I'm trying to understand why do we have a resolution $0 \to O(-3)^{\oplus 2} \to O(-2)^{\oplus 3} \to I_{p_1,p_2,p_3} \to 0$ of the ideal sheaf.
I understand that when I have a complete intersection, for example two points in $\mathbb{P}^2$ I can just take the map $\mathcal{O}(-2) \oplus \mathcal{O}(-1) \to \mathcal{O}$ given by polynomial multiplication $(f \ g)$, and so we can complete the resolution by defining another map $0 \to \mathcal{O}(-3) \overset{(-g \ f)^t}{\to} \mathcal{O}(-2) \oplus \mathcal{O}(-1) $.
But I don't know how to find such resolutions when dealing with 3 or more points. Can anyone explain or give me a hint about what is happening in this case?
Thank you!
If you take a $2 \times 3$ matrix of linear forms on $\mathbb{P}^2$, it defines a map $\mathcal{O}(-3)^{\oplus 2} \to \mathcal{O}(-2)^{\oplus 3}$, and the minors of the map are three quadrics that define a map $\mathcal{O}(-2)^{\oplus 3} \to \mathcal{O}$. The composition is zero by the Cramer's rule. This gives a complex $$ 0 \to \mathcal{O}(-3)^{\oplus 2} \to \mathcal{O}(-2)^{\oplus 3} \to \mathcal{O} \to 0 $$ whose cohomology sits in the rightmost term and is equal to the structure sheaf of the corank-2 degeneration locus of the first map, which happens to be a subscheme of length~3.