Advanced Algebraic Curves problem

[mothergoose64]

[Question]
Let p1, p2 and p3 be 3 distinct points in PC2( Projective space, ie
(z0,z1,z2) belong to PC2) Find the dimension of the linear system of
cubics containing these 3 points.


I have solved it for the non collinear case, by taking a projective
transformation of the 3 points to [1,0,0],[0,1,0] and [0,0,1]
respectively.
And substituting those values into the equation of a cubic, to get
that there are 6 coefficients remaining, therefore the dimension of
the linear system is 6 by definition.

But I am stuck on the collinear case (or maybe it can be shown generally?), thanks in advance for any help
that can be given.

 

[jvicens]

Hello there!

First of all, great job on solving the non-collinear case. You are correct in your approach and your answer of 6 is also correct.

For the collinear case, let's consider the points p1=[1,0,0], p2=[0,1,0] and p3=[0,0,1]. These points are collinear and they form a line in PC2. Now, let's take a general cubic equation f(x,y,z) and substitute these points into it. We get:

f(1,0,0) = a + b(1) + c(0) + d(0)^2 + e(0)^3 = a
f(0,1,0) = a + b(0) + c(1) + d(0)^2 + e(0)^3 = c
f(0,0,1) = a + b(0) + c(0) + d(1)^2 + e(1)^3 = d + e

Since these points lie on the same line, the values of a, c and d+e must be equal. Therefore, we only have 3 independent coefficients left (b, d and e). This means that the dimension of the linear system for the collinear case is 3.

To generalize this result, we can say that if we have n collinear points in PC2, then the dimension of the linear system of cubics containing these points would be 3n-3. This can be proven using a similar approach as above.

Hope this helps! Let me know if you have any further questions.