# rational points and closed subvariety

Let $k$ is an algebraically closed field of characteristic zero. Assume that $Γ$ is a finite group, acting on an affine scheme $X = \mathrm{Spec}(A)$ by automorphisms (Where $A$ is commutative associative $k$-algebra). Let $X_{rat}$ denote the set of $k$-rational points of $X$ and $\mathcal{O}(X)$ be the structure sheaf of $X$.

1) Given a finite subset $\mathbf{x} ⊆ X_{rat}$ with this property that $y \notin Γ · x$ for distinct $x, y \in \mathbf{x}$. Let $Z = ‎\cup_{x \in ‎\mathbf{x}} ‎\Gamma ‎\cdot x$. Proof that $Z$ is a $Γ$-invariant closed subvariety and the $\mathcal{O}(X) \rightarrow \mathcal{O}(Z)$ is surjective?

2) We say that $X$ is an affine variety if $A$ is finitely generated and reduced. Proof that $X_{rat}$ is $Γ$-invariant and that $X_{rat} = X$ if $X$ is an affine variety.

thank you.

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In the question 2), your $X$ should be the set of maximal ideals of $A$. Otherwise $X_{rat}=X$ holds only when $\dim X=0$. – user18119 Feb 15 '13 at 11:13
Please explain that how $\Gamma$ induces an isomorphism on the residue fields. Thanks – Masoud Feb 15 '13 at 21:53
Any $\gamma\in \Gamma$ induces an isomorphism of local rings $O_{X, x}\to O_{X, \gamma x}$, so it induces an isomorphism of residue fields. – user18119 Feb 15 '13 at 22:07
Thanks for your help! – Masoud Feb 17 '13 at 3:55

1) $\Gamma$ maps a rational point to a rational (hence closed) point because it induces an isomorphism on the residue fields of $x$ and $\gamma(x)$ for any $\gamma\in\Gamma$. So $Z$ is a finite subset consisting in closed points. This implies $Z$ itself is closed (finite union of closed subsets). As $X$ is affine, $\mathcal O(X)\to \mathcal O(Z)$ is thus surjective. Finally $Z$ is invariant by $\Gamma$ by construction (it is a union f orbits.
2) As said above, $\Gamma$ maps a rational point to a rational, so $\Gamma(X_{rat})=X_{rat}$. The equality $X=X_{rat}$ is just the weak form of Nullstellensatz for finitely generated $k$-algebras.