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Aug 17, 2016 at 15:49 vote accept DCT
Aug 17, 2016 at 8:39 answer added Francesco Polizzi timeline score: 5
Aug 17, 2016 at 8:32 comment added Francesco Polizzi @JasonStarr: I think that the original work of Severi already answers the question about the possibility of independently smoothing the singularity of nodal curves. In fact, the proof of this result only involves checking the smoothness of some deformation space, and Severi's proof was in this aspect correct (as far I can remember). Harris solved the much harder problem of proving its irreducibility, for which Severi's proof was flawed.
Aug 17, 2016 at 8:21 history edited Francesco Polizzi CC BY-SA 3.0
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Aug 15, 2016 at 22:42 comment added Jason Starr For curves that are worse than nodal, this map is not surjective. For instance, consider the "hypercuspidal curve", $C=\text{Zero}(Y^d - X Z^{d-1})$. In this case, $\textit{Ext}^1 ( \Omega_C, \mathcal{O}_C )$ is a torsion sheaf supported at the unique singular point $[X,Y,Z] = [1,0,0]$ with length $(d-1)(d-2)$. On the other hand $k[X,Y,Z]_d$ has length $(d+2)(d+1)/2$. For $d\geq 9$, the map cannot be surjective. For the nodal case, you might check Harris's paper solving the Severi conjecture -- I remember something about this in that paper.
Aug 15, 2016 at 21:17 history edited DCT CC BY-SA 3.0
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Aug 14, 2016 at 22:50 history edited DCT
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Aug 14, 2016 at 22:30 history asked DCT CC BY-SA 3.0