Timeline for Flat module and torsion-free module
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Jul 4, 2022 at 22:41 | history | edited | YCor | CC BY-SA 4.0 |
formatting
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| Jul 4, 2022 at 21:12 | answer | added | Bruno Kahn | timeline score: 1 | |
| Jul 24, 2012 at 1:41 | history | edited | David White | CC BY-SA 3.0 |
Fixed typos, since it was on the front-page anyway.
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| Jul 24, 2012 at 1:12 | answer | added | Philipp Rothmaler | timeline score: 5 | |
| Jan 4, 2011 at 20:36 | comment | added | Qing Liu | Another source of free-torsion algebras comes from blowing-ups. Take $R=k[x,y]$ and $A=k[x, y/x]\subset k(x,y)$. | |
| Jan 4, 2011 at 19:08 | answer | added | Sándor Kovács | timeline score: 8 | |
| Jan 4, 2011 at 17:08 | answer | added | Pete L. Clark | timeline score: 17 | |
| Jan 4, 2011 at 15:02 | comment | added | Karl Schwede | Keerthi's example is right, and it is certainly the ``simplest'' example of a torsion-free non-flat module over the coordinate ring of a smooth variety. Generally speaking, EVERY coordinate ring of an algebraic variety of dimension > 1 will have torsion free modules which are not flat (the ideal of any closed point is exactly such a module). | |
| Jan 4, 2011 at 14:14 | comment | added | Keerthi Madapusi | Take $R=k[x,y]$ and take $M=(x,y)\subset R$. This is torsion-free, but not flat. | |
| Jan 4, 2011 at 14:08 | history | edited | Liu Hang | CC BY-SA 2.5 |
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| Jan 4, 2011 at 14:08 | comment | added | Liu Hang | @Theo: all rings in this question are integral, thanks! | |
| Jan 4, 2011 at 14:01 | history | edited | Liu Hang | CC BY-SA 2.5 |
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| Jan 4, 2011 at 13:21 | comment | added | Theo Buehler | @liu: What exactly is your definition of torsion-freeness? Usually one says that an $R$-module $M$ is torsion-free if $rm = 0$ implies $r = 0$ or $m = 0$. But then your ring had better be a domain because otherwise the projective (hence flat) module $R$ would be a counterexample to the first sentence of your question. | |
| Jan 4, 2011 at 12:32 | vote | accept | Liu Hang | ||
| Jan 4, 2011 at 12:31 | comment | added | Liu Hang | @Johannes: It is true for Prüfer domain, but not for general commutative rings. | |
| Jan 4, 2011 at 12:27 | history | edited | Liu Hang | CC BY-SA 2.5 |
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| Jan 4, 2011 at 11:16 | comment | added | Johannes Hahn | @liu hang: One of the equivalent definitions wikipedia lists is exactly "Every torsion-free R-modul is flat." Isn't your first question already answered with that? | |
| Jan 4, 2011 at 10:46 | answer | added | Georges Elencwajg | timeline score: 21 | |
| Jan 4, 2011 at 9:00 | history | asked | Liu Hang | CC BY-SA 2.5 |