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This question probably follows from standard geometric invariant theory. If true I'd to know a reference for it.

Given a projective scheme $X\rightarrow S$ over the base $S$. Let's assume a finite group $G$ is acting on $X$ and its quotient is an $S$-scheme $X//G$.

Is the quotient projective or at least proper? (I have seen versions of this over fields but not for arbitrary base.)

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I think the reference you want is:

Seshadri, C. S., Geometric reductivity over arbitrary base. Advances in Math. 26 (1977), no. 3, 225–274.

In particular, I think you will find conditions for a positive answer in Theorem 4 and Remark 10.

Here is part of Remark 10:

"We have been so far principally working over affine base schemes [$S$] for the sake of simplicity. Theorems 3 and 4 generalize immediately to arbitrary base (cf. Remark 4)...Let $X = \mathrm{Proj} B$, and $Y = \mathrm{Proj} B^G$ and $\varphi$ the canonical rational morphism $\varphi: X\to Y$ induced by $B^G \subset B$. Then Theorem 4 remains true in this case and of course to say that $Y$ is projective over $S$ we have to assume that $S$ is locally of finite type over a universally Japanese ring."

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  • $\begingroup$ Isn't this a bit of an overkill? Our group here is finite, whereas Seshadri works with any reductive group. It'd be nice to have a simpler proof for finite groups. $\endgroup$
    – Marco Flores
    Commented Feb 1, 2022 at 13:42

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