Is there a general geometric characterization for polynomials to be linearly dependent?

Question

Consider $P$ the complex projective plane, and fix a line $L$ in $P$

I had a conjecture, that prof. I. Dolgachev showed me how to prove, that $3$ quadratic polynomials depending on a variable $z \in L$, say $p_1$, $p_2$ and $p_3$, are linearly dependent over $\mathbb{C}$ if and only if:

there exist $4$ points $A$, $B$, $C$ and $D$ in $P \setminus L$, such that the intersections:

$AB$ with $L$, $CD$ with $L$

$AC$ with $L$, $BD$ with $L$

$AD$ with $L$, $BC$ with $L$

are precisely the sets of roots of $p_1$, $p_2$ and $p_3$ respectively. Strictly speaking, one should allow for limiting points too, and so allow for $A$, $B$, $C$ and $D$ to degenerate.

Dolgachev's argument is elegant, and uses the fact that $4$ points in the plane in general position define a pencil of quadrics, namely the family of quadrics which pass through these points. Such a family contains $3$ completely reducible quadrics, which are the $3$ pairs of lines mentioned above. Those are the key ingredients. I can provide more details if needed.

My question is, is there a characterization for $n$ polynomials of degree $n-1$ in one variable $z \in L$ (with $L$ being a complex projective line) to be linearly dependent over $\mathbb{C}$, which is along the same lines as the characterization above for $n=3$? For instance, is there a generalization of the criterion above for $n>3$, using algebraic geometry?

Edit: for $n=4$, the best I could find is the following statement. Consider $7$ points $P_a$, $1 \leq a \leq 7$, in general position in the complex projective plane, and not on $L$. Consider a partition of the $7$ points into a set $S$ of $5$ points, and a set $T$ of $2$ points.

There is a unique quadric passing through all the points in $S$, and this quadric intersects $L$ at two points, counting multiplicity. On the other hand, there is unique line passing through the two points of $T$, which intersects $L$ at one point. So each partition of the $7$ points into two sets as above gives a divisor of degree $3$ on $L$ ($2$ points coming from $S$ and $1$ point coming from $T$).

If I am not mistaken, if we choose $4$ different partitions of the $7$ points, then this leads to $4$ linearly dependent (over $\mathbb{C}$) divisors of degree $3$ on $L$, and I believe the converse is true.

Answer 1

For an arbitrary $n \geq 3$, consider a collection $X$ of $n(n-1)/2 + 1$ points in general position in $P^2(\mathbb{C})$. Subdivide this collection of points into a set $S$ with $n(n-1)/2 - 1$ points and a set $T$ with the remaining $2$ points.

There is a unique planar curve of degree $n-2$ passing through all the points in $S$, and a unique line passing through the two points in $T$. By intersecting with a fixed line $L$ in $P^2(\mathbb{C})$, this gives in total $n-2 + 1 = n-1$ points on $L$, counting multiplicity.

Moreover, by counting the dimension of the linear family of degree $n-1$ curves passing through the full collection $X$, it is clear that if one chooses $n$ different partitions of the same full collection $X$, then the corresponding $n$ divisors of degree $n-1$ on $L$ are linearly dependent over $\mathbb{C}$.

Notice that the family of possible $X$s is $n(n-1) + 2$ dimensional, while the family of $n$ linearly dependent divisors of degree $n-1$ on $L$ is $n(n-1) - 1$ dimensional, so that the family of possible $X$s has a larger dimension than the family of $n$ linearly dependent divsors of degree $n-1$.

Ok! I think I found a suitable generalization.