Let $R$ be a finitely generated $\mathbb{Z}$-algebra with an [edit: linear algebraic] action of $G(\mathbb{Z})$ where $G$ is a split simply-connected semisimple group.


Then for any prime $p$ we have a map $R^{G(\mathbb{Z})} \otimes \mathbb{F}\_p \rightarrow (R \otimes \mathbb{F}\_p)^{G(\mathbb{F}\_p)}$. Is this map necessarily surjective for sufficiently large $p$?


Comments:
(1) The simply-connectedness assumption may seem weird; it is made to ensure that $G(\mathbb{Z}) \rightarrow G(\mathbb{F}\_p)$ is surjective so that there is a map at all.


(2) If $G$ is a finite group, then the answer is yes by an averaging argument.


(3) If $G$ is unipotent, then the answer is no. For example, take $x \mapsto x+1$ acting on $k[x]$; there are many invariants in positive characteristic (Artin-Schreier covers!).



Behavior of invariants under reduction mod p
No.
Let $G=SL\_n$, acting on its defining representation $V$, with $n\geq2$.
Let $R=\mathbb{Z}[X\_1,\dots,X\_n]$ be the obvious $\mathbb{Z}$-form of the ring
of polynomial functions on $V$. Let $p$ be a prime. For any $f\in R/pR$ the
product over all $g\in SL\_n(\mathbb{F}\_p)$ of $f\circ g$ is invariant under
$SL\_n(\mathbb{F}\_p)$. But there are no nontrivial $G(\mathbb{Z})$-invariants in $R$
because $G(\mathbb{Z})$ has a Zariski dense orbit in $V$.


