$G$- Fixed Point Scheme explicitly

Question

Let $G$ be an abstract finite group acting on a separated $k$-scheme $X$. ($k$ a field; note we can canonically promote $G$ to a $k$- scheme). Then a result by Demazure and Grothendieck (in "Schémas en Groupes") shows that under these assumptions ( indeed one can replace $k$ by more general base $S$, but for my purposes $k$ as base suffice) the $G$-fixed point functor

$$ X^G(T) := \{ x \in X(T) \mid\text{$G(T')$-action on $X(T') $ fixes $x$ for all $T$-schemes $T'$}\} $$

($T$ $k$-scheme) is representable by a closed subscheme $X^G \subset X$.

Question: The result is in general setting given abstractly. Is it possible in "simple" situations to describe $X^G$ as subscheme explicitly? Say $X=\operatorname{Spec}(R)$, where $R:= k[X_1,...,X_n]/I= k[r_1,...,r_n]$ for approp ideal $I$ and $G := \Bbb Z/d =\langle g \rangle $ acting on $R$.
Can we explicitly conceptionally describe the defining ideal $J \subset R$ defining closed $X^G = V(J) \subset X$?

"Morally"/ intuitively this should be in appropriate sense be "dual" to GIT quotient ( ...let assume it exists) which then it given as spec of $R^G$, so tends to expect that this should be somehow related to it.
If I would have beeing asked to make a guess about $J$ I would tend to conjecture that $J = \langle r_1 - g(r_1),..., r_ n-g(r_n) \rangle$, but I not know to justify it precisely. Any idea?

Answer 1

Your guess is right. Note that a functor of points is determined by its value on affine schemes, so we can assume that $T = \operatorname{Spec} A$ and $T'= \operatorname{Spec} B$ is affine.

Then $X^G(T)$ consists of homomorphisms $f\colon R \to A$ such that the $G(B)$ action on homomorphisms $R \to B$ fixes $f$ for all $A$-algebras $B$. You want to check these are exactly the homomorphisms that factor through $R/J$.

It suffices to check that the homomorphisms that factor through $R/J$ are exactly the homomorphism $R \to A$ fixed by the $G(A)$-action, since $R \to A$ factors through $R/J$ if and only if all compositions $R \to A\to B$ factor through $R/J$.

It even suffices to check that the homomorphisms that factor through $R/J$ are exactly the homomorphisms $R \to A$ fixed by $g$, since everything fixed by $G(A)$ is fixed by $g$ and every element of $G(A)$ becomes after passing to some ring $B$ a power of $g$, and again if a homomorphism factors through $R/J$ then its composition with $A \to B$ factors through $R/J$.

But this is clear: $g$ sends the homomorphism $f$ that takes $r_i$ to $f(r_i)$ to the homomorphism $f \circ g$ that takes $r_i$ to $f(g(r_i))$, and these are equal if and only if $f( r_i-g(r_i))=0$ for all $i$.