A NOTE ON LOCAL DUALITY
RICHARD BELSHOFF CAMERON WICKHAM
A
Let (R, G) be a local Noetherian ring. We show that if R is complete, then an R-module M satisfies local duality if and only if the Bass numbers µi(G, M) are finite for all i. The class of modules with finite Bass numbers includes all finitely generated, all Artinian, and all Matlis reflexive R-modules. If the ring R is not complete, we show by example that modules with finite Bass numbers need not satisfy local duality. We prove that Matlis reflexive modules satisfy local duality in general, where R is any local ring with a dualizing complex.
1. Introduction In this paper all rings are commutative Noetherian rings with identities, and all modules are unitary. For a ring S and an S-module M, ES(M) (or simply E(M) when the ring is understood) will denote the injective hull of M. Let R be a local ring with maximal ideal G, and let E denote E(R\G). Recall that for an R-module M,
! n HG(M) lox?MQG xl0 for some natural number nq, ! i and that the right derived functors of HG are the local cohomology functors HG with support in the maximal ideal. For an R-module M, Soc M will denote the Socle of M, and M G will denote the Matlis dual HomR(M, E) of the R-module M. In this situation, a module M is called Matlis reflexie if the natural map M ! MGG is an isomorphism. The term complete will always mean complete with respect to the G-adic topology. The ith Bass number of an R-module M with respect to G will be denoted µi(G, M), i or simply µi(M), and is the vector space dimension of ExtR(R\G, M) over R\G. Equivalently, if 0 ! M ! Id is a minimal injective resolution of M, then µi(M) is the number of times E appears as a direct summand in Ii (see [1, Section 2]). A dualizing complex for a ring S of dimension n is a complex ! Dd:0,- D ,-(,-Dn,- 0 of injective S-modules such that the following two conditions are satisfied: i (i) D % & E(S\J) for each i, where the sum is taken over all J ? Spec S such that i l nkdim R\J; (ii) Hi(Dd) is finitely generated for each i. If such a complex exists, we say S has a dualizing complex Dd.IfSis a Gorenstein ring, then the minimal S-injective resolution of S provides a dualizing complex for S.
Received 16 May 1995; revised 31 July 1995. 1991 Mathematics Subject Classification 13D45. Bull. London Math. Soc. 29 (1997) 25–31 26 Suppose now that R has a dualizing complex Dd and dim R l n. By Proposition (2.3) of [2], there is a natural transformation of functors (from the category of all R-modules to itself)
! ! n G φ : HG(k) ,- [H (HomR(k, Dd))] . When R is Gorenstein,
! ! n G φ : HG(k) ,- Ext (k, R) . ! This φ can be incorporated into a uniquely determined homomorphism of (strongly) connected sequences of functors
i i n−i G Φ l (φ )i&!:(HG(k))i&! ,- ([H (HomR(k, Dd))] )i&!. For R Gorenstein,
i i n−i G Φ l (φ )i&!:(HG(k))i&! ,- (Ext (k, R) )i&!. i When Mh is a finitely generated R-module, φMh is an isomorphism for all i & 0.
D 1. Let R be a local ring with a dualizing complex, and let the maps φi be given as above. For any R-module M, we shall say that M satisfies local duality i if and only if φM is an isomorphism for all i & 0.
Grothendieck’s Local Duality Theorem, as given in [4]or[2], states that finitely generated modules satisfy local duality. In Section 2 we shall show that Matlis reflexive modules satisfy local duality. In Section 3 we consider the special case in which R is a complete local ring, and show that an R-module M satisfies local duality if and only if the Bass numbers µi(M) are finite for all i. The class of modules with finite Bass numbers includes all finitely generated, all Artinian, and all Matlis reflexive modules. If the ring R is not complete, we show by example that modules with finite Bass numbers need not satisfy local duality.
2. Matlis reflexie modules In this section, we shall prove that all Matlis reflexive R-modules satisfy local duality, where R is any local ring with a dualizing complex. We begin by recalling some general facts about Matlis reflexive modules. First, any module of finite length is Matlis reflexive. An R-module M is Matlis reflexive if and only if M G is Matlis reflexive. Also, submodules and quotients of Matlis reflexive modules are Matlis reflexive, and finite direct sums of Matlis reflexive modules are Matlis reflexive (see, for example, [7, Section 3.4]). If RV denotes the G-adic completion of R, then by Theorem 3.7 of [5],
HomR(E, E) % RV . Hence the ring R is a Matlis reflexive R-module if and only if R is complete. In this case, finitely generated and Artinian R-modules are Matlis reflexive. Also, M is an Artinian (Noetherian) R-module if and only if M G is Noetherian (Artinian); see [5, Corollary 4.3], for example. In particular, RG % E and, if R is complete, E G % R, and hence both are Matlis reflexive. 27 An example of a Matlis reflexive R-module that is neither finitely generated nor Artinian is given by E(R) where R is a complete Gorenstein ring of dimension one. To see this, observe that there is an exact sequence 0 ,- R ,- E(R) ,- E ,- 0. Since R and E are Matlis reflexive, so is E(R). If R is complete, then Matlis reflexive modules are characterized by the following proposition.
P 1. If R is complete, then M is Matlis reflexie if and only if M contains a finitely generated submodule S such that M\S is Artinian.
This proposition was proved by E. Enochs in [3, Proposition 1.3], and also by H. Zo$ schinger in [8] (see also [7, Theorem 3.4.13]). In fact, the proof of the ‘only if’ part, as given in [3]or[7], does not use the completeness assumption. So we shall actually use the following.
L 1. Let R be any local ring. If M is a Matlis reflexieR-module, then there is a short exact sequence 0 ,- S ,- M ,- M\S ,- 0 with S finitely generated and complete, and M\S Artinian.
For the rest of this section we assume that R is an n-dimensional local ring with j maximal ideal G which has a dualizing complex Dd. Let D (k) denote the functor j H (HomR(k, Dd)).
i L 2. Let M be an Artinian R-module. Then D (M) l 0 for all i n, and n G ! ! n G D (M) l M . Furthermore, the map φM: HG(M) ! D (M) is the natural map M ! M GG.
Proof. The first part follows directly from the definition of a dualizing complex and the fact that HomR(M, E(R\J)) l 0ifJis a non-maximal prime ideal. Recall that ! i if M is Artinian, then H (M) l M and H (M) l 0 for i " 0. The second part then G ! G follows from the definition of φM.
C 1. LetMbeanR-module which contains a submodule S such that M\S is an Artinian module. If j: S M is the inclusion map, then the induced maps
i i i n−i G n−i G n−i G HG( j):HG(S),- HG(M) and D ( j) : D (S) ,- D (M) are isomorphisms for i " 1.
Proof. For any i " 1 we have the following commutative diagram.
i i–1 i H G( j ) i i ,! H G (M/S) ,! H G(S) ,,,! HG(M) ,! H G(M/S) ,! ,! ,! ,! ,! i–1 i i i φ M/S φ S φ M φ M / S D n – i( j ) ∨ ,! D n – i + 1(M / S) ∨ ,! D n – i(S) ∨ ,,,! D n–i(M)∨ ,! D n – i(M/S) ∨ ,! 28 " Since M\S is Artinian, Hi− (M\S) and Hi (M\S) are zero, and by Lemma 2, " " G G Dn−i+ (M\S)G and Dn− (M\S)G are zero. The conclusion then follows.
T 1. LetMbeanR-module which contains a finitely generated submodule S such that M\S is an Artinian, reflexieR-module. Then the maps
i i n−i G φM: HG(M) ,- D (M) are isomorphisms for all i & 0. In particular, if M is a Matlis reflexie module, then M satisfies local duality.
i i n−i G n−i G Proof. By Corollary 1, HG(S) % HG(M) and D (S) % D (M) for i " 1. i i n−i G Since S is finitely generated, the maps φS: HG(S) !D (S) are isomorphisms for all i i n−i G i. Thus φM: HG(M) ! D (M) are isomorphisms for i " 1. For the other two isomorphisms, the exact sequence 0 ! S ! M ! M\S ! 0 gives rise to the following commutative diagram.
0 0 0 1 1 0 ,! H G (S) ,! H G(M) ,! H G(M/S) ,! H G(S) ,! H G(M) ,! 0 ,! ,! ,! ,! ,! 0 0 0 1 1 φ S φ M φ M / S φ S φ M 0 ,! D n(S) ∨ ,! D n(M) ∨ ,! D n(M/S) ∨ ,! D n – 1(S) ∨ ,! D n – 1(M) ,! 0
Since M\S is Artinian, by Lemma 2 the map
! ! n G φM/S: HG(M\S) ,- D (M\S) is the natural map M\S ,- (M\S)GG. ! Now M\S is assumed to be reflexive, so the map φM S is an isomorphism. Since S is ! " / ! " finitely generated, φS and φS are isomorphisms. Thus φM and φM are also isomorphisms. 3. Modules with finite Bass numbers
L 3. Let R be a local ring, and let M be an R-module such that µi(M) !_ i for all i. Then HG(M) is Artinian for all i.
Proof. Suppose that µi B µi(M) is finite for all i, and let M ! Id be the minimal ! ! injective resolution of M. Using the fact that HG(E) l E and HG(E(R\J)) l 0ifJis ! i a non-maximal prime ideal of R, it follows that H (I ) l E µi, where E µi denotes the ! G direct sum of µi copies of E. Then HG(Id) is the complex
d! di−" di 0 ,- E µ! ,- (,-Eµi,- (.
Every module in this complex is Artinian. Hence ker (di), im (di−") and their quotient i HG(M) are all Artinian.
Now let R be a complete local Gorenstein ring with dim R l n, and let M be an R-module. Since R is complete, there is a natural isomorphism R n−i Torn−i(M, E)G % ExtR (M, R). 29 n i R Since R is Gorenstein, HG(R) l E and HG(M) % Torn−i(M, E) for all i, functorially in M (see 9.1.4 and 10.1.10 of [7], for example). Therefore, for any i & 0, the map
i i n−i G φM: HG(M) ,- ExtR (M, R) is an isomorphism if and only if the natural map R R Torn−i(M, E) ,- Torn−i(M, E)GG i is an isomorphism. So M satisfies local duality if and only if HG(M) is Matlis reflexive.
P 2. For a complete local Gorenstein ring R, an R-module M satisfies local duality if and only if µi(M) !_ for all i.
i Proof. Assume that M satisfies local duality. Then each HG(M) is Matlis reflexive by the remarks above. Since a Matlis reflexive module cannot contain an ! infinite direct sum (see [7], the paragraph just before 3.4.3 on p. 41), dimk(Soc H (M)) ! G is finite, where k l R\G. But Soc HG(M) l Soc M % HomR(k, M), and so ! dimk(Soc HG(M)) l µ!(M) is finite. To see that µi(M) !_ for i & 1 consider the short exact sequence 0 ,- M ,- I ,- C ,- 0, ! where I l E(M) and C l I\M. Then µi(M) l µi "(C) for i & 1. Since H (M) and ! − G HG(I) are Matlis reflexive (see the proof of Lemma 3), it follows from the long exact ! i i+" sequence of local cohomology modules that HG(C) and HG(C) % HG (M) for i & 1 are Matlis reflexive. Therefore µ"(M) l µ!(C) is finite. Repeating the argument with C in place of M shows that µi(M) !_ for all i. i Conversely, if µi(M) is finite for all i, then HG(M) is Artinian by Lemma 3. Hence i HG(M) is Matlis reflexive, and so M satisfies local duality.
Recall that if S is a Gorenstein ring, then the minimal S-injective resolution Id of S is a dualizing complex for S.IfTis a homomorphic image of S, then HomS(T, Id) is a dualizing complex for T. So any homomorphic image of a Gorenstein ring has a dualizing complex (see, for example, [6]). Now let S be a complete local ring. By the Cohen structure theorem, S l R\I where R is a regular local ring and I is an ideal of R. Since a regular local ring is Gorenstein, a complete local ring has a dualizing complex, namely Dd l HomR(S, Id), where Id is the minimal R-injective resolution of the Gorenstein ring R.
T 2. Let S be a complete local ring, and let M be an S-module. Then M satisfies local duality if and only if µi(M) !_ for all i.
Proof. We shall assume that S l R\I, and let the notation be as in the paragraph just before the statement of the theorem. By a standard reduction (see, for example, various proofs of the local Lichtenbaum–Hartshorne theorem) we can assume that S and R have the same Krull dimension. Now let M be an S-module. We have
HomS(M, Dd) l HomS(M, HomR(S, Id)) % HomR(M, Id). 30 n−i n−i Thus for any i we have H (HomS(M, Dd)) l ExtR (M, R) and n−i n−i [H (HomS(M, Dd))]G l ExtR (M, R)G. i i Also, HG(k) and HGS(k) are naturally isomorphic functors from S-modules to R-modules (see Proposition 4.1.4 of [7]). The local rings (R, G) and (S, GS) have the same residue field k,soµi(M)l i dimk Ext (k, M) does not depend on whether we regard M as an S-module or as an R-module. By the previous proposition, the R-linear maps i φM i n−i G HG(M) ,,- ExtR (M, R) are isomorphisms for all i & 0 if and only if µi(M) !_ for all i. Therefore the S-linear maps i φM i n−i G HGS(M) ,,- [H (HomS(M, Dd))] are isomorphisms for all i & 0 if and only if µi(M) !_ for all i.
A natural question to ask is whether one can drop the completeness assumption in Theorem 2. The next proposition answers the question and gives us the following example.
E. The Artinian module E does not satisfy local duality if R is a Gorenstein ring that is not complete.
P 3. Let R be a local Gorenstein ring. The Artinian module E satisfies local duality if and only if R is complete.
Proof.IfRis complete, then E satisfies local duality by Proposition 2. For the converse, we shall show that if E satisfies local duality, then R is complete. i ! Since E is Artinian, we have HG(E) l 0 for i " 0, and HG(E) l E. By the remarks before Proposition 2, E satisfies local duality if and only if the natural map E ! E GG is an isomorphism. But E l RG, and if RG is reflexive then R is reflexive. Hence R is complete.
A. We thank Professor Rodney Sharp for some helpful comments, and in particular for suggesting Definition 1 to us after a careful reading of an earlier version of this paper.
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Department of Mathematics Southwest Missouri State University Springfield, MO 65804 USA