I am currently studying quotients of varieties. In the book "Algebraic Geometry" by J. Harris (http://userpage.fu-berlin.de/aconstant/Alg2/Bib/Harris_AlgebraicGeometry.pdf), Harris wants to show that a quotient always exists in the case where we have a finite group $G$ acting on an affine variety $X$. For that, he shows that the invariant ring $A(X)^G$ (where $A(X)$ is the coordinate ring of $X$) is a finitely generated $K$-algebra, so it's actually the coordinate ring of some affine variety $Y$, now Harris wants to show that $Y$ is indeed the desired quotient, which means that there must be a surjective morphism $\pi :X\rightarrow Y$ which is constant on orbits, so $\pi(x) = \pi(g(x)), \forall g\in G, x\in X$. Since we have the trivial inclusion $A(Y) = A(X)^G \rightarrow A(X)$, this induces a map $\pi : X \rightarrow Y$, which looks like a good candidate for our quotient map. However, I don't quite understand the proof (page 125):
"We have to see that the points of $Y$ correspond to orbits of $G$ on $X$. Suppose that two points $p,q \in X$ are not in the same orbit under $G$. To see that $\pi(p) \neq \pi(q)$, observe that we can find a function $f \in A(X)$ vanishing at $p$ but not at $g(q)$ for any $g \in G$; the product $\prod g^*(f)$ of the images of $f$ under $G$ is then a function $f \in A(Y)$ vanishing at $\pi(p)$ but not at $\pi(q)$."
The highlighted text is the step that troubles me. Intuitively, this sounds fine, however I couldn't really come up with a mathematical explanation. For example, in the case where our field $K$ is $\mathbb{R}$, one can easily see that $f(x_1, ... , x_m) = (x_1 - p_1)^2 + ... + (x_m - p_m)^2$ would be such a polynomial, where $p = (p_1, ... , p_m)$. But how does it work for the general case? Is there maybe a nice way to construct such a polynomial, or how else can this statement be shown?
After that, he seems to show surjectivity of $\pi$:
"To see that $\pi$ is surjective, suppose that $\mathfrak{m} = (h_1,... ,h_k)$ is an ideal in $A(Y)$; We have to show that $\mathfrak{m}\cdot A(X) \neq A(X)$ unless $\mathfrak{m} = A(Y)$ is itself the unit ideal in $A(Y)$. But now ..."
Here I am completely lost. How does the highlighted statement imply surjectivity?
Edit:
I finally understood it. For the first question: Let $\{g(q) : g \in G \} = \{q_1,...,q_n\}$. Let $f \in I(\{p\})$. For the sake of contradiction, assume that there is an $i \in \{ 1, ... ,n\}$ s.t. $f(q_i) = 0$, then we have $I(\{p\}) \subseteq \cup_{i=1}^n I(\{q_i\})$. However, using the prime avoidance lemma, we can see that that $I(\{p\}) \subseteq I(\{q_i\})$ for some $i$. Taking the vanishing set on both sides implies $\{q_i\} \subseteq \{p\} \implies p = q_i$, which is a contradiction.
For the second question: The points of $X$ correspond to maximal ideals in $A(X)$ (same thing for $Y$). We want to show that every point of $Y$ has a point of $X$ as its preimage, but that's equivalent to saying that the image of any maximal ideal $\mathfrak{m} $ in $A(Y)$ is a maximal ideal in $A(X)$. If we have shown that $\mathfrak{m} \cdot A(X) \neq A(X)$, then we know that there exists some maximal ideal $\mathfrak{n}$ s.t. $\mathfrak{m} \cdot A(X) \subseteq \mathfrak{n}$. Intersection with $A(Y)$ yields: $\mathfrak{m} \subseteq A(Y) \cap \mathfrak{n}$, but since $\mathfrak{m}$ is maximal, we get $\mathfrak{m} = A(Y) \cap \mathfrak{n}$, which implies the claim.