Suppose we have an algebraic subgroup $G \subset GL(V)$, where $V$ is a finite dimensional vector space over the field of complex numbers $\mathbb{C}$. I'm trying to prove the following:
If $G$-orbits in $V$ can be separated by $G$-invariant polynomials, then $G$ is finite.
I've found out that it's important that the field is $\mathbb{C}$ (I think because $\mathbb{C}$ is closed). Over $\mathbb{R}$ we have a counterexample $SO_1 \subset GL(\mathbb{R}^3)$. But the requirement for subgroup to be algebraic prevented me to construct something of this kind for $\mathbb{C}$. So this condition is also crucial.
I have a feeling it's an easy question, but I didn't succeed in making all those conditions work together. Any hint would be appreciated.
We can take homogeneous generators $f_1, \dots, f_n$ of $\mathbb{C}[V]^G$ and consider a map $f : V \to \mathbb{C}^n$, $f(x) = (f_1(x), \dots, f_n(x))$. Since invariant polynomials separate $0$ from other orbits, $f^{-1}(0) = 0$. This condition together with homogeneity implies that $f$ has finite fibers (it is actually finite, When does a homogeneous morphism have only finite fibers?). Hence all orbits are finite. Hence $G$ is finite.