Linear homogenous polynomials that generates one quadratic polynomial

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

Assume that for every $i$ and for every $j$ the polynomial $f$ belongs to the ideal $\langle P_i, Q_j \rangle$.

Is it true that the rank of $\{P_1, \ldots, P_m \}$ (in the vector space of all linear polynomials $\mathbb{C}[x_0,\ldots,x_n]$) or the rank of $\{Q_1, \ldots Q_k \}$ is less than some constant?

(I think the answer of this question can help to solve this question)

UPD(First I asked this question for $f=x_0^2$ and Zach Teitler answered to it.)

I think I have understood my question better:).

It is not difficult to see that $f \in \langle P_i, Q_j \rangle \Leftrightarrow$ for every $x$ in subspace $\{t\in \mathbb{C}^n | P_i(t)=Q_j(t)=0\}$, $f(x)=0$.

So, we can ask some another question.

Let $M$ be a quadratic surface (the zeros of $f$). Assume that $M$ contains some finite set of subspaces of codimension $2$. What can we say about this set of subspaces? Can this set be large?

• all the non-degenerate quadrics in $\mathbb{P}^n$ have the same geometry - so yes, you will be lots of subspaces on it. In the degenerate cases it's also true, as you will have the kernel and the complement, and you build subspaces... – Dima Pasechnik Dec 19 '17 at 20:29
• @DimaPasechnik Do you mean that I can consider $f = x_0^2 + \ldots x_n^2$ consider subspaces that contained this $f$ and then ask is it possible to set these subspace by some $P_1, \ldots P_m$ and $Q_1, \ldots, Q_k$? – Alexey Milovanov Dec 19 '17 at 20:54
• the classical example in $\mathbb{P}^3$ is given by the equation $x_0x_1=x_2x_3$. – Dima Pasechnik Dec 19 '17 at 21:00
• @DimaPasechnik yes, but I need to consider $\mathbb{P}^n$ since I want to prove or disprove that some dimension is less than constant. – Alexey Milovanov Dec 19 '17 at 21:37
• I meant to say that in $\mathbb{P}^3$ you have a nondegenerate example with infinitely many codimension 2 subspaces. Now you can build examples in any dimension $3+k.$ Just add $k$ more variables, and keep the same equation. I.e. your quadric is still of rank 3. – Dima Pasechnik Dec 20 '17 at 22:16

To answer the updated bottom part of the question. The nondegenerate quadrics in $\mathbb{P}^{2m-1}$ (i.e. in the case $n=2m$) are all equivalent to the quadric given by $f_{2m-1}:=\sum_{k=0}^{m-2}x_{k}x_{2m-1-k}=0$, and in $\mathbb{P}^{2m}$ (i.e. in the case $n=2m+1$) they are equivalent to the quadric given by $f_{2m}=f_{2m-1}+x_{2m}^2$. The maximal dimension of a subspace on such a quadric $f$ equals $m$, and there is infinitely many of them: the orthogonal group $O(f)$, which is isomorphic to the matrix group $\{A\in GL(n)\mid AA^\top=I\}$ has either one or two orbits on such subspaces (called maximal totally isotropic subspaces). Each subspace on the quadric lies in a maximal totally isotropic subspace.

Moreover, any quadric $M$ in $\mathbb{P}^{n+1}$, then it is equivalent to a quadric given by either by $g:=f_{2m}$ or by $g:=f_{2m-1}$ for some $m<n/2$. The maximal subspaces on it are direct sums of the kernel of $M$ and a maximal totally isotropic subspace of $g$. Thus in particular as soon as there is one codimension 2 subspace in $M$, there is infinitely many of them there.

There are many sources where this can be found, e.g. the classic book La géométrie des groupes classiques by Dieudonné (translated in many languages).

Each ideal $\langle P_i, Q_j \rangle$ is a radical ideal. So in fact $x_0 \in \langle P_i, Q_j \rangle$. That means that $x_0 \in \operatorname{span}(P_i,Q_j)$. This holds for each $i,j$.

Case 1. Every $P$ is a scalar multiple of $x_0$, or every $Q$ is, or both. Then clearly the $P$'s have rank at most $1$, or the $Q$'s do, or both.

Case 2. Without loss of generality $P_1$ is linearly independent of $x_0$ and so is $Q_1$. Then every $Q$ is in the span of $P_1$ and $x_0$, so the $Q$'s have rank at most $2$; and every $P$ is in the span of $Q_1$ and $x_0$, so the $P$'s also have rank at most $2$. But they can't both have rank $2$. If they did, the $P$'s plus $x_0$ would have the same span as the $Q$'s plus $x_0$. Say this $2$-dimensional subspace has basis $x=x_0$ and $y$. Then $y$ is one of the $P$'s, or a linear combination of them anyway. And it is also a linear combination of the $Q$'s. But $x_0 \notin \langle y,y \rangle$.

No doubt there's a smoother argument, but in any case it follows that the $P$s or the $Q$s (or both) have rank $1$.

Update. I'm not sure if this idea is right, but: If $f \in \langle P_i, Q_j \rangle$ for every $i,j$, then it means $V(P_i,Q_j) \subset V(f)$. The union of those codimension $2$ subspaces is in fact $V(P_1 \dotsm P_m, Q_1 \dotsm Q_k)$! So by Nullstellensatz, there is some power $f^\ell \in \langle P_1 \dotsm P_m, Q_1 \dotsm Q_k \rangle$. Now that should be a radical ideal, if the $V(P_i,Q_j)$s are pairwise distinct subspaces, but it's very late at night here, so I don't know, I might be overlooking something. If it's right that $f$ is in that ideal, then in order to generate a quadratic form it must be the case $m \leq 2$ or $k \leq 2$. So... are you assuming much of anything about the $P$s and $Q$s? You haven't stated explicitly that they are pairwise distinct, let alone any other properties like any $n+1$ are linearly independent, etc. Is it possible that some $P_i=Q_j$, or anything like that?

We will assume that $f$ is irreducible (if $f$ is not irreducible then in fact the argument of Zach Teitler's answer works).

Consider $M:= f \cap P_1$ (I mean the intersection of the zeros $f$ and $P_1$).

This set is the zeros of a quadratic form in plane $P_1$ of codimension $1$ (it can not be $P_1$ since $f$ is irreducible). For every $i$ the intersection of $Q_i$ and $f$ must contain a subspace of codimension $2$. Hence $M$ is the union of one or two subspace of codimension $2$. So, there exists at most $2$ subspace of codimension $2$ such that every $Q_i$ must contain at least one of them. Now, it is not difficult to see that rank of $\{Q_1, \ldots, Q_k\}$ is bounded by $4$. The similar argument works for the rank of $\{P_1, \ldots, P_m\}$.