Space Curves as Determinantal Varieties - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T14:00:29Z http://mathoverflow.net/feeds/question/45406 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/45406/space-curves-as-determinantal-varieties Space Curves as Determinantal Varieties Richard Montgomery 2010-11-09T05:46:31Z 2010-11-09T07:28:49Z <p>I read in a paper of Goryunov (`Functions on space curves', Journal of The London Mathematical Society, vol. 61 (2000), 807-822; available on his home page) that every space curve can be defined as the vanishing of the N minors of some N by N+1 matrix with entries functions of x, y, z. How does one prove this? (I can do it for the rational normal curve, thanks to Harris's book.)</p> http://mathoverflow.net/questions/45406/space-curves-as-determinantal-varieties/45411#45411 Answer by Denis Serre for Space Curves as Determinantal Varieties Denis Serre 2010-11-09T06:28:04Z 2010-11-09T06:28:04Z <p>Not exactly an answer to the question, but some related information. The following article deals with the real case. It states that every reduced projective plane curve defined over the real field has an equation of the form $\det(xA+yB+zC)=0$ where $A,B,C$ are Hermitian. In addition, if the curve contains a set of $[n/2]$ ovals totally ordered by inclusion, then one may choose $A,B,C$ such that a linear combination of them be positive definite. This yields the solution of a famous problem raised by P. Lax.</p> <blockquote> <p>Vinnikov, V. Selfadjoint determinantal representations of real plane curves. Math. Ann. 296 (1993), no. 3, 453–479</p> </blockquote> http://mathoverflow.net/questions/45406/space-curves-as-determinantal-varieties/45413#45413 Answer by Hailong Dao for Space Curves as Determinantal Varieties Hailong Dao 2010-11-09T07:07:23Z 2010-11-09T07:28:49Z <p>Let $I\subset R = k[x,y,z]$ be the defining ideal of your curve. Then $R/I$ has dimension one and no embedded components, so has projective dimension $2$ by the Auslander-Buchsbaum formula. Therefore $I$ itself has projective dimension $1$, and so can be fit into a short exact sequence:</p> <p>$$0 \to F \to G \to I \to 0$$</p> <p>with $F,G$ free (a minor point: one needs that projective modules are free here, it is easy if you assume $I \subset (x,y,z)$, since you may as well look at the local ring at the origin). If $N=\text{rank} F$, then $\text{rank} G=N+1$, and the matrix representing the map from $F$ to $G$ is what you want. This is known as the Hilbert-Burch theorem, and details can be found in Chapter 20 of Eisenbud's book "Commutative Algebra with a view..."</p>