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S Mar 1, 2014 at 22:52 history bounty ended CommunityBot
S Mar 1, 2014 at 22:52 history notice removed CommunityBot
Feb 22, 2014 at 2:39 comment added Vladimir Baranovsky Ok, if you say that the tangent spaces are $H^0(\mathcal{N}_{Z|\mathbb{P}^n})$ then you are deforming $Z$ and $Z'$ as subschemes of $\mathbb{P}^n$ not of $X$, as I misunderstood. Then taking the $m$-power of the deforming ideal does not quite work.
S Feb 21, 2014 at 21:21 history bounty started user46578
S Feb 21, 2014 at 21:21 history notice added user46578 Draw attention
Feb 21, 2014 at 18:21 comment added user46578 @Baranovsky: May be I am getting this wrong. $X \otimes Spec(k[\epsilon]/(\epsilon^2)$ is not a first order deformation of $X$. May be you meant $X \times Spec(k[\epsilon]/(\epsilon^2)$ which is the trivial deformation. It is also not guarenteed that a first deformation of $Z$ comes from a first order deformation of $X$. Could you elaborate on your argument a bit more?
Feb 21, 2014 at 18:10 comment added Vladimir Baranovsky If your first order deformation of $Z$ is a subscheme of $X \otimes Spec (k[\varepsilon]/\varepsilon^2)$ with ideal sheaf $J$ then it seems that $J^m$ will define a subscheme deforming $Z'$. Unless I misunderstood the question.
Feb 21, 2014 at 18:01 comment added user46578 @Baranovsky: I do not understand the question. "What happens" is a very broad question. Is there any specific property you are inquisitive about?
Feb 20, 2014 at 21:52 comment added Vladimir Baranovsky If your deformation of Z is a subscheme with ideal sheaf J, what happens when you consider the ideal J^m?
Feb 19, 2014 at 21:14 history asked user46578 CC BY-SA 3.0