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May 21, 2019 at 21:17 comment added Asvin Thanks! Let me think about it a little.
May 21, 2019 at 21:12 comment added François Brunault I see, then the restriction of the defining polynomial to that line will be completely determined (its roots will correspond to the points you took), then you can add any multiple of the equation of the line. So the dimension of your linear subspace will be the dimension of the space of homogeneous polynomials of degree $d-1$. So I think you have codim $d$ in this case.
May 21, 2019 at 21:03 history edited Asvin CC BY-SA 4.0
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May 21, 2019 at 21:02 comment added Asvin In fact, I think I would like to assume that the points all line on a line. So as far from generic as possible.
May 21, 2019 at 20:59 comment added Asvin Nine = number...
May 21, 2019 at 20:59 comment added Asvin Right, but I think the fact that the nine of points is smaller than the dimension of the space of curves should make it so that it doesn't really matter which number I care about. I am not sure...
May 21, 2019 at 20:55 comment added François Brunault I forgot that your $r$ points are rational, so there are only finitely many possibilities for them and genericity doesn't make much sense. I was thinking geometrically (say over the algebraic closure). So the number of curves will depend on the position of these points; which number do you want to estimate? You have $r$ equations, so the codimension is at most $r$, but can be less e.g. if your points satisfy collinearity conditions. For conics ($5$-dim parameter space) passing through 5 points, "generic" means "no 4 points collinear", but in general I don't know.
May 21, 2019 at 20:47 comment added Asvin Right, that would be ideal. Also, why does genericity imply the right codimension?
May 21, 2019 at 17:47 comment added François Brunault The space of plane curves of degree $d$ is a projective space, and passing through $r$ points gives you linear conditions on the coefficients of the equation, so what you get is a linear subspace, which has codimension $r$ if the points are in generic position. But maybe you also want to have control in every case?
May 21, 2019 at 11:14 history edited Asvin CC BY-SA 4.0
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May 21, 2019 at 11:14 comment added Asvin Yes, I am okay with assuming they are rational.
May 21, 2019 at 10:00 comment added François Brunault Are your points rational over $\mathbb{F}_q$? If not, then I guess you want to assume that your set of points is stable under Galois.
May 21, 2019 at 6:36 history asked Asvin CC BY-SA 4.0