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I am working on Exercise 2.9 in Eisenbud's Geometry of Syzygies:

Let $X$ be a set of $n\leq 2r+1$ points in $\mathbb{P}^r$ in linearly general position. Show that $X$ imposes independent conditions on quadrics (for every $p\in X$ there is a quadric not vanishing at $p$ but vanishing at all other points of $X$).

Could somebody please help me solve this problem? I have never done an exercise involving independent conditions before. This is not homework.

Chris Suk
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Given $p\in X$, we split the remaining points of $X$ into $p_1,\ldots,p_r$ and $q_1,\ldots,q_r$. Let $H_1$ be the hyperplane spanned by $p_1,\ldots,p_r$ and $H_2$ be the hyperplane spanned by $q_1,\ldots,q_r$. Since $X$ is in linear general position, we know that $p$ does not lie on $H_1$ or $H_2$.

Finally, we can let our quadric $Q$ be $H_1\cup H_2$.

(I'm curious if anybody knows if $2r+1$ is the best we can do given linear general position, since we haven't used all of the quadrics available)

DCT
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  • Yeah, I guess $2r+1$ bound is the best we have. For example, given $5$ general points in $\mathbb P^2$, we can draw two lines passing through all but one point. But for six general points, we'll always miss two points when we attempt to draw lines. Of course, this is not rigorous. (btw +1) – cqfd Nov 25 '20 at 17:09
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    Good point, I think your comment can be made into a counterexample in the case of six points in $\mathbb{P}^2$. Namely, we can just pick six points on a conic that are in linear general position (no two coincide, no three colinear). But we can still ask whether the bound is sharp for $r\geq 3$. – DCT Nov 26 '20 at 16:30
  • The same carries over to higher dimensions, I guess. If there are $2r+2$ points, then the argument given in your answer won't work, right? – cqfd Nov 27 '20 at 02:29