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.)
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asked Nov 9, 2010 at 5:46
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2 Answers
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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:
$$0 \to F \to G \to I \to 0 $$
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..."
answered Nov 9, 2010 at 7:07
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1$\begingroup$ @Richard: you are welcome. Please feel free to make comments here if you have related questions, I think I left out a few details (it was late at night). $\endgroup$ Nov 9, 2010 at 18:04
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$\begingroup$ @Hailong: Could you elaborate on why it's okay to apply Auslander-Buchsbaum here? R certainly isn't local, and I don't understand how you can reduce to the local case. $\endgroup$ Nov 9, 2010 at 18:35
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3$\begingroup$ @Daniel: Sure, locally $R/I$ has projective dimension at most $2$, so if you let $F$ be a second sygyzy, $F$ has to be locally free, hence projective, hence free. $\endgroup$ Nov 9, 2010 at 18:58
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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.
Vinnikov, V. Selfadjoint determinantal representations of real plane curves. Math. Ann. 296 (1993), no. 3, 453–479
answered Nov 9, 2010 at 6:28