Quotient variety as a quotient set

Question

In Shafarevich's text Basic Algebraic Geometry, Ch. 1 section 2.3, he describes the quotient of an affine variety $X$ by the action of a finite group $G$ (it is the algebraic set corresponding to the invariant subring of the coordinate ring $R$), and gives a proof that it is equal to the quotient set. Call the quotient map $\varphi$. His proof that $\varphi(x)=\varphi(gx)$ makes sense to me, but his proof of the converse does not. His argument is as follows: Suppose $x\neq gy$ for each $g\in G$. Then, Shafarevich claims, $R$ contains an element $f$ such that $f(gx)=1$ and $f(gy)=0$ for each $g\in G$. Granting this, it is easy to see that the symmetrization of $f$ is $G$-invariant and hence witnesses that $\varphi(x)\neq\varphi(y)$, but I don't see why such a function should exist in the first place. Why is this? Or, if it does not, how can we show that $\varphi(x)\neq\varphi(y)$?

Answer 1

This can be solved using the generalized Chinese remainder theorem. Pick some finite set of distinct closed points $\{x_1,\cdots,x_n\}\subset X$ with corresponding maximal ideals $m_1,\cdots,m_n\subset R$. Then $R\to R/(m_1m_2\cdots m_n)$ is surjective, and as $m_1,\cdots,m_n$ are all pairwise comaximal, we have that $R/(m_1m_2\cdots m_n)\cong (R/m_1)\times\cdots(R/m_n)$. So there's some element in $R$ which maps to $(1,0,\cdots,0)\in (R/m_1)\times\cdots(R/m_n)$, and any such element can play the role of our function $f$.