• mothergoose64
In summary, the dimension of the linear system of cubics containing 3 distinct points in PC2 is 6 for the non-collinear case and 3 for the collinear case. For n collinear points, the dimension is 3n-3.
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.

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.

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