IULIU CRIVEI
Received 23 September 2004 and in revised form 9 January 2005
A submoduleAof a rightR-moduleBis calleds-pure if f ⊗R1Sis a monomorphism for every simple leftR-moduleS, wheref :A→Bis the inclusion homomorphism. We estab-lish some properties ofs-pure submodules and uses-purity to characterize commutative rings with every maximal ideal idempotent.
1. Introduction
Purity in module categories and its different generalizations have been present in the literature since the 1960s. Apart from its special role in module theory, the importance of purity increased during the last two decades through the developments of model theory and the study of locally finitely presented categories (e.g., [2,6]). In the module theory context, that will be our framework, a purity of a particular interest is purity with respect to a class of cyclic modules, that was introduced by Stenstr¨om [7]. Connected to that, the present paper deals with purity with respect to the class of simple modules, previously considered by the author in [3,4].
We set some notation and terminology. Throughout the paperRdenotes an associative ring with nonzero identity and Mod-Rdenotes the category of right R-modules. The injective hull of a rightR-moduleAwill be denoted byE(A).
Consider a short exact sequence in Mod-R
0−→A−−→f B−→g C−→0. (1.1)
If f ⊗R1S:A⊗RS→B⊗RSis a monomorphism for every simple leftR-moduleS, it is said that the sequence (1.1) iss-pure. If f is the inclusion homomorphism and (1.1) is s-pure, thenAis said to be ans-pure submoduleofB[7]. A rightR-moduleAis called absolutelys-pureif it iss-pure in every rightR-moduleBthat containsAas a submodule [3].
Some characterizations ofs-pure submodules and absolutelys-pure submodules were established in [3,4]. In this paper we investigate further properties of these notions. For a commutative ringR, we show that every maximal ideal ofRis idempotent if and only
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if every maximal submodule of anR-module iss-pure. Rings with the previous property generalize commutative von Neumann regular rings.
2.s-purity
This section contains some further properties ofs-pure submodules and absolutelys-pure modules. The following characterization of absolutelys-pure modules will be frequently used.
Proposition2.1. LetAbe a rightR-module. Then the following statements are equivalent: (i)Ais absolutelys-pure;
(ii)Aiss-pure inE(A).
Proof. It follows by [5, Proposition 1.7]. Example 2.2. It is known that a ringRis von Neumann regular if and only if every right R-module is absolutely pure [8, Theorem 37.6]. Hence if Ris a von Neumann regular ring, every rightR-module is absolutelys-pure.
Corollary 2.3. Let R be right hereditary. Then the class of absolutely s-pure right R-modules is closed under homomorphic images.
Proof. LetAbe an absolutelys-pure rightR-module and letDbe a submodule ofA. The class of short exact sequences
0−→L−→α M−→β N−→0 (2.1)
of rightR-modules, where Imαis ans-pure submodule ofM, is a proper class of short exact sequences in the terminology of [7]. Then, sinceAiss-pure inE(A),A/Diss-pure inE(A)/D[7, Proposition 2.1]. ButRis right hereditary, henceE(A)/Dis injective. Now byProposition 2.1, it follows thatA/Dis absolutelys-pure. Denote byᏹthe class of rightR-modulesAwith the property that: for everys-pure right idealIofR,Ais injective with respect to the inclusion homomorphism f :I→R. It is clear thatA∈ᏹif and only if Ext1R(R/I,A)=0 for everys-pure right idealIofR. Proposition2.4. ᏹis closed under extensions.
Proof. Let 0→A→B→C→0 be a short exact sequence withA,C∈ᏹand letIbe an s-pure right ideal ofR. We get the induced exact sequence
Ext1R(R/I,A)−→Ext1R(R/I,B)−→Ext1R(R/I,C). (2.2)
The first and the last Ext are zero by hypothesis, hence Ext1R(R/I,B)=0, showing that
B∈ᏹ.
Proof. (i)⇒(ii). LetB∈ᏹand letAbe a submodule ofB. LetIbe ans-pure right ideal ofR. The exact sequence 0→I→R→R/I→0 induces the exact sequence
Ext1R(I,A)−→Ext2R(R/I,A)−→Ext2R(R,A). (2.3)
The first and the last Ext are clearly zero, hence Ext2R(R/I,A)=0. Now the exact sequence 0→A→B→B/A→0 induces the exact sequence
Ext1R(R/I,B)−→Ext1R(R/I,B/A)−→Ext2R(R/I,A). (2.4)
Since the first and the last Ext are zero becauseB∈ᏹand by what we have showed above, Ext1R(R/I,B/A)=0. ThusB/A∈ᏹ. Therefore the classᏹis closed under homomorphic images.
(ii)⇒(i). LetIbe ans-pure right ideal ofRand letAbe a rightR-module. Then the exact sequence 0→A→E(A)→E(A)/A→0 induces the exact sequence
Ext1RR/I,E(A)/A−→Ext2R(R/I,A)−→Ext2RR/I,E(A). (2.5)
By hypothesis we have the first Ext zero, and clearly the last one is zero, hence Ext2R(R/I, A)=0. Now the exact sequence 0→I→R→R/I→0 induces the exact sequence
Ext1R(R,A)−→Ext1R(I,A)−→Ext2R(R/I,A). (2.6)
The first and the last Ext are zero, hence we get Ext1R(I,A)=0, showing thatIis projective.
Corollary2.6. LetRbe right hereditary. Thenᏹis closed under homomorphic images. Now we recall a basic characterization ofs-pure submodules.
Proposition2.7 [7, page 170]. LetAbe a submodule of a rightR-moduleB. Then the following statements are equivalent:
(i)Aiss-pure inB;
(ii)AM=A∩BMfor every maximal left idealMofR.
Corollary2.8. LetI⊆J(R)be an idempotent right ideal ofR, whereJ(R)is the Jacobson radical ofR. ThenIiss-pure inR.
Proof. LetMbe a maximal left ideal ofR. Then we haveI⊆MandI=I2⊆IM⊆I∩M=I.
HenceIM=I∩M. NowIiss-pure inRbyProposition 2.7.
3. The commutative case
Throughout this section the ringRis assumed to be commutative. The following charac-terization ofs-pure short exact sequences will be very useful.
Proposition3.1 [3, Corollary 4.5]. Consider the short exact sequence (1.1). Then the fol-lowing statements are equivalent:
(i)the sequence (1.1) iss-pure;
Theorem3.2. LetAbe a submodule of anR-moduleB. Then the following statements are equivalent:
(i)Aiss-pure inB;
(ii)A/Dis a direct summand ofB/Dfor each maximal submoduleDofA.
Proof. (i)⇒(ii). Suppose thatAiss-pure inBand letDbe a maximal submodule ofA. ThenA/Diss-pure inB/D[7, Proposition 2.1]. Also,A/Dis simple, hence byProposition 3.1A/Dis a direct summand ofB/D.
(ii)⇒(i). Assume that (ii) holds. Let Sbe a simpleR-module and let p:A→Sbe a homomorphism. Without loss of generality, we may assume that pis an epimorphism. Let f :A→Bandj: Kerp→Abe the inclusion homomorphisms,i:A/Kerp→B/Kerp the inclusion homomorphism induced by f, andw:A→A/Kerp andv:B→B/Kerp the natural homomorphisms. Theniw=v f. Since Kerpis a maximal submodule ofA, A/Kerpis a direct summand ofB/Kerpby hypothesis. Letq:B/Kerp→A/Kerpbe the canonical projection. Sincew j=p j=0, there exists an isomorphismh:S→A/Kerp such thathp=w. Thus we obtain the following commutative diagram in Mod-Rwith the first row and the column exact:
0
Kerp j
0 A f
w p
B v
S h A/Kerp i B/Kerp
0
(3.1)
Letg=h−1qv. Theng f =h−1qv f=h−1qiw=h−1w=p. HenceSis injective with respect
tof. ByProposition 3.1,Aiss-pure inB.
Corollary3.3. LetAbe anR-module such thatA/Dis injective for every maximal sub-moduleDofA. ThenAis absolutelys-pure.
Proof. Under the hypothesis,A/D is a direct summand ofE(A)/D for every maximal submoduleDofA. ByTheorem 3.2,Aiss-pure inE(A). Now byProposition 2.1,Ais
absolutelys-pure.
Corollary3.4. LetRbe hereditary and letAbe anR-module. Then the following state-ments are equivalent:
(i)Ais absolutelys-pure;
Proof. (i)⇒(ii). Suppose thatAis absolutelys-pure. LetDbe a maximal submodule ofA. SinceRis hereditary,E(A)/Dis injective. ByProposition 2.1andTheorem 3.2,A/Dis a direct summand ofE(A)/D, hence injective.
(ii)⇒(i). ByCorollary 3.3.
Theorem3.5. Consider the short exact sequence (1.1). Then the following statements are equivalent:
(i)the sequence (1.1) iss-pure;
(ii)for every commutative diagram inMod-R
B g
v
C α
E(S) p E(S)/S
(3.2)
whereSis a simpleR-module and p:E(S)→E(S)/Sis the natural homomorphism, there exists a homomorphismγ:C→E(S)such thatpγ=α.
Proof. (i)⇒(ii). Suppose that the sequence (1.1) iss-pure and consider the commutative diagram (3.2). It can be completed to the following commutative diagram in Mod-Rwith exact rows:
0 A f
β
B g
v
C α
0
0 S h E(S) p E(S)/S 0
(3.3)
whereh:S→E(S) is the inclusion homomorphism. ByProposition 3.1,Sis injective with respect to f. Hence there exists a homomorphismδ:B→Ssuch thatδ f =β. Then there exists a homomorphismγ:C→E(S) such thatpγ=α[8, Lemma 7.16].
(ii)⇒(i). Assume that (ii) holds. LetSbe a simpleR-module and letβ:A→Sbe a ho-momorphism. Leth:S→E(S) be the inclusion homomorphism and letp:E(S)→E(S)/S be the natural homomorphism. By the injectivity ofE(S), there exists a homomorphism v:B→E(S) such thatv f =hβ. Thus we can construct a commutative diagram (3.3) with exact rows. By hypothesis, there exists a homomorphismγ:C→E(S) such thatpγ=α. Then there exists a homomorphismδ:B→Ssuch thatδ f =β[8, Lemma 7.16]. Now by
Proposition 3.1, the sequence (1.1) iss-pure.
For a commutativeRthe result ofCorollary 2.8can be strengthened as follows. Proposition3.6. Every idempotent ideal ofRiss-pure.
Proof. LetI be an idempotent ideal ofRand letM be a maximal ideal ofR. IfIM, thenI+M=R, henceIM=I∩M. IfI⊆M, thenI=I2⊆IM⊆I∩M=I, henceIM=
I∩M. NowIiss-pure inRbyProposition 2.7.
Proposition3.7 [1, Corollary 1.4.3]. The following statements are equivalent for a right R-moduleE:
(i)Eis injective with respect to every inclusion homomorphismM→R, whereM is a maximal right ideal ofR;
(ii)Eis injective with respect to every inclusion homomorphismA→B, whereA is a submodule of a rightR-moduleBandB/Ais semiartinian.
Theorem3.8. The following statements are equivalent: (i)every maximal ideal ofRis idempotent;
(ii)every maximal ideal ofRiss-pure inR;
(iii)ifAis a submodule of anR-moduleBsuch thatB/Ais semiartinian, thenAiss-pure inB;
(iv)every maximal submoduleAof anR-moduleBiss-pure inB. Proof. (i)⇒(ii). ByProposition 3.6.
(ii)⇒(i). Assume that (ii) holds. LetMbe a maximal ideal ofR. ThenMiss-pure inR, henceM2=MbyProposition 2.7.
(ii)⇒(iii). Assume that (ii) holds. LetMbe a maximal ideal ofR,Sa simpleR-module, and f :M→Rthe inclusion homomorphism. ThenMiss-pure inR. ByProposition 3.1, Sis injective with respect to f. ByProposition 3.7,Sis injective with respect to every inclusion homomorphismA→B, whereAis a submodule of anR-moduleBsuch that B/Ais semiartinian. Now apply againProposition 3.1.
(iii)⇒(iv). Assume that (iii) holds. LetAbe a maximal submodule of anR-moduleB. ThenB/Ais clearly semiartinian. NowAiss-pure inBby hypothesis.
(iv)⇒(ii). Clear.
Remark 3.9. Clearly,Theorem 3.8holds for a commutative von Neumann regular ring. Corollary3.10. LetRbe semiartinian such that every maximal ideal ofRis idempotent. ThenRis von Neumann regular.
Proof. LetI be an ideal ofR. ThenR/I is semiartinian. ByTheorem 3.8,I iss-pure in R. ByProposition 3.1, every simpleR-moduleSis injective with respect to the inclusion homomorphismI→R. HenceSis injective. It follows thatRis von Neumann regular.
Example 3.11. LetR=KΛfor some fieldKand some infinite setΛ. ThenRis von
Neu-mann regular and not semiartinian. Since in a commutative von NeuNeu-mann regular ring every prime ideal is maximal, there exists a nonzero idealIofRsuch thatR/Iis not semi-artinian. ButIiss-pure inR. Hence the converse of condition (iii) ofTheorem 3.8does not hold.
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Iuliu Crivei: Department of Mathematics, Technical University of Cluj-Napoca, Str. C. Daicoviciu 15, 400020 Cluj-Napoca, Romania
