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Timeline for Flatness and local freeness

Current License: CC BY-SA 3.0

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Dec 31, 2015 at 19:05 comment added Daniel Litt @user26857: Added a link to the original example in Ravi's notes--thanks!
Dec 31, 2015 at 19:04 history edited Daniel Litt CC BY-SA 3.0
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Dec 31, 2015 at 18:46 comment added user26857 The link to the example is dead. (It can be replaced by a reference: Lam, Exercises in Modules and Rings, Exercise 4.17. Also here.)
Jul 21, 2011 at 20:49 comment added Daniel Litt (The link above should be math.stanford.edu/~vakil/216blog ...I think MO screwed up the html.)
Jul 21, 2011 at 17:51 comment added Ravi Vakil Also, that example in your UPDATE is great. Does anyone know a published reference with an argument? (I now mention it as an aside in Exercise 25.4.E in the July ~21 notes <a href="math.stanford.edu/~vakil/216blog/">here</a>.)
Jul 20, 2011 at 17:08 comment added Daniel Litt Yeah, this is kind of an amazing fact :).
Jul 20, 2011 at 16:36 comment added Ravi Vakil @Daniel --- as I now mention in my "answer", finite presentation can be weakened to finitely generated. (Your argument is the slickest one I know in the finitely presented case though.)
Jul 27, 2010 at 16:49 history edited Daniel Litt CC BY-SA 2.5
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Jul 27, 2010 at 16:02 history edited Daniel Litt CC BY-SA 2.5
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Jul 27, 2010 at 15:46 history edited Daniel Litt CC BY-SA 2.5
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Jul 27, 2010 at 15:46 comment added Keenan Kidwell Okay, yeah, the Tor sequence. Thanks.
Jul 27, 2010 at 15:41 history edited Daniel Litt CC BY-SA 2.5
Added counterexample.
Jul 27, 2010 at 15:37 comment added Daniel Litt @Keenan: The exactness of that sequence is exactly where I'm using the flatness of $M$. Consider the LES of Tor. And finite flat over a Noetherian local ring implies free, so obviously projective.
Jul 27, 2010 at 15:33 comment added Keenan Kidwell That should be "Noetherian local ring."
Jul 27, 2010 at 15:32 comment added Keenan Kidwell Aren't you saying that the sequence obtained $0\rightarrow K\rightarrow A^n\rightarrow M\rightarrow 0$ by lifting a basis for $M/\mathfrak{m}M$ remains exact upon tensoring with the residue field? I don't understand why this is the case, maybe 'cause I don't see where you're using the flatness of $M$. Is it true that $M$ is necessarily projective, i.e., does finite flat over a Noetherian ring imply projective?
Jul 27, 2010 at 15:26 comment added ashpool I agree with Akhil's answer wholeheartedly. But the question is whether we can relax the condition of finite presentation with finite generation.
Jul 27, 2010 at 15:17 history answered Daniel Litt CC BY-SA 2.5