I'm sure this is a silly question, but for some reason I'm stuck in it.
The $k$-vector space generated by the monomials $x,y,z$ give all possible equations for lines in $\Bbb{P}^2$ and has projective dimension $3-1=2$. So two points in $\Bbb{P}^2$ determine a line.
Analogously, the space of $x^2,xy,xz, y^2,yz,z^2$ gives the equations of conics and has projective dimension $6-1=5$. So five points determine a conic.
For cubics, the space is $x^3,x^2y,x^2z,xy^2,xz^2,xyz,y^3, y^2z, yz^3, z^3$ with projective dimension $10-1=9$. Then shouldn't nine points determine a cubic?
Take for example a cubic $C$ and three concurrent lines $L_1,L_2,L_3$ meeting at $P\notin C$, each $L_i$ meeting $C$ simply. (for example, take $C:yz^2-(x^2-3z^2)x=0$, $L_1:y=0$, $L_2:y-z=0$ and $L_3:y+z=0$).
The intersection $C\cap L_1L_2L_3$ consists of $9$ distinct points, but $C$ and $L_1L_2L_3$ are distinct cubic through them.
What am I missing?
Best Answer
9 general points determine a curve, but 9 specific points might determine lots of curves. It’s similar to how a $9\times9$ matrix is usually invertible, but if you’re not careful about choosing your 9 rows, they could be linearly dependent and have lots of solutions.
In particular, once you choose 8 points, there’s still a degree of freedom, so there’s lots of curves. However, by Cayley–Bacharach, there’s a specific point which all of them go through. If you choose that 9th point in a special way (for example by choosing your 9 points in the first place by intersecting cubics), it doesn’t provide any additional restrictions so you can still get multiple cubics. Any other generic point will restrict you down to a unique cubic.