If $G$ is a finite subgroup of $GL_n(C)$ then there is a polynomial $f\in C[x_1,...,x_n]$ such that $f(gv)=f(v)$ Problem. Let $G$ be a finite subgroup of $\textrm{GL}_n(\Bbb C)$ and $a \in \Bbb C^n-\{0\}$. Show that there is some polynomial $f \in \Bbb C[x_1,~\cdots,x_n]$ that satisfies the following conditions. 
(1) $f(gv) = f(v)$ for all $g\in G$ and $v\in \Bbb C^n$.
(2) $f(0) = 0$ and $f(a) = 1$.
I'm trying to solve this problem. The condition $f(0)=0$ means that $f$ should have no constant term. Also to be $f(a)=1$, it suffices to have $f(a)\neq 0$, because we can normalize at the end. I got stuck here. I can't see how to use the assumption that $G$ is a finite subgroup of $GL_n(C)$. What I know about it is, that every $g\in G$ should be diagonalizable because $g$ has finite order, and so the minimal polynomial of $g$ splits. But I'm not sure that this fact is useful. Any hints?
 A: Probably the easiest construction is as follow: Let $h(v)$ be any non-trivial polynomial, and then let
$$f(v) = \prod_{g \in G} h(gv).$$
It's both non-constant and invariant by $G$. If $h$ was homogenous of degree one, then $f$ is homogenous of degree $|G|$.
More generally, if $V$ is the representation of $G$ on the linear terms, then one can study the ring of  $G$-invariants $k[V]^{G}$. Hilbert proved that this ring is always finitely generated. It's also easy to show it has transcendence degree $n$. This goes under the name "invariant theory".
A: Let $X$, $Y$ disjoint Zariski closed subsets of $\mathbb{C}^n$.  The system 
$f=0$, $f\in I(X)$
$g=0$, $g \in I(Y)$ 
has not solution. By Hilbert nulstellensatz, the sum of the ideals $I(X)$, $I(Y)$ is $1$. So there exists $f \in I(X)$, $g\in I(Y)$ so that 
$$f+ g = 1$$  Therefore the polynomial $f$ is $0$ on $X$ and $1$ on $Y$.
Assume moreover that $X$, $Y$ are $G$ invariant. Consider $f^G$ the $G$ average of $f$.  We have $f^G$ is $G$ invariant, $f^G=0$ on $X$ and $=1$ on $Y$. 
