trivial, so isomorphic to the trivial module, and at $x=1,y=0$, that ideal is the maximal ideal of a a PID, thus isomorphic to the trivial module. | |
| Dec 27, 2012 at 21:14 | history | closed |
Fernando Muro Martin Brandenburg Eric Wofsey Andreas Blass Will Sawin |
too localized | |
| Dec 27, 2012 at 21:05 | answer | added | Steven Landsburg | timeline score: 6 | |
| Dec 27, 2012 at 18:00 | comment | added | user30230 | Thanks to Fernando Muro, it seems that my question has negative answer! I am not familiar with line bundles. Can you explain more or provide a more accessible (for me) example !? | |
| Dec 27, 2012 at 17:41 | history | edited | user30230 | CC BY-SA 3.0 |
corrected speling
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| Dec 27, 2012 at 13:32 | comment | added | Steven Landsburg | Meta-thread here: tea.mathoverflow.net/discussion/1500/locally-isomorphic-modules | |
| Dec 27, 2012 at 13:24 | answer | added | Steven Landsburg | timeline score: 6 | |
| Dec 27, 2012 at 11:32 | comment | added | Martin Brandenburg | mathoverflow.net/questions/24361/… | |
| Dec 27, 2012 at 11:30 | comment | added | Sasha | Adding to Fernando's comment --- an example of an affine variety with nontrivial Picard group is the complement to a hypersurface of degree $d \ge 2$ in $P^n$. | |
| Dec 27, 2012 at 10:22 | comment | added | Fernando Muro | The module $M$ of sections of a non-trivial line bundle on an affine variety $X=\operatorname{Spec} R$ satisfies $R_{\mathfrak{P}}\cong M_{\mathfrak{P}}$ for any prime ideal $\mathfrak{P}\subset R$, but $R\ncong M$. | |
| Dec 27, 2012 at 9:26 | history | asked | user30230 | CC BY-SA 3.0 |