Let $X/k$ be a scheme of finite type over a field $k$, $G/k$ be a finite group scheme and suppose $G$ acts on $X$, i.e we have a $k$-morphism $ \mu : G \times_k X \rightarrow X$ satisfies some conditions, see Mumford's book "Abelian Varieties", p. 108.
Suppose that $k$ is algebraically closed and consider the map $\mu$ induced on closed points $G(k) \times X(k) \rightarrow X(k)$, which gives $G(k) \rightarrow \mathrm{Aut}(X(k))$. Under the assumption that each $x \in X$ has an open affine neighborhood $U$ such that $U$ is invariant under $G$, one can form the quotient $X/G$. In particular, quasi-projective varieties have this property. This assumption reduces the construction from general variety to affine case.
For a general field $k$, if one can cover $X$ by open affine subset $U_i$ such that the image of $ G \times_k U_i$ under $\mu$ is contained in $U$, then one reduces the construction to the affine case. But I couldn't figure out if a quasi-projective variety $X$ over $k$ always has such a covering. By working with base change to the algebraic closure of $k$, one gets a $G$-invariant open affine covering of $X_{\overline{k}}$. The images of these open affine subsets of $X_{\overline{k}}$ under the projection to $X$ is still $G$-invariant, but not necessary affine. So for quasi-projective varieties over $k$, do we have the existence of the quotient $X/G$?
Another question is that if it exists, then if it is also quasi-projective? What I know is that, in the classical case ( over algebracially closed field and giving finite group action $H \rightarrow \mathrm{Aut}_k (X))$, if the quotient exists and $X$ is complete, then $X/H$ is also complete.
Best Answer
Let $x\in X$. The morphism $G\to \mathrm{Spec}\;k$ is finite, hence so is the projection $p:G\times X\to X$. In particular it is quasi-finite and hence $p^{-1}(x)$ is a finite set. Consequently so is $\mu(p^{-1}(x))$ which could be called the "naive orbit" of $x$. Any finite set of points in a quasi-projective variety is contained in an affine open subscheme. This is an easy exercise and one doesn't need the ground field to be algebraically closed. You can now proceed as in SGA3, V.5.b) on page 270 f. to produce an open affine neighborhood of $x$ which is invariant under $G$ in the sense you mentioned.