GOURMET'S GUIDE TO GORENSTEINNESS PETER JRGENSEN AND JAMES J. ZHANG 0. Introduction If we were very clever, we might be able to solve the following pr

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1 GOURMET'S GUIDE TO GORENSTEINNESS PETER JRGENSEN AND JAMES J. ZHANG 0. Introduction If we were very clever, we might be able to solve the following problem: Let A be a noetherian N-graded connected k-algebra given by generators and relations. Now determine, by looking at the relations, if A is AS-Gorenstein. It must, however, be admitted that the problem as stated seems rather hard. A more modest goal is hence the following: Find some criteria by which it can be decided whether some naturally occuring algebras are AS-Gorenstein. That is what we shall do in this manuscript. We prove generalizations of several well-known theorems from commutative algebra, notably Watanabe's theorem, [2, thm ], and Stanley's theorem, [11, thm. 4.4]. The main results are the following (note that throughout this manuscript, \algebra" means N-graded connected k-algebra): Theorem 4.7. (Watanabe.) Let A be a noetherian AS-regular algebra with dual A #, let G be a group, and let : G ;! GrAut(A) : G ;! GrAut(A) resp. be a homomorphism resp. an anti-homomorphism satisfying det (g) = det (g) for each g 2 G (we dene the determinant map in section 3). Then G acts on A A # through the induced homomorphism G ;! GrAut(A A # ) g 7;! (g) (g) suppose that its image is nite, of order coprime to char(k). Then the xed ring (A A # ) G is AS-Gorenstein Mathematics Subject Classication. 16W50, 16E30. Key words and phrases. AS-Gorenstein algebra, Watanabe's theorem, xed ring of nite group, Stanley's theorem. 1

2 2 PETER JRGENSEN AND JAMES J. ZHANG Theorem 6.2. (Stanley.) Let A be a noetherian AS-Cohen-Macaulay domain which satises one (or both) of the following conditions: A has enough normal elements. A is quotient of an Auslander-regular algebra. Then A is AS-Gorenstein precisely if the Hilbert series of A satises the functional equation H A (t) =t ;m H A (t ;1 ) as rational functions over Q. There are also several corollaries and related results, some of which solve problems posed in [3], to which this manuscript can be viewed as a sequel. For instance, Theorem 3.6. Let A be a noetherian AS-Gorenstein algebra. 1 ) If h ;1 2 k, and hj`, then the Veronese subalgebra A (h) is AS- Gorenstein. 2 ) If h6 j`, and dim A d > 1 for all d > 0, then A (h) has innite injective dimension over itself, and so is not AS-Gorenstein. This generalizes [3, prop. 6.5]. Theorem 4.7 builds on a theory of traces and determinants which extends the methods described in [3]. We thus continue the line laid out in [3] of using traces for invariant theory, even though the underlying rings are non-commutative (in the commutative case, traces are a classical tool for invariant theory). Theorem 6.2 builds on methods for working with dualizing modules akin to those of [7], and on a generalization of the Similar Submodule Condition from [17]. The manuscript is structured thus: Sections 1 to 3 study traces and determinants. Section 4 uses these methods to prove (among other things) the Watanabe theorem. Section 5 introduces the generalizations of the Similar Submodule Condition and studies its properties. And section 6 uses it to prove (among other things) the Stanley theorem. To close the introduction, let us run through some notation and conventions. Throughout this manuscript, k is a xed eld, and the word \algebra" means N-graded, connected, locally nite k-algebra (an algebra is not necessarily noetherian). We denote the group of graded k-algebra automorphisms of A by GrAut(A). The word \module" means graded module. We always think of A- modules as being left-modules, and obtain A-right-modules as A -leftmodules. The category of (graded left-) A-modules and homomorphisms of degree zero is denoted GrMod(A), while the subcategory of nitely generated modules is denoted grmod(a). The degree of the

3 GOURMET'S GUIDE TO GORENSTEINNESS 3 lowest non-vanishing piece of the module M is denoted by i(m) if M is not left-bounded, then i(m) =;1. AmoduleM is called locally nite if each graded piece M i is a nitedimensional vector space over k. For a locally nite module, we introduce the Hilbert function H(M ;) dened by H(M i) = dim k M i : If M is also right{ resp. left-bounded, we introduce the Hilbert series X H M (t) = H(M i)t i i it is a Laurent series in t ;1 resp. t. Shifting of modules is dened by M(i) j = M i+j. Matlis dual is dened by (M 0 ) i =(M ;i ) 0, where the right-hand prime denotes taking the dual vector space. Matlis dual sends left-modules to right-modules and vice versa. If M 2 GrMod(A) and 2 GrAut(A), then the module M is determined by having the same underlying vector space as M, butmultipli- cation dened by a m := (a)m. Since A is connected, the graded bi-module A=A 1 is one-dimensional over k. We usually denote it simply by k, and call it the trivial module. We denote \graded Ext" by Ext, so for M N 2 GrMod(A), we have Ext i A(M N) = M m Ext i GrMod(A)(M N(m)) where Ext GrMod(A) is the Ext of the category GrMod(A). If m = A 1 is the graded maximal ideal of A, we use the notation H m for the local cohomology functors, so H i m(m) =lim ;! Ext i A(A=A m M): There are certain classes of particularly interesting algebras: Denition 0.1. Let A be an algebra. We say that A is AS-Cohen- Macaulay if there exists an integer n such that i 6= n ) H i (A) m =Hi m(a) =0: AS stands for Artin and Schelter. Next two denitions were in fact introduced by Artin and Schelter. Denition 0.2. Let A be an algebra. We say that A is AS-Gorenstein if the following conditions are satised: We have id A (A) =id A (A) =n<1 where injective dimension is measured in the categories of graded modules.

4 4 PETER JRGENSEN AND JAMES J. ZHANG There is an integer ` such that Ext i A(k A) = Ext i A (k A) = ( 0 for i 6= n k(`) for i = n: Denition 0.3. Let A be an algebra. We say that A is AS-regular if it is AS-Gorenstein and it has nite left and rigth global dimension. If A is AS-regular, then the trivial module k has a minimal free resolution of the form 0! F n ;!;! F 0 ;! k! 0 where each F i is nitely generated, and where F n = A(;`). The integers n and ` appearing here agree with the n and ` appearing in denition 0.2. Note also the following: Important convention: Throughout the paper, when a theorem/lemma/proposition begin \Let A be an AS-Gorenstein (resp. AS-regular) algebra", the proof of the result uses n and ` to mean the integers appearing in denition 0.2. Acknowledgements. The second author was supported by the NSF and by a Sloan Research Fellowship. The rst author wishes to thank S. Paul Smith for invitations to visit the University ofwashing- toninjune97 and February The trace of a -linear map A classical tool in invariant theory is trace functions. The study of non-commutative xed rings can be viewed as a non-commutative generalization of invariant theory. So it is natural to try applying trace functions to xed rings. This was done in [3]. The present manuscript continues the application of traces this section sets up a few denitions, including the various forms of rationality thatwe will use a great deal. Denition 1.1. Let A be an algebra, and let 2 GrAut(A). M N 2 GrMod(A). A k-linear graded map f : M ;! N will be called a -linear map if it gives a homomorphism of graded A- modules, f : M ;! N i.e. if f(am) =(a)f(m). (1) Let

5 GOURMET'S GUIDE TO GORENSTEINNESS 5 Note that : A ;! A is itself -linear. Note also: If f : M ;! M is -linear, then f : M ;! M is A- linear. So if M ;! E is an injective resolution (of M in the category GrMod(A)), then f lifts to an A-linear chain map E ;! E, that is, to a -linear map E ;! E. In other words, -linear maps lift to -linear chain maps, and hence the -linear map f induces -linear maps H i m(f) :H i m(m) ;! H i m(m): Denition 1.2. Let A be an algebra. A module M 2 GrMod(A) is called ;-nite if it satises the following two conditions: We have Hm(M) i =0for i 0. Each Hm(M) i is locally nite and right-bounded. Denition 1.3. Let A be an algebra, and let 2 GrAut(A). M 2 GrMod(A), and let f : M ;! M be -linear. We write Tr M (f t) = X d tr(f j Md )t d whenever this is well-dened. When M is left-bounded and locally nite, we interpret Tr M (f t) as an element of k((t)), the fraction eld of the power series ring k[[t]]. When M is right-bounded and locally nite, we interpret Tr M (f t) as an element of k((t ;1 )), the fraction eld of the power series ring k[[t ;1 ]]. When M is ;-nite, we also write Br M (f t) = X i (;1) i Tr H i m(m)(h i m(f) t) interpreting this as an element of k((t ;1 )). When M is left-bounded, locally nite, and ;-nite, we say that f is rational over k if it satises the conditions: Tr M (f t) resp. Br M (f t) are rational functions over k (inside k((t)) resp. k((t ;1 ))). As rational functions over k, we have Tr M (f t) =Br M (f t): Denition 1.4. Let A be an algebra, and let M 2 GrMod(A). If M is left-bounded and locally nite, we can interpret the Hilbert series H M (t) as an element of k((t)), the fraction eld of the power series ring k[[t]]. When M is ;-nite, we introduce X B M (t) = (;1) i H H m(m)(t) i i Let

6 6 PETER JRGENSEN AND JAMES J. ZHANG and we can interpret this as an element of k((t ;1 )). When M is left-bounded, locally nite, and ;-nite, we say that M is rational over k if it satises the conditions: H M (t) and B M (t) are rational functions over k (inside k((t)) resp. k((t ;1 ))). As rational functions over k, we have H M (t) =B M (t): Similarly, when M is left-bounded, locally nite, and ;-nite, we can interpret H M (t) resp. B M (t) as elements of Q((t)) resp. Q((t ;1 )). We say that M is rational over Q of it satises the conditions: H M (t) and B M (t) are rational functions over Q (inside Q((t)) resp. Q((t ;1 ))). As rational functions over Q, we have H M (t) =B M (t): 2. The determinant of an element in GrAut(A) Recall the \Important convention" of the introduction: When a lemma starts \Let A be an AS-Gorenstein algebra", n and ` always mean the integers determined by equation (1). When A is an AS-Gorenstein algebra (see denition 0.2), it turns out that one can dene a group homomorphism, det : GrAut(A) ;! k which we call the determinant (k is the multiplicative group of the eld k). This generalizes the case A = k[x 1 ::: x n ], where graded automorphisms are given by elements of the group GL n (k), on which the usual (matrix) determinant is dened. If G is a nite subgroup of GrAut(A), then the restriction detj G is important for questions concerning properties of the xed ring A G. The present section sets up the homomorphism det. First a lemma about (not necessarily noetherian) AS-Gorenstein algebras. Lemma 2.1. Let A be an AS-Gorenstein algebra. Then Proof. Let H i m(a) = ( 0 for i 6= n AA 0 (`) for i = n: 0! A ;! E 0 ;!;! E n! 0 be a minimal injective resolution of A as an object of the category GrMod(A).

7 GOURMET'S GUIDE TO GORENSTEINNESS 7 Let j n. If E j has a non-zero torsion-element e, then we can also nd a non-zero element e 1 2 E j such that A 1 e 1 = 0. This yields k(; deg(e 1 )),! E j by 1 7! e 1. But k is a simple module, so the condition of minimality on E shows that the composition k ;! E j ;! E j+1 is zero, so Ext j A(k A) 6= 0,so j = n. In other words, E 0 ::: E n;1 are torsion-free. So since Ext n A(k A) = k(`), we must have k(`),! E n, and hence k(`)'s injective envelope is a direct summand in E n, E n = A0 (`) I and here I must be torsion-free (otherwise dim k Ext n A(k A) would be larger than 1). This implies the statement on H m(a). 2 We now aim to dene the determinant of an automorphism of an AS-Gorenstein algebra. Lemma 2.2. Let A be an AS-Gorenstein algebra, let 2 GrAut(A), and consider the map of right-modules ;1 : A A (;`) ;! A A (;`) which is ;1 -linear (in the equation, we really should write ;1 (;`) rather than just ;1, but we will be sloppy on this point). The Matlis dual map ( ;1 ) 0 : A A 0 (`) ;! A A 0 (`) is -linear. We now have: There exists a scalar c 2 k such that the -linear map H n m() :H n m(a) ;! H n m(a) is equal to c( ;1 ) 0 : A A 0 (`) ;! A A 0 (`): Proof. First note that by lemma 2.1, the module Hm(A) n is equal to AA 0 (`), so the statement of the lemma makes sense. Now, the lemma really tells us to consider the A-linear map : A A ;! AA and apply H n m, getting the A-linear map H n m() :H n m(a) ;! H n m( A)= H n m(a): Using lemma 2.1, we may write this as H n m() : A A 0 (`) ;! AA 0 (`):

8 8 PETER JRGENSEN AND JAMES J. ZHANG But using Matlis duality, it is trivial to see that the only A-linear maps from A A 0 (`) to AA 0 (`) are of the form where c 2 k. So c( ;1 ) 0 H n m() =c( ;1 ) 0 for some c 2 k. Finally we want to see c 6= 0. But that is easy, since H n m()h n m( ;1 )=H n m( ;1 )=id H n m (A) whence the map H n m() =c( ;1 ) 0 cannot be zero. 2 This enables us to give the denition of the determinant: Denition 2.3. Let A be an AS-Gorenstein algebra. For each 2 GrAut(A), we have by lemma 2.2 that and write H n m() =c( ;1 ) 0 det := c ;1 calling this the determinant of. In this way, we have dened a map, det : GrAut(A) ;! k : Remark 2.4. Note that in denition 2.3, the map is equal to the identity in graded degree zero that is, in its lowest non-vanishing degree. The map ( ;1 ) 0 : A 0 (`) ;! A 0 (`) is therefore equal to the identity in graded degree ;`, that is, in its highest non-vanishing degree. The map H n m() = (det ) ;1 ( ;1 ) 0 is therefore, in its highest non-vanishing degree (that is, degree `), just multiplication with the scalar (det ) ;1. 2 As claimed in the beginning of this section, the map det has the crucial property of being a group homomorphism: Proposition 2.5. Let A be an AS-Gorenstein algebra. Then the determinant map of denition 2.3 satises det() = det()det() for any 2 GrAut(A), and is thus a group homomorphism det : GrAut(A) ;! k :

9 GOURMET'S GUIDE TO GORENSTEINNESS 9 Proof. Let 2 GrAut(A). We have a commutative triangle of A-linear _ so we get A A H n m(a) Hn m () >H n H n @R _ H n m ( ) H n m( A): Now we use remark 2.4: It states that in graded degree ;`, the map H n m() is multiplication with (det ) ;1, the map H n m() is multiplication with (det ) ;1, whereas the map H n m() is multiplication with (det ) ;1. But then our diagram of H n m's clearly means that (det ) ;1 =(det) ;1 (det ) ;1 which implies the proposition. 2 Let us nish the section with a proof of another key property of the determinant, which links it to the trace of graded automorphisms. Lemma 2.6. Let A be an AS-Gorenstein algebra, and let 2 GrAut(A). If is k-rational in the sense of denition 1.3, then the rational function Tr A ( t) has the form Tr A ( t)=(;1) n (det ) ;1 t ;` + lower terms, when we write it out as a Laurent series in t ;1. Proof. When is k-rational, we have Tr A ( t) (a) = Br A ( t) (b) = (;1) n Tr H n m(a)(h m () t) (c) = (;1) n (det ) ;1 t ;` + lower terms, where \(a)" is by denition of rationality, \(b)" is by denition of Br, and \(c)" is by remark Among other things, this lemma enables us to motivate the name \determinant" for the map det by proving the claim made at the beginning

10 10 PETER JRGENSEN AND JAMES J. ZHANG of the section: If A = k[x 1 ::: x n ] is a commutative polynomial algebra, and the automorphism 2 GrAut(A) is given by the invertible matrix C 2 GL n (k), then our det coincides with the usual (matrix) determinant of C. To see this, note that by [3, (1-1)], we have Tr A ( t)= 1 det(i ; Ct) = 1 (;1) n det(c)t n + lower terms =() and writing this out as a Laurent series in t ;1, () =(;1) n (det C) ;1 t ;n + lower terms: Comparison with the formula of lemma 2.6 then gives det C = det (for the polynomial algebra, n = `). 3. The invariant ring of a finite group Suppose that B is an AS-Gorenstein algebra, and that G is a nite subgroup of GrAut(B). It turns out that the restriction detj G is connected to the structure of the dualizing module of the xed ring B G. The connection results in one of our two principal tools for proving Gorensteinness, theorem 3.3, which says that if B if noetherian and detj G trivial, then B G is AS-Gorenstein. In this section, x the following notation: B is an algebra, and G is a nite subgroup of GrAut(B) for which jgj ;1 2 k. We write A = B G for the invariant ring of G it is again an algebra. We write n = B 1 and m = A 1. It is standard that B is a graded A-bi-module, and that as an A-bimodule, B splits as B = A C. The map F : B ;! B F (b) = 1 jgj X 2G (b) is a projection of B onto the summand A. If B is (left-)noetherian, so is A, and in this case, B is (left-)nitely generated over A. For proofs of these statements, see [9, cor and cor. 5.9]. If B is noetherian and satises condition, then A also satises condition by the remark following [1, prop. 8.7]. If B is noetherian and has a balanced dualizing complex, then A also has a balanced dualizing complex by [15, prop. 4.17]. We begin by computing the dualizing module of the xed ring, A. Lemma 3.1. Suppose that B is noetherian AS-Gorenstein. Then

11 GOURMET'S GUIDE TO GORENSTEINNESS 11 1 ) When i 6= n, we have Hm(A) i =H i m(a) =0. Consequently, A is AS-Cohen-Macaulay. 2 ) We have as A-left-modules H n m(a) =fx 2 B 0 (`) j (det ) ;1 ( ;1 ) 0 (x) =x for all 2 Gg (and of course, there is a similar formula of A-right-modules). Proof. First note that B satises condition, by [16, cor. 4.3(1)]. As observed, the same is the case for A, and since B is module-nite over A, [15, lem. 4.13] implies that H i m = H i n and H i = m Hi n for all i. 1 : We have H i n(b) =H i m(b) =H i m(a C) =H i m(a) H i m(c): By lemma 2.1, we have that H i n(b) = 0 for i 6= n, so we must have H i m(a) =0for i 6= n. Of course, one can do the same computation for the opposite rings. 2 :TheA-bi-module homomorphism F introduced at the beginning of this section is a splitting of the inclusion of A into B. So the A-bimodule-homomorphism H n m(f ):H n m(b) ;! H n m(b) is a splitting of the inclusion of H n m(a) into H n m(b). So the module H n m(a) can be obtained as the set of xed points inside H n m(b) of the map H n m(f ), H n m(a) =fz 2 H n m(b) j H n m(f )z = zg: But the group G acts on H n m(b) through the maps H n m(), and H n m(f )= 1 jgj X 2G H n m(): Now let z 2 H n m(b). On one hand, suppose that z is invariant under all maps H n m() for 2 G. Then it is clear that H n m(f )z = z, and hence z 2 H n m(a). On the other hand, suppose that z 2 H n m(a). Then H n m(f )z = z, and hence for each 2 G, we get H n m()(z) = H n m()(h n m(f )z) = H n m() X 2G H n m()(z) = X 2G H n m()(z) = H n m(f )z = z:

12 12 PETER JRGENSEN AND JAMES J. ZHANG So inside H n m(b), one gets H n m(a) by taking the module of elements which are xed under all the maps H n m(): H n m(a) =fx 2 H n m(b) j H n m()(x) =x for all 2 Gg: Moreover, we know that H n m = H n n, so actually H n m(a) =fx 2 H n n (B) j H n n ()(x) =x for all 2 Gg: Replacing H n n (B) with B 0 (`), and using H n n () = (det ) ;1 ( ;1 ) 0,we get the lemma's statement. 2 Gorensteinness of an AS-Cohen-Macaulay algebra can be read o its dualizing module: Lemma 3.2. Let C be a noetherian AS-Cohen-Macaulay algebra with a balanced dualizing complex. Write o = C 1. If C satises H n o (C) =C 0 (`) as C-left-modules, then C is AS-Gorenstein. Proof. There is a convergent spectral sequence, E pq 2 = Ext p C(k Ho(C)) q ) Ext p+q (k C) see [5, prop. 1.1]. It implies that for any algebra satisfying the lemma's assumptions, ( 0 for i 6= n Ext i C(k C) = k(`) for i = n and using [6, thm. 4.5] this implies that id C (C) < 1 (here we used the existence of a balanced dualizing complex for C, which is enough to ensure the validity of the proof of [6, thm. 4.5]). So C is left AS- Gorenstein. And then by [4, cor. 4.6], we know that C is also right AS-Gorenstein, so all in all, C is AS-Gorenstein. 2 The following theorem generalizes a result by Watanabe, [2, thm ]. It will serve as one of our two key tools for proving Gorensteinness (the other one being Stanley's theorem, theorem 6.1). Theorem 3.3. Suppose that B is noetherian AS-Gorenstein. If we have det =1 for each 2 G, then the xed ring A = B G is AS-Gorenstein. Proof. First note that B has a balanced dualizing complex, by [14, cor. 4.14] (the proof of which works for AS-Gorenstein algebras). Then note that by the remarks at the beginning of this section, A is C

13 GOURMET'S GUIDE TO GORENSTEINNESS 13 noetherian, and has a balanced dualizing complex. Note also that by lemma 3.1, part 1,thealgebra A is AS-Cohen-Macaulay. However, by lemma 3.1, part 2, we may obtain the A-left-module H n m(a) as the submodule of B 0 (`) which is xed under all homomorphisms (det ) ;1 ( ;1 ) 0, for 2 G. Under the present assumption of det() = 1 for all 2 G, we therefore get H n m(a) as the submodule of B 0 (`) xed under all homomorphisms ( ;1 ) 0, for 2 G. And since A is the xed ring of G, it is clear that this submodule is A 0 (`), that is H n m(a) =A 0 (`): By lemma 3.2, A is then AS-Gorenstein. 2 As an illustration of how theorem 3.3 might be used, let us write down an immediate corollary. Corollary 3.4. Suppose that B is noetherian AS-Gorenstein. Under any one of the following conditions, the xed ring A = B G is AS-Gorenstein: 1 ) G = h 1 ::: r i, and det i =1for each i. 2 ) G =[G G]. 3 ) G is non-abelian simple. 4 ) For any 2 G we have n = e, but k contains no non-trivial n'th root of unity. 5 ) jgj is odd, and k = Q. Proof. Under any of the assumptions, it is clear that the homomorphism det : G ;! k is trivial, in other words that det =1for any 2 G. Hence A is AS-Gorenstein by theorem The following corollary is less striking than corollary 3.4, but will serve us in the next section, when we prove our generalization of Watanabe's theorem. Corollary 3.5. Suppose that B is noetherian AS-Gorenstein, and suppose that the group G is generated bytheelements 1 ::: r. Suppose that the following two conditions are satised: If M is a nitely generated B-module, and 1 i r, and f : M ;! M is a i -linear map, then f is rational over k. If 1 i r, then Tr B ( i t)=(;1) n t ;` + lower terms. Then the xed ring A = B G is AS-Gorenstein. Proof. Let 1 i r. By the rst assumption in the corollary, the map i is k-rational. Using lemma 2.6, we have that Tr B ( i t)=(;1) n (det i ) ;1 t ;` + lower terms :

14 14 PETER JRGENSEN AND JAMES J. ZHANG Combining this with the second assumption in the corollary, we see that det i =1: But by corollary 3.4, part 1, we then get that A = B G is AS- Gorenstein. 2 Let us round o with an application of corollary 3.4, provingageneralization of [3, prop. 6.5]. Theorem 3.6. Let C be a noetherian AS-Gorenstein algebra. 1 ) If h ;1 2 k, and hj`, then the Veronese subalgebra C (h) is AS- Gorenstein. 2 ) If h6 j`, and dim C d > 1 for all d > 0, then C (h) has innite injective dimension over itself, and so is not AS-Gorenstein. Proof. Part 2 follows from lemma 3.1 and the last part of [3, proof of prop. 6.5]. We now prove part 1, dividing the proof into three steps. Step 1. First we assume that the eld k contains a primitive h'th root of unity,. We can dene a graded automorphism of C by (x) = deg(x) x. We write G = hi for the (nite cyclic) subgroup of GrAut(C) generated by. It is clear that C G = C (h) when we think of the Veronese algebra as graded by ( C (h) 0 for h6j i i = for hji: C i It is easy to see that det = ;`. If hj`, then det = 1. But then C (h) = C G is AS-Gorenstein by corollary 3.4, part 1. Step 2. Now we suppose that k is arbitrary. We write F = k() for the extension eld obtained by adjoining a primitive h'th root of unity. The F -algebra C k F is noetherian, since k F is a nite eld extension. There is an obvious \change of scalars" functor, ;F : GrMod(C) ;! GrMod(C F ) which is exact (we omit the subscript k on from here on), and for M N 2 GrMod(C), there are natural isomorphisms, Ext i CF (M F N F ) = Exti C(M N) F: This easily implies that C F is AS-Gorenstein with the same values of n and ` as C. So step 1 above applies to C F, and tells us that the algebra (C F ) (h) = C (h) F

15 GOURMET'S GUIDE TO GORENSTEINNESS 15 is AS-Gorenstein. Step 3. We repeat the trick from step 2, observing that for M N 2 GrMod(C (h) ), we have natural isomorphisms Ext i C(h) F (M F N F ) = Ext i C(h)(M N) F: When C (h) F is AS-Gorenstein, this equation easily yields that C (h) itself is AS-Gorenstein Watanabe's Theorem This section proves our generalizations of Watanabe's theorem (theorems 4.7 and 4.8). These results generalize [3, thm. 6.6]. Most of the section consists of technicalities related to traces and rationality these will bring us in a position to use corollary 3.5 in the cases we are interested in. Lemma 4.1. Let A be an algebra, let 2 GrAut(A), let 0! M 1 ;! M 2 ;! M 3! 0 be an exact sequence in GrMod(A), and let f i : M i ;! M i be -linear maps which are compatible with the maps in the exact sequence. Then 1 ) If two of M 1 M 2 M 3 are ;-nite, so is the third. 2 ) We have P i(;1) i Tr Mi (f i t)=0 P i(;1) i Br Mi (f i t)=0 Proof. This is elementary. 2 Proposition 4.2. Let A be an AS-regular algebra. Let M be a nitely generated A-module which has a nite free resolution (that is, a free resolution terminating after nitely many steps, and consisting of nitely generated modules). Let 2 GrAut(A), and suppose that f : M ;! M is -linear. Then f is rational over k. Proof. Observe that since A itself in ;-nite (see denition 1.2), by working our way along the nite free resolution of M, we can see that M is ;-nite. We prove the theorem in four steps. Step 1: Suppose that M = A(s) and f =. Note that Tr M (f t) = Tr A(s) ( t) is a rational function over k, by [3, thm. 2.3(3)]. We have

16 16 PETER JRGENSEN AND JAMES J. ZHANG H n m(a) =A 0 (`), and can compute as follows: Br M (f t) = Br A(s) ( t) = (;1) n Tr H n m(a(s))(h n m() t) (a) = (;1) n Tr A0 (s+`)((det ) ;1 ( ;1 ) 0 t) = (;1) n (det ) ;1 t ;s;` Tr A 0(( ;1 ) 0 t) = (;1) n (det ) ;1 t ;s;` Tr A ( ;1 t ;1 ) (b) = (;1) n (det ) ;1 t ;s;`at m Tr A ( t) = bt m;`t ;s Tr A ( t) = bt m;` Tr A(s) ( t) = bt m;` Tr M (f t) where b = (;1) n (det ) ;1 a is a scalar. Here \(a)" is by lemma 2.2 and denition 2.3, and \(b)" is by [3, thm. 3.1]. The integer m which appears in \(b)" is dened by m =deg(p (t)), where p (t) isthepolynomial from [3, thm. 3.1]. So we get the equation Br M (f t) =bt m;` Tr M (f t): (2) Note that b and m do not depend on s. Step 2: Suppose that M is nitely generated free, f an arbitrary - linear map. We will prove by induction on the rank of M that Tr M (f t) is rational over k, and that equation (2) is satised. To start the induction, assume rank(m) = 1. Then M equals A(s) for some integer s, and f is itself, so the assertion follows from step 1. Now assume rank(m) 2. We can use an argument from [3, proof of lem. 2.2] to get ashortexact sequence 0! A(s) ;! M ;! M 00! 0 where A(s) is a direct summand in M, where f preserves A(s), and where rank(m 00 ) = rank(m) ; 1. Using the inductional assumption, and lemma 4.1, we learn that Tr M (f t) is rational over k. We alsohave Br M (f t) = Br M 00(f t)+br A(s) (f t) (a) = bt m;` Tr M 00(f t)+bt m;` Tr A(s) (f t) = bt m;` Tr M (f t) where \(a)" is by the inductional assumption on M 00 and step 1. So M and f satisfy equation (2).

17 GOURMET'S GUIDE TO GORENSTEINNESS 17 Step 3: Now look at the general case where M has a nite free resolution, and where f is an arbitrary -linear map. We want to prove thatf satises equation (2). We can lift f : M ;! M to a chain map dened on the nite free resolution of M, and by step 2, each homomorphism which is a component of this chain map has rational trace, and satises equation (2). Therefore f has the same properties. Step 4: To nish the proof, we just need to check that the constant b from equation (2) equals 1, and that the integer m ; ` from equation (2) equals 0. But that follows from setting M = k and f =id k. 2 Corollary 4.3. Let A be a noetherian AS-regular algebra, and let M 2 grmod(a). Let 2 GrAut(A), and let f : M ;! M be -linear. Then f is rational over k. Proof. Since A is noetherian AS-regular, M has a nite free resolution. Now use proposition Lemma 4.4. Let A and B be noetherian algebras with balanced dualizing complexes. Suppose that A is a graded subalgebra of B, in a way such that B A and A B are nitely generated. Let 2 GrAut(B) be such that (A) = A, let M 2 grmod(b), and let f : M ;! M be -linear. Then 1 ) H i n(m) =H i m(m). 2 ) Br B M(f t) =Br A M(f t). 3 ) If each j A -linear map on a nitely generated A-module is rational over k, then each -linear map on a nitely generated B-module is rational over k. Proof. 1 : Since A has a balanced dualizing complex, it satises condition by [13, thm. 6.3]. And B is module nite over A, so the lemma's part 1 can be read in [15, lem. 4.13]. 2 : The isomorphism in 1 is functorial, so f induces the same maps on H m(m) respectively H n (M). This implies 2. 3 : Any nitely generated B-module can be viewed as a nitely generated A-module. And the trace of f does not depend on whether M is considered to be an A{ or a B-module. So this follows from 2. 2 For any algebra A, we can equip the dual A # = Ext A(k k) with the Yoneda product. With this product, A # is a priori an ungraded algebra. We need to view it both as a Z 2 -graded algebra, with A # ij =Ext j A(k k) i

18 18 PETER JRGENSEN AND JAMES J. ZHANG and as an N-graded algebra, with A # m =Ext A(k k) ;m : Note the change of sign between the two gradings which makes A # i = A # ;i : The context will make it clear which grading we think of. Note that if A is a Koszul algebra, then A # is equal to the Koszul dual A! of A, ifwe equip A # with the N-grading described above (cf. [3, p. 367]). One might have expected a generalization of the Watanabe theorem to deal with a Koszul algebra A, and to involve A!, but as we shall see, the Koszul condition is superuous if we use A # rather than A!. If A is AS-regular, then [3, lem. 4.2] tells us that any 2 GrAut(A) induces an automorphism of A #. The denition of makes it clear that it respects both the above gradings on A #. The map 7;! is an anti-homomorphism by [3, lem. 4.2], that is, it satises GrAut(A) ;! GrAut(A # ) = : If we use the N-grading on A #, the tensor algebra A A # also becomes N-graded, and induces an element, 2 GrAut(A A # ): But note that the map 7;! is not a group homomorphism (nor an anti-homomorphism), so a priori, G does not act on A A #. Corollary 4.5. Let A be a noetherian AS-regular algebra. Let 2 GrAut(A), and 2 GrAut(A # ), and consider the induced automorphism 2 GrAut(A A # ): Let M 2 grmod(a A # ), and let f : M ;! M be -linear. Then f is rational over k. Proof. We have A = A k,! A A # and AA # is module-nite over A, since dim k A # < 1. Clearly, when we view A as a subalgebra of A A # in this way, it is preserved by any. Also, both A and A A # have balanced dualizing complexes, so we are in the situation of lemma 4.4.

19 GOURMET'S GUIDE TO GORENSTEINNESS 19 By part 3 of that lemma, the present corollary will follow ifwecan see that any ( )j A -linear map is rational over k, for 2 G. But ( )j A =, sothe statement we need follows from corollary Note that the following lemma is in a sense the \dual" of lemma 2.6. Lemma 4.6. Let A be a noetherian AS-regular algebra, and let 2 GrAut(A). Then Tr A #( t) = det()t` + lower terms when we write the trace as Laurent series in t ;1. Proof. First observe that as A # ij = Ext j A(k k) i, we have by [12, prop. 3.1(3)] that A # ;` = A # ;` n = k: So and tr( <;` )=0 (3) tr( ;` q ) = 0 for q 6= n: (4) Now we have equations of Laurent series in t ;1, Tr A #( t ;1 ;1) ;1 (a) = Tr A ( t) (b) = (;1) n (det ) ;1 t ;` +lower terms where \(a)" is by [3, (4-2)], while \(b)" is by lemma 2.6, whence Tr A #( t ;1 ;1) = (;1) n det()t` +lower terms: On the other hand, by denition X Tr A #( t ;1 ;1) = tr(j A # pq )t ;p (;1) q p q see [3, p. 366]. Comparing the two formulae for Tr A #( t ;1 ;1) and using the knowledge of equations (3) and (4) shows that tr( ;` n )=det (5) and combining equations (3), (4), and (5) yields that as Laurent series in t ;1, Tr A #( t ;1 1) = det()t` +lower terms: But then as Laurent series in t ;1, Tr A #( t) =Tr A #( t ;1 1) = det()t` +lower terms as claimed. Note how the rst \=" here involves the sign inversion between the two gradings of A #, changing a t into a t ;1. 2

20 20 PETER JRGENSEN AND JAMES J. ZHANG With all the above machinery in place, we can nally prove the main results of this section, the non-commutative generalizations of the Watanabe theorem. Since G does not a priori act on A A #, one has to think of some recipe to get a group which does there are (at least) two ways of doing so, hence two dierent generalizations of Watanabe (theorems 4.7 and 4.8). Theorem 4.7. (Watanabe I.) Let A be a noetherian AS-regular algebra, let G be a group, and let : G ;! GrAut(A) : G ;! GrAut(A) resp. be a homomorphism resp. an anti-homomorphism satisfying det (g) = det (g) for each g 2 G. Then G acts on A A # through the induced homomorphism G ;! GrAut(A A # ) g 7;! (g) (g) suppose that its image is nite, of order coprime to char(k). Then the xed ring (A A # ) G is AS-Gorenstein. Proof. AA # is AS-Gorenstein of injective dimension n, by [3, cor. 6.3(1)]. Denoting the image of G in GrAut(A A # ) by G 0, we want to use corollary 3.5 on A A #,soneed to check its two conditions on A A # and each (g) (g). 1 : Follows immediately from corollary : For each g 2 G, we can compute Tr AA #((g) (g) t) (a) = Tr A ((g) t)tr A #((g) t) (b) = ((;1) n (det (g)) ;1 t ;` + lower terms ) (det (g)t` + lower terms ) (c) = (;1) n t 0 + lower terms where \(a)" is by [3, (2-4)], \(b)" is by lemmas 2.6 and 4.6, and \(c)" is because det (g) = det (g) for each g. 2 This theorem contains the commutativewatanabe theorem, [2, thm ], as a special case: Let V be a nite-dimensional vector space over k, and let A = k[v ]. Let G be a nite subgroup of GL n (k) = GrAut(A)

21 GOURMET'S GUIDE TO GORENSTEINNESS 21 for which jgj ;1 2 k. We set equal to the inclusion of G in GrAut(A), and equal to the anti-homomorphism obtained by composing the inclusion with transposition of matrices. The dual of A is the exterior algebra A # = V (V 0 ), and one can check that the action of G on A A # = k[v ] V (V 0 )given by the recipe of theorem 4.7, g 7;! (g) (g) is the action appearing in [2, thm ]. But now theorem 4.7 says that the xed algebra (A A # ) G =(k[v ] ^(V 0 )) G is AS-Gorenstein, and that is exactly the statement of [2, thm ]. Theorem 4.8. (Watanabe II.) Let A be a noetherian AS-regular algebra, let G be anitesubgroup of GrAut(A), and consider the elements 2 GrAut(A A # ) for 2 G. They generate a subgroup of GrAut(A A # ) which we call G 0. Assume that the order of G 0 is coprime to char(k) (note that G 0 is necessarily nite, as it consists of elements of the form, for 2 G). Then (A A # ) G0 is AS-Gorenstein. Proof. This is almost the same as the proof of theorem 4.7: A A # is AS-Gorenstein of injective dimension n, by[3, cor. 6.3(1)]. We want to use corollary 3.5 on A A #,soneed to check its two conditions on A A # and a chosen set of generators of G 0. Naturally, wechoose the generators. 1 : Follows immediately from corollary : For each 2 G, we can compute Tr AA #( t) (a) = Tr A ( t)tr A #( t) (b) = ((;1) n (det ) ;1 t ;` + lower terms ) (det()t` + lower terms ) = (;1) n t 0 + lower terms where \(a)" is by [3, (2-4)], and \(b)" is by lemmas 2.6 and The Similar Submodule Condition The present section introduces a condition, the Similar Submodule Condition, and shows that an algebra satisfying this condition has a number of benecial properties. In the next section, they will help us

22 22 PETER JRGENSEN AND JAMES J. ZHANG to check that various algebras satisfy the conditions which will appear in the core form of our generalized Stanley theorem, theorem 6.1. First we generalize [17, p. 397]: Denition 5.1. Let A be an algebra, and let M N 2 GrMod(A). We say that M and N are similar, abbreviated to M N, if they satisfy the following conditions: (S1) M = N as graded k-vectorspaces, and (S2) H i m(m) = H i m(n) as graded k-vectorspaces, for each integer i. If A is noetherian AS-Gorenstein, the local duality theorem [14, thm. 4.18] implies that this denition coincides with the denition of similarity given in [17, p. 397]. Denition 5.2. Let A be an algebra, and let M 2 GrMod(A). We say that M has a proper similar submodule if: There exists an M 0 2 GrMod(A) such that M 0 is similar to M, and an integer ` > 0 such that M 0 (;`) is isomorphic to a proper graded submodule of M. If A is an algebra such that any torsionfree graded module M has a graded submodule N, such that N has a proper similar submodule, we say that A satises the Similar Submodule Condition, abbreviated to (SSC). If A is noetherian AS-Gorenstein, the present condition (SSC) coincides with condition (SSC) introduced on AS-Gorenstein algebras in [17, p. 398]. The following two results exhibit the basic good behaviour which results from condition (SSC). They generalize [17, lem. 2.2 and thm. 3.1]. The notion of multi-polynomial functions is used in the rst of the results see [17, p. 399] for the denition. Lemma 5.3. Let A be anoetherian algebra satisfying condition (SSC), and let M 2 grmod(a). Then 1 ) The Hilbert function H(M ;) is multi-polynomial for large values of the argument, and GKdim(M) =deg(h(m ;)) + 1 < 1: 2 ) Kdim(M) GKdim(M). 3 ) If the graded module N is isomorphic to M as graded k-vector spaces, and there is an integer m>0, and an exact sequence 0! K 1 ;! N(;m) ;! M ;! L ;! K 2! 0 then we have GKdim(M) GKdim(L)+1. And if K 1 = K 2 =0, we even have GKdim(M) =GKdim(L)+1 Proof. Use the proof of [17, lem. 2.2]. 2

23 GOURMET'S GUIDE TO GORENSTEINNESS 23 Theorem 5.4. Let A be anoetherian algebra satisfying conditions (SSC) and. Then 1 ) A has a balanced dualizing complex, and satises the generalized Auslander condition. It is also graded Auslander GKdim-Macaulay (see [15] for denitions of the generalized Auslander condition and the gradedauslander GKdim-Macaulay condition). Moreover, GKdim(A) = lcd(a) =lcd(a ). 2 ) For any nitely generated graded module, M, we have GKdim(M) = Kdim(M) < 1: Proof. Use the proof of [17, thm. 3.1]. 2 We go on to show some properties of algebras with (SSC) which will be handy for the next section. The key results are propositions 5.5 and 5.7. Proposition 5.5. Let A be a noetherian algebra. Suppose that A satisfeis one (or both) of the following conditions: A satises (SSC) and. A is quotient of a noetherian AS-regular algebra. Then any M 2 grmod(a) is rational over Q, in the sense of denition 1.4. Proof. The case where A satises (SSC) and. First note that we are in the situation of theorem 5.4. In particular, lcd(a) < 1 and A satises, so B M (t) iswell-dened. We also know GKdim(M) < 1. We use induction on d = GKdim(M). For d = 0, rationality of M over Q is obvious, since M has nite length. So assume d > 0, and suppose that rationality over Q holds for any L 2 grmod(a) whichhas GKdim(L) <d. Since H M (t) andb M (t) are both additive with respect to short exact sequences in M, a standard noetherian argument shows that we just need to produce a non-zero graded submodule N in M such thath N (t) and B N (t) are rational over Q, and such that the equation H N (t) = B N (t) holds for N. If M has torsion, we can use N = (M). If M is torsion-free, we use condition (SSC) to choose N M such that N has a proper similar submodule: There is N ~ such that N N, ~ and such that N(;m) ~ N for some m>0. So there is a short-exact sequence, Lemma 5.3, part 3, tells us that 0! ~N(;m) ;! N ;! N 00! 0: (6) GKdim(N 00 ) = GKdim(N) ; 1 GKdim(M) ; 1=d ; 1

24 24 PETER JRGENSEN AND JAMES J. ZHANG so by induction, we have that N 00 is rational over Q. In other words, H N 00 and B N 00 are rational functions, and H N 00(t) =B N 00(t): (7) However, the sequence (6) clearly gives us the equation (1 ; t m )H N (t) =H N 00(t) (8) and using the fact that N ~ N, whence H i m(n) = H i m( ~ N) for all i, the sequence (6) also gives the equation (1 ; t m )B N (t) =B N 00(t): (9) By equations (8) and (9), H N (t) andb N (t) are rational over Q. And combining equations (7), (8), and (9), we have the desired equality of rational functions, H N (t) =B N (t): The case where A is quotient of an AS-regular algebra. Let B be the AS-regular algebra of which A is quotient. By [15, lem. 4.13], local cohomology of A-modules can be computed equally well over A and B. It is therefore enough to show that any nitely generated B-module is rational over Q. First note that both H M (t) and B M (t) are additive with respect to shortexact sequences in M. Since each nitely generated module has a nite free resolution consisting of direct sums of shifts of B, itisenough to see that H B (t) and B B (t) are rational, and that H B (t) =B B (t). But this follows easily from [12, thm. 2.4(2)], and the remark preceding that theorem. 2 Lemma 5.6. Let A be anoetherian Auslander-Gorenstein algebra. Then A has an Artinian ring of quotients. Proof. This follows from [15, thm. 6.23]. 2 Proposition 5.7. Let A be a noetherian AS-Cohen-Macaulay algebra which has a balanced dualizing complex given by K[n], where K is the dualizing module. Suppose also that A satises one (or both) of the following conditions: A satises (SSC) and. A is quotient of a noetherian Auslander-regular algebra. If s > ; i(k), then the algebra A K(;s) has an Artinian ring of quotients (the product in this algebra is given by (a k)(a 1 k 1 ) = (aa 1 ak 1 + ka 1 )).

25 GOURMET'S GUIDE TO GORENSTEINNESS 25 Proof. Generally speaking, according to [7, prop. 1.5], the algebra A K(;s) is AS-Gorenstein when s>; i(k). Now, in the case where A satises (SSC) and, according to theorem 5.4, part 1, the algebra A satises the generalized Auslander condition. And in the case where A is quotient of a noetherian Auslanderregular algebra, A also satises the generalized Auslander condition, by [15, cor. 4.15(2)]. And AK(;s) is module-nite over A, so again by[15,cor. 4.15(2)], the algebra A K(;s) satises the generalized Auslander condition in either of the two cases we are considering. Combining the properties of A K(;s), we see that A K(;s) is Auslander-Gorenstein when s>; i(k). But then lemma 5.6 says that it has an Artinian ring of quotients. 2 Finally, let us note that (SSC) is satised for many practically occuring algebras. Lemma 5.8. Let A be an algebra which has enough normal elements, in the sense of [17, p. 392]. Then 1 ) A satises condition (SSC) of denition ) A satises condition. Proof. 1 : Let M 2 grmod(a) be torsion-free. As in [17, proof of prop. 2.3(2)], we may choose N 6= 0 as a graded M-submodule, such that p = Ann(N) is a prime ideal in A, and such that N is fully faithful over A=p. Since M is torsion-free, we have p 6= m, so choose a homogeneous regular normal element x 2 (A=p) 1. Write m = deg(x), and let be the graded automorphism of A=p such that ax = x(a) for all a 2 A=p. Look at N as an A=p-module. It contains the graded submodule xn, and in fact as A=p-modules we have So in particular, we have xn = N(;m): xn = N(;m) as graded vector-spaces, that is, condition (S1) of denition 5.1 is satised for xn and N(;m). We also get, for each integer i, an isomorphism over A=p, H i m=p(xn) = H i m=p( N(;m)) = (H i m=p(n(;m))) so in particular, as graded vector-spaces, H i m=p (xn) = H i m=p (N(;m)). But A satises, and by [15, lem. 4.13] wethenhave H i m=p (L) = H i m(l)

26 26 PETER JRGENSEN AND JAMES J. ZHANG for any L 2 grmod(a=p). All in all, H i m(xn) = H i m(n(;m)) as graded vector-spaces, that is, condition (S2) of denition 5.1 is satised for xn and N(;m). So (xn)(m) N, and N has a submodule isomorphic to xn, that is, to the ;m'th shift of (xn)(m). 2 : This is [1, cor. 8.12(2)] Stanley's Theorem This section proves our generalizations of Stanley's theorem. There is both an \abstract" version, theorem 6.1, which contains some rather complicated conditions on the algebra in question, a more \concrete" version with more comprehensible conditions, theorem 6.2, and nally a version for xed rings, theorem 6.4. Theorem 6.1. (Stanley.) Let A be a noetherian AS-Cohen-Macaulay algebra which is a domain, and satises the following conditions. 1 ) A has a balanced dualizing complex. By the AS-Cohen-Macaulay condition, the dualizing complex is equal to K[n] for some bimodule K we call K the dualizing module. 2 ) There is an integer s > ; i(k) such that the algebra A K(;s) has an Artinian ring of quotients. 3 ) The Hilbert series H A (t) and H K (t) are both rational functions over Q, and as rational functions they satisfy H K (t) =H A (t ;1 ): Then A is AS-Gorenstein if and only if the equation of rational functions is satised for some integer m. H A (t) =t ;m H A (t ;1 ) (10) A comment on the conditions 1 to 3 in the theorem: We shall see below, using the results from section 5, that they are all satised whenever A is AS-Cohen-Macaulay, and satises conditions (SSC) and. Proof. (of theorem 6.1.) Suppose on one hand that A is AS-Gorenstein. Then the dualizing module K is just A(;m), so we have the equation H A (t) =t ;m H K (t) (a) = t ;m (H A (t ;1 )) = t ;m H A (t ;1 ) where \(a)" is by condition 3. This shows that equation (10) holds.

27 GOURMET'S GUIDE TO GORENSTEINNESS 27 Suppose on the other hand that A satises equation (10). We write B = A K(;s) by condition 2, there is a choice of s such that B has an Artinian ring of quotients. We can then argue in the same way as the proof of [7, thm. 2.1], and obtain the same statement as in [7, thm. 2.1(4)]: If x 2 A is a non-zero-divisor of A, then x is regular on K, both from the left and from the right. Since A is a domain by assumption, this means that any non-zero x 2 A is regular on K, both from the left and from the right. Combining equation (10) and condition 3, we have the equation H K (t) =H A (t ;1 )=(t m H A (t)) = t m H A (t) which is valid when viewed as an equation of power series with rational coecients, and clearly the plus sign must apply, so H K (t) =t m H A (t) hence the Hilbert series of K is just a shift of the Hilbert series of A. Let y be a non-zero homogeneous element of minimal degree in K we have deg(y) = m. Look at the left-sub-module Ay K. It was established above that any non-zero x 2 A is regular on K, and so H Ay (t) =t m H A (t). But then H Ay (t) =t m H A (t) =H K (t) and we conclude that Ay = K, that is, K is a free module with one generator, y. Thus we see that id A (A) =id A (K) < 1: The same device, from the right, can be employed to see that id A (A) =id A (K) < 1: And A and A satisfy by condition 1 and [13, thm. 6.3], so by [16, thm. 4.2(3)] the algebra A is AS-Gorenstein. 2 As an application, we will show that Stanley's theorem is valid in certain practically occuring cases. Theorem 6.2. (Stanley's theorem for rings with enough normal elements and quotients of Auslander-regular algebras.) Let A be an AS- Cohen-Macaulay algebra which is a domain. Suppose that A satises one (or both) of the following conditions: A has enough normal elements in the sence of [17, p. 392]. A is quotient of an Auslander-regular algebra.

28 28 PETER JRGENSEN AND JAMES J. ZHANG Then A is an AS-Gorenstein algebra if and only if the equation of rational functions H A (t) =t ;m H A (t ;1 ) is satised for some integer m. Proof. It is clear that we canprove thisbychecking that the algebra A satises conditions 1 to 3 from theorem 6.1. The case where A has enough normal elements. First note that by lemma 5.8, A satises both (SSC) and. Condition 1 : A satises condition, and by theorem 5.4, part 1, we also have lcd(a) < 1. The same statements hold for A, since the assumption of enough normal elements is left/right-symmetric. By [13, thm. 6.3], A has a balanced dualizing complex, so condition 1 holds. Condition 2 : If s>; i(k), we know from proposition 5.7 that the algebra A K(;s) has an Artinian ring of quotients. Condition 3 : Every nitely generated A-module is rational over Q by proposition 5.5. The case where A is quotient of an Auslander-regular algebra. Note again that A satises condition, by [16, cor. 4.3(1)]. Condition 1 : A has a balanced dualizing complex by [14, thm. 4.14]. Condition 2 : This works just as when A has enough normal elements, by appealing to proposition 5.7. Condition 3 : Note that an Auslander-regular algebra is AS-regular by [8, thm. 6.3]. Now use proposition 5.5 to see that every nitely generated A-module is rational over Q. 2 This theorem contains Stanley's original result, [11, thm. 4.4], as a special case, since any commutative noetherian graded k-algebra is quotient of a polynomial algebra k[x 1 ::: x n ] (where the generators are not necessarily of degree one). As another application of Stanley's theorem in the form 6.1, we will give a necessary and sucient condition for a xed subring of an Auslander-regular ring to be AS-Gorenstein. First a lemma. Lemma 6.3. Let B be a noetherian algebra with a balanced dualizing complex. let G be a nite subgroup of GrAut(B) with jgj ;1 2 k, and set A = B G. If every -linear map on every nitely generated B-left-module is rational over k, then every nitely generated A-left-module is rational over k.

29 GOURMET'S GUIDE TO GORENSTEINNESS 29 Proof. We recall from section 3 that as A-bi-modules, B = A C, and that the projection onto the A-summand is the map F = 1 X : jgj 2G Let M 2 grmod(a), and set M 0 = B A M. We have M 0 =(A C) A M = M (C A M): Clearly, the projection onto the M-summand is the map F 0 = F id M : It is also clear that we have the fact When G acts on M 0 by m b 7;! M (b), the xed module is precisely M. (11) Using the same observation as in the proof of lemma 3.1, we nowsee that H i m( A M 0 )=H i m(m) H i m(c A M) and that we have the fact The projection onto the H i m(m)-summand is H i m(f 0 )= 1 jgj We can now compute with Laurent series over k: B A M(t) (a) P 2G H i m(): (12) = Br A M 0(F 0 t) = 1 X Br jgj A M 0( id t) (b) = 1 jgj (c) = 1 jgj 2G X 2G X 2G (d) = Tr (M 0 ) G(id t) (e) = Tr M (id t) = H M (t): Br B M 0( id t) Tr M 0( id t) \(a)" simply expresses that the dimension of a space is equal to the trace of a projection mapping to that space, cf. (12). \(b)" is by lemma 4.4, part 3. \(c)" is by our assumption on B. \(d)" is by an obvious generalization of the formula in [3, lem. 5.2]. And \(e)" is by (11).