Null cone of a $G$-module is $0$ when $G$ is finite Let $G$ a finite group and $W$ a finite dimensional $G$ module. Define the null-cone $$ N_W := \{ w \in W \mid f(w) = 0, \ \forall f \in \mathbb C[W]_+^G \} $$ Here $\mathbb C[W]_+^G$ are the $G$ invariant polynomials with zero constant term. 
I want to show $N_W = 0$ for $G$ finite. Since $N_W$ is a cone and is Zariski closed, it is enough to show it has dimension zero. But I have no ideas how to prove it rigourously. Of course, $N_W$ is the fiber at the origin of the map $\pi : W \to \mathbb C^k, w \to (f_1(w), \dots, f_k(w))$, where $f_i$ generate $\mathbb C[W]^G$. Intuitively this map is a quotient map which parametrize the orbits, and since $G$ has dimension $0$ the dimension of the orbits should be also $0$. But I don't know how to prove it rigourously. 
Thanks in advance ! 
 A: You can show that $\Bbb C[W]$ is a finite $\Bbb C[W]^G$ module: Take any $f\in\Bbb C[W]$ and consider the polynomial 
$$
P(t):=\prod_{g\in G} (t-g.f)
$$
which has coefficients in $\Bbb C[W]^G$ and satisfies $P(f)=0$. This shows that $\Bbb C[W]$ is integral over $\Bbb C[W]^G$, hence module-finite. Therefore, the map $W\to\Bbb C^k$ that you describe is a finite morphism onto its image, which has finite fibers.
A: Here's a direct argument for the requested fact:
Let $w \in W$, $w \not = 0$. To show that $w$ is not in the Null cone, it suffices to find a $G$ invariant form that vanishes at $0$ but not at $w$.
Let $L$ be a linear form : $L = \Sigma a_i x_i$, where $x_i$ is the dual basis, and $a_i$ are scalars. Then $\Pi gL$ is $G$ invariant, and its zero locus is the $G$ orbit of the hyperplane cut out by $L$. 
I claim that if we chose $L$ generically, this $G$ orbit does not contain $w$. 
Intuitively, this is clear - if for some choice of $L$ the $G$ orbit contained $w$, find (intersection of) planes that intersected $w$, and nudge them off of $w$. The rest of the planes couldn't have swung back to catch $w$ quickly enough, so for small nudges the new form doesn't vanish on $w$.
Formally, we can write this out:
The condition that $w$ is in the zero locus of $\Pi gL$ is $\{ \text{ w in zero locus of gL for some $g$}\} = \{ \text{ for some g, gw is in the zero locus of L} \}$. 
So it suffices to pick a $L$ that does not vanish on the $G$ orbit of $w$. But it is always possible to find a hyperplane that misses a finite set of points, since not vanishing at each point imposes a Zariski open condition on the space of hyperplanes. In fact, this shows that generic $L$ works.
