Let $X$ be $ \operatorname{Proj}(A)$ for some graded ring A, and let $G$ be a finite group acting on $A$ with morphisms of graded rings; consequently $G$ acts on $X$.
I know I can write $X = \bigcup_{f\in A^+}X_f$ where $X_f=\{p \in X \mid f \notin p\}$.
Is it true that the quotient $X/G$, as a topological space, can be written as $X/G = \bigcup_{f\in A^G}p(X_f)$ ? With $p:X \to X/G$ I intend the projection, while $A^G $ is the subring of $G$ invariants element of $A$.
I tried to prove it, but no idea on how to proceed. Any help/hint would be appreciated!
Yes; in fact, $X$ itself can already be written as $\bigcup_{f\in (A^G)^+}X_f$. Suppose $p\in X$ and $p$ is not in $X_f$ for any $f\in (A^G)^+$, i.e. $p$ contains all of $(A^G)^+$. Let $f$ be any element of $A^+$. Enumerate the elements of $G$ as $g_1,\dots,g_n$ and observe that every elementary symmetric polynomial in $g_1f,\dots,g_nf$ is in $(A^G)^+$, and thus in $p$. Thus, the polynomial $q(t)=(t-g_1f)\cdots(t-g_nf)$ has all coefficients except the leading coefficient in $p$. Taking this polynomial mod $p$, then, it reduces to just $t^n$. But $f$ is a root of $q(t)$, so this means $f^n$ is $0$ mod $p$. Since $p$ is prime, this means $f\in p$. Since $f\in A^+$ was arbitrary, this means $p$ is all of $A^+$, which is a contradiction.