1 Introduction

In view of [12], the following theorem will be the key.

Theorem 1

Let k be a noetherian commutative ring, and let G be a finite flat group scheme over k. There is a positive integer n that annihilates \(H^i(G,M)\) for all \(i>0\) and all G-modules M.

Remark 1

If k does not contain \(\mathbb Z\), then one may clearly take n to be the additive order of \(1\in k\).

Remark 2

If G is a constant group scheme, then it is well known that one may take n to be the order of the group [15, Theorem 6.5.8]. (A proof is also implicit in the proof of Theorem 1 below.)

Theorem 2

(Friedlander-Suslin theorem over noetherian base ring) Let k be a noetherian commutative ring, and let G be a finite flat group scheme over k. Let G act rationally on a finitely generated commutative k-algebra A. Then, the cohomology algebra \(H^*(G,A)\) is a finitely generated k-algebra.

As usual, cf. [11, Theorem 1.5], this implies

Corollary 3

If further M is a finitely generated A-module with a compatible G-action, then \(H^*(G,M)\) is a finitely generated \(H^*(G,A)\)-module.

Remark 3

By invariant theory, \(A^G=H^0(G,A)\) is a finitely generated k-algebra [13, Theorem 8, Theorem 19].

Remark 4

If G is a constant group scheme, then Theorem 2 follows from [2, Theorem 8.1]. The original proof in [2] is much more efficient than a proof that relies on [12].

Remark 5

If k is a field of finite characteristic, then Theorem 2 is implicit in [4]. (If G is finite reduced, see Evens. If G is finite and connected, take \(C = A^G\) in [4, Theorem 1.5, 1.5.1]. If G is not connected, one finishes the argument by following [2] as on pages 220–221 of [4].)

Remark 6

The flatness assumption is essential for this kind of representation theory. One needs it to ensure that taking invariants is left exact [6, I 2.10 (4)] and that the category of comodules is abelian [6, I 2.9].

2 Discussion

While the present paper is short, the full proof of Theorem 2 is of book length. Presently, it uses [3, 4, 10, 11], and [12], with ideas taken from [1, 2, 5, 7, 9] ...

3 Conventions

The coordinate ring k[G] is a Hopf algebra [6, Part I, Chapter 2]. The dual Hopf algebra M(G) is the algebra of measures on G. Recall that over a noetherian commutative ring finite flat modules are finitely generated projective. Thus, both k[G] and M(G) are finitely generated projective k-modules.

Following [6, Part I, Chapter 8], we denote by \(M(G)^G_\ell \) the k-module of left invariant measures. Any G-module V may be viewed as a left M(G)-module, and one has \(M(G)^G_\ell V\subseteq V^G=H^0(G,V)\). But in [6, Part I, Chapter 8], it is assumed that k is a field, so we will refer to Pareigis [8] for some needed facts.

We say that G acts rationally on a commutative k-algebra A, if A is also a G-module and the multiplication map \(A\otimes _k A\rightarrow A\) is a map of G-modules.

An abelian group L is said to have bounded torsion if there is a positive integer n so that \(nL_{\textrm{tors}}=0\), where \(L_{\textrm{tors}}\) is the torsion subgroup of L. By [12, Theorem 10.5], bounded torsion is intimately related to finite generation of cohomology algebras.

4 Proofs

Proof of Theorem 2 assuming Theorem 1

First, embed G into some \(\mathop { GL }\nolimits _N\), as follows. As in [6, I 2.7], let \(\rho _r\) denote right regular representation on k[G]. Let \(1\in G(k)\) denote the unit element. Observe that \(\rho _r(g)(f)(1)=f(g)\) for \(g\in G(k)\), \(f\in k[G]\). It follows that \(\rho _r\) is faithful. Furthermore, k[G] is a finitely generated projective k-module, so there is a k-module Q so that \(k[G]\oplus Q\) is free, of rank N say. We let G act trivially on Q and get a faithful action of G on \(k[G]\oplus Q\cong k^N\).

Lemma 4

This defines a closed embedding \(G\subset \mathop { GL }\nolimits _N\).

Proof of Lemma

Choose algebra generators \(f_1,\cdots ,f_r\) of k[G]. If R is a k-algebra then any \(g\in G(R)\) is determined by its values \(f_i(g)\). Let \(\mathcal {G}(R)\) be the subset of \(R^r\) consisting of the \(a=(a_1,\cdots ,a_r)\) such that \(f_i\mapsto a_i\) extends to an algebra homomorphism, denoted \(g_a\), from k[G] to R. In other words, \(\mathcal {G}\) is the subfunctor associated with the closed embedding \(G\subset {\mathbb {A}}^r\) given by the \(f_i\). Note that \(g_a\in G(R)\) for \(a\in \mathcal {G}(R)\).

We seek equations on \(\mathop { GL }\nolimits _N\) that cut out the image of G in \(\mathop { GL }\nolimits _N\). It turns out to be more convenient to cut out an intermediate subscheme X. Let \(L\in \mathop { GL }\nolimits _N(R)\), viewed as an R-linear automorphism of \(R[G]\oplus (R\otimes Q)\). If L is of the form \(\rho _r(g)\oplus \mathop {\textrm{id}}\nolimits _{R\otimes Q}\) for some \(g\in G(R)\), then it satisfies

  • \(L(R[G])=R[G]\),

  • If \(a=(L(f_1)(1),\cdots ,L(f_r)(1))\), then \(a\in \mathcal {G}(R)\),

  • \(g=g_a\).

The first two properties define a k-closed subscheme X of \(\mathop { GL }\nolimits _N\), and the last property shows that G is a retract of X. Therefore, the embedding is closed.

Alternative proof of the Lemma: If k[G] and Q are free k-modules, then one may ignore Q and use the proof of [14, Theorem 3.4], taking \(V=A\). Now use that \(G\rightarrow \mathop { GL }\nolimits _N\) is a closed embedding if there is a cover by affine opens of \({\textrm{Spec}}(k)\) over which it is a closed embedding.\(\square \)

We may thus view \(\mathop { GL }\nolimits _N\) as a group scheme over k with G as k-subgroup scheme. Notice that \(\mathop { GL }\nolimits _N/G\) is affine [6, I 5.5(6)] so that \({\textrm{ ind}}_G^{\mathop { GL }\nolimits _N}\) is exact [6, Corollary I 5.13]. Thus by [6, I 4.6] we may rewrite \(H^*(G,A)\) as \(H^*(\mathop { GL }\nolimits _N,{\textrm{ ind}}_G^{\mathop { GL }\nolimits _N}(A))\), with \({\textrm{ ind}}_G^{\mathop { GL }\nolimits _N}(A)\) a finitely generated k-algebra, by invariant theory. As \(A^G\) is noetherian, it has bounded torsion, and by Theorem 1\(H^{>0}(\mathop { GL }\nolimits _N,{\textrm{ ind}}_G^{\mathop { GL }\nolimits _N}(A))=H^{>0}(G,A)\) also has bounded torsion. Theorem 2 now follows from Theorem 10.5 in [12].\(\square \)

Remains to prove Theorem 1.

Proof of Theorem 1

(\(\bullet \)) Observe that the problem is local in the Zariski topology on \({\textrm{Spec}}(k)\), by the following Lemma.

Lemma 5

Let \(\mathcal M\) be a collection of k-modules. Let \(f_1,\cdots , f_s \in k\) and let \(n_1,\cdots ,n_s\) be positive integers, such that \(n_i (M\otimes k[1/f_i])_{\textrm{tors}}=0\) for all \(M\in \mathcal M\) and all i. If the \(f_i\) generate the unit ideal, then \(n_1\cdots n_s M_{\textrm{tors}}=0\) for all \(M\in \mathcal M\).

Proof of Lemma

Recall that the \(f_i\) generate the unit ideal if and only if the principal open subsets \(D(f_i)= {\textrm{Spec}}(k[1/f_i])\) cover \({\textrm{Spec}}(k)\). Take \(M\in \mathcal M\) and \(m\in M_{\textrm{tors}}\). The annihilator of \(n_1\cdots n_s m\) contains a power of \(f_i\) for each i. These powers generate the unit ideal.\(\square \)

Let H be the Hopf algebra k[G]. In the notations of Pareigis [8], we have a rank 1 projective k-module \(P(H^*)\) that is a direct summand of the k-module \(H^*=M(G)\) [8, Lemma 2, Proposition 3]. If that projective module is free, then Pareigis shows that \(M(G)^G_\ell \) is a direct summand of \(H^*=M(G)\), free of rank one [8, Lemma 3]. By the observation (\(\bullet \)), we may and shall assume that \(P(H^*)\) is indeed free. Take a generator \(\psi \) of \(M(G)^G_\ell \). By Remark 1, we may also assume that k contains \(\mathbb Z\), so that tensoring with \(\mathbb Q\) does not kill everything.

We claim that \(\psi (1)\) is now a unit in \(k_1=\mathbb Q\otimes k\). It suffices to check this at a geometric point \(x={\textrm{Spec}}(F)\) of \({\textrm{Spec}}(k_1)\). As F is an algebraically closed field of characteristic zero, G is a constant group scheme at x by Cartier’s Theorem [14, 11.4, 6.4]. The coordinate ring F[G] is now the F-algebra of maps from the finite group G(F) to F. Evaluation at an element g of G(F) defines a Dirac measure \(\delta _g:F[G]\rightarrow F\), and \(\psi \) is a nonzero scalar multiple of the sum \(\psi _0=\sum _{g\in G(F)}\delta _g\) of the Dirac measures. Evaluating \(\psi _0\) at 1 yields the order of G(F), which is indeed invertible in F.

Put \(\psi _1=(\psi (1))^{-1}\psi \) in \(\mathbb Q\otimes M(G)^G_\ell \). Then, \(\psi _1(1)=1\), and we conclude that \(1\in \mathbb Q\otimes k\psi (1)\). Then, there is \(a\in k\) so that \(a\psi (1)\) is a positive integer n. Put \(\phi =a\psi \). We now observe that \(\phi - n\) annihilates k in k[G] and thus annihilates all invariants in G-modules. And for any G-module M, we have \(\phi (M)\subseteq M^G\) because \(\phi \) is left invariant.

Consider a short exact sequence of G-modules

$$0\rightarrow M'\rightarrow M{\mathop {\rightarrow }\limits ^{\pi }}M''\rightarrow 0.$$

If \(m''\in M''^G\), let \(m\in M\) be a lift. One has \(0=(\phi -n)m''=\pi \phi (m)-nm''\). As \(\phi (m)\in M^G\), we conclude that n annihilates the cokernel of \(M^G\rightarrow M'^G\). Taking M injective, we see that n annihilates \(H^1(G,M')\). This applies to arbitrary G-modules \(M'\). By dimension shift, we get Theorem 1.\(\square \)

Remark 7

One does not need to use (\(\bullet \)), because actually \(\mathbb Q\otimes M(G)^G_\ell \) maps onto \(k_1\), even when \(M(G)^G_\ell \) is not free over k. Indeed, consider the map \(v:\mathbb Q\otimes M(G)^G_\ell \rightarrow k_1\) induced by \(\chi \mapsto \chi (1):M(G)^G_\ell \rightarrow k\). To see that v is surjective, it suffices again to check at the arbitrary geometric point \(x={\textrm{Spec}}(F)\) of \({\textrm{Spec}}(k_1)\).

In fact, \(\mathbb Q\otimes M(G)^G_\ell \) is always free over \(k_1=\mathbb Q\otimes k\).