# Thunv

## The answer for an interesting Question!

trên 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 $(x_1,\cdots,x_n)\ne (0,\cdots,0)$?

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

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

Similarly, we also have

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 😦

• thunv nói:

Quoc Anh doc ma khong hieu thi con ai hieu duoc day!

2. Dino nói:

May ra co thay Thai cua ong hieu thoi.