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Oct 16, 2012 at 16:58 history edited Choa CC BY-SA 3.0
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Oct 16, 2012 at 16:44 comment added Matt This can be made totally rigorous. There are a ton of details to check, but I think it is fairly straightforward when (as you point out) you know what the target of $f$ is supposed to be. Let $P$ be the Hilbert polynomial of $\mathcal{F}_x$. The image is $Hilb^P(X)$ (the scheme representing the functor $S\to X \mapsto $ flat quotients of $\mathcal{O}_{S\times X}$ with Hilbert polynomial $P$). Commutativity of the appropriate diagram and identification of tangent spaces with the Ext groups using deformation theory shows that $df$ is exactly $\kappa (x)$.
Oct 16, 2012 at 16:44 answer added Jonathan Wise timeline score: 4
Oct 16, 2012 at 12:25 history edited Choa CC BY-SA 3.0
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Oct 15, 2012 at 14:02 history edited Choa CC BY-SA 3.0
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Oct 15, 2012 at 5:16 history edited Choa CC BY-SA 3.0
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Oct 15, 2012 at 1:14 history edited Choa CC BY-SA 3.0
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Oct 14, 2012 at 15:58 history asked Choa CC BY-SA 3.0