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The answer for an interesting Question!

on Tháng Hai 5, 2009

We have a question:  Let Q_1,\cdots,Q_n be  homogeneous polynomials of same degree d>0 in \mathbb C[x_1,\cdots,x_n]. Prove that if the Jacobian \frac{\partial (Q_1,\cdots,Q_n)}{\partial (x_1,\cdots,x_n)}\equiv 0 on \mathbb C^n , then Q_1,\cdots,Q_n have a common zero (x_1,\cdots,x_n)\ne (0,\cdots,0)?

This is an answer:
Let S be a variety defined by S=\{x=(x_1,\cdots,x_n)\in \mathbb C^{n}|Q_3(x)=\cdots=Q_n(x)=0\}. If \dim S> 2, then \dim\{x=(x_1,\cdots,x_n)\in \mathbb C^{n}|Q_j(x)=0, 1\leq j\leq n\}>0. Therefore, Q_1,\cdots,Q_n have a common zero (x_1,\cdots,x_n)\ne (0,\cdots,0). So, we may assume that \dim S=2. Denote by \tilde S the set of all smooth points in S. Let us consider an arbitrary smooth point x_0\in \tilde S. Without loss of generality, we may assume that there exist a neighborhood U of x_0 in S and a homeomorphism h: \mathbb C^2\to U\cap S such that (U,h) is a chart on S and \frac{\partial (Q_3,\cdots,Q_n)}{\partial (x_3,\cdots,x_n)}\ne 0 on U\cap S. It means that Q_j(h(u,v))=0 for every (u,v)\in \mathbb C^2 and for every 3\leq j\leq n.
Let \tilde Q_1(u,v):=Q_1(h(u,v)) and \tilde Q_2(u,v):=Q_2(h(u,v)) for every (u,v)\in \mathbb C^2. Using Euler’s formula and note that
\frac{\partial \tilde Q _l}{\partial u}=\sum_{k=1}^{n}\frac{\partial Q_l}{\partial x_k}\frac{\partial h_k}{\partial u}
, \frac{\partial \tilde Q _l}{\partial v}=\sum_{k=1}^{n}\frac{\partial Q_l}{\partial x_k}\frac{\partial h_k}{\partial v}, \sum_{k=1}^{n}\frac{\partial Q_j}{\partial h_k}\frac{\partial x_k}{\partial u}=0 and \sum_{k=1}^{n}\frac{\partial Q_j}{\partial x_k}\frac{\partial h_k}{\partial v}=0 for 1\leq l\leq 2, 3\leq j\leq n, it follows from \frac{\partial (Q_1,\cdots,Q_n)}{\partial (x_1,\cdots,x_n)}\equiv 0 on \mathbb C^n
that

hinh0

Since \frac{\partial (Q_3,\cdots,Q_n)}{\partial (x_3,\cdots,x_n)}\ne 0 on U\cap S,

hinh1

Similarly, we also have

hinh2
Therefore, there exists a  constant C such that \tilde Q_1(u,v) =C \tilde Q_2(u,v) for every (u,v)\in \mathbb C^2, that is, Q_1(x) =CQ_2(x)  for all x\in U\cap S. Since x_0 is an abitrary smooth point in S, Q_1(x) =CQ_2(x) for all x\in\tilde S. Hence,  Q_1 =C Q_2 on S since \tilde S is dense in S. Consequently, there is a common zero (x_1,\cdots,x_n)\ne (0,\cdots,0) of  Q_1,\cdots,Q_n. This completes the proof.

PS. We would like to thank Prof.  SIU Yum Tong  for showing the way to prove this problem!

Open Problem: are Q_1,\cdots,Q_n linearly independent?  (it is true for case n=2)


3 responses to “The answer for an interesting Question!

  1. Ngô Quốc Anh nói:

    Thú thực là em đọc chẳng hiểu gì cả😦

  2. Dino nói:

    May ra co thay Thai cua ong hieu thoi.

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