# Linear homogenous polynomials that generates several quadratic polynomials

This is a generalization of this question.

Let $P_1, \ldots, P_m$, $Q_1, \ldots, Q_k \in \mathbb{C}[x_0,\ldots,x_n]$ be linear homogenous polynomials. Let $f_1, \ldots, f_s$ be a homogenous quadratic irreducible polynomials of degree $2$.

Assume that for every $i$ and for every $j$ the ideal $\langle P_i, Q_j \rangle$ contains some $f_l$.

Assume also that the dimenstion of the span of $\{f_1, \ldots, f_s \}$ (in the vector space of all quadratic homogenous polynomials in $\mathbb{C}[x_0,\ldots,x_n]$) is equal to some constant $c$ .

Question: Is it true that the dimension of the span of $\{P_1, \ldots, P_m \}$ or the dimension of the span of $\{Q_1, \ldots Q_k \}$ is less than some constant (i.e. some function from $c$)?

I can affirmatively answer this question if $s$ (the number of quadratic polynomials) is bounded by a constant:

Consider those polynomials in $\{f_1, \ldots, f_s\}$ that belongs to $\langle P_1, Q_j \rangle$ for some $j$. W.l.o.g. we can assume that this set is $\{f_1, f_2,, \ldots, f_{s'} \}$ for some $s' \le s$.

Consider $M_i:= f_i \cap P_1$ (I mean the intersection of the zeros $f_i$ and $P_1$) for some $i \le s'$.

This set is the zeros of a quadratic form in plane $P_1$ with codimension $1$ (it can not be $P_1$ since $f$ is irreducible). For some $j$ the intersection $Q_j \cap f$ must contain a subspace of codimension $2$. Hence $M_i$ is the union of one or two subspace of codimension $2$. So, there exists at most $2s'$ subspaces of codimension $2$ such that every $Q_j$ must contain at least one of them. Now, it is not difficult to see that the dimension of the span of $\{Q_1, \ldots, Q_k\}$ is bounded by $4s' \le 4 s$. The similar argument works for the dimension of the span of $\{P_1, \ldots, P_m\}$.

Yes.

1) A quadratic homogenous polynomial $f$ (over $\mathbb{C}$) is irreducible iff $\text{rk}(f) \ge 3$. Here $\text{rk}(f)$ is the rank of $f$ as a quadratic form. Indeed, if $\text{rk}(f) < 3$ then it is obvious that $f$ is not irreducible. To prove that in other cases $f$ is irreducivle it is enough to show that polynomial $x^2 + y^2 + z^2$ is irreducible (see, for example https://math.stackexchange.com/questions/486668/x2-y2-z2-is-irreducible-in-mathbb-c-x-y-z).

2) Consider some $f_l$. For some $i$ and $j$ the ideal $\langle P_i, Q_j \rangle$ contains $f_l$. Hence, the intersection $P_i \cap f_l$ is a quadratic form with rank at most $2$. Therefore, $\text{rk}f_l \le 3$. Combine this result with 1) we conclude that $\text{rk}f_l = 3$ for every $l$.

3) Consider the largest linear independent subset of $\{P_1, \ldots, P_m\}$. W.l.o.g. we may assume that this set is $\{P_1, \ldots, P_t\}$ for some $1 \le t \le m$. We will show that $t \le 3c$. (Remind that $c$ is the dimension of the span of $\{f_1, \ldots, f_s\}$). This will give us what we want.

4) Add new $P'_{m+1},\ldots, P'_n$ such that $\{P_1, \ldots, P_t, P'_{t+1},\ldots, P'_n\}$ is a basis. Consider the (symetric) matrixes $A_1, \ldots, A_s$ of quadratic forms $f_1, \ldots, f_s$ in the dual basis of $\{P_1, \ldots, P_t, P'_{t+1},\ldots, P'_n\}$.

5) The span of $\{A_1, \ldots, A_s\} = c$, the rank of every $A_i$ is equal to $3$. Hence, there exist $3c$ numbers of rows such that other rows are linearly depend from these in every matrix $A_i$. The same is true for columns since these matrixes are symmetric.

6) For every $P_j$ the exists $f_i$ such that $f_i \cap P_j$ is a quadratic form of rank $2$. Hence for every $i = 1, \ldots, t$ there exists a matrix $A_j$ such that matrix $A_{j,i}$ (that obtained from $A$ be deleting the $i$th row and the $i$th column) has rank $2$. But from 5) it follows that there are at most $3c$ such numbers $i$. Therefore $t \le 3c$.