Suppose that $R = \mathbb{C}[x_1, \dots, x_n]/I$ is a finitely generated commutative $\mathbb{C}$-algebra which is an integral domain, and suppose that $G \leqslant \text{GL}(n, \mathbb{C})$ is a finite group acting faithfully on $R$. Is it possible for every maximal ideal of $R$ to have a nontrivial stabiliser?
Obviously, by the Nullstellensatz, every maximal ideal has the form $( x_1- \alpha_1, \dots, x_n - \alpha_n)/I$, where $I \subseteq ( x_1- \alpha_1, \dots, x_n - \alpha_n)$. I feel like there's some easy density argument that I'm missing, along the lines of the $\alpha_i$ having to satisfy a polynomial equation.
I originally thought that $S_2$ acting naturally on $\mathbb{C}[x,y]/(x-y)$ gave a counterexample, but here $S_2$ is actually acting trivially! The integral domain hypothesis might not even be necessary, but I thought that I had a counterexample when $R$ wasn't a domain.
Let $G$ be a finite group acting faithfully by regular maps on an irreducible complex algebraic affine variety $X$. Then some closed point in $X$ has trivial stabilizer.
Proof: For $g \in G$ other than the identity, the subvariety $\{x \in X \mid g.x=x\}$ is a proper subvariety (a priori, not necessarily irreducible) in $X$, so it has strictly smaller dimension than $X$. An irreducible complex variety cannot be a finite union of smaller dimension subvarieties. So some point in $X$ is outside the union of these fixed point subvarieties.