By Audun Holme, Robert Speiser

This quantity offers chosen papers because of the assembly at Sundance on enumerative algebraic geometry. The papers are unique learn articles and focus on the underlying geometry of the subject.

**Read or Download Algebraic Geometry Sundance 1986: Proceedings of a Conference held at Sundance, Utah, August 12–19, 1986 PDF**

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**Extra info for Algebraic Geometry Sundance 1986: Proceedings of a Conference held at Sundance, Utah, August 12–19, 1986**

**Example text**

O b s e r v e t h a t w h e n w e pull t h e s e loci b a c k to t h e ( r , s ) - p l a n e , t h e l o c u s of c u r v e s w i t h t h r e e n o d e s is g i v e n in b r a n c h 1) b y r = 0, t h e locus of c u r v e s w i t h a t a c n o d e b y s 2 = 4r; and that these have intersection multiplicity b r a n c h 2) t h e s e t w o loci a r e g i v e n b y t h e e q u a t i o n s respectively, and have intersection number s2 = - 4 r , s2 = 4r 2; a n d in b r a n c h 3) b y r -- 0 a n d again having intersection multiplicity m u l t i p l i c i t y of t h e s e t w o loci is t h u s r = 0 and 2; s i m i l a r l y in 2.

O b s e r v e t h a t w h e n w e pull t h e s e loci b a c k to t h e ( r , s ) - p l a n e , t h e l o c u s of c u r v e s w i t h t h r e e n o d e s is g i v e n in b r a n c h 1) b y r = 0, t h e locus of c u r v e s w i t h a t a c n o d e b y s 2 = 4r; and that these have intersection multiplicity b r a n c h 2) t h e s e t w o loci a r e g i v e n b y t h e e q u a t i o n s respectively, and have intersection number s2 = - 4 r , s2 = 4r 2; a n d in b r a n c h 3) b y r -- 0 a n d again having intersection multiplicity m u l t i p l i c i t y of t h e s e t w o loci is t h u s r = 0 and 2; s i m i l a r l y in 2.

As sets of points a r e a l w a y s aritmetically Cohen-Macaulay, this example shows that the analogue of the problem has a negative solution for sets of points. N o w if one of these bad sets of points were the hyperplane section of a projectively normal curve that was cut out scheme-theoretically by quadrics, then the solution to the problem above would be negative. 4 that no such curve can exist. ) In addition to thanking Joe Harris, we would like to t h a n k Jee Koh and Michael Stillman, in conversations w i t h w h o m we first considered t h e problems a t t a c k e d in this note, and Bill Lang and David Morrison, who provided first aid for o u r K3 b u m p s a n d bruises.