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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.

Bombyx mori
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    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 $C$. –  Jun 04 '13 at 22:18
  • Note that once Bezout's theorem has been introduced, you can use it to justify the second point more succinctly. –  Jun 04 '13 at 22:20
  • I see. Maybe I can just skip this until he prove Bezout's theorem later. – Bombyx mori Jun 05 '13 at 00:00
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    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! –  Jun 05 '13 at 13:58
  • @user6487 Very clever !!!! – Jam Aug 19 '21 at 20:39

2 Answers2

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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 $Q_3\cap L\subset C$.

Puzzled
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Let a homogeneous polynomial of degree 2 on $\mathbb{P}^3$ be written as $$ a_{00}z_0^2 + a_{01}z_0z_1 + \cdots + a_{33}z_3^2$$ and $p=[p_0:p_1 : p_2: p_3], q=[q_0:q_1:q_2:q_3] \in \{p_1, \dots, p_7\} \subset C$. Since to give a quadric hypersurface some $a_{ij}$ must be non-zero, and since $F,\lambda F=0$ define the same quadric hypersurface in $\mathbb{P}^3$, we have a projective space $\mathcal{P}$ of dimension $\binom{4}{2} + 4-1=9$ of quadric hypersurfaces in $\mathbb{P}^3$ as stated above.

The condition that a given quadric contain $p$ is given by $H_p := \sum_{i,j}a_{ij}p_ip_j=0$ which is linear in $a_{ij}$ (as the $p_i,p_j$ are constants), so gives a linear hyperplane in $\mathcal{P} \simeq \mathbb{P}^9$. Since the $\{p_i\}_{i=1}^7$ lie in general position, we may choose $\{a_{ij}\}$ in each $H_p$ so that $H_p, H_q$ intersect transversely for all $p,q\in \{p_i\}_{i=1}^7$. Therefore, the subspace of $\mathcal{P}$ given by homogeneous quadratic polynomials vanishing at all $\{p_i\}_{i=1}^7$ has projective dimension $9-7=2$, as stated above. But this subspace is already spanned by $Q_0,Q_1,Q_2$ defining $C$ as stated above, since their coefficients lie in general position in $\mathcal{P}$.

[there is not much novelty in my answer beyond previous responses, but this is my understanding of the details]

locally trivial
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