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The Hilbert Scheme of 11 Points in \(\mathbb{A}^{3}\) Is Irreducible

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Combinatorial Algebraic Geometry

Part of the book series: Fields Institute Communications ((FIC,volume 80))

Abstract

We prove that the Hilbert scheme of 11 points on a smooth threefold is irreducible. In the course of the proof, we present several known and new techniques for producing curves on the Hilbert scheme.

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Acknowledgements

This article was initiated during the Apprenticeship Weeks (22 August–2 September 2016), led by Bernd Sturmfels, as part of the Combinatorial Algebraic Geometry Semester at the Fields Institute. The authors wish to thank the Fields Institute, the organizers of the Thematic Program on Combinatorial Algebraic Geometry in Fall 2016, and the organizers of the Apprenticeship Weeks which took place during the program. We are very grateful to Mark Huibregtse, Anthony Iarrobino, Gary Kennedy, Greg Smith, Bernd Sturmfels, and several anonymous referees for numerous helpful comments. This work was supported by a grant from the Simons Foundation (#354574, Zach Teitler). JJ was supported by Polish National Science Center, project 2014/13/N/ST1/02640. BIUN was supported by NRC project 144013.

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Douvropoulos, T., Jelisiejew, J., Nødland, B.I.U., Teitler, Z. (2017). The Hilbert Scheme of 11 Points in \(\mathbb{A}^{3}\) Is Irreducible. In: Smith, G., Sturmfels, B. (eds) Combinatorial Algebraic Geometry. Fields Institute Communications, vol 80. Springer, New York, NY. https://doi.org/10.1007/978-1-4939-7486-3_15

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