Timeline for A projective module over a domain that is not faithfully flat?
Current License: CC BY-SA 4.0
21 events
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| S Oct 11, 2021 at 15:06 | history | bounty ended | CommunityBot | ||
| S Oct 11, 2021 at 15:06 | history | notice removed | CommunityBot | ||
| S Oct 3, 2021 at 13:18 | history | bounty started | Dick Johnson | ||
| S Oct 3, 2021 at 13:18 | history | notice added | Dick Johnson | Draw attention | |
| Sep 28, 2021 at 16:00 | history | edited | Tim Montegue | CC BY-SA 4.0 |
edited title
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| Sep 28, 2021 at 15:19 | comment | added | Jeremy Rickard | @Wojowu Of course, you might argue that if one can’t even be bothered to be a domain, then one doesn’t deserve to have any torsion free modules. | |
| Sep 28, 2021 at 15:10 | comment | added | Jeremy Rickard | @Wojowu For rings that are not domains, I think that “torsion free” usually means that no nonzero element is annihilated by a non zero divisor. Otherwise no nonzero module for a ring that is not a domain would be torsion free. Appeal to authority: en.m.wikipedia.org/wiki/Torsion-free_module | |
| Sep 28, 2021 at 15:02 | comment | added | Wojowu | @JeremyRickard I don't think your example is torsion-free: the element $(1,0)\in R$ annihilates the element $(0,1)\in \{0\}\times\mathbb Z$. | |
| Sep 28, 2021 at 13:43 | comment | added | Tim Montegue | @Fernando: Thank you for the information, I have removed the redundent assumption. | |
| Sep 28, 2021 at 13:42 | history | edited | Tim Montegue | CC BY-SA 4.0 |
Edit: Removed the assumption of torsion-freeness since it was redundent.
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| Sep 28, 2021 at 13:14 | comment | added | Fernando Muro | If you ring is a domain, protective implies torsion free. Also, it's kind of unfair to change your questions after getting correct answers. | |
| Sep 28, 2021 at 12:03 | comment | added | Tim Montegue | So I guess I was also assuming that $R$ has no zero-divisors. This rules out the counter example $R = S \times \mathbb{Z}$. Perhaps there are still trivial examples? | |
| Sep 28, 2021 at 12:01 | history | edited | Tim Montegue | CC BY-SA 4.0 |
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| Sep 28, 2021 at 12:01 | history | undeleted | Tim Montegue | ||
| Sep 28, 2021 at 11:01 | history | deleted | Tim Montegue | via Vote | |
| Sep 28, 2021 at 10:57 | history | edited | YCor | CC BY-SA 4.0 |
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| Sep 28, 2021 at 10:56 | comment | added | Jeremy Rickard | There are still trivial counterexamples. For example, if $R=S\times\mathbb{Z}$ then $0\times\mathbb{Z}$ is projective and torsion-free, but not faithfully flat. | |
| Sep 28, 2021 at 10:51 | comment | added | Tim Montegue | @Jeremy: I had assumed that $\mathcal{N}$ is non-zero without writing it. I have now edited to make this clear. | |
| Sep 28, 2021 at 10:50 | history | edited | Tim Montegue | CC BY-SA 4.0 |
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| Sep 28, 2021 at 10:47 | comment | added | Jeremy Rickard | Is that really what you meant to ask? The zero module is projective and torsionfree, but not faithfully flat. | |
| Sep 28, 2021 at 10:40 | history | asked | Tim Montegue | CC BY-SA 4.0 |