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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?

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  • $\begingroup$ @saulspatz sure, thank you. $\endgroup$
    – rmdmc89
    Jun 3, 2021 at 19:17
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    $\begingroup$ The points should be in "general positions". For example, if we have five points on the same line, then they don't uniquely determine a conic. $\endgroup$
    – WhatsUp
    Jun 3, 2021 at 19:22
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    $\begingroup$ I think this can be viewed as conditions for solution to a linear system. If you write the equation for a cubic, you have 10 monomials, so you have 10 coefficients of the monomials. Now, for each point you want on the curve you put in the values of $x,y,z$ you get one linear equation in the coeffcients. Since you're working projective, you eliminate the $z$ and remain with 9 coefficients multiplied by a number, and a constant term. Then after 9 points you have to check the determinant of the $9x9$ matrix to see if it's non-zero. $\endgroup$
    – user932138
    Jun 3, 2021 at 19:34

1 Answer 1

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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.

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