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AUTOMORPHISM GROUPS AND PICARD GROUPS OF ADDITIVE FULL SUBCATEGORIES

NAOYA HIRAMATSU and YUJI YOSHINO

Abstract

We study category equivalences between additive full subcategories of module categories over commutative rings. And we are able to define the Picard group of additive full subcategories. The aim of this paper is to study the properties of the Picard groups and show that the automorphism group of an additive full subcategory is a semi-direct product of the Picard group with the group of algebra automorphisms of the ring.

1. Introduction

Let (A,) be a (commutative) Cohen-Macaulay local ring and CM(A)be the category consisting of maximal Cohen-Macaulay modules overAandA- module homomorphisms. It is known (e.g. [12]) that the structure of CM(A) reflects the properties of the singularity of Spec(A). To understand CM(A) well, we consider in this paper the problem how big (or small) is the auto- morphism group of the category CM(A).

Our main theorem in the paper is the following (see Theorem 6.2).

Theorem1.1. LetAbe a Cohen-Macaulay localk-algebra with dimen- siond, wherek is a commutative ring. Suppose thatAhas only an isolated singularity (i.e.,Ais a regular local ring for each prime idealexcept).

LetAutk(CM(A))denote the group ofk-linear automorphisms of the category CM(A). Then we have the following isomorphisms of groups.

Autk(CM(A))∼=

Autk-alg(A) (d =2) Autk-alg(A)C(A) (d =2),

where Autk-alg(A)denotes the group of k-algebra automorphisms ofAand C(A)denotes the divisor class group ofA.

Motivated by this result, we generalize in this paper these isomorphisms to hold for much wider classes of additive full subcategoriesᑝofA-Mod, and we shall show a certain structure theorem for Autk()in the most generality.

Received 27 November 2008.

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Our primary setting is the following: Letk be a commutative ring andA be a commutativek-algebra. Assume thatᑝis an additive full subcategory of the module category overAsuch thatᑝcontainsAas an object. SinceAis ak-algebra, every additive full subcategoryis ak-category. We say that a covariant functorᑝ → ᑝis ak-linear automorphism ofᑝif it is ak-linear functor giving an auto-equivalence of the categoryᑝ. Then the set of all the isomorphism classes of k-linear automorphisms of ᑝ, which we denote by Autk(), forms a group by defining the multiplication to be the composition of functors. The final goal of this paper is to prove the isomorphism of groups (1) Autk()∼=Autk-alg(A)Pic(ᑝ),

under some subtle assumptions onᑝ. In this isomorphism, Pic(ᑝ)is the Picard group ofᑝwhich we shall define in this paper below. Note that this type of isomorphism is known for the classical Picard group of the ring. (Compare with [2, Proposition 5.4, Chapter 2] and its non-commutative analogue in [11].) However, an original point of this paper is that we can define the Picard group for any additive full subcategory of module category overAwhenever it containsAas an object.

In Section 2, we shall prove some results on the presentation of functors which give equivalences between additive full subcategories. See Theorems 2.5 and 2.7. These results can be regarded as part of a generalization of Morita theory. There are many variants and generalizations of Morita theory. See Rickard [9], [10] for other variants of Morita theory to derived categories.

We should emphasize that the commutativity of the ringAis essential in our argument in Section 2.

Based on the results in Section 2, we give the definition of the Picard group for an additive full subcategory in Section 3. See Definition 3.2.

In Section 4, we shall give our computational results for the Picard groups of various additive full subcategories. It is worth noting that ifAis a Krull domain and ifᑝis the category of reflexiveA-lattices, then we shall have the divisor class groupC(A)as Pic(ᑝ), cf. Propositions 4.1 and 4.2.

Section 5 is devoted to prove the above mentioned isomorphism (1) under the assumption that the category is stable under Autk-alg(A). See Defin- ition 5.2 for the stability under Autk-alg(A). As a result, the Picard group Pic(ᑝ)is naturally a normal subgroup Autk()and the residue class group Autk()/Pic(ᑝ)is isomorphic to Autk-alg(A), cf. Corollary 5.10.

In Section 6, we concentrate upon the category of Cohen-Macaulay mod- ules over a Cohen-Macaulay localk-algebra. We prove the above mentioned Theorem 1.1 in Theorem 6.2.

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An application to the duality of an additive full subcategory is given in Section 7.

Convention.Throughout the paper,kalways denotes a commutative ring which we fix. LetAbe a commutativek-algebra. Then we denote the category of allA-modules andA-homomorphisms byA-Mod. When we say thatᑝis a subcategory ofA-Mod, we always assume thatᑝis closed under isomorphisms, i.e., ifXandY are isomorphicA-modules and ifXis an object ofᑝ, then so is Y. Since we only consider full subcategories ofA-Mod in this paper, we often identify the full subcategoryofA-Mod with the class of objects of. Hence we simply writeX∈ᑝto indicate thatXis an object ofᑝ. A full subcategory ᑝofA-Mod is said to be additive if it is closed under finite direct sums, i.e., ifX, Y ∈ᑝthenXY ∈ᑝ.

2. Equivalences of additive full subcategories

Let A and B be commutative k-algebras and suppose we are given an ad- ditive full subcategory(resp. ) ofA-Mod (resp. B-Mod). And suppose that there is an additive covariant functor F from ᑝ to ᑞ. We say that F is ak-linear functor if it induces a k-linear mapping from HomA(X, Y ) to HomB(F (X), F (Y )) for any objectsX, Y ∈ ᑝ. We denote the set of all the natural transformationsFF by End(F ). Note that End(F )naturally has structure of abelian group, sinceF :ᑝ→ᑞis an additive functor of additive categories. Moreover End(F )is a ring by defining the composition of natural transformations as the multiplication.

SinceAis a commutative ring, taking an elementaAwe have the natural transformationα(a):FFdefined byα(a)(X)=F (aX):F (X)F (X) for eachX ∈ ᑝ, whereaXdenotes the multiplication map bya on an object X∈ᑝ. (Note thataX:XXis anA-module homomorphism, sinceAis a commutative ring.) In such a way we have the ring homomorphism

α:A→End(F ); aF (a( )).

If F is a k-linear functor, then it is easy to see that End(F ) naturally has structure ofk-algebra and thatαis ak-algebra homomorphism.

Now suppose thatB∈ᑞand there is an objectN ∈ᑝsuch thatF (N )∼=B.

Note in this case that the mappingB →EndB(F (N ))which sendsbB to the multiplication mappingbF (N )bybonF (N )is an isomorphism of rings.

Furthermore ifF is ak-linear functor, then this isomorphism is a k-algebra isomorphism. We often identify EndB(F (N )) with B by this isomorphism.

Then we can define a ring homomorphism

β: End(F )→EndB(F (N ))∼=B; ϕϕ(N ).

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IfF is ak-linear functor, thenβ is ak-algebra homomorphism.

We observe the following lemma holds.

Lemma 2.1. Let A andB be commutativek-algebras and let and be additive full subcategories ofA-Mod and B-Mod respectively such that A∈ᑝandB ∈ ᑞ. Then the following hold for ak-linear covariant functor F :ᑝ→ᑞ.

(1) IfFis a faithful functor, then thek-algebra homomorphismαis injective.

(2) IfFis a full functor and it has a right adjoint functor, thenαis surjective.

(3) Assume that there is an objectN ∈ ᑝsuch thatF (N ) ∼= B. Then the k-algebra homomorphismβ is surjective.

(4) Adding to the assumption of(3), ifF is a full functor, thenβis injective as well.

Proof. (1) Suppose α(a) = 0 for aA. Since A is an object of ᑝ and sinceF is a faithful functor, a natural mappingϕA : A = EndA(A)→ EndB(F (A)) defined by ϕA(a) = F (aA) is injective. Note that there is a commutative diagram

()

A −−−−−−−→α End(F )

=

evA

A−−−−−−−→ϕA EndB(F (A)),

where the right vertical arrow evA is the evaluation map at Awhich maps γ ∈ End(F ) toγ (A). This implies thatϕA(a)= F (aA) = evA(α(a)) = 0, which forcesa=0.

(2) SinceFis full, the natural mapϕA :A=EndA(A)→EndB(F (A))is surjective. If the mapping evAin the diagram of()is injective, then it follows from the commutativity of()thatαis surjective. Thus we only have to show that evAis injective.

LetGbe a right adjoint functor ofF and we have a commutative diagram End(F )=Hom(F, F ) −−−−−−→= Hom(1, GF )

evA

evA

EndB(F (A))=HomB(F (A), F (A))−−−−−−→= HomA(A, GF (A)).

Therefore it is enough to show that evA: Hom(1, GF )→HomA(A, GF (A))is an injective map.

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To prove this, assume evA(ψ ) = ψ (A) = 0 for ψ ∈ Hom(1, GF ).

For any X ∈ ᑝ and any xX, we define an A-module homomorphism hx:AXbyhx(a)=ax(a∈A). Sinceψis a natural transformation from 1toGF, there is a commutative diagram

A−−−−−−−−−→ψ (A) GF (A)

hx

GF (hx) X −−−−−−−−−→ψ (X) GF (X),

which implies that ψ (X)(x) = 0 for all xX, since ψ (A) = 0. Hence ψ (X)=0 for allX∈ᑝ, and we conclude thatψ =0.

(3) For anybBand anyX∈ᑝ, we denote byϕb(X)the multiplication map byb onF (X). SinceF is a functor, for anyf ∈ HomA(X, Y )where X, Y ∈ᑝ,F (f )is aB-module homomorphism fromF (X)toF (Y ), therefore we have a commutative diagram

F (X)−−−−−−−−→F (f ) F (Y )

ϕb(X)

ϕb(Y ) F (X)−−−−−−−−→F (f ) F (Y ).

Thusϕbis a natural transformation ofF, henceϕb ∈ End(F )for allbB.

Clearly the equalityβ(ϕb)=bholds, henceβis surjective.

(4) To prove the injectivity ofβ, letβ(ψ )=ψ (N )=0 forψ ∈End(F ).

We have to show thatψ (X) :F (X)F (X)is a zero map for eachX∈ᑝ. For anyyF (X), we define aB-module homomorphismhy : BF (X) by hy(b) = by (b ∈ B). Since F is a full functor and since F (N ) ∼= B, there exists an fy ∈ HomA(N, X) such that F (fy) = hy. Then there is a commutative diagram

F (N )−−−−−−−−−−→F (fy)=hy F (X)

ψ (N )

ψ (X) F (N )−−−−−−−−−−→F (f

y)=hy

F (X),

and hence it follows thatψ (X)(y)= 0. Sincey is any element ofF (X), we haveψ (X)=0.

The lemma holds for a contravariant functor. Actually we can prove the following lemma whose proof will be left to the reader.

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Lemma 2.2. Let A andB be commutativek-algebras and let and be additive full subcategories ofA-Mod and B-Mod respectively such that A ∈ ᑝ and B ∈ ᑞ. Then the following hold for a k-linear contravariant functorF :→ᑞop.

(1) IfFis a faithful functor, then thek-algebra homomorphismαis injective.

(2) IfFis a full functor and it has a right adjoint functor, thenαis surjective.

(3) Assume that there is an objectN ∈ ᑝsuch thatF (N ) ∼= B. Then the k-algebra homomorphismβ is surjective.

(4) Adding to the assumption of(3), ifF is a full functor, thenβis injective as well.

Recall that we say thatᑝandᑞare equivalent to each other ask-categories if there is ak-linear functor→ᑞwhich gives an equivalence of categories.

Corollary2.3. LetAandBbe commutativek-algebras and letandbe additive full subcategories ofA-ModandB-Modrespectively such that A∈ ᑝandB ∈ ᑞ. Suppose the categoriesandare equivalent to each other ask-categories. ThenAis isomorphic toBas ak-algebra.

Proof. This is a consequence of Lemma 2.1, since thek-linear equivalence F :ᑝ→ᑞsatisfies all the necessary conditions assumed in the lemma.

It is well known that there exist examples of non-commutative rings for which Corollary 2.3 does not hold. For example, letAbe a commutative ring andB be anm×m full matrix algebra over A. Then, AandB are Morita equivalent, hence there is an equivalence of categories A-Mod ∼= B-Mod.

However,AareBare not isomorphic as algebras ifm2.

Next we argue about the form of a functor which gives an equivalence of categories.

Definition 2.4. LetAand B be commutative k-algebras and letN be anA-module. Given ak-algebra homomorphismσ : BA, we define an (AkB)-moduleNσbyNσ =Nas an abelian group on which the ring action is defined by

(ab)·n=aσ (b)n

for anyabAkBandnN. In such a case, we can define a functor HomA(Nσ,):A-ModB-Mod.

Note that theB-module structure on HomA(Nσ, X)forXA-Mod is defined by(b·f )(n)=f ((1b)·n)forf ∈HomA(Nσ, X),bBandnN.

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Theorem2.5. LetAandBbe commutativek-algebras and letand be additive full subcategories ofA-Mod and B-Mod respectively such that A∈ᑝandB∈ᑞ. Suppose thatandare equivalent ask-categories and letF : → ᑞbe ak-linear covariant functor which gives an equivalence between them. Then there is ak-algebra isomorphismσ :BAsuch thatF is isomorphic to the functorHomA(Nσ,)|, whereN ∈ᑝis chosen so that F (N )∼=Bin.

Proof. SinceF is an equivalence, there exists anA-moduleN ∈ ᑝsuch thatF (N ) ∼= B. By virtue of Lemma 2.1, thek-algebra homomorphismsα andβare defined well and they are isomorphisms. Now we define ak-algebra isomorphismσ :BAas the composition ofα1andβ1:

B−−−→= EndB(F (N ))−−−→β−1 End(F ) −−−→α−1 A

b −−−→ bF (N ) −−−→β1(bF (N ))−−−→σ (b)=α11(bF (N ))).

Note from the definition ofαandβ, the equalityF (σ (b)N)=bF (N )holds for anybB.

SinceF is ak-linear equivalence, we have isomorphisms ofk-modules for eachX∈ᑝ;

HomA(Nσ, X)−−−−−−→= HomB(F (N ), F (X))−−−−−−→= F (X) f −−−−−−→ F (f ) −−−−−−→F (f )(1), whose composition we denote byϕX. We claim thatϕXis aB-module homo- morphism for eachX∈ᑝ. For this, we show thatF (b·f )=bF (f )for any bB and anyf ∈ HomA(Nσ, X). Since the equality(b·f )(n)= f ((1b)·n)=f (σ (b)n)= (σ (b)f )(n)= σ (b)X(f (n))holds for anynN, we should note thatb·f =σ (b)Xf, hence we haveF (b·f )=F (σ (b)X)F (f ).

Therefore it is enough to show the equalityF (σ (b)X)=bF (X)for anybB andX ∈ᑝ. To prove this, foryF (X), lethy :BF (X)be aB-module homomorphism defined byhy(b) = by (b ∈ B). SinceF (N ) ∼= B and F is full, there existsgy ∈ HomA(N, X)such thathy = F (gy). Sincegy is an A-module homomorphism, we have the equalityσ (b)X·gy=gy·σ (b)N for bB. Thus we haveF (σ (b)X)·F (gy) = F (gy)·F (σ (b)N). This implies F (σ (b)X)·hy = hy·F (σ (b)N) = hy ·bF (N ) = bF (X)·hy, hence in partic- ular, we haveF (σ (b)X)(y) = bF (X)(y)for allyF (X). This shows that F (σ (b)X)=bF (X)as desired.

As the final step of proof, we show that theB-module isomorphismϕX : HomA(Nσ, X)F (X) is functorial in the variable X so that it induces the isomorphism of functorsϕ : HomA(Nσ,)F. To prove this letf :

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XY be any morphism inᑝ. We want to show the equalityF (f )·ϕX = ϕY·HomA(Nσ, f ). But evaluating the both ends atg∈HomA(Nσ, X), we have F (f )(ϕX(g))=F (f )(F (g)(1)), and(ϕY·HomA(Nσ, f ))(g)=ϕY(f·g)= F (f ·g)(1)=F (f )(F (g)(1)), hence the equality above holds.

It is worth noting the following corollary as a direct consequence of the theorem.

Corollary2.6. LetAandBbe commutativek-algebras and letandbe additive full subcategories ofA-ModandB-Modrespectively such that A∈ᑝandB ∈ᑞ. Suppose thatF :ᑝ→ᑞis ak-linear covariant functor which gives an equivalence between them and that F (A) ∼= B in . Then there is ak-algebra isomorphismσ :BAsuch thatF is isomorphic to the functorHomA(Aσ,)|.

In particular, if F : ᑝ → ᑝ is a k-linear auto-equivalence of and if F (A)∼=A, thenF is isomorphic to the identity functor on.

The contravariant version of the theorem also holds, whose proof will go through similarly to the proof of Theorem 2.5, but using Lemma 2.2 instead of Lemma 2.1.

Theorem2.7. LetAandBbe commutativek-algebras and letandbe additive full subcategories ofA-ModandB-Modrespectively such thatA∈ᑝ andB ∈ᑞ. Suppose there is ak-linear contravariant functorF :→ᑞop which gives a duality between them. Then there is ak-algebra isomorphism σ : BAsuch thatF is isomorphic to the functor HomA(, Nσ)| for N ∈ᑝwithF (A)∼=B.

Note that for any X ∈ ᑝ, the B-module structure on HomA(X, Nσ) is defined as follows:

(b·f )(x)=(1b)·f (x)=σ (b)f (x), forxX,f ∈HomA(X, Nσ)andbB.

3. Picard groups of additive full subcategories

In this section, we study the group of all theA-linear automorphisms of an additive full subcategory ofA-Mod.

Let A be a commutative ring and let be an additive full subcategory of A-Mod such that A ∈ ᑝ. We denote by AutA() the group of all the isomorphism classes ofA-linear automorphisms overᑝ, i.e.,

AutA()=

F :ᑝ→ᑝ

F is anA-linear covariant functor that gives an equivalence of the categoryᑝ

∼=.

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Since the identity mappingAAis the uniqueA-algebra automorphism of A, Theorem 2.5 implies the following.

Corollary3.1. For any element[F]∈AutA(), there is an isomorphism of functorsF ∼=HomA(N,)|for someN ∈ᑝ.

Taking this corollary into consideration, we make the following definition.

Definition3.2. LetAbe a commutative ring and letᑝbe an additive full subcategory ofA-Mod such thatA∈ᑝ. We define Pic(ᑝ)to be the set of all the isomorphism classes ofA-modulesM ∈ᑝsuch that HomA(M,)|gives an equivalence of the category. That is,

Pic(ᑝ)= {M ∈ᑝ|HomA(M,)|gives

an (A-linear) equivalenceᑝ→ᑝ}/∼=. We define the group structure on Pic(ᑝ) as follows: Let [M] and [N] be in Pic(ᑝ). Since the composition HomA(M,)|◦HomA(N,)|is also an A-linear equivalence, it follows from Corollary 3.1 that there exists anL∈ᑝ such that

HomA(L,)|∼=HomA(M,)|◦HomA(N,)|. We define the multiplication in Pic(ᑝ)by [M]·[N]=[L]. Note that

HomA(M,)|◦HomA(N,)|∼=HomA(MAN,)|

∼=HomA(N,)|◦HomA(M,)|, we see that [M]·[N]=[N]·[M]. In such a way Pic()is an abelian group with the identity element [A]. We call Pic(ᑝ)the Picard group ofᑝ.

Note from Yoneda’s lemma that objectsM andN inᑝare isomorphic as A-modules if and only if the functors HomA(M,)| and HomA(N,)|

are isomorphic to each other. Therefore the multiplication in Pic()is well- defined. We should note that the multiplication is not necessarily defined to be MAN in the usual way, sinceMANis not always an object ofᑝ.

Furthermore, the mapping Pic(ᑝ) → AutA() which sends [M] to HomA(M,)|is an isomorphism of groups by Corollary 3.1.

Lemma3.3. There is an isomorphismPic(ᑝ)∼=AutA()as groups.

In the following we shall give some examples of the Picard groups of ad- ditive full subcategories.

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LetAbe a commutative ring. We denote by PicAthe ordinary Picard group of the ringA. Recall that anA-moduleM is called invertible if there is anA- moduleMsuch that MAM ∼= AasA-modules. Then PicAconsists of all the isomorphism classes of invertibleA-modules, and the multiplication in PicAis defined by tensor product overA. See [2, §5, Chapter 2] for more details. Recall the following lemma holds for invertible modules. (Cf. [3].)

Lemma3.4. LetAbe a commutative ring and letM be an invertible A- module. ThenM is a finitely generated projectiveA-module.

Adding to the lemma, we note that ifM is an invertibleA-module, then there is an isomorphism HomA(M,) ∼= MA− as functors onA-Mod, where M denotes HomA(M, A). In fact, there is a natural transformation MA − → HomA(M,) of functors that is an isomorphism if M is a finitely generated projectiveA-module. Note also thatMAM ∼=Aholds for an invertibleA-moduleM.

Recall that an additive full subcategory is called additively closed if it is closed under summands of finite direct sums. Note that ifᑝis an additively closed full subcategory ofA-Mod and if A ∈ ᑝ, then all finitely generated projectiveA-modules belong to, since they are summands of finite direct sums of copies ofA. We note the following lemma holds.

Lemma3.5. LetAbe a commutative ring and letbe a full subcategory ofA-Modthat is additively closed andA∈ ᑝ. Then the classes of invertible A-modules are elements of Pic(ᑝ), and hencePicAis naturally a subgroup of Pic(ᑝ).

Proof. Let [M] ∈ PicA. ThenM ∈ ᑝas remarked above. Assume that M is a summand of the free module An of rank n. Then, for anyX ∈ ᑝ, HomA(M, X) is a summand of HomA(An, X) ∼= Xn which is an object of ᑝ. Therefore we have HomA(M, X) ∈ ᑝfor any X ∈ ᑝ. This shows that HomA(M,)defines a functorᑝ→ᑝ. LetMbe anA-module withMA

M∼=A. Then by the same reason as above, HomA(M,)defines a functor ᑝ → ᑝas well. Since HomA(M,)|◦HomA(M,)| ∼= HomA(MA

M,)|∼=1, we see that [M]∈Pic(ᑝ).

A full subcategoryᑝofA-Mod is said to be closed under tensor products ifMAN ∈ᑝfor allM, N ∈ᑝ.

Lemma 3.6. Let Abe a commutative ring and let be an additive full subcategory ofA-Modthat is closed under tensor products andA∈ᑝ. Then Pic()is naturally a subgroup ofPicA.

Proof. Let [M]∈Pic(). Then there is anM∈ᑝsuch that HomA(MA

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M,)|∼=1. SinceMAM∈ᑝ,Yoneda’s lemma implies thatMAM ∼= AasA-modules, henceM is an invertibleA-module. Therefore [M]∈PicA.

Combining Lemmas 3.5 and 3.6 above, we have the following proposition.

Proposition3.7. LetAbe a commutative ring and letbe an additively closed full subcategory ofA-Modwhich is closed under tensor products and A∈ᑝ. Then we have the equality

Pic(ᑝ)=PicA.

Note in the proposition that the equality means an equality of subsets of the set of all the isomorphism classes ofA-modules.

Example3.8. LetAbe a commutative ring. We denote byA-mod the full subcategory ofA-Mod consisting of all finitely generatedA-modules. We also denote by Proj(A)(resp. proj(A)) the full subcategory ofA-Mod consisting of all projectiveA-modules (resp. all finitely generated projectiveA-modules).

Since all of these full subcategories are additively closed and closed under tensor products, we have the equalities

Pic(A-Mod)=Pic(A-mod)=Pic(Proj(A))=Pic(proj(A))=PicA.

4. Further examples of Picard groups

In this section we shall give numerous examples of Picard groups for additive subcategories.

LetAbe a Krull domain. We denote byC(A)the divisor class group of A which is a group consisting of all the isomorphism classes of divisorial ideals of A. Recall that an A-module M is called an A-lattice if there are freeA-modulesF andFwith the same finite rank such thatFMF. AnA-lattice M is called reflexive if the natural mapping MM∗∗ is an isomorphism, where()denotes HomA(, A). A divisorial ideal is nothing but a reflexiveA-lattice of rank one. Note that the multiplication inC(A)is defined by [I]·[J]=[(I ⊗AJ )∗∗]. See [3] or [6] for the details.

Proposition4.1. LetAbe a Krull domain and we denote byRef(A)the full subcategory ofA-Modconsisting of all reflexiveA-lattices. Then we have an equality

Pic(Ref(A))=C(A).

Proof. For any [M] ∈ Pic(Ref(A)), there exists an [M] ∈ Pic(Ref(A)) such that 1Ref(A)∼=HomA(MAM,)|Ref(A)as functors on Ref(A). Since

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anyA-homomorphism fromMAMto a reflexive module factors through the naturalA-homomorphismMAM(MAM)∗∗, we have isomorphisms of functors

1Ref(A)∼=HomA(MAM,)|Ref(A)∼=HomA((MAM)∗∗,)|Ref(A). Since(MA M)∗∗ is again a reflexiveA-lattice, we conclude that(MA

M)∗∗ ∼= Aby Yoneda’s lemma. This implies that the reflexiveA-latticeM has rank one, thusM is a divisorial ideal and [M]∈C(A).

On the other hand, we shall show that HomA(I,)|Ref(A)is a well-defined automorphism for any divisorial fractional ideal I of A, hence that [I] ∈ Pic(Ref(A)).

First of all we remark from Bourbaki [3, Chapter VII, §2] or Fossum [6, Chapter 1, §5] that an A-lattice M is reflexive if and only if the equality M =

H (A)Mholds, whereH (A)is the set of all prime ideals of height one. Secondly we note that that the equality

HomA(X, Y )=

H (A)

HomA(X, Y)

holds for X, Y ∈ Ref(A). In fact, any f

H (A)HomA(X, Y) maps X toY for allH (A), hencef (X)

H (R)Y = Y, and thusf ∈ HomA(X, Y ). Combining these two claims one can see that HomA(I, X)is a reflexive lattice for allX∈Ref(A). Hence HomA(I,)yields a functor from Ref(A)to itself. SinceI is a divisorial ideal, there exists a divisorial idealJ with [J]= −[I] inC(A), i.e.,(IAJ )∗∗∼=A. Then there are isomorphisms of functors on Ref(A);

HomA(J,HomA(I,))|Ref(A)∼=HomA(IAJ,)|Ref(A)

∼=HomA((IAJ )∗∗,)|Ref(A)

∼=1Ref(A).

This shows that Hom(I,−)|Ref(A)is an automorphism on Ref(A)as desired.

In a Noetherian case we have the following example.

Proposition4.2. Let Abe a Noetherian normal domain. We denote by ref(A)the full subcategory ofA-Mod consisting of all finitely generated re- flexiveA-modules. Then we have the following equality.

Pic(ref(A))=C(A).

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Proof. In fact, ifAis a Noetherian Krull domain, then all reflexive A- lattices are finitely generated, hence we have Ref(A)=ref(A).

LetAbe an integral domain. The torsion partt (M)of anA-moduleM is defined to be

t (M)= {mM |am=0 for some non-zero elementaA}. AnA-module M is called torsion-free ift (M) = 0. Note that M/t (M)is always torsion-free for anyA-moduleM.

Proposition4.3. LetAbe a commutative integral domain. We denote by Tf(A)(resp.tf(A)) the full subcategory ofA-Mod consisting of all torsion- freeA-modules (resp. all finitely generated torsion-freeA-modules). Then we have the equalities

Pic(Tf(A))=Pic(tf(A))=PicA.

Proof. Since Tf(A)(resp. tf(A)) is additively closed, the inclusion PicA⊆ Pic(Tf(A))(resp. PicA⊆Pic(tf(A))) holds by Lemma 3.5.

To prove the other inclusion, assume [M] ∈ Pic(Tf(A)) (resp. [M] ∈ Pic(tf(A))). We have only to show thatM is an invertibleA-module. Take a torsion-freeA-module [M]∈Pic(Tf(A))(resp. [M]∈Pic(tf(A))) such that HomA(MAM,)∼= 1 as functors on Tf(A)(resp. tf(A)). We define the torsion-free tensor product by

M ¯⊗AM:=(MAM)/t (MAM).

It is easy to see that everyA-homomorphism fromMAMto a torsion-free A-module factors through the natural surjectionMAMM¯⊗AM. Hence the above isomorphism of functors induces an isomorphism

HomA(M ¯⊗AM,)∼=1,

as functors on Tf(A)(resp. tf(A)). SinceM¯⊗AM is torsion-free, it follows from Yoneda’s lemma thatM ¯⊗AM ∼= AasA-modules. Note from this iso- morphism thatMhas rank one.

Now we prove the following claim.

Claim. Assume thatM¯⊗AM∼=Afor torsion-freeA-modulesM andM. Then we haveM∼=M ∼=Afor any prime idealofA.

In fact, take an elementr

i=1mimiMAMwhich maps to 1 by the natural epimorphism

π :MAMM¯⊗AM∼=A.

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Localizing atᒍ, we have theA-homomorphism

π:MAM ∼=(MAM)(M¯⊗AM)∼=A.

SinceA is a local ring, we may assume thatm1/1m1/1MA M is mapped to a unit element ofA by π. Letf be anA-homomorphism fromA to M defined by f (a) = a·(m1/1)(a ∈ A). Taking the tensor product ofM with f and making the composition with π, we obtain an A-homomorphism

M=MAA−−−−−−−→Mf MAM −−−−→π A,

which we denote byg. Note that g is an epimorphism, sinceg(m1/1) = π(m1/1m1/1)is a unit ofA. Therefore we see thatM ∼= AK as A-modules, whereK =ker(g). SinceMhas rank one,Kmust be a torsion A-module. HoweverMis a torsion-freeA-module, hence we haveK=0.

ThusM∼=A. This completes the proof of the claim.

Now from the claim we have that

(MAM)∼=MAM ∼=A,

which is a torsion-freeA-module. It follows from this thatt (MAM)=0.

Since this equality holds for any prime idealᒍ, we have thatt (MAM)=0.

Therefore, we haveMAM ∼= M¯⊗AM ∼= A, which shows thatM is an invertibleA-module.

Let (A,)be a Noetherian local ring. We consider the full subcategory d1(A)ofA-Mod which consists of all the finitely generatedA-modulesM with depthM ≥1. Denote byH0(M)the 0th local cohomology module of an A-moduleM, i.e.,

H0(M)= {xM |ᒊnx =0 for some natural numbern}.

Note that a finitely generatedA-moduleMis an object ofd1(A)if and only if H0(M)= 0. Moreover, for any finitely generatedA-moduleM,M/H0(M) is always an object ofd1(A).

Proposition4.4. LetAbe a Noetherian local ring withdepthA >0. Then Pic(d1(A))is a trivial group.

Proof. Let [M] ∈ Pic(d1(A)), and we shall show thatM ∼= Aas A- modules. Take an element [M]∈Pic(d1(A))such that

1d≥1(A)∼=HomA(MAM,)|d≥1(A).

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We defineM ˜⊗AMas the residue moduleMAM/H0(MAM). Then M˜⊗AMd1(A)and it is easy to see that everyA-homomorphism from MAMto an object ofd1(A)factors through the natural surjectionMA

MM˜⊗AM. Hence the above isomorphism of functors induces an iso- morphism

1d≥1(A)∼=HomA(M˜⊗AM,)|d≥1(A).

SinceM˜⊗AMd1(A), it follows from Yoneda’s lemma that there is an isomorphismM˜⊗AM ∼= AasA-modules. ThusMAM is isomorphic to AH0(MAM). We denote byπthe natural projection fromMAMto A. Note that ker(π )∼=H0(MAM)is of finite length, hence the localized mappingπ : MA M ∼= (MAM)Ais an isomorphism ofA- modules for any prime idealᒍ=ᒊofA. ThusMis an invertibleA-module.

SinceAis a local ring, it forces thatM∼=AasA-modules.

Letn

i=1mimibe an element ofMAMsuch thatπn

i=1mimi = 1. SinceAis a local ring, we may assume thatπ(m1m1)is a unit element inA. We define anA-homomorphismf :AMbyf (a)=am1(a∈A).

Taking the tensor product withM and making the composition withπ, we have anA-homomorphism

M −−−−−−→Mf MAM−−−−→π A,

which we denote byg. Thengis an epimorphism, sinceg(m1)=π(m1m1) is a unit ofA. For any prime idealᒍwithᒍ=ᒊ, since the localized mapping g : MA is surjective and since M ∼= A, we can see that g is an isomorphism ofA-modules. This implies that ker(g)is anA-module of finite length. Since depthM ≥1, the submodule ker(g)ofM of finite length must be 0. Thereforeg:MAis an isomorphism ofA-modules.

Now let(A,)be a Noetherian complete local ring, and we denote byE the injective envelope of theA-moduleA/. AnA-moduleMis called Matlis- reflexive if the natural mappingMM∨∨is an isomorphism, where() denotes HomA(, E). It is well known by [7] that every Noetherian or Artinian A-module is Matlis-reflexive. We denote by Mat(A)the full subcategory of A-Mod consisting of all the Matlis-reflexiveA-modules. Note from [5] that an A-moduleMbelongs to Mat(A)if and only if there is a short exact sequence

0−−−→N−−−→M −−−→L−−−→0,

whereNis a NoetherianA-module andLis an ArtinianA-module. It is easy to see that any submodule and any quotient module of a Matlis-reflexive module is again Matlis-reflexive. Hence the subcategory Mat(A)is an abelian category.

It is also easy to see that Mat(A)is additively closed.

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Proposition4.5. Let (A,)be a Noetherian complete local ring. Then Pic(Mat(A))is a trivial group.

Proof. Let [M] ∈ Pic(Mat(A))be an arbitrary element. It is enough to show thatMis an invertibleA-module. Take an inverse element [N]=[M]1 in Pic(Mat(A))withN ∈Mat(A). SettingL=MAN, from the definition we have an isomorphism of functors

HomA(L,)|Mat(A)∼=1Mat(A).

We have only to show thatL∼=Afrom this isomorphism. SinceE∈Mat(A), we have an isomorphismL=HomA(L, E)∼=E. Since the natural mapping LL∨∨is always an injection, we have an embeddingLL∨∨ ∼=E ∼= A. Therefore Lis isomorphic to an ideal ofA, hence it is finitely generated.

On the other hand, we have isomorphisms ofA-modules:

HomA(LAA/, E)∼=HomA(A/, L)

∼=HomA(A/, E)∼=A/. SinceLAA/ᒊis finitely generated, we have

LAA/∼=(LAA/)∨∨∼=(A/)∼=A/.

This shows thatL is generated by a single element by Nakayama’s lemma.

Hence we may writeL∼=A/I for some idealI ofA. SinceA∈Mat(A), we also have an isomorphism ofA-modules:

HomA(L, A)∼=A.

SinceI annihilates the left hand side, we have I A = 0, henceL ∼= A as desired.

Remark 4.6. Any subgroups of PicA (resp. C(A)) will appear as the Picard groups of some additive full subcategories. In fact, it is not difficult to see that the following claims hold. (We leave their proofs to the reader.)

(1) LetAbe a commutative ring. For any additive full subcategoryᑝof proj(A)withA∈ᑝ, Pic(ᑝ)is naturally a subgroup of PicA. Conversely, for any subgroupGof PicA, there exists an additive full subcategoryᑝof proj(A) such that Pic(ᑝ)=G.

(2) LetAbe a Krull domain. For any additive full subcategoryᑝof Ref(A) such thatA∈ᑝ, Pic(ᑝ)is naturally a subgroup ofC(A). Conversely, for any subgroupGofC(A), there exists an additive full subcategoryᑝof Ref(A) such that Pic()=G.

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5. Structure of the group ofk-linear automorphisms

Letkbe a commutative ring and letAbe a commutativek-algebra. We denote by Autk-alg(A) the group of all the k-algebra automorphisms of A. For an automorphism σ ∈ Autk-alg(A), we can define a covariantk-linear functor σ : A-ModA-Mod as in the following manner. For eachA-moduleM, we defineσ(M)to beσM =Mas an abelian group on which theA-module structure is given by

am=σ1(a)m (aA, mM).

For anA-homomorphismf :MNofA-modules, we defineσf :σMσNto be the same mapping asf. Note thatσfis anA-homomorphism, since the equalitiesf )(am)=f (σ1(a)m)=σ1(a)f (m)=aσf (m)hold foraAandmM. It is easily verified that the equality(σ τ )=στholds forσ, τ ∈Autk-alg(A). In fact, we see thatam=(σ τ )1(a)mon(σ τ )M, on the other handam = τ11(a))monστM. Sinceσ1)is the identity functor onA-Mod, thek-linear functorσ: A-ModA-Mod is an equivalence of the category.

Letᑝbe any subcategory ofA-Mod. We denote by Autk()the group of all the isomorphism classes ofk-linear automorphism of, i.e., an element of Autk()is the isomorphism class [F] of ak-linear equivalenceF :ᑝ→ᑝ, and the multiplication in Autk()is defined to be the composition of functors.

Under these notations we have the following lemma.

Lemma5.1. There is an injective group homomorphism: Autk-alg(A)→ Autk(A-Mod)which mapsσto the isomorphism class ofσ.

Proof. It is easy to see from the observation above that is a group homomorphism. we have only to show that it is injective. For this, it is enough to prove the following claim.

Claim. If σ is isomorphic to the identity functor on A-Mod, then the k-algebra automorphismσ is the identity mapping.

In fact, lettingϕ : 1A-Modσbe an isomorphism of functors, for any aAwe have the following commutative diagram, from the definition ofσ.

A−−−−−−→ϕ(A) σA A

aA

σ(aA)

aA

A−−−−−−→ϕ(A) σAA,

whereaA denotes the multiplication mapping byaonA. Thus it follows the equalityϕ(A)(a)=a·ϕ(A)(1)as an element ofσA=A. On the other hand,

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sinceϕ(A)is anA-homomorphism, we must haveϕ(A)(a)=aϕ(A)(1)= σ1(a)·ϕ(A)(1). Therefore we have a = σ1(a). Since this holds for all aA, we conclude thatσ is the identity mapping.

It is worth noting that may not be surjective.

Definition5.2. Letᑝbe an additive full subcategory ofA-Mod and let σ ∈Autk-alg(A). Thenis calledσ-stable ifσgives an equivalence→ᑝ, i.e.,σ(M)∈ᑝand1)(M)∈ᑝwheneverM ∈ᑝ. We also say thatᑝis stable under Autk-alg(A)ifᑝisσ-stable for allσ ∈Autk-alg(A).

Example 5.3. The following subcategories of A-Mod are stable under Autk-alg(A).

(1) A-mod, proj(A)and Proj(A). (See Example 3.8.)

(2) Ref(A)in caseAis a Krull domain, and ref(A)in caseAis a Noetherian normal domain. (See Propositions 4.1 and 4.2.)

(3) Tf(A)and tf(A)in caseAis an integral domain. (See Proposition 4.3.) (4) the full subcategorydi(A)which consists of all the finitely generated A-modulesMwith depthMiwhenAis a Noetherian local ring with depthAi, whereiis any natural number. (Cf. Proposition 4.4.) (5) Mat(A) in caseAis a Noetherian complete local ring. (See Proposi-

tion 4.5.)

(6) CM(A)in caseAis a Cohen-Macaulay local ring. (See §5 below.) Notice that an additive full subcategory ofA-Mod is not necessarily stable under Autk-alg(A).

RegardingAas anA-algebra, we can consider the group AutA()for an additive full subcategory ᑝ of A-Mod. Since any A-linear functor is a k- linear functor in an obvious way, there is a group homomorphism AutA()→ Autk()whose kernel consists of a single element that is the class of the identity functor on. In such a way, AutA()is naturally a subgroup of Autk(A).

If the additive full subcategoryᑝcontains Aas an object, then we have shown in Lemma 3.3 that Pic(ᑝ)is isomorphic to AutA(). Thus we obtain the following lemma.

Lemma5.4.LetAbe a commutative algebra over a commutative ringk, and letbe an additive full subcategory ofA-ModwithA∈ᑝ. ThenPic(ᑝ) is naturally a subgroup ofAutk()by mapping[M] ∈Pic(ᑝ)to the class of automorphismHomA(M,)|.

We should remark that Pic()is an abelian group while Autk()is not.

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Definition5.5. In the rest of the paper, we always assume that the additive full subcategoryᑝis stable under Autk-alg(A). Then by definition there is a group homomorphism

|: Autk-alg(A)→Autk()

which mapsσto the class ofσ|. We denote the image of|by Ik(). Note that Ik()is a subgroup of Autk(). We call Ik()the subgroup ofk-linear automorphisms ofᑝinduced byk-algebra automorphisms.

We should note the following lemma.

Lemma 5.6. Let be an additive full subcategory of A-Mod which is stable underAutk-alg(A)andA ∈ ᑝ. Then the group homomorphism| : Autk-alg(A) → Autk() is injective. In particular, Autk-alg(A) ∼= Ik() as groups.

Proof. To show the injectivity of|, it is enough to prove thatσ∼= 1 as functors onᑝonly ifσis the identity mapping onA. But the proof of this will be the same as that of the claim in the proof of Lemma 5.1.

In this way we obtain the subgroups AutA()and Ik()in Autk()which are isomorphic as groups to Pic(ᑝ)and Autk-alg(A)respectively. For [M]∈Pic(ᑝ) andσ ∈Autk-alg(A), it is easy to see thatσM defines an element of Pic(ᑝ), hence the group Autk-alg(A)acts on Pic(ᑝ). In fact we can show the following lemma.

Lemma5.7. Let [M] ∈ Pic(ᑝ) and let σ ∈ Autk-alg(A)as above. Then there is an isomorphism of functors on

HomAM,)|∼=σ|◦HomA(M,)|1)|. Proof. It is enough to show that there is a natural isomorphism

HomAM, X)∼=σHomA(M, (σ1)X)

asA-modules for anyX∈ᑝ. But the both sides are the set of homomorphisms of abelian groupsf :MXsatisfyingf (σ1(a)m)=af (m)for allaA and mM, on which theA-module structure is given as (af )(m) = f (σ1(a)m)foraAandmM.

This lemma shows that under the isomorphism Pic(ᑝ) ∼= AutA() in Lemma 3.3, an element [σM]∈Pic(ᑝ)corresponds to the product of functors σ|◦HomA(M,)|1)|.

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