3
$\begingroup$

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 $\{P_1, \ldots, P_m\}$.

$\endgroup$
  • $\begingroup$ what do you mean by "rank of $\{f_1,...,f_s\}$ ? Do you mean the ranks of the quadratic forms $f_k$, $1\leq k\leq s$? Or something else? $\endgroup$ – Dima Pasechnik Dec 30 '17 at 22:20
  • $\begingroup$ @DimaPasechnik I mean they do not linearly independent (if $c<s$) $\endgroup$ – Alexey Milovanov Dec 30 '17 at 22:28
  • $\begingroup$ @DimaPasechnik I mean the vector space of all quadratic homogenous polynomials over $\mathbb{C}$ $\endgroup$ – Alexey Milovanov Dec 30 '17 at 22:29
  • $\begingroup$ in such a case one writes "dimension of the span" rather than "rank". $\endgroup$ – Dima Pasechnik Dec 31 '17 at 7:07
  • $\begingroup$ @DimaPasechnik thank you! I have corrected it $\endgroup$ – Alexey Milovanov Dec 31 '17 at 9:46
0
$\begingroup$

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 $t \le 3c$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.