This is a generalization of [this](https://mathoverflow.net/questions/288640/linear-homogenous-polynomials-that-generates-one-quadratic-polynomial) 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 $\{P\_1, \ldots, P\_m\}$. 



Linear homogenous polynomials that generates several quadratic polynomials
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 $t \le 3c$. 


