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Power reductivity over an arbitrary base

Our starting point is Mumford's conjecture, on representations of Chevalley groups over fields, as it is phrased in the preface of"Geometric Invariant Theory". After extending the conjecture appropriately, we show that it holds over an arbitrary commutative base ring. We thus obtain the first fundamental theorem of invariant theory (often referred to as Hilbert's fourteenth problem) over an arbitrary Noetherian ring. We also prove results on the Grosshans graded deformation of an algebra in the same generality. We end with tentative finiteness results for rational cohomology over the integers.


Introduction
The following statement may seem quite well known: Theorem 1. Let k be a Dedekind ring and let G be a Chevalley group scheme over k. Let A be a finitely generated commutative k-algebra on which G acts rationally through algebra automorphisms. The subring of invariants A G is then a finitely generated k-algebra. 1 LMJL -Laboratoire de Mathématiques Jean Leray, CNRS/Université de Nantes. The author acknowledges the hospitality and support of CRM Barcelona during the tuning of the paper. 2 The project started in Nantes, the author being the first visitor in the MATPYL program of CNRS' FR 2962 "Mathématiques des Pays de Loire".

Power surjectivity
To deal with the strong form of integrality we encounter, we find it convenient to make the following definition.
Definition 4. A morphism of k-algebras: φ : S → R is power-surjective if every element of R has a power in the image of φ. It is universally powersurjective if for every k-algebra A, the morphism of k-algebras A ⊗ φ is powersurjective, that is: for every k-algebra A, for every x in A⊗R, there is a positive integer n so that x n lies in the image of A ⊗ φ.
If k contains a field, one does not need arbitrary positive exponents n, but only powers of the characteristic exponent of k (compare [20, Lemma 2.1.4, Exercise 2. 1.5] or Proposition 41 below). Thus if k is a field of characteristic zero, any universally power-surjective morphism of k-algebras is surjective.

Consequences
We start by listing consequences of power reductivity, as defined in the introduction (Definition 2).
Convention 5. An algebraic group over our commutative ring k is always assumed to be a flat affine group scheme over k. Flatness is essential, as we tacitly use throughout that the functor of taking invariants is left exact.
Proposition 6 (Lifting of invariants). Let k be a ring and let G be a powerreductive algebraic group over k. Let A be a finitely generated commutative k-algebra on which G acts rationally through algebra automorphisms. If J is an invariant ideal in A, the map induced by reducing mod J: For an example over Z, see 2.3.2.
Remark 7. Let G be power reductive and let φ : A → B be a power-surjective G-map of k-algebras. One easily shows that A G → B G is power-surjective.
Theorem 8 (Hilbert's fourteenth problem). Let k be a Noetherian ring and let G be an algebraic group over k. Let A be a finitely generated commutative k-algebra on which G acts rationally through algebra automorphisms. If G is power-reductive, then the subring of invariants A G is a finitely generated kalgebra.
Proof. We apply [20, p. 23-26]. It shows that Theorem 8 relies entirely on the conclusion of Proposition 6, which is equivalent to the statement [20, Lemma 2.4.7 p. 23] that, for a surjective G-map φ : A → B of k-algebras, the induced map on invariants A G → B G is power-surjective. To prove that power reductivity implies this, consider an invariant b in B, take for L the cyclic module k.b and for M any submodule of A such that φ(M ) = L. We conclude with a commuting diagram: Theorem 9 (Hilbert's fourteenth for modules). Let k be a Noetherian ring and let G be a power-reductive algebraic group over k. Let A be a finitely generated commutative k-algebra on which G acts rationally, and let M be a Noetherian A-module, with an equivariant structure map A ⊗ M → M . If G is power-reductive, then the module of invariants M G is Noetherian over A G .
Proof. As in [14, 2.2], consider either the symmetric algebra of M on A, or the 'semi-direct product ring' A ⋉ M as in Proposition 57, whose underlying G-module is A ⊕ M , with product given by (a 1 , m 1 )(a 2 , m 2 ) = (a 1 a 2 , a 1 m 2 + a 2 m 1 ).

2.3.1
Let k = Z. Consider the group SL 2 acting on 2 × 2 matrices a b c d by conjugation. Let L be the line of homotheties in M := M 2 (Z). Write V # to indicate the dual module Hom Z (V, Z) of a Z-module V . The restriction: Taking SL 2 -invariants: the trace t = a + d is sent to 2λ, so λ does not lift to an invariant in M # . The determinant D = ad − bc is sent to λ 2 however, illustrating power reductivity of SL 2 .

2.3.2
Similarly, the adjoint action of SL 2 on sl 2 is such that u(a) := 1 a 0 1 sends X, H, Y ∈ sl 2 respectively to X + aH − a 2 Y, H − 2aY, Y . This action extends to Documenta Mathematica · Extra Volume Suslin (2010) 171-195 the symmetric algebra S * (sl 2 ), which is a polynomial ring in variables X, H, Y . Take k = Z again. The mod 2 invariant H does not lift to an integral invariant, but H 2 + 4XY is an integral invariant, and it reduces mod 2 to the square H 2 in F 2 [X, H, Y ]. This illustrates power reductivity with modules that are not flat, and the strong link between integral and modular invariants.

2.3.3
Consider the group U of 2 × 2 upper triangular matrices with diagonal 1: this is just an additive group. Let it act on M with basis {x, y} by linear substitutions: u(a) sends x, y respectively to x, ax + y. Sending x to 0 defines M → L, and since (S * M ) U = k[x], power reductivity fails.

Equivalence of power reductivity with property (Int)
Following [14], we say that a group G satisfies (Int) if (A/J) G is integral over the image of A G for every A and J with G action. Note that if (A/J) G is a Noetherian A G -module (compare Theorem 9), it must be integral over the image of A G . As explained in [14,Theorem 2.8], when k is a field, the property (Int) is equivalent to geometric reductivity, which is equivalent to power-reductivity by [20,Lemma 2.4.7 p. 23]. In general, property (Int) is still equivalent to power-reductivity. But geometric reductivity in the sense of [19] looks too weak.
Proposition 10. An algebraic group G has property (Int) if, and only if, it is power-reductive.
Proof. By Proposition 6, power reductivity implies property (Int). We prove the converse. Let φ : M → L be as in the formulation of power reductivity in Definition 2. Choose a generator b of L. Property (Int) gives a polynomial t n + a 1 t n−1 + · · · + a n with b as root, and with a i in the image of S * (ϕ) : (S * M ) G → S * L. As b is homogeneous of degree one, we may assume a i ∈ S i ϕ((S i M ) G ). Write a i as r i b i with r i ∈ k. Put r = 1 + r 1 + · · · r n . Then rb n = 0, and r (n−1)! annihilates b n! . Since a n!/i i lies in the image of S n! ϕ : (S n! M ) G → S n! L, the cokernel of this map is annihilated by r n!/i i . Together r (n−1)! and the r n!/i i generate the unit ideal. So the cokernel vanishes.
Example 11. Let G be a finite group, viewed as an algebraic group over k. Then A is integral over A G , because a is a root of g∈G (x − g(a)). (This goes back to Emmy Noether [18].) Property (Int) follows easily. Hence G is power reductive.
Theorem 12 (Mumford conjecture). A Chevalley group scheme is powerreductive for every base.
By a Chevalley group scheme over Z, we mean a connected split reductive algebraic Z-group G Z , and, by a Chevalley group scheme over a ring k, we mean an algebraic k-group G = G k obtained by base change from such a G Z . We want to establish the following: Property. Let k be a commutative ring. Let L be a cyclic k-module with trivial G-action. Let M be a rational G-module, and let ϕ be a G-module map from M onto L. Then there is a positive integer d such that the d-th symmetric power of ϕ induces a surjection:

Reduction to local rings
We first reduce to the case of a local ring. For each positive integer d, consider the ideal in k formed by those scalars which are hit by an invariant in (S d M ) G , and let: be its radical. Note that these ideals form a monotone family: if d divides d ′ , then I d is contained in I d ′ . We want to show that I d (k) equals k for some d.
To that purpose, it is enough to prove that for each maximal ideal M in k, the localized I d (k) (M) equals the local ring k (M) for some d. Notice that taking invariants commutes with localization. Indeed the whole Hochschild complex does and localization is exact. As a result, the localized I d (k) (M) is equal to the ideal I d (k (M) ). This shows that it is enough to prove the property for a local ring k.
For the rest of this proof, k denotes a local ring with residual characteristic p.

Reduction to cohomology
As explained in Section 3.5, we may assume that G is semisimple simply connected. Replacing M if necessary by a submodule that still maps onto L, we may assume that M is finitely generated. We then reduce the desired property to cohomological algebra. To that effect, if X is a G-module, consider the evaluation map on the identity id X : Hom k (X, X) # → k (we use V # to indicate the dual module Hom k (V, k) of a module V ). If X is k-free of finite rank d, then S d (Hom k (X, X) # ) contains the determinant. The determinant is G-invariant, and its evaluation at id X is equal to 1. Let b a k-generator of L and consider the composite: Its d-th power S d ψ sends the determinant to b d . Suppose further that ψ lifts to M by a G-equivariant map. Then, choosing d to be the k-rank of X, the d-th power of the resulting map S d (Hom k (X, X) # ) → S d M sends the determinant to a G-invariant in S d M , which is sent to b d through S d ϕ. This would establish the property.
The existence of a lifting would follow from the vanishing of the extension group: Ext 1 G ((Hom k (X, X) # , Kerϕ), or, better, from acyclicity, i.e. the vanishing of all positive degree Ext-groups. Inspired by the proof of the Mumford conjecture in [6, (3.6)], we choose X to be an adequate Steinberg module. To make this choice precise, we need notations, essentially borrowed from [6,2].

Notations
We decide as in [11], and contrary to [12] and [6], that the roots of the standard Borel subgroup B are negative. The opposite Borel group B + of B will thus have positive roots. We also fix a Weyl group invariant inner product on the weight lattice X(T ). Thus we can speak of the length of a weight. For a weight λ in the weight lattice, we denote by λ as well the corresponding one-dimensional rational B-module (or sometimes B + -module), and by ∇ λ the costandard module (Schur module) ind G B λ induced from it. Dually, we denote by ∆ λ the standard module (Weyl module) of highest weight λ. So ∆ λ = ind G B + (−λ) # . We shall use that, over Z, these modules are Z-free [11, II Ch. 8]. We let ρ be half the sum of the positive roots of G. It is also the sum of the fundamental weights. As G is simply connected, the fundamental weights are weights of B. Let p be the characteristic of the residue field of the local ring k. When p is positive, for each positive integer r, we let the weight σ r be (p r − 1)ρ. When p is 0, we let σ r be rρ. Let St r be the G-module ∇ σr = ind G B σ r ; it is a usual Steinberg module when k is a field of positive characteristic.

3.4
We shall use the following combinatorial lemma: Lemma 13. Let R be a positive real number. If r is a large enough integer, for all weights µ of length less than R, σ r + µ is dominant.
So, if r is a large enough integer to satisfy the condition in Lemma 13, for all G-modules M with weights that have length less than R, all the weights in σ r ⊗ M are dominant. Note that in the preceding discussion, the G-module M is finitely generated. Thus the weights of M , and hence of Kerϕ, are bounded. Thus, Theorem 12 is implied by the following proposition.
Proposition 14. Let R be a positive real number, and let r be as in Lemma 13 . For all local rings k with characteristic p residue field, for all G-module N with weights of length less than R, and for all positive integers n: Proof. First, the result is true when k is a field. Indeed, we have chosen St r to be a self-dual Steinberg module, so, for each positive integer n: Vanishing follows by [6, Corollary (3.3')] or the proof of [6, Corollary (3.7)]. Suppose now that N is defined over Z by a free Z-module, in the following sense: We then use the universal coefficient theorem [4, A.X.4.7] (see also [11, I.4.18]) to prove acyclicity in this case. Specifically, let us note Y Z := Hom Z ((St r ) Z , (St r ) Z ) ⊗ N Z , so that, using base change (Proposition 16 for λ = σ r ): This cohomology is computed [7, II.3] (see also [11, I.4.16]) by taking the homology of the Hochschild complex C(G, Y Z ⊗ V ). This complex is isomorphic to the complex obtained by tensoring with V the integral Hochschild complex C(G Z , Y Z ). Since the latter is a complex of torsion-free abelian groups, we deduce, by the universal coefficient theorem applied to tensoring with a characteristic p field k, and the vanishing for the case of such a field, that: H n (G Z , Y Z ) ⊗ k = 0, for all positive n. We apply this when k is the residue field of Z (p) ; note that if p = 0 the residue field k is just Q. Since the cohomology H n (G Z , Y Z ) is finitely generated [11, B.6], the Nakayama lemma implies that: ) is a complex of torsion-free Z (p) -modules, we thus can apply the universal coefficient theorem to tensoring with V . The vanishing For the general case, we proceed by descending induction on the highest weight of N . To perform the induction, we first choose a total order on weights of length less than R, that refines the usual dominance order of [11,II 1.5]. Initiate the induction with N = 0. For the induction step, consider the highest weight µ in N and let N µ be its weight space. By the preceding case, we obtain vanishing for ∆ µ Z ⊗ Z N µ . Now, by Proposition 21, ∆ µ Z ⊗ Z N µ maps to N , and the kernel and the cokernel of this map have lower highest weight. By induction, they give vanishing cohomology. Thus Hom k (St r , St r ) ⊗ N is in an exact sequence where three out of four terms are acyclic, hence it is acyclic.
This concludes the proof of Theorem 12.

Reduction to simply connected group schemes
Let Z Z be the center of G Z and let Z be the corresponding subgroup of G. It is a diagonalisable group scheme, so M Z → L is also surjective. We may replace M with M Z and G with G/Z, in view of the general formula M G = (M Z ) G/Z , see [11, I 6.8(3)]. So now G has become semisimple, but of adjoint type rather than simply connected type. So choose a simply connected Chevalley group schemeG Z with centerZ Z so thatG Z /Z Z = G Z . We may now replace G with G.
Remark 15. Other reductions are possible, to enlarge the supply of power reductive algebraic groups. For instance, if G has a normal subgroup N so that both N and G/N are power reductive, then so is G (for a proof, use Remark 7). And if k → R is a faithfully flat extension so that G R is power reductive, then G is already power reductive. So twisted forms are allowed, compare the discussion in [19, p. 239].

Generalities
This section collects known results over an arbitrary base, their proof, and correct proofs of known results over fields, for use in the other sections. The part up to subsection 4.3 is used, and referred to, in the previous section.

Notations
Throughout this paper, we let G be a semisimple Chevalley group scheme over the commutative ring k. We keep the notations of Section 3.3. In particular, the standard parabolic B has negative roots. Its standard torus is T , its unipotent radical is U . The opposite Borel B + has positive roots and its unipotent radical is U + . For a standard parabolic subgroup P its unipotent radical is R u (P ).

4.2
We first recall base change for costandard modules.
Proposition 16. Let λ be a weight, and denote also by λ = λ Z ⊗ k the Bmodule k with action by λ. For any ring k, there is a natural isomorphism: In particular, ind G B λ is nonzero if and only if λ is dominant.
Proof. First consider the case when λ is not dominant. Then ind G B λ vanishes when k is a field [11, II.2.6], so both ind G Z B Z λ Z and the torsion in R 1 ind G Z B Z λ Z must vanish. Then ind G B λ vanishes as well for a general k by the universal coefficient theorem. In the case when λ is dominant, [11,II 8.8(2)]. Thus, by [11, I.4 Proposition 17 (Tensor identity for weights). Let λ be a weight, and denote again by λ the B-module k with action by λ. Let N be a G-module. There is a natural isomorphism: Remark 18. The case when N is k-flat is covered by [11, I.4.8].
We warn the reader against Proposition I.3.6 in the 1987 first edition of the book.
Indeed, suppose we always had ⊗ λ ⊗ N triv , given in non-functorial notation by: To make these maps into G-module maps, one must use the given G-action on N as the action on N triv . So B acts on N , but not N triv , and for G it is the other way around. In particular, if M has highest weight λ, then there is a natural map from ∆ λ Z ⊗ Z M λ to M , its kernel has lower weights, and λ is not a weight of its cokernel.
Proof. By the tensor identity Proposition 17: . If M has highest weight λ, then M λ = Hom B + (λ, M ). Tracing the maps, the second part follows from Proposition 20.

Notations
We now recall the notations from [13, §2.2]. Let the Grosshans height function ht : X(T ) → Z be defined by: the Grosshans filtration, and we call its associated graded the Grosshans graded gr M of M . We put: hull ∇ (gr M ) = ind G B M U + . Let A be a commutative k-algebra on which G acts rationally through k-algebra automorphisms. The Grosshans graded algebra gr A is given in degree i by:

Erratum
When k is a field, one knows that gr A embeds in a good filtration hull, which Grosshans calls R in [10], and which we call hull ∇ (gr A): When k is a field of positive characteristic p, it was shown by Mathieu [16,Key Lemma 3.4] that this inclusion is power-surjective: for every b ∈ hull ∇ (gr A), there is an r so that b p r lies in the subalgebra gr A. Frank Grosshans has pointed out that the proof of this sublemma is not convincing beyond the reduction to the affine case. Later A. J. de Jong actually gave a counterexample to the reasoning. The result itself is correct and has been used by others. As power surjectivity is a main theme in this paper, we take the opportunity to give a corrected treatment. Mathieu's result will be generalized to an arbitrary base k in Section 5. Suppose we can show that for every i, j we have m i,j so that b p m i,j j ∈ A + p i B. Then for every i we have m i so that b p m i ∈ A + p i B for every b ∈ B. Then b p m 1 +···ms ∈ A + p 1 · · · p s B for every b ∈ B. As p 1 · · · p s is nilpotent, one finds m with b p m ∈ A for all b ∈ B. The upshot is that it suffices to prove the sublemma for the inclusion A/p i ⊂ B/p i B. [It is an inclusion because there is a prime ideal q i in B with A ∩ q i = p i .] Therefore we further assume that A is a domain. Let r denote the nilradical of B. If we can show that for all b ∈ B there is m with b p m ∈ A + r, then clearly we can also find a u with b p u ∈ A. So we may as well replace A ⊂ B with A ⊂ B/r and assume that B is reduced. But then at least one component of Spec(B) must map onto Spec(A), so bijectivity implies there is only one component. In other words, B is also a domain. Choose t so that the field extension Frac(A) ⊂ Frac(AB p t ) is separable. (So it is the separable closure of Frac(A) in Frac(B).) As X → Spec(AB p t ) is also bijective, we have that Spec(AB p t ) → Spec(A) is bijective. It clearly suffices to prove the proposition for A ⊂ AB p t . So we replace B with AB p t and further assume that Frac(B) is separable over Frac(A). Now X → Y has a degree which is the degree of the separable field extension. There is a dense subset U of Y so that this degree is the number of elements in the inverse image of a point of U . [Take a primitive element of the field extension, localize to make its minimum polynomial monic over A, invert the discriminant.] Thus the degree must be one because of bijectivity. So we must now have that Frac(B) = Frac(A). Let c = { b ∈ B | bB ⊂ A } be the conductor of A ⊂ B. We know it is nonzero. If it is the unit ideal then we are done. Suppose it is not. By induction applied to A/c ⊂ B/c (we need the induction hypothesis for the original problem without any of the intermediate simplifications) we have that for each b ∈ B there is an m so that b p m ∈ A + c = A.

4.5
This subsection prepares the ground for the proof of the theorems in Section 5. We start with the ring of invariants k[G/U ] of the action of U by right translation on k[G].
Lemma 23. The k-algebra k[G/U ] is finitely generated.
Proof. We have: By Proposition 16, this equals the sum ⊕ λ ∇ λ over dominant weights λ only. When G is simply connected, every fundamental weight is a weight, and the monoid of dominant λ is finitely generated. In general, some multiple of a fundamental weight is in X(T ) and there are only finitely many dominant weights modulo these multiples. So the monoid is still finitely generated by Dickson's Lemma [5, Ch. 2 Thm. 7]. The maps ∇ λ ⊗ ∇ µ → ∇ λ+µ are surjective for dominant λ, µ, because this is so over Z, by base change and surjectivity for fields [11,II,Proposition 14.20]. This implies the result.
In the same manner one shows: Lemma 24. If the k-algebra A U is finitely generated, so is hull ∇ gr A = ind G B A U + .
Proof. Use that A U + is isomorphic to A U as k-algebra.
Lemma 25. Suppose k is Noetherian. If the k-algebra A with G action is finitely generated, then so is A U .

Now apply Lemma 23 and Theorem 3.
Lemma 26. If M is a G-module, there is a natural injective map Proof. By Lemma 19, the weights of M U + are dominant. If one of them, say λ, has Grosshans height i, the universal property of Weyl modules (Proposition 21) shows that (M U + ) λ is contained in a G-submodule with weights that do not have a larger Grosshans height. So the weight space (M U + ) λ is contained in M ≤i , but not M <i . We conclude that the T -module ⊕ i (gr i M ) U + may be identified with the T -module M U + . It remains to embed gr i M into ind G B (gr i M ) U + .
The T -module projection of gr i M onto (gr i M ) U + may be viewed as a Bmodule map, and then, it induces a G-module map gr i M → ind G B ((gr i M ) U + ), which restricts to an isomorphism on (gr i M ) U + by Proposition 20. So its kernel has weights with lower Grosshans height, and must therefore be zero.
In the light of Lemma 26, one may write: Definition 27. A G-module M has good Grosshans filtration if the embedding of gr M into hull ∇ (gr M ) is an isomorphism.
It is worth recording the following characterization, just like in the case where k is a field.
Proposition 28 (Cohomological criterion). For a G-module M , the following are equivalent.
i. M has good Grosshans filtration.
ii. Conversely, suppose that M does not have good Grosshans filtration. Choose i so that M <i has good Grosshans filtration, but M ≤i does not. Choose λ so that Hom(∆ λ , hull(gr i M )/ gr i M ) is nonzero. Note that λ has Grosshans height below i. As Hom(∆ λ , hull(gr i M )) vanishes, Ext 1 is nonzero as well. Now use that Hom(∆ λ , M/M ≤i ) vanishes, and conclude that H 1 (G, M ⊗ k[G/U ]) does not vanish.

5.1
When G is a semisimple group over a field k, Grosshans has introduced a filtration on G-modules. As recalled in Section 4.3, it is the filtration associated to the function defined on X(T ) by: ht γ = α>0 γ, α ∨ . Grosshans has proved some interesting results about its associated graded, when the G-module is a kalgebra A with rational G action. We now show how these results generalize to an arbitrary Noetherian base k, and we draw some conclusions about H * (G, A). All this suggests that the finite generation conjecture of [13] (see also [14]) deserves to be investigated in the following generality.
Problem. Let k be a Noetherian ring and let G be a Chevalley group scheme over k. Let A be a finitely generated commutative k-algebra on which G acts rationally through algebra automorphisms. Is the cohomology ring H * (G, A) a finitely generated k-algebra?
Let k be an arbitrary commutative ring.
Theorem 29 (Grosshans hull and powers). The natural embedding of gr A in hull ∇ (gr A) is power-surjective.
This will then be used to prove: Theorem 30 (Grosshans hull and finite generation). If the ring k is Noetherian, then the following are equivalent.
i. The k-algebra A is finitely generated; ii. For every standard parabolic P , the k-algebra of invariants A Ru(P ) is finitely generated; iii. The k-algebra gr A is finitely generated; iv. The k-algebra hull ∇ (gr A) is finitely generated.
Remark 31. Consider a reductive Chevalley group scheme G. As the Grosshans height is formulated with the help of coroots α ∨ , only the semisimple part of G is relevant for it. But of course everything is compatible with the action of the center of G also. We leave it to the reader to reformulate our results for reductive G. We return to the assumption that G is semisimple.
Theorem 32. Let A be a finitely generated commutative k-algebra. If k is Noetherian, there is a positive integer n so that: n hull ∇ (gr A) ⊆ gr A.
In particular H i (G, gr A) is annihilated by n for positive i. This is stronger than the next result.
Theorem 33 (generic good Grosshans filtration). Let A be a finitely generated commutative k-algebra. If k is Noetherian, there is a positive integer n so that A[1/n] has good Grosshans filtration. In particular Remark 34. Of course A[1/n] may vanish altogether, as we are allowed to take the characteristic for n, when that is positive.
Theorem 35. Let A be a finitely generated commutative k-algebra. If k is Noetherian, for each prime number p, the algebra map gr A → gr(A/pA) is power-surjective.

5.2
We start with a crucial special case. Let k = Z. Let λ ∈ X(T ) be dominant. Let S ′ be the graded algebra with degree n part: Let us view ∆ λ as a submodule of ∇ λ with common λ weight space (the 'minimal admissible lattice' embedded in the 'maximal admissible lattice'). Let S be the graded subalgebra generated by ∆ λ in the graded algebra S ′ . If we wish to emphasize the dependence on λ, we write S ′ (λ) for S ′ , S(λ) for S.
given by the 'linear system' ∇ λ on G/B. The projective scheme Proj(S ′ ) corresponds with its image, which, by direct inspection, is isomorphic to G/P , where P is the stabilizer of the weight space with weight −λ of ∇ # λ . Indeed Proof. As explained in [EGA II, 2.8.1], the domain G(φ) contains the principal open subset D + (s) of Proj(S ′ ) for any s ∈ S 1 . Consider in particular a generator s of the λ weight space of ∇ λ . It is an element in S 1 , and, by Lemma 20, it generates the free k-module Γ(P/P, L(λ)). Thus, the minimal Schubert variety P/P is contained in D + (s). We then conclude by homogeneity: s is also U + invariant, so in fact the big cell Ω = U + P/P is contained in D + (s), and the domain G(φ) contains the big cell Ω. Then it also contains the Weyl group translates wΩ, and thus it contains all of G/P . Lemma 37. The graded algebra S ′ is integral over its subalgebra S.
Proof. We also put a grading on the polynomial ring S ′ [z], by assigning degree one to the variable z.

5.3
We briefly return to power surjectivity for a general commutative ring k.
Definition 40. Let t be a positive integer and let f : Q → R a k-algebra homomorphism. We say that f is t-power-surjective if for every x ∈ R there is a power t n with x t n ∈ f (Q).
Proposition 41. Let f : Q → R be a k-algebra homomorphism and Y a variable.
is power-surjective, then Q → R/pR is ppower-surjective for every prime p; • Assume t is a positive integer such that tR ⊂ f (Q). If Q → R/pR is p-power-surjective for every prime p dividing t, then f is universally power-surjective.
is power-surjective. Let x ∈ R/pR. We have to show that x p n lifts to Q for some n.
is also power-surjective. Choose n prime to p and m so that (x + Y ) np m lifts to Q[Y ]. Rewrite (x + Y ) np m as (x p m + Y p m ) n and note that the coefficient nx p m of Y (n−1)p m must lift to Q. Now use that n is invertible in k/pk. Next suppose tR ⊂ f (Q) and Q → R/pR is p-power-surjective for every prime p dividing t. Let C be a k-algebra. We have to show that f ⊗C : Q⊗C → R⊗C is power-surjective. Since f ⊗ C : Q ⊗ C → R ⊗ C satisfies all the conditions that f : Q → R does, we may as well simplify notation and suppress C. For x ∈ R we have to show that some power lifts to Q. By taking repeated powers we can get x in f (Q) + pR for every prime p dividing t. So if p 1 ,. . . ,p m are the primes dividing t, we can arrange that x lies in the intersection of the f (Q) + p i R, which is f (Q) + p 1 · · · p m R. Now by taking repeated p 1 · · · p m -th powers, one pushes it into f (Q) + (p 1 · · · p m ) n R for any positive n, eventually into f (Q) + tR ⊆ f (Q).

5.4
We come back to the k-algebra A, and consider the inclusion gr A ֒→ hull ∇ (gr A), as in Theorem 29.
Notations 42. Let λ be a dominant weight and let b ∈ A U + be a weight vector of weight λ. Then we define ψ b : S ′ (λ) ⊗ k → hull ∇ (gr A) as the algebra map induced by the B-algebra map S ′ (λ) ⊗ k → A U + which sends the generator (choose one) of the λ weight space of ∇ λ to b.
Lemma 43. For each c in the image of ψ b , there is a positive integer s so that c s ∈ gr A.
Proof. The composite of S ⊗ k → S ′ ⊗ k with ψ b factors through gr A, so this follows from Lemma 39.
Lemma 44. Suppose k is Noetherian. If hull ∇ (gr A) is a finitely generated k-algebra, so is gr A.
Proof. Indeed, hull ∇ (gr A) is integral over gr A by Theorem 29. Then it is integral over a finitely generated subalgebra of gr A, and it is a Noetherian module over that subalgebra.
Lemma 45. If gr A is finitely generated as a k-algebra, then so is A.
Proof. Say j 1 , . . . , j n are nonnegative integers and a i ∈ A ≤ji are such that the classes a i + A <ji ∈ gr ji A generate gr A. Then the a i generate A.
Lemma 46. Suppose k is Noetherian. If A U is a finitely generated k-algebra, so is A.
Lemma 47. Let P be a standard parabolic subgroup. Suppose k is Noetherian. Then A is a finitely generated k-algebra if and only if A Ru(P ) is one.
Proof. Let V be the intersection of U with the semisimple part of the standard Levi subgroup of P . Then U = V R u (P ) and A U = (A Ru(P ) ) V . Suppose that A is a finitely generated k-algebra. Then A U = (A Ru(P ) ) V is one also by Lemma 25, and so is A Ru(P ) by Lemma 46 (applied with a different group and a different algebra). Conversely, if A Ru(P ) is a finitely generated k-algebra, Lemma 25 (with that same group and algebra) implies that A U = (A Ru(P ) ) V is finitely generated, and thus A is as well, by Lemma 46.
Proof of Theorem 32. Let k be Noetherian and let A be a finitely generated k-algebra. By Theorem 30, the k-algebra hull ∇ (gr A) is finitely generated. So we may choose b 1 ,. . . ,b s , so that ψ b1 ⊗ · · · ⊗ ψ bs has image hull ∇ (gr A). By extending Lemma 38 to tensor products, we can argue as in the proof of Lemma 43 and Theorem 29, and see that there is a positive integer n so that n hull ∇ (gr A) ⊆ gr A. Now, hull ∇ (gr A) ⊗ k[G/U ] is acyclic by Proposition 28, and thus its summand hull ∇ (gr A) is acyclic as well. It follows that H i (G, gr A) is a quotient of H i−1 (G, hull ∇ (gr A)/ gr A), which is annihilated by n.
Proof of Theorem 33. Take n as in Theorem 32, and use that localization is exact.
Proof of Theorem 35. It suffices to show that the composite: is power-surjective. It coincides with the composite gr A → hull ∇ (gr(A)) → hull ∇ (gr(A/pA)). Now A U + → (A/pA) U + is p-power-surjective by a combination of Theorem 12, Proposition 6, Proposition 41 and the transfer principle [9, Ch. Two] as used in 25. After inducing up, hull ∇ (gr(A)) → hull ∇ (gr(A/pA)) is still p-powersurjective, indeed the same p-power is sufficient. And gr A → hull ∇ (gr(A)) is power-surjective by Theorem 29.

Finiteness properties of cohomology algebras
In this section we study finiteness properties of H * (G, A), primarily when the base ring k is Z. We shall always assume that the commutative algebra A is finitely generated over the ring k, with rational action of a Chevalley group scheme G. Further, M will be a noetherian A-module with compatible Gaction. Torsion will refer to torsion as an abelian group, not as an A-module. We say that V has bounded torsion if there is a positive integer that annihilates the torsion subgroup V tors .
Lemma 48. A noetherian module over a graded commutative ring has bounded torsion.
Recall that we call a homomorphism f : R → S of graded commutative algebras noetherian if f makes S into a noetherian left R-module. Recall that CFG refers to cohomological finite generation. The main result of this section is the following.
• Every H m (G, M ) is a noetherian A G -module.
• If H * (G, A) is a finitely generated algebra, then H * (G, M ) is a noetherian H * (G, A)-module.
• H * (G, gr A) is a finitely generated algebra.
• If H * (G, A) has bounded torsion, then the reduction H even (G, A) → H even (G, A/pA) is power-surjective for every prime number p.
• If H even (G, A) → H even (G, A/pA) is noetherian for every prime number p, then H * (G, A) is a finitely generated algebra.
Remark 50. Note that the first statement would fail badly, by [11, I 4.12], if one replaced G with the additive group scheme G a . This may explain why our proof is far from elementary.
We hope to show in the future that H even (G, A) → H even (G, A/pA) is noetherian for every prime number p. The theorem suggests to ask: Problem. Let k be a Noetherian ring and let G be a Chevalley group scheme over k. Let A, Q be finitely generated commutative k-algebras on which G acts rationally through algebra automorphisms. Let f : A → Q be a power-surjective equivariant homomorphism. Is H * (G, A) → H * (G, Q) power-surjective?
We will need the recent theorem of Touzé [ We leave it to the reader to try a limit argument to deal with the case where k is essentially of finite type over a field. First let the ring k be noetherian. We are going to imitate arguments of Benson-Habegger [3]. We thank Dave Benson for the reference.
Proof. We may assume m is prime. By the Chinese Remainder Theorem we may then also assume that n is a power of that same prime. (If n is prime to m the Lemma is clear.) Let x ∈ H even (G, A/nA). We show that some power x m r of x lifts. Arguing as in the proof of Proposition 41 we may assume x is homogeneous. Let I be the kernel of A/mnA → A/nA. Note that m annihilates I, hence also H * (G, I). Theorem 49 has been proven.