# Degrees of syzygies of points in $\mathbb P^2$

Let $$X$$ be a collection of points in $$\mathbb P^2$$ over the complex numbers. Let $$I_X$$ be the defining ideal. I am interested in knowing when:

The syzygies of $$I_X$$ contains no linear forms. Since we are in $$\mathbb P^2$$, this just says that the Hilbert-Burch matrix contains no (non-zero) linear entries. $$(*)$$

One obvious case when this happens is when we take two general curves $$F,G$$ of degrees $$a,b\geq 2$$ and let $$X=V(F)\cap V(G)$$. Then the syzygies is just the Kozsul relation and has degrees $$a,b$$. I don't know further examples and would like to know if there are interesting geometric conditions that would imply $$(*)$$.

One obvious necessary condition is that the generators of $$I_X$$ have degree at least $$2n-2$$, where $$n$$ is the number of generators.

Also, perhaps if you fix the degree $$d$$ of $$X$$, then $$(*)$$ defines a closed subscheme (? I am not sure about this) of the Hilbert scheme $$\mathbb P^{2[d]}$$. If so, then knowing it's dimension would be nice.

• If you take sufficiently many points in general position, shouldn't the condition be satisfied? – Angelo May 31 at 8:23
• @Angelo: ideal of general points tend to have linear entries in the H-B matrix, but it's not clear when. – Hailong Dao May 31 at 15:18
• I see. This seems quite subtle. – Angelo Jun 1 at 11:19

I asked David Eisenbud and was told about Exercises 12 and 13 of Chapter 3 in his book "Geometry of Syzygies". Putting together, they show that for a generic set of $$n$$ points $$X$$ in $$\mathbb P^2$$, the syzygies of $$I_X$$ has no linear forms if and only if $$n= \binom{2s+1}{2}+s$$, for some positive integer $$s$$. It is unclear if a complete characterization can be found.