# Lifting $G$-invariants from characteristic $p\gg 0$ to characteristic 0 for a reductive algebraic group $G$

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.

• 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. Jan 17 '20 at 10:53
• Yes, thanks! I am certainly fine with making the assumptions you suggested.
– Akk
Jan 17 '20 at 17:38
• 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. Jan 18 '20 at 3:31
• Jan 18 '20 at 9:11
• Indeed it is! Thank you very much!
– Akk
Jan 18 '20 at 16:46

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.

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