Let $S\subset \mathbb{C}$ be a finitely generated ring, let $R$ be a finitely generated commutative ring over $S$. Let $G$ be a linear algebraic group over $S$, such that $G\_{\mathbb{C}}$ is reductive. Suppose that Spec$(R)$ is equipped with a $G$-action over $S$.


In this setting, I hope that the following statement holds.


For any large enough prime $p$, given a base change 
$S\to \bf{k}$ to an algebraically closed field of characteristic $p,$ then $G\_{\bf{k}}$-invariants of $R\_{\bf{k}}$ are generated by 
the image of $G$-invariants of $R.$


It feels that the above statement is either explicitly well-known, or at least should follow from a well-known result. Any suggestions or references will be greatly appreciated.



Lifting $G$-invariants from characteristic $p\gg 0$ to characteristic 0 for a reductive algebraic group $G$
We offer two facts and a Theorem.


Let $S$ be a commutative noetherian ring containing $\mathbb Z$ and let $G=G\_S$ be 
 reductive over $S$ in the sense of SGA3. That is, $G$ is smooth over $S$
 with geometric fibers that are connected reductive.
 Let $R$ be a finitely generated commutative $S$-algebra.
 Suppose that $\mathrm{Spec}(R)$ is equipped with a right
$G$-action over $S$.


**Fact 1** Let us be given a base change $S\to \bf k$ to a field of positive characteristic $p$.
For every $x\in (R\otimes\_S {\bf k})^G$ there is an $r\geq1$ so that $x^{p^r}$ lies in the $\bf k$-span of the image 
of $R^G$.


**Fact 2** For almost all primes $p$ the map $R^G\to (R/pR)^G$ is surjective.


Remark. We do not need to distinguish between $(R/pR)^G$ and $(R/pR)^{G\_{S/pS}}$. Base change of the group
is unnecessary when computing invariants or cohomology. (See 1.13. Restriction in our survey 
[Reductivity properties over an affine base](https://arxiv.org/abs/1910.13749).)


**Theorem** Assume further that $R$ is flat over $S$ and that $S$ is of finite global homological dimension.
Then there is an integer $n\geq1$ so that if $S[1/n]\to \bf k$ is a base change to a field, then the map 
$R^G\otimes\_S{\bf k}\to (R\otimes\_S{\bf k})^{G\_{\bf k}}$ is surjective.


To prove Fact 1, let $D$ be the image of $S$ in $\bf k$. As $D\to \bf k$ is flat, we have $(R\otimes\_S {\bf k})^G=(R\otimes\_S D)^G\otimes\_D {\bf k}$,
so it suffices to show that for every $x\in (R\otimes\_S D)^G$ there is an $r\geq1$ so that $x^{p^r}$ lies in the the image 
of $R^G$. Now $G$ is power reductive, so we may apply Proposition 41 
of our paper 
[Power Reductivity over an Arbitrary Base](https://www.math.uni-bielefeld.de/documenta/vol-suslin/franjou_vanderkallen.pdf).
See also my survey 
[Reductivity properties over an affine base](https://arxiv.org/abs/1910.13749).


To prove Fact 2, recall from Theorem 10.5 of 
[Good Grosshans filtration in a family](https://arxiv.org/abs/1109.5822)
 that $H^1(G,R)$ is a finitely generated module over $R^G$. 
 This module is also $\mathbb Z$-torsion. 
 To see this, first take an fppf base change to reduce to the case that $G$ is split over $S$ (see SGA3).
Then $G\_{\mathbb Q}$ makes sense and 
$H^1(G,R)\otimes\_{\mathbb Z}{\mathbb Q}=H^1(G\_{\mathbb Q},R\otimes\_{\mathbb Z}{\mathbb Q})=0$.
Choose $n\geq1$ so that $n$ annihilates the generators of $H^1(G,R)$. Choose $m\geq1$ so that $m$
annihilates the generators of the $\mathbb Z$-torsion ideal of $R$.
Now if $p$ does not divide $mn$, then $\partial$ vanishes in the exact sequence
$0\to R^G\stackrel{\times p}\to R^G\to (R/pR)^G\stackrel\partial\to H^1(G,R)$.


Now we turn to the proof of the Theorem. If $G$ is split over $S$ then we may apply 
Remark 31 and Theorem 33 of 
[Power Reductivity over an Arbitrary Base](https://www.math.uni-bielefeld.de/documenta/vol-suslin/franjou_vanderkallen.pdf)
to obtain $n$ so that $H^i(G,R[1/n])$ vanishes for $i\geq1$.
If $G$ is not yet split the same result is true. Indeed we may by SGA3 do an fppf base change $S\to T$ so that $G\_T$ is split over $T$.
Then we may arrange that $H^i(G,R[1/n])\otimes\_ST=H^i(G\_T,R[1/n]\otimes\_ST)$ vanishes for $i\geq1$. This implies that $H^i(G,R[1/n])$ vanishes for $i\geq1$.


Having chosen $n$ this way we now claim that for every $S$-module $N$ with trivial $G$ action, the 
$H^i(G,R[1/n]\otimes\_SN)$ also vanish for $i\geq1$. This is clear if $N$ is free and then it follows by induction on the projective dimension of the $S$-module $N$.
(If $0\to N'\to F \to N\to0$ is exact, with $F$ free, consider the long exact sequence for $G$-cohomology associated with the exact sequence
$0\to R[1/n]\otimes\_SN'\to R[1/n]\otimes\_SF \to R[1/n]\otimes\_SN\to0$.)


Now let us be given a base change $S[1/n]\to \bf k$ to a field. Let $N$ be the kernel of $S[1/n]\to \bf k$ and let $D$ be its image.
Note that $0\to R\otimes\_SN\to R\otimes\_SS[1/n] \to R\otimes\_SD\to0$ is exact and that $R\otimes\_SN=R[1/n]\otimes\_SN$.
As $H^1(G,R[1/n]\otimes\_SN)=0$ we have a surjection $(R\otimes\_SS[1/n])^G\to (R\otimes\_SD)^G$.
As $D\to \bf k$ is flat, $(R\otimes\_S{\bf k})^{G\_{\bf k}}=(R\otimes\_SD)^G\otimes\_D{\bf k}$.
We see that $(R\otimes\_SS[1/n])^G\otimes\_S{\bf k}$ maps onto $(R\otimes\_S{\bf k})^{G\_{\bf k}}$. But $(R\otimes\_SS[1/n])^G\otimes\_S{\bf k}$ equals
$R^G\otimes\_SS[1/n]\otimes\_S{\bf k}=R^G\otimes\_S{\bf k}$. The result follows.


