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.

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

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