INJECTIVE COVERS OVER COMMUTATIVE ... - MathNet Korea

Bull. Korean Math. Soc. 40 (2003), No. 1, pp. 167–176

INJECTIVE COVERS OVER COMMUTATIVE NOETHERIAN RINGS WITH GLOBAL DIMENSION AT MOST TWO Edgar E. Enochs, Hae-sik Kim and Yeong-moo Song Abstract. In [3], Del Valle, Enochs and Mart´ınez studied flat envelopes over rings and they showed that over rings as in the title these are very well behaved. If we replace flat with injective and envelope with the dual notion of a cover we then have the injective covers. In this article we show that these injective covers over the commutative noetherian rings with global dimension at most 2 have properties analogous to those of the flat envelopes over these rings.

1. Introduction R will denote a commutative ring with identity. Let X be a class which is closed under isomorphism, direct summands, and finite direct sums. An X -envelope of a module M is a linear map φ : M → X with X ∈ X such that the following two conditions hold; 0

0

0

(1) HomR (X, X ) → HomR (M, X ) → 0 is exact for any X ∈ X . (2) Any f : X → X with f ◦ φ = φ is an automorphism of X. If φ : M → X satisfies (1), and perhaps not (2), φ is called an X preenvelope of M . Dually, an X -precover of M is a linear map ψ : X → 0 0 M with X ∈ X such that HomR (X , X) → HomR (X , M ) → 0 is exact 0 for any X ∈ X and if an X -precover ψ : X → M of M satisfies the condition that any f : X → X with ψ ◦ f = ψ is an automorphism of X, then ψ : X → M is called an X -cover of M . Note that an X -envelope (an X -cover, respectively) of a module is unique up to isomorphism, if it exists. By convention, (pre)envelopes and (pre)covers are named according to the name of the class X . For Received May 4, 2002. 2000 Mathematics Subject Classification: 16D50, 16E30. Key words and phrases: injective cover.

168

Edgar E. Enochs, Hae-sik Kim and Yeong-moo Song

example, an injective cover is an X -cover where X is the class of injective modules. We recall that for a class X of R-modules, X ⊥ consists of all Rmodules K such that Ext1R (X, K) = 0 for all X ∈ X and ⊥ X consists of all R-modules G such that Ext1R (G, X) = 0 for all X ∈ X (See [7, p.29]). Wakamatsu’s lemma ([7, Lemma 2.1.1] or [6]) says that if ψ : X → M is an X -cover of an R-module M and if X is closed under extensions, then Kerψ ∈ X ⊥ . Conversely, if ψ : X → M is a surjection with X ∈ X and 0 0 0 Kerψ ∈ X ⊥ , then for any X ∈ X , HomR (X , X) → HomR (X , M ) → 0 Ext1R (X , Kerψ) = 0 is exact. So ψ : X → M is an X -precover. Such a precover is called a special X -precover of M . In studying injective covers, the modules C such that HomR (E, C) = 0 and Ext1R (E, C) = 0 for all injective modules E play an important role (because of Wakamatsu’s lemma). If the ring is k[[x, y]] with k a field, this class contains all direct summands of products of modules of finite length (Theorem 2.9). One objective of this paper is to study modules C with HomR (E, C) = 0 and Ext1R (E, C) = 0 for all injective modules E over a noetherian ring R with global dimension at most 2.

2. Rings with global dimension at most 2 From results in [2, Proposition 8.1], we know that if φ : E → M is an ¯ K) = 0 and Ext1 (E, ¯ K) = injective cover with kernel K, then HomR (E, R ¯ Conversely, given such a module K, if 0 for all injective modules E. K ⊂ E is an injective envelope, then the natural map ψ : E → E/K is an injective precover. Moreover, Lemma 2.1. The diagram ψ

/ E/K z= z z z zz ² zz ψ

E

E

can only be completed to a commutative diagram by idE .

Injective covers over commutative Noetherian rings

169

Proof. If f and g both complete the diagram ψ

/ E/K = zz f g zzz z ² zz ψ

E

E

then the map f − g : E → E has its image in K. But Hom(E, K) = 0, so f − g = 0. Since f = idE is one possibility, this is the only one. Definition 2.2. An X -cover ψ : X → M for some class X will be called a rigid X -cover if ψ ◦ f = ψ for f : X → X implies f = idX . A rigid X -envelope is defined in an analogous manner. ¯ K) = 0 and Lemma 2.3. Let K be an R-module with HomR (E, 1 ¯ ¯ ExtR (E, K) = 0 for all injective modules E and let K ⊂ E be an injective envelope of K. Then E/K has no nonzero injective submodules. ¯ be an injective submodule of E/K. Then E/K = E ¯ ⊕N Proof. Let E for some N . Let ψ : F → N be an injective cover of N . Then σ : ¯ → N ⊕E ¯ is also an injective cover. Since the natural map F ⊕E ¯ ∼ E → E/K is a rigid injective cover by Lemma 2.1, so F ⊕ E = E. ¯ ¯ = 0. Moreover, Kerσ = Kerψ, and so K ⊂ F . Since F ∩ E = 0, K ∩ E ¯ = 0. But K ⊂ E is an injective envelope, and thus E Definition 2.4. ([7]) Let X be a class of R-modules and let M be an R-module. Then an element ξ ∈ Ext1R (L, M ) where L ∈ X is said to generate Ext1R (X , M ) (or ξ is a generator of Ext1R (X , M )) if for any ¯ ∈ X and ξ¯ ∈ Ext1 (L, ¯ M ), there is a linear map f : L ¯ → L such that L R 1 ¯ ExtR (f, M )(ξ) = ξ. Diagrammatically this says we have a commutative diagram 0

/M

/G ¯

/L ¯

²

²

/0

f

0

/M

/G

/L

/ 0,

where the rows represent the extensions ξ¯ and ξ. Moreover, ξ ∈ Ext1R (L, M ) is called a minimal generator if it is a generator and if any f ∈ HomR (L, L) such that Ext1R (f, M )(ξ) = ξ is necessarily an automor¯ M) phism of L. Then we see that if ξ ∈ Ext1R (L, M ) and ξ¯ ∈ Ext1R (L, 1 ¯ L) are both minimal generators of ExtR (X , M ), then any f ∈ HomR (L, 1 such that ExtR (f, M )(ξ) = ξ¯ is an isomorphism.

170


¯ = If ξ ∈ Ext1R (L, M ) is a generator of Ext1R (X , M ) and Ext1R (f, M )(ξ) 1 ¯ ¯ ¯ ¯ ξ for ξ ∈ ExtR (L, M ) and L ∈ X , then ξ is also a generator. Theorem 2.5. If R is a noetherian ring, then every R-module M has a generator of Ext1R (E, M ), where E is the class of injective R-modules. Proof. We will only sketch this argument (which is due to Naveed Zaman). Since R is noetherian, there is a representative set of indecomposable R-modules. Then for the given R-module M , we see that there is a family 0 → M → Gi → Ei → 0(i ∈ I) of short exact sequences where Ei is indecomposable and injective such that any short exact sequence 0 → M → G → E → 0 with E indecomposable and injective is isomor¯ be phic over M to one of the sequence 0 → M → Gi → Ei → 0. Let G the direct sum of the Gi amalgamated over M . Then we have an exact ¯ → ⊕Ei → 0. Now let 0 → M → H → E → 0 sequence 0 → M → G be an exact sequence with E an arbitrary injective module. Then E is 0 the direct sum of indecomposable injective modules. If E is any indecomposable injective submodule of E, we have the short exact sequence 0 0 0 → M → H → E → 0 which we derive from 0 → M → H → E → 0 0 0 (H is the preimage of E ). By construction there is a commutative diagram ; 0

/M

0

/M

/

H

0

² /G ¯

/

E

0

² / ⊕Ei

/0

/ 0.

But then using the fact that E is the direct sum of some set of such 0 E ⊂ E we see that we have the following commutative diagram 0

/M

/H

/E

/0

0

/M

² /G ¯

² / ⊕Ei

/ 0.

¯ → ⊕Ei → 0 is a generator of Hence the exact sequence 0 → M → G 1 ExtR (E, M ). Proposition 2.6. If 0 → M → K → E → 0 is an exact sequence with E injective and K ∈ E ⊥ , where E is the class of injective R-modules, then M → K is an envelope for the class E ⊥ . Proof. Let G ∈ E ⊥ . Then HomR (K, G) → HomR (M, G) → Ext1R (E, G) = 0 is exact. So M → K is an E ⊥ -preenvelope. Now if 0 → M →


171

¯ →E ¯ → 0 is exact with E ¯ ∈ E, then HomR (K, ¯ K) → HomR (M, K) → K 1 ¯ ExtR (E, K) = 0 is exact, so the diagram M

/K ¯

M

² /K

can be completed to a commutative diagram. This shows that 0 → M → K → E → 0 is a generator of Ext1R (E, M ). Since E is closed under direct limits, 0 → M → K → E → 0 is a minimal generator by [7, Theorem 2.2.2]. So the diagram /K M ² ~

K

can be completed only by automorphisms. Thus M → K is an E ⊥ envelope. Note that if 0 → M → K → K/M → 0 is an exact sequence with M → K an E ⊥ -envelope, then K/M is an injective R-module. Corollary 2.7. If R is a noetherian ring with gl.dim.R ≤ 2, then every R-module has a rigid E ⊥ -envelope. Proof. By Theorem 2.5 and [7, Theorem 2.2.2], every R-module M has a minimal generator for Ext1R (E, M ) where E is the class of injective R-modules. So if 0 → M → K → E → 0 is a minimal generator of Ext1R (E, M ), then K ∈ E ⊥ by [7, Proposition 2.2.1]. Thus M → K is an E ⊥ -envelope by the proposition above. Now if f, g both complete the diagram /M /K / K/M /0 0 f g

²

/K M then f −g = 0 by the argument in the proof of Lemma 2.1. So f = g.

When we say that (R, M, k) is a local ring, we mean that R is a local ring, M is the unique maximal ideal of R and k is the residue field of R. For any module M , the injective envelope of M is denoted by E(M ). Example 2.8. If (R, M, k) is a complete local ring of depthR ≥ 2, ∼ ˆ ∼ then by [1], Ext1R (k, R) = 0. But R = E(k), kˆ ∼ = k and Ext1R (k, R) =

172


ˆ (this uses the fact that the elements of Ext1 (M, N ) can be ˆ k) Ext1R (R, R identified with equivalence classes of short exact sequences 0 → N → U → M → 0). So for P ∈ SpecR, if P 6= M, let t 6∈ P and t ∈ M. Then multiplication by t on E(R/P ) is an isomorphism and is zero on k. From this we can get HomR (E(R/P ), k) = 0 and Ext1R (E(R/P ), k) = 0. If P = M, the Matlis duality gives HomR (E(k), k) = 0 and Ext1R (E(k), k) ˆ = Ext1 (k, R) ˆ = 0. For example, if = 0. This is because HomR (k, R) R k[[x, y]] R = k[[x, y]], then k ∼ has the above properties. = (x, y) The next theorem shows that there is a plentiful supply of modules C over R = k[[x, y]] with HomR (E, C) = 0 and Ext1R (E, C) = 0 whenever E is an injective R-module. Theorem 2.9. If a module C over R = k[[x, y]] with k a field is a direct summand of a product of modules of finite length, then C has the property that HomR (E, C) = 0 and Ext1R (E, C) = 0 for all injective R-modules E. Proof. If we use induction on the length of C, it is not hard to argue that any module C of finite length over R has the desired properties. And we easily see that the class of modules C with these properties is closed under products and summands. Definition 2.10. ([2]) Let R be a noetherian ring and M an Rmodule. The complex · · · → E1 → E0 → M → 0 with the requirement that E0 → M , E1 → Ker(E0 → M ) and En+1 → Ker(En → En−1 ) for n ≥ 1 are injective precovers is called an injective resolvent of M . If the maps En+1 → Ker(En → En−1 ) are furthermore injective covers, then we call our sequence a minimal injective resolvent. Note that the minimal injective resolvent is unique up to isomorphism and finite sums of minimal resolvents are minimal resolvents. For arbitrary class X , it is not in general true that the product of X -covers is an X -cover (even if X is closed under products). The next result shows this is the case in our situation. Theorem 2.11. Let R be a noetherian ring with gl.dim.R ≤ 2. Q Q (1) If Q φi : Ei → Mi are injective covers for i ∈ I, then φi : Ei → Mi is also an injective cover. (2) If φi : Ei → Mi are injective covers for i = 1, 2, · · · , n, then so is ⊕ni=1 φi : ⊕ni=1 Ei → ⊕ni=1 Mi .


173

Q Q Q Q Proof. (1) Since Ei is injective, φi : Ei → Mi is an injective precover by [7, Theorem 1.2.9]. From [2, Proposition 8.1], we see that if R is noetherian and gl.dim.R ≤ 2, then the minimal resolvent of any R-module M is of the form 0

/0

/0

/E

φ

/M

0

/0. 0

But thisQmeansQthat Hom QR (E , Kerφ) = 0 for any injective module E . To get φi : Ei → Mi is an injectiveQcover, we need only show that if Ci = Kerφi for each i ∈ I, then Ci has no nonzero injec¯ is an injective submodule of Q Ci , then tive submodules. Now if E Q ¯ Q Ci ) ∼ ¯ Ci ) = 0. So E ¯ = 0. HomR (E, = HomR (E, (2) This follows from [2, Proposition 4.1] when we take Em = 0 for m > n. Theorem 2.12. If (R, M, k) is a Gorenstein local ring of Krull dimension ≥ 2, then Ext1R (E(k), R) = 0. Proof. From [1], we know that if 0 → R → E 0 (R) → · · · → E h (R) → · · · is a minimal injective resolution of R, then for each h ≥ 0, E h (R) = ⊕htP =h E(R/P ). To get Ext1R (E(k), R) = 0 we only need to argue that HomR (E(k), E(R/P )) = 0 when htP = 1 by [5]. If γ ∈ M and γ 6∈ P , then the multiplication E(R/P ) → E(R/P ) by γ is an isomorphism. But for x ∈ E(R/P ), γ n x = 0 for some n ≥ 1. If f : E(k) → E(R/P ) is linear, then 0 = f (γ n x) = γ n f (x). This implies f (x) = 0 by the above. So we get Ext1R (E(k), R) = 0. 3. An equivalence of module categories In this section we assume R is a commutative noetherian ring with global dimension at most 2. We find an equivalence between the category of modules C we have been considering and another category of modules. Let C be the category of R-modules C such that HomR (E, C) = 0 and Ext1R (E, C) = 0 for all injective modules E and let D be the category of R-modules D such that D is isomorphic to a quotient of an injective R-module (or equivalently the injective cover E → D is surjective) and D has no nonzero injective submodules. We can define a functor G : D → C as follows: for each D ∈ obj(D), define G(D) = the kernel of the injective cover of D. Then HomR (E, G(D)) = 0 and Ext1R (E, G(D)) = 0 for all injective R-modules

174


E. So G(D) ∈ obj(C). For the given D-morphism D1 → D2 with the injective covers E1 → D1 and E2 → D2 , we can complete uniquely to the following commutative diagram E1

/ D1

²

² / D2 .

E2

So from this way we get a well-defined functor G : D → C. Also, it is easy to see that G is an additive functor. Remark 3.1. (1) Since an injective cover of D is unique up to iso0 morphism, if we make a different choice of injective cover E → D for 0 each D and get another functor G : D → C, then it is not hard to see 0 that G and G would be naturally isomorphic functors. (2) It is easy to see that D is an object D if and only if its injective cover E → D is surjective and if Ker(E → D) ⊂ E is an injective envelope. We know from Lemma 2.1 and Lemma 2.3 that for each C ∈ obj(C), E(C)/C has no nonzero injective submodules and E(C) → E(C)/C is the injective cover. If f : C1 → C2 is a C-morphism, then there is a linear f¯ : E(C1 ) → E(C2 ) such that the following diagram commutes 0

/ C1

f¯

f

²

0

/ E(C1 )

/ C2

α

/ E(C1 )/C1

/0

g

²

/ E(C2 )

²

β

/ E(C2 )/C2

/ 0.

So if f ∈ HomC (G(D1 ), G(D2 )), then there exists an extension f¯ : E(G(D1 )) → E(G(D2 )) such that f¯|G(D1 ) = f . Thus there is a map g : E(G(D1 ))/G(D1 ) → E(G(D2 ))/G(D2 ) such that g ◦ α = β ◦ f¯. Hence we get the following proposition. ∆

Proposition 3.2. HomD (D1 , D2 ) → HomC (G(D1 ), G(D2 )) is always bijective.


175

Proof. We only need to ensure that ∆ is injective. From the following diagram ; / E1 / G(D1 ) / D1 /0 0

0

² / G(D2 )

² / E2

² / D2

/0

Ker ∆ = {f ∈ HomD (D1 , D2 )| G(f ) ∼ = 0} = {f ∈ HomD (D1 , D2 )| G(f ) can be factored through an injective module E1 } by [4, 13.2] = {f ∈ HomD (D1 , D2 )| f can be factored through an injective module E2 } by [4, 13.8] = {f ∈ HomD (D1 , D2 )| f ∼ = 0} = [0] . RemarkQ3.3. (1) We Q know that the functor G “preserves products”, that is, G( Di ) ∼ G(Di ) naturally. It is probably not true that = ∼ G(⊕Di ) = ⊕G(Di ) in general. (2) If gl.dim.R ≤ 1 (so then gl.dim.R ≤ 2), then every injective cover is just of the form E ,→ M , where E is the largest injective submodule of M . So D = 0 and C = 0. References [1] H. Bass, On the ubiquity of Gorenstein rings, Math. Z. 82 (1963), 8–28. [2] E. Enochs, Injective and flat covers, envelopes, and resolvents, Israel J. Math. 39 (1981), 189–209. [3] E. Enochs, J. Mart´ınez Hern´ andez and A. Del Valle, Coherent rings of finite weak global dimension, Proc. Amer. Math. Soc. 126 (1998), 1611–1620. [4] P. Hilton, Homotopy theory and duality, Gordon and Breach, New York, 1965. [5] E. Matlis, Injective modules over Noetherian rings, Pacific J. Math. 8 (1958), 511–528. [6] T. Wakamatsu, Stable equivalence of self-injective algebras and a generalization of tilting modules, J. Algebra 134 (1990), 298–325. [7] J. Xu, Flat Covers of Modules, Lecture Notes in Math. 1634, Springer-Verlag, Berlin, 1996. Edgar E. Enochs, Department of Mathematics, University of Kentucky, Lexington, KY 40506-0027, USA E-mail : [email protected] Hae-sik Kim, Department of Mathematics, Kyungpook National University, Taegu 702-701, Korea E-mail : [email protected]

176


Yeong-moo Song, Department of Mathematics Education, Sunchon National University, Sunchon 540-742, Korea E-mail : [email protected]