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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.

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    $\begingroup$ The hypothesis that $G_{\mathbb{C}}$ is reductive looks too weak. One would prefer to have $G$ reductive over $S$ in the sense of SGA3. That is, one wants $G$ to be smooth over $S$ with geometric fibers that are connected reductive. $\endgroup$ Jan 17 '20 at 10:53
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    $\begingroup$ Yes, thanks! I am certainly fine with making the assumptions you suggested. $\endgroup$
    – Akk
    Jan 17 '20 at 17:38
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    $\begingroup$ Perhaps this should be viewed as a question about representation theory of $G$; here much but not everything is known, especially for $p$ large enough. $\endgroup$ Jan 18 '20 at 3:31
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    $\begingroup$ Related to mathoverflow.net/questions/176393/git-over-integers/… $\endgroup$ Jan 18 '20 at 9:11
  • $\begingroup$ Indeed it is! Thank you very much! $\endgroup$
    – Akk
    Jan 18 '20 at 16:46
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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 .)

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 . See also my survey Reductivity properties over an affine base .

To prove Fact 2, recall from Theorem 10.5 of Good Grosshans filtration in a family 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 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.

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  • $\begingroup$ Thanks, this is great! $\endgroup$
    – Akk
    Jan 27 '20 at 4:07

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