# Does the Bertini Theorem imply that there exists $k$ points such that passing through them imposes linearly independent conditions?

Consider the space of all homogeneous degree $d$ polynomials in three variables $[X,Y,Z]$, i.e. $$f[X,Y,Z] = f_{d00} X^d + f_{d10} X^{d-1}Y + \ldots .$$ This can be thought of as a section of the line bundle

$$\gamma_{\mathbb{P}^2} ^{* d} \rightarrow \mathbb{P}^2$$

Upto scaling this is the projective space $\mathbb{P}^{\delta_{d}}$, where $\delta_{d} = \frac{d(d+3)}{2}.$
Note that passing through a point '' imposes a linear condition on the coefficients $f_{ijk}$. Fix any number $k \leq \delta_{d}$. Is there an abstract way to show that there exist $k$ points $[X_1, Y_1, Z_1]\ldots, [X_k, Y_k, Z_k]$ such that passing through those $k$ points imposes linearly independent conditions on the coefficients? More precisely, the conditions will not be linearly independent implies that a certain determinant involving the $[X_i,Y_i, Z_i]$ is zero. How do I know that this determinant is not $\textit{always}$ zero?
One can explicitly write out the matrix whose determinant should not be zero and try to produce some explicit choice of points for which the determinant is not zero. But is there a way to avoid doing that? Does the Bertini theorem (or some modification of that) imply that there exists $k$ points so that passing through $k$ points imposes linearly independent conditions? In any case what is the simplest way to prove this seemingly obvious statement? Everything is over the complex numbers.

-
Well, this is just equivalent to the fact that a general matrix has maximal rank. If you want a "geometric" proof, you can just reason by induction. Assume that the propery you want holds for some collection $\mathcal{P}$ of $k-1$ points $p_1, \ldots, p_{k-1}$. Now the base locus $B$ of the linear system of curves of degree $d$ passing through $\mathcal{P}$ is a Zariski closed subset of $\mathbf{P}^2$. Choose $p_k$ outside $B$ and you are done. –  Francesco Polizzi Oct 17 '11 at 19:16
Of course, the base of the induction process is the trivial fact that two points always impose independent conditions. –  Francesco Polizzi Oct 17 '11 at 19:25
This is not any different than what Francesco is suggesting, but one possible shortcut is to argue by choosing coefficients for your points that generate a sufficiently transcendental extension. –  Thierry Zell Oct 18 '11 at 2:23

It is quite easy to give $k$ points imposing independent conditions. Just choose the points $p_1,\dots p_k$ one at a time as follows:

After choosing the first $j$ points for some $j<k$, take any curve $C$ passing through these $j$ points. Pick any point not lying on $C$. Call this point $p_{j+1}$ and continue. The effect is that the sets of curves passing through the first $j$ points are strictly decreasing sets as $j$ increases, which is all you that you want to achieve.

-
Actually, this is just what Francesco said above. –  Peter Kronheimer Oct 21 '11 at 21:12
Actually the question I had in mind is this one mathoverflow.net/questions/78505/… This is merely a special case of that (for which there seems to be a simple answer). –  Ritwik Oct 21 '11 at 21:32

The simplest solution of the problem is very-well known in approximation theory. The desired construction is called Berzolari-Radon: One takes $d+1$ lines $l_0,\dots,l_d$. Take any $d+1$ points on $l_0$. Next take $d$ points on $l_1$ not lying on $l_0$, take $d-1$ points on $l_2$ not lying on $l_0$ and $l_1$, and so on, finally taking $1$ point on $l_d$ not lying on any other previous lines. Thus we get maximal number of $d$-independent points. Of course any of its subsets (of $k$ points) is independent as well.

-