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?)

  • $\begingroup$ For the first one, I definitely get confused in the situation that $G$ doesn't have enough $\mathbb Z$-points. Say $\mathbb G_m$ acts on $Spec\ \mathbb Z[x,y]$ with weights $+1,-1$ on the coordinates. What does $A^G$ mean here, where the $\mathbb Z$-points of $\mathbb G_m$ are just $\pm 1$? Do we first think of the ring $A$ as an algebraic variety of infinite type, on which $G$ is acting algebraically? $\endgroup$ Jul 17, 2014 at 21:20
  • 2
    $\begingroup$ @AllenKnutson You can define it algebraically. The group action gives a map $\phi: A \to A \otimes \mathcal{G}$. The invariants are the elements $a$ which satisfy $\phi(a) = a \otimes 1$. $\endgroup$ Jul 17, 2014 at 23:57
  • 1
    $\begingroup$ @AllenKnutson: Similar issues arise to make useful sense of $A^G$ when $G$ is an infinitesimal group scheme in characteristic $p > 0$ (e.g., $\mu_p$), or more generally a non-smooth group (in which case invariance by field-valued points is too limited to be a useful notion). $\endgroup$
    – user27920
    Jul 18, 2014 at 17:18
  • $\begingroup$ For #2, if $G$ is semisimple but not adjoint type and acts on $A$ through its adjoint quotient then $\theta=1$ and $M^{\theta,f}$ is empty but $M//_{\theta}G$ is generally not empty. Please clarify. Do you mean to ask if $M^{\theta,f}$ is a $G$-torsor over an open subspace of $M//_{\theta} G$? In other words, since $M^{\theta,f}$ is a $G$-torsor over the algebraic space $M^{\theta,f}/G$, is #2 asking whether the natural map $M^{\theta,f}/G\rightarrow M//_{\theta}G$ is an open immersion? $\endgroup$
    – user27920
    Jul 19, 2014 at 3:36

2 Answers 2


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 $S$.


  • 2
    $\begingroup$ Dear user52824, I'd like to use your answer in a paper and would like to properly acknowledge your input. If that's okay, could you please reveal your identity (say, by sending an e-mail to [email protected]) so I can mention your name in acknowledgments? Otherwise we'll need to thank an anonymous mathoverflow user. $\endgroup$
    – fnklberg
    Jul 20, 2014 at 7:50
  • $\begingroup$ As for the second question, your version of the question is what I need. In fact, in the situation I'm interested in (construction of quiver varieties in positive characteristic) the action on the set of stable points is free, $X^{f,\theta} = X^\theta$. By the way, I used that formulation in the original question because I thought that for irreducible X the subset $X^{\theta,f}$ is either empty or equals $X^\theta$, isn't that right? $\endgroup$
    – fnklberg
    Jul 20, 2014 at 7:52
  • $\begingroup$ I am sceptical about this claim. Consider for example $G=SO_3$ over $\mathbb Z$ acting on its Lie algebra $M=so_3$ by the adjoint action. Then the invariant ring is $\mathbb Z[c_2]$ where $c_2$ is the coefficient of the char. polynomial of antisymmetric matrices which has degreee 2 in the variables. However after the base change $\mathbb Z\to \mathbb F_2$ the element $c_2$ does not generate the invariant ring. Namely, the determinant is the square of the pfaffian, and the invariant ring now is $\mathbb F_2[c_2,\pi_1]$ where $\mathbb\pi_1$ is... $\endgroup$ Sep 19, 2022 at 19:56
  • $\begingroup$ ... the coefficient of the "pfaffian polynomial" 𝑝𝑓(π‘‡β‹…πΌπ‘‘βˆ’π‘€) which has degreee 1 in the variables. See Chaput and Romagny, On the adjoint quotient of Chevalley groups over arbitrary base schemes, JIMJ 2010. $\endgroup$ Sep 19, 2022 at 19:57
  • $\begingroup$ @MatthieuRomagny, just to have your own remark from elsewhere recorded here, your comment was based on missing the "almost" in the quantifier "almost all primes $p$". $\endgroup$
    – LSpice
    Jun 22, 2023 at 20:20

The answer to the first question is positive. One does not need normality of the generic fiber. By the universal coefficient theorem [Prop I 4.18 (a) in Jantzen Representations of algebraic groups], it suffices to know that $H^1(G,A)$ vanishes generically on $Spec(R)$. But this is implied by theorem 33 of my paper with Franjou Power reductivity over an arbitrary base.

Using further suggestions by user52824 we see that the right setting has a Dedekind domain $D$ as base ring. We localize by inverting a nonzero element of $D$. We first localize to get $A$ flat over $D$, then we localize further to make $H^1(G,A)$ flat over $D$. This uses Theorem 10.5 of Good Grosshans filtration in a family as the big hammer. Then the universal coefficient theorem implies compatibility with any base change. Notice that we no longer try to kill $H^1(G,A)$. Just its $D$-torsion.

  • $\begingroup$ Thanks for the reference. Had I imposed "geometric normality" on the generic fiber (equivalent in generic char. 0) then the argument in my answer would have applied verbatim in positive generic characteristic (say assuming the Dedekind domain to be excellent, depending on which reference is used for the finite generation of the algebra of invariants). Does your Theorem 33 adapt to give an analogous vanishing result on H$^1$ for positive generic characteristic? $\endgroup$
    – user27920
    Jul 19, 2014 at 18:59
  • $\begingroup$ I am not sure what the question is. If your Dedekind domain $R$ is of finite characteristic $p$, then the result just amounts to the trivial fact that you kill $H^1$ when inverting $p$. Our result that you may kill $H^1$ by inverting some integer is of interest only when $R$ itself does not have this property that you may kill it that way. $\endgroup$ Jul 19, 2014 at 19:53
  • $\begingroup$ The formulation of the Theorem as in my answer makes sense (and is non-vacuous) when $K$ has characteristic $p > 0$, and I briefly indicated in my preceding comment that the method used in my answer gives such a result for positive characteristic when assuming geometric normality over $K$. So my question was whether one can expect a version of your Theorem 33 (for positive characteristic) where instead of inverting a nonzero integer, one inverts a nonzero element of the Dedekind domain over which everything is given. $\endgroup$
    – user27920
    Jul 19, 2014 at 20:19
  • $\begingroup$ One may expect it. By inverting a nonzero element one may first kill the $R$ torsion in $A$, then hope to kill the $R$ torsion in $H^1(G,A)$. One hopes that $H^1(G,A)$ is finitely generated as an $A^G$ module. But I only seem to know that when $R$ contains a field, cf. arXiv:1109.5822 $\endgroup$ Jul 19, 2014 at 22:54
  • $\begingroup$ Ah, Theorem 10.5 of the arxiv link gives that over a noetherian base ring $\mathbf{k}$, the cohomology $H := {\rm{H}}^1(G,A)$ is a noetherian $A^G$-module. Taking $\mathbf{k}$ to be a domain, for every $c \in \mathbf{k}$ the $c$-torsion $H[c]$ is an $A^G$-submodule of $H$. Clearly $H[c], H[c'] \subset H[cc']$ for nonzero $c, c' \in \mathbf{k}$. So by the usual acc argument, the torsion submodule of $H$ (over $\mathbf{k}$!) coincides with $H[c]$ for sufficiently divisible nonzero $c \in \mathbf{k}$. So $H[1/c]$ is torsion-free over $\mathbf{k}[1/c]$. Have I understood Theorem 10.5 correctly? $\endgroup$
    – user27920
    Jul 19, 2014 at 23:21

Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.