Question

Let $S$ be a base scheme. Let $X$ be a scheme over $S$ and let $G$ be a group scheme over $S$ acting on $X$ via $\sigma: G \times_S X \to X$. Suppose that we have a scheme $Y$ over $S$ together with $\varphi: X \to Y$ such that $\varphi \circ \sigma = \varphi \circ p_2$ (where $p_2: G \times_S X \to X$), $\varphi$ is surjective and the image of $(\sigma,1):G \times_S X \to X \times_S X$ is equal to $X \times_Y X$.

Why is this condition equivalent to saying that "the geometric fibres of $\varphi$ are precisely the orbits of the geometric points of $X$, for geometric points over an algebraically closed field of sufficiently high transcendence degree"?

(Source: GIT by Mumford...)

Answer 1

As far as I recall from Mumford's book there are two more conditions an a "geometric quotient", namely that $\varphi$ is submersive and that ${\mathcal O}_Y \to \varphi_*({\mathcal O}_X)^G$ is an isomorphism. Moreover, the condition on geometric fibers is only equivalent to

(*) surjectivity of $\varphi$ and the image of $(\sigma,1)$ is $X \times_Y X$,

under the assumption that $\varphi \circ \sigma = \varphi \circ p_2$ (this condition ensures that $(\sigma,1)$ factors through $X \times_Y X$).

To see this equivaleny it suffices to combine the following two obervations (the first is purely formal, the second is not difficult to prove - and can be found in EGA1).

1. Let $T$ be any $S$-scheme and let $\varphi_T\colon X(T) \to Y(T)$ the map induced by $\varphi$ on $T$-valued points, and similarly for other morphisms. Then the map (of sets) $\varphi_T$ is surjective and the map $(\sigma,1)_T$ has image $(X \times_Y X)(T)$ if and only if the fibers of $\varphi_T$ (again as a map of sets) are precisely the $G(T)$-orbits on $X(T)$ under $\sigma_T$.

2. A morphism of schemes $f\colon X \to Y$ is surjective if and only if for every field $K$ and for every $K$-valued point $y \in Y(K)$ there exists a field extension $L$ of $K$ and $x \in X(L)$ such that $f_L(x) = y_L$, where $y_L$ is the image of $y$ under $Y(K) \to Y(L)$.