0
$\begingroup$

Let $Q_1, \ldots, Q_k$ and $P_1,\ldots, P_m$ be irredicable homogenous polynomials in $\mathbb{C}[x_0,\ldots, x_n]$ such that $V(Q_1, \ldots, Q_k) \subseteq \cup_i V(P_i)$. Here $V$ is projective variety.

Question 1: Is it true that for some $j$: $V(Q_1, \ldots, Q_k) \subseteq V(P_j)$?

I believe that the answer is "No". So, I also ask:

Question 2: What is the answer for the Question 1 for polynomials $Q_1, \ldots, Q_k$ and $P_1,\ldots, P_m$ of degree at most $2$?

Improve this question
$\endgroup$
2
  • 5
    $\begingroup$ For $n$ equal to $3$, for $Q_1 = x_0x_3-x_1x_2$ and for $Q_2=x_0x_3+x_1x_2$, the common zero locus is contained in the union $\text{Zero}(x_0)\cup \text{Zero}(x_1)\cup\text{Zero}(x_2)\cup \text{Zero}(x_3)$, yet it is not contained in any one of these hyperplanes. $\endgroup$
    – Jason Starr
    Commented Dec 1, 2017 at 21:14
  • 1
    $\begingroup$ Your question is whether $V(Q_1,\dotsc,Q_k)$ is an irreducible variety. Even if the $Q_i$ are each irreducible, the variety that they cut out is not necessarily irreducible. Jason's example is a reducible intersection of two irreducible quadric surfaces in $\mathbb{P}^3$. In the plane, the intersection of two irreducible conics is (generally) a set of $4$ points, which is reducible. $\endgroup$
    – Zach Teitler
    Commented Dec 1, 2017 at 22:03

0

Reset to default

You must log in to answer this question.