# Why 7 points on a twisted cubic is enough to fix a quadratic?

From Joe Harris, Algebraic Geometry, Page 10.

Show that if seven points $p_{1},\cdots,p_{7}$ on a twisted cubic curve, then the common zero locus of the quadratic polynomials vanishing at the $p_{i}$ is the twisted cubic.

I am looking for a hint to solve this problem (and the more general case of rational normal curves). I know $p_{i}$ must be in general position, such that any four of them span $\mathbb{P}^{3}$. But this alone does not give an answer immediately. Here is an argument based on Harris' argument on page 7.

Consider $p_{1},p_{2},p_{3}$. It must span a hypersurface of dimension 2 in $\mathbb{P}^{3}$. And similarly is $p_{4},p_{5},p_{6}$. If $p_{7}$ lies on all quadratic polynomial passing through them, then $p_{7}$ has to lie on the intersection of two hypersurfaces - therefore must be lying on at least one hyperplane spanned by the above 2 sets. So we can write $p_{7}$ to be the linear combination of $p_{1},p_{2},p_{3}$ without loss of generality. But this contradicts the fact that they lying on general position (such that $p_{1},p_{2},p_{3},p_{7}$ span $\mathbb{P}^{7}$). So $p_{7}$ must not lie on all quadratic polynomials passing through them.

Since we know every quadratic in $\mathbb{P}^{3}$ is determined by $4+6-1=9$ coefficients, two quadratics pass through the same 6 points should give us 3 dimension "wiggle room" left. But then I am lost as what to proceed.

• This is one of those cases where the appropriate answer depends on what methods you are "allowed" to use. Here's my explanation. First, the twisted cubic $C$ is cut out in P^3 by quadratic polynomials. So the common zero locus of all the quadratics vanishing at $p_1,\ldots,p_7$ is at most $C$. On the other hand, if you have a quadratic vanishing at $p_1,\ldots,p_7$, you can compose it with the map $P^1 \rightarrow C$ to get a degree-6 homog. poly. on $P^1$ which vanishes at 7 points; that's impossible unless the quadratic is identically zero on $C$. – user64687 Jun 4 '13 at 22:18
• Note that once Bezout's theorem has been introduced, you can use it to justify the second point more succinctly. – user64687 Jun 4 '13 at 22:20
• I see. Maybe I can just skip this until he prove Bezout's theorem later. – Bombyx mori Jun 5 '13 at 0:00
• But I explained in my comment above that you really don't need Bezout's theorem to solve this problem (just that one part of the argument becomes very short when you use that theorem). There's no reason to skip it! – user64687 Jun 5 '13 at 13:58

Let $C$ be the twisted cubic. The space $V$ of quadric surfaces containing $p_1,...,p_7$ has projective dimension $9-7 = 2$. Now, any quadric $Q$ containing $p_1,...,p_7$ intersects $C$ in $7 > 2deg(C) = 6$ points. Therefore $C\subset Q$.
We have three independet quadrics $Q_1,Q_2,Q_3\in V$ such that $C\subset Q_i$ for any $i = 1,2,3$. Now, $Q_1\cap Q_2 = C\cup L$ where $L$ is a line. Therefore $Q_1\cap Q_2\cap Q_3 = C$ because $Q_3\cap L\subset C$.