1. Let $G$ be a reductive algebraic group over ${\mathbb Z}$ (or a finite
localization of a ring of integers $R$ in a number field) acting on an
affine scheme of finite type $M=Spec(A)$ over $R$. 
We can take the invariants $A^G$, and then reduce
modulo a prime $p$: $A^G\otimes\_{\mathbb Z}{\mathbb F}\_q$. We can first
reduce modulo $p$, and then take invariants: 
$(A\otimes\_{\mathbb Z}{\mathbb F}\_q)^{G\_p}$. Is it true that for almost all
primes $p$ the results coincide?
2. Let $M^\theta\subset M$ be the open subset of $\theta$-stable
points for a choice of a character $\theta$ of $G$. 
Let $M^{\theta,f}\subset M^\theta$ be the open subset of $M^\theta$ formed by
all the points with trivial stabilizers. Let $M//\_\theta G$ be the GIT quotient
(the projective spectrum of the ring of $\theta$-semiinvariants of the 
$G$-action on $A$). Is it true that the natural projection  

$M^{\theta,f}\to M//\_\theta G$ is a $G$-torsor? (\'etale locally trivial?)



GIT over integers
The answer to the first question is affirmative if we assume the generic fiber is normal. 


First, some general setup (so it is well-posed to speak of "compatibility with any base change" below). Let $S = {\rm{Spec}}(D)$ be an affine scheme and $X = {\rm{Spec}}(A)$ a finitely presented affine $S$-scheme equipped with an $S$-groupoid $R \rightarrow X \times\_S X$ for which the two maps $R \rightrightarrows X$ are flat. For example, we could have $R = G \times\_S X$ with $R \rightarrow X \times\_S X$ given by $(g,x) \mapsto (gx,x)$ for a flat finitely presented affine $S$-group $G$ equipped with a left action on $X$. We define the "naive quotient" $X/R := {\rm{Spec}}(A^R)$ where
$$A^R = \{a \in A\,|\,p\_1^{\ast}(a) = p\_2^{\ast}(a)\}$$
with equality of pullbacks in the coordinate ring of $R$. For instance, in the "action by a flat affine group of finite type" this recovers Speyer's comment and when working with connected reductive groups over an infinite field it coincides with what Knutson's comment seems to refer to (since in such groups the locus of rational points in schematically dense, due to unirationality over the ground field).


For any affine $S$-scheme $S' = {\rm{Spec}}(D')$ and associated base change $R' \rightarrow X' \times\_{S'} X'$, there is an evident "base change morphism" of $S'$-schemes $X'/R' \rightarrow (X/R)\_{S'}$, and we say that the formation of $X/R$ "commutes with any base change" when these are always isomorphisms. It is clear that the base change map is an isomorphism when $S'$ is $S$-flat (e.g., a localization), as $A^R$ is expressed as a kernel of a linear map between $D$-modules in an evident manner (with the formation of that map compatible with base change in an evident manner, etc.).


We will use a hammer provided by Seshadri to prove:


$\mathbf{Theorem}$: Assume $D$ is a Dedekind domain with fraction field $K$ of characteristic 0, $X\_K$ is normal, and $R$ arises from the action on $X$ by a smooth affine $D$-group $G$ with connected reductive fibers. Then $X/R$ is finite type over $D$, and for some nonzero $d \in D$ the formation of $X/R$ over $D[1/d]$-algebras commutes with any base change on $D[1/d]$ and has geometrically normal fibers.


Localize a bit on $D$ so that the finite type affine $X$ is $D$-flat. Then by standard saturation arguments over Dedekind domains (or see Proposition 3 in section 1 of Seshadri's 1977 paper "Geometric reductivity over arbitrary base" in Advances in Math vol. 26), we can equivariantly realize $X$ as closed in an affine space over $D$ equipped with a linear $G$-action. Since $K$ has characteristic 0, so $D$ is excellent, it follows by Theorem 2 in section 4 of Seshadri's paper that $X/R$ is finite type over $D$. 


By localizing $D$ a bit, we can arrange that all $D$-fibers of $A$ are geometrically normal (since true for generic fiber, as $K$ has characteristic 0, so EGA IV$\_3$, 9.9.5(iii) applies). Since orbit closures in $X\_K$ lie in distinct connected components, the topological description of $X\_K/R\_K$ in terms of $X\_K$ on $\overline{K}$-points (see GIT, or Theorem 3 of section 4 of Seshadri's paper), there is no harm in assuming $X\_K$ is connected, so by some localization on $D$ we may assume that the normal $X\_K$ is connected, hence integral, so $A\_K$ is a domain and thus $A^R$ is a domain. In particular, $A\_K \cap {\rm{Frac}}(A\_K^{R\_K}) = A\_K^{R\_K}$, so $A\_K^{R\_K}$ inherits normality from $A\_K$. This is geometric normality since $K$ has characteristic 0, so by localizing a bit on $D$ we may arrange that the $D$-flat finite type $X/R$ has geometrically normal fibers.


Now $R$ is $D$-flat (as $A$ is), so we see that the $D$-module cokernel $A/A^R$ is torsion-free over $D$, hence is $D$-flat. Thus, for any $D$-algebra $D'$, the natural map $(A^R)\_{D'} \rightarrow A' := A\_{D'}$ obtained by scalar extension is injective. Letting $R' := R\_{D'}$, we therefore have $(A^R)\_{D'} \subset {A'}^{R'}$ for any $D'$. By limit considerations in $D'$ as a $D$-module, the proof of equality is reduced to the case when $D$ is a discrete valuation ring and $D'$ is its residue field $k$ which we may moreover assume (by appropriate flat local dvr base change) is algebraically closed.


Theorem 3 of section 4 in Seshadri's paper shows that the natural map $X \rightarrow X/R$ is surjective and describes the induced map on $k$-points in terms that are *intrinsic* to the $G\_k$-action on $X\_k$. It follows from this that the induced map $f:X\_k/R\_k \rightarrow (X/R)\_k$ is bijective on $k$-points, the analogue for the geometric generic fiber being obvious. A closed set in $X\_k/R\_k$ "is" a $G\_k$-invariant closed set in $X\_k$ (by Theorem 3(iii) in Seshadri's paper applied over the base $k$), yet $X\_k$ is closed inside $X$ and likewise $(X/R)\_k$ is closed in $X/R$, so by that Theorem 3(iii) applied over the base $D$ we see that any $G\_k$-invariant closed set in $X\_k$ has closed image in $(X/R)\_k$. In other words, $X\_k/R\_k \rightarrow (X/R)\_k$ is closed. 


Likewise, if $P\_0$ is a polynomial ring over $k$ in finitely many variables and $P$ is the corresponding polynomial ring over $D$ then $f\_{P\_0}$ is the natural map $X\_{P\_0}/R\_{P\_0} = (X\_k/R\_k)\_{P\_0} \rightarrow ((X/R)\_k)\_{P\_0}$ under which a closed set in the source "is" a $G\_{P\_0}$-stable closed set in $X\_{P\_0}$ and hence is a $G\_P$-stable closed set in $X\_P$, so its image in $X\_P/R\_P$ is closed by Seshadri's Theorem 3(iii) applied over the base ring $P$. But $X\_P/R\_P = (X/R)\_P$ because $P$ is $D$-flat, so we conclude that $f\_{P\_0}$ is closed for all such polynomial rings $P\_0$. In other words, $f$ is "universally closed against affine spaces" (over $k$), so $f$ is closed after base change against curves, so it is proper. Thus, $f$ is a finite radiciel surjection. The *module-finite inclusion* $(A^R)\_k \hookrightarrow A\_k^{R\_k}$ is between normal $k$-algebras (as $A\_k^{R\_k}$ inherits normality from $A\_k$), so the radiciel property implies that this is a direct product of normalization maps arising from some purely inseparable finite extensions of fields. Thus, it is an equality *provided* that it is birational.


Now we return to the situation *before* we passed to local $D$, and have to show that after a bit of localization on $D$ we can arrange that $X\_s/R\_s \rightarrow (X/R)\_s$ is birational for all but finitely many $s \in S$. We arranged harmlessly that $A$ is a domain. The dominant map $X \rightarrow X/R$ corresponding to an injection between domains of finite type over $D$ is generically flat, so by localizing a bit on $D$ we can arrange that there exists some nonzero $a\_0 \in A^R$ such that $(A^R)[1/a\_0] \rightarrow A[1/a\_0]$ is faithfully flat. Some more localization on $D$ ensures that $A/a\_0 A$ is $D$-flat. Since $A^R/a\_0 A^R \rightarrow A/a\_0 A$ is injective, it follows that $A^R/a\_0 A^R$ is also $D$-flat. Thus, $a\_0$ is fiberwise not a zero divisor on $X$ or $X/R$ over $S$. Thus, for all $s \in S$ we get an inclusion $A\_s^{R\_s} \subset (A\_s^{R\_s})[1/a\_0] \cap A\_s$
inside $A\_s[1/a\_0]$ (algebras over the residue field at $s$) and it is an equality.
By inverting $a\_0$ on $A\_s$, we conclude that $X\_s[1/a\_0]/R\_s[1/a\_0] = (X\_s/R\_s)[1/a\_0]$. In other words, the open locus $\{a\_0 \ne 0\}$ is fiberwise-dense in $X/R$ and $X$ over which the formation of $X/R$ commutes with passage to geometric fibers over $S$. 


QED


