TITLE: Ideal of 8 general points in $\mathbb{P}^2$
QUESTION [6 upvotes]: I am working through chapter 3 of Eisenbud's Geometry of Syzygies. In the first example he makes the claim that the ideal of 8 general points in $\mathbb{P}^2$ is generated by two cubics and a quartic.
Q: How can I see that this is the case?
I am familiar with Bezout's theorem and the Cayley-Bacharach theorem, but I have never done one of these types of arguments on my own before.

REPLY [2 votes]: The linear forms, quadratics, cubics, and quartics on the plane are vector spaces of dimensions 3,6,10,15 respectively.
If we choose each point P generally, imposing the extra condition of vanishing at P will decrease each of the dimensions above by 1 (unless the dimension is already 0). If this helps, we are just saying that, if we have chosen $P_1$ through $P_i$, we can choose the next point to not lie on the common zero locus of the degree d forms that vanish on our first $i$ points, unless the zero polynomial is the only such form of degree d.
Therefore, after choosing 8 general points, the dimensions are now 0,0,2,7. 
So there are two cubics that vanish on the 8 points and one additional quartic that is not generated by the 2 cubics. The cubics will intersect in a subscheme of length 9. If we add the quartic, we must decrease the length, so the length must become 8, so we have found a generating set.