Hypersurface in $\mathbb P^n$ containing a linear subspace of dimension $r \geq n/2$ has singular points

Question

I'm trying to prove that if I have a hypersurface $X = Z(F)$ (where $F \in K[x_0, \dots, x_n]_{d>1}$) which contains a linear subspace of dimension $r \geq n/2$ then there exists singular points on $X$.

What I have thought so far:

A point $p=[y_0, \dots, y_n]$ is singular for $Z(F)$ iff $$ \left( \dfrac{\partial F}{\partial x_1} (p), \dots, \dfrac{\partial F}{\partial x_n} (p) \right) = 0.$$ $Z(F)$ contains a subspace of dimension $r$ (let's assume without loss of generality it has equations $x_{r+1} = 0, \dots, x_{n} = 0$), so we can rewrite $$ F = x_{r+1}G_{r+1}+ \dots + x_n G_n,$$ where $G_i \in K[x_0, \dots x_n]_{d-1}0$. The condition for $p$ to be singular becomes $$ \begin{cases} \sum_{i=r+1}^n y_i \partial_0 G_i (p) = 0,\\ \vdots \\ \sum_{i=r+1}^n y_i \partial_r G_i(p)=0,\\ G_{r+1} + \sum_{i=r+1}^n y_i \partial_{r+1} G_i (p) = 0,\\ \vdots \\ G_n + \sum_{i=r+1}^n y_i \partial_n G_i (p) = 0.\end{cases}$$

Why this system of equations should have a non-zero solution $[y_0, \dots, y_n]$ if $r \geq n/2$?

Answer 1

It is a known fact that, if $s \leq k$, $s$ homogeneous polynomials in $K[x_0, \dots, x_k]$ have a common zero in $\mathbb P^k$ (see for example Shafarevich, Basic Algebraic Geometry, Corollary 5, p.71). Taking $s:=n-r$, $k:=r$ gives the thesis.

Answer 2

The statement you want to prove is false: Consider $X := \{ (x_0 :...:x_4) \in \mathbb P^4: x_0 = 0 \}$, $Y := \{ (x_0 :...:x_4) \in \mathbb P^4: x_1 = 0 \}$ two 3-dimensional hyperplanes. The intersection

$$Z:= X \cap Y =\{ (x_0 :...:x_4) \in \mathbb P^4: x_0 = 0, x_1 = 0\} \cong \mathbb P^2$$

is a 2-dimensional subspace of X. We have $n = 4, r = 2 \geq 4/2$, but X is regular, q.e.d.