Timeline for Is there a Whitney theorem type theorem for projective schemes?
Current License: CC BY-SA 2.5
9 events
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Feb 6, 2010 at 16:04 | comment | added | Pete L. Clark | @Liu: We try to avoid the "s word" here on MO, even applied to oneself. :) | |
Feb 5, 2010 at 21:37 | comment | added | Qing Liu | Yes, you are right. My remark is stupid. | |
Feb 5, 2010 at 21:37 | comment | added | Qing Liu | Yes you are right. My remark is stupid. | |
Feb 4, 2010 at 22:16 | comment | added | David E Speyer | I didn't make any specific claim about the embedding dimension of singular varieties. I said "I think there should be a result like this" and then explained the smooth case. That said, I don't buy your example. There is no reason I have to embed by a complete linear system. It looks to me like your example embeds in P^3: just take a generic n+1 points in P^3 and join them up by lines. | |
Feb 4, 2010 at 22:06 | comment | added | Qing Liu | David:regarding the embedding dimension of singular varieties, the local embedding dimensions are not enough. Take a chain of n+1 projective lines, then the tangent spaces have dimension at most 2, but any ample divisor D on this curve has degree at least n, and L(D) has dimension at least n+2 (up to $\epsilon$, I didn't check carefully), so the curve can not be embedded in $\mathbb P^{n+1}$. Maybe you mean irreducible varieties. | |
Feb 4, 2010 at 20:02 | history | edited | David E Speyer | CC BY-SA 2.5 |
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Feb 4, 2010 at 19:55 | comment | added | Ryan Eberhart | Sorry--I did mean smooth. It was in the curve case but as Pete said and you noticed I forgot to explicitly include it in the hypothesis of my question. | |
Feb 4, 2010 at 19:45 | comment | added | Pete L. Clark | @DS: OK. I was thinking about the case of a smooth variety (Ryan said "smooth projective curve"; I didn't notice that he didn't repeat "smooth" after that). | |
Feb 4, 2010 at 19:42 | history | answered | David E Speyer | CC BY-SA 2.5 |