9 points on a quadric hypersurface

Question

I found on some notes the statement

9 points in $\mathbb{P}^3$ always lie on a quadric hypersurface $Q$.

Actually I can't understand if this problem is easy to solve using the -elementary- techniques I got from a course in projective geometry I attended at the first year of my bachelor degree, or it can be solved only by using more advanced tools.

It's just curiosity, I tried some attacks without any success. I'd like to see why this should works, and in general how to attack statements like this. Thanks in advance.

Answer 1

The number of degree monimals in $P^3$ are $x_ix_j$ where $0\le i \le j\le 3$ so there are $10$ of them. Each quadric is linear combination of them so it is parametrized by 10 variables. Now suppose you are given 9 points. To find the suitable quadric, you need to solve the coefficients and you get a linear system of equations. Since the number of variables is more than the number equations, you always get a nonzero solution.

Answer 2

It's also important that for ''generic'' $9$ points there exists a unique quadric containing them.

Fact: if in $k^N$ we have some vectors such that every $N$ of these vectors are linearly dependent, then $\it{all}$ of these vectors lie in a subspace of $N-1$. Indeed, the span of these vectors is of dimension $\le N-1$. From here we conclude ( contrapositive):

If $N$ function $\phi_{i} \colon X \to k$, are linearly independent, then, there there exist $N$ points $(x_j)_{j}$ such that the vectors $(\phi_{i}(x_{j}))_{i,j}$ are linearly independent, there exist $N$ points $(x_{j})$ such that $$ \det (\phi_{\alpha}(x_{\beta}) )\ne 0$$

Consider now $(\phi_{\alpha})$ $N$ polynomials that are linearly independent as polynomials. They they are also independent as polynomial functions ( assume $k$ infinite, or take an infinite extension $\bar k$ of $k$). Therefore, there exist $N$ points $(x_{\beta})$ such that the determinant $\det (\phi_{\alpha} ( x_{\beta}) \ne 0$. This determinant as a function of $N$ points $(x_{\beta})$, is a polynomial function, not $0$. We also see that for $k$ generic points $(x_{\beta})$, $1\le k \le N$, the matrix $(\phi_{\alpha} ( x_{\beta})$ has rank $k$

Let's consider now $\phi_{i}$ $N$ homogenous polynomials in variable $x$, of same degree $d$, that are linearly independent. For $N-1$ generic points $(x_{\beta})$ the matrix $(p_{\alpha}(x_{\beta}))$ has rank $N-1$. Since the polynomials are homogenous of same degree, a rescaling of the points will just multiply the rows of the matrix by constants. Therefore, we can say: for $N-1$ generic points in the projective space, there exists a unique ( up to a multiplication by a constant) linear combination of $\sum c_{\alpha} \phi_{\alpha}$ that is zero at these points. The equation equation can be obtained by considering an $N\times N$ determinant with entries $\phi_i(x_j)$, $\phi_i(x)$.