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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