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:



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> 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. $(\*)$
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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. 



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


