Automorphism-invariant modules

Automorphism-invariant modules

ADELALAHMADI(*) - ALBERTOFACCHINI(**) - NGUYENKHANHTUNG(***)

ABSTRACT- A moduleMis calledautomorphism-invariantif it is invariant under automorphisms of its injective envelope. In this paper, we study the en- domorphism rings of automorphism-invariant modules and their injective en- velopes. We investigate some cases where automorphism-invariant modules are quasi-injective and a connection between automorphism-invariant modules and boolean rings.

MATHEMATICSSUBJECTCLASSIFICATION(2010). 16D50; 16U60, 16W20.

KEYWORDS. Automorphism-invariant modules, injective envelopes.

1. Introduction

A moduleMis calledautomorphism-invariantif it is invariant under automorphisms of its injective envelope, that is, if W(M)M for every W2Aut(E(M)) (equivalently, if W(M)ˆM for every W2Aut(E(M))). In [6, Theorem 16], it was shown that automorphism-invariant modules are

(*) Indirizzo dell'A.: Department of Mathematics, King Abdulaziz University, Jeddah 21589, Saudi Arabia.

E-mail: [email protected]

(**) Indirizzo dell'A.: Dipartimento di Matematica, UniversitaÁ di Padova, 35121 Padova, Italy.

E-mail: [email protected]

The second author is grateful to UniversitaÁ di Padova (Progetto di ricerca di Ateneo CPDA105885/10 ``Differential graded categories'' and Progetto ex 60%

``Anelli e categorie di moduli'') and Fondazione Cassa di Risparmio di Padova e Rovigo (Progetto di Eccellenza ``Algebraic structures and their applications'') for the financial support.

(***) Indirizzo dell'A.: Dipartimento di Matematica, UniversitaÁ di Padova, 35121 Padova, Italy.

E-mail: [email protected]

preciselypseudo-injective modules, where a moduleMis calledpseudo- injectiveif, for any submoduleAof M, every monomorphismf:A!M can be extended to an element of End(M). In this paper, we show that if M is an automorphism-invariant module, then

End(M)=J(End(M)) turns out to be a rationally closed subring of

End(E(M))=J(End(E(M))):

Both the rings End(M)=J(End(M)) and End(E(M))=J(End(E(M))) are von Neumann regular [11, Proposition 1]. We consider in particular the case of automorphism-invariant modules of finite Goldie dimension or indecomposable. Notice that automorphism-invariant modules have the exchange property [11], so that indecomposable automorphism-invariant modules have a local endomorphism ring. Moreover, idempotents can be lifted modulo every right ideal both in End(M) and in End(E(M)) [16].

We then study the connection between automorphism-invariant mod- ules and boolean rings. The existence of such a connection was suggested to us by the results in Section 5 of [19], where VaÂmos considers modules whose endomorphism ring (or endomorphism ring modulo the Jacobson radical) is a boolean ring. He studies modules in which the identity en- domorphism is the sum of two automorphisms. This condition is related to the existence of factors of the endomorphism ring isomorphic to the field F2with two elements [12]. Notice that ifMis an automorphism-invariant right R-module and End(M) has no factor isomorphic toF₂, then M is quasi-injective [10, Theorem 3].

Every automorphism-invariant module is the direct sum of a quasi-in- jective module and a square-free module [6, Theorem 3]. This leads us to study, for an automorphism-invariant square-free moduleM, the relation betweenMbeing quasi-injective and the existence of factors isomorphic to F₂ in End(M) and in End(E(M)).

Throughout, all rings have identity element and modules are right unital. For a moduleM,E(M) denotes the injective envelope ofM.

2. Notation, definitions, and some properties of automorphism-invari- ant modules

LetRbe a ring. For every pair of rightR-modulesMandN, letD(M;N) denote the set of all module morphismsf:M!Nwhose kernel ker (f) is an

essential submodule ofM. In [11, Proposition 1], it was shown that ifMis an automorphism-invariant module, then the Jacobson radical J(End(M)) of End(M) consists exactly of all the endomorphisms of M with essential kernel, that is,J(End(M)ˆD(M;M). Moreover, End(M)=J(End(M) is von Neumann regular and idempotents lift moduloJ(End(M)).

Recall that the Jacobson radicalJ of a preadditive categoryCis the ideal defined, for everyA;B2Ob(C), byJ(A;B):ˆ ff 2HomC(A;B)j1A gf has a left inverse (equivalently, a two-sided inverse) for every morphism g:B!AinC g. LetAbe the full subcategory of Mod-Rwhose objects are all automorphism-invariant rightR-modules. It is easily seen thatDis an ideal ofAcontained in the Jacobson radicalJ ofA, because iff 2D(M;N), then ker (gf) is essential inMfor everyg. Hencegf 2J(End(M)) for every g [11, Proposition 1]. Thus for the idealsDandJ ofA, we have thatD J. Nevertheless the idealDcan be properly contained inJ. For instance, Z=2Z and Z=4Z are automorphism-invariant Z-modules, the mono- morphism f:Z=2Z!Z=4Z does not have an essential kernel, but for every morphism g:Z=4Z!Z=2Zthe composite mapping gf is the zero mapping. Hence gf has an essential kernel, and so gf belongs to J(End(Z=2Z)). ThereforeD(Z=2Z;Z=4Z) J(Z=2Z;Z=4Z).

A ring morphismW:R!Sislocalif, for everyr2R,W(r) invertible in Simpliesrinvertible inR. We will denote byU(R) the group of units, i.e., invertible elements, of a ringR. Arationally closedsubring of a ringSis a subringRofSsuch that the embeddingR!Sis a local morphism, that is, a subringRofSsuch thatU(R)ˆR\U(S) [4].

Recall that a moduleMsatisfies Condition (C₁) if every submodule ofM is essential in a direct summand ofM. An automorphism-invariant module satisfies (C1) if and only if it is quasi-injective [13]. Every automorphism- invariant moduleMsatisfies Condition (C2); that is, every submodule ofM isomorphic to a direct summand ofMis itself a direct summand ofM[5], [6]. Moreover, every automorphism-invariant module M satisfies Condi- tion (C₃); that is, ifN₁andN₂are direct summands ofMwithN₁\N₂ˆ0, thenN1N2is also a direct summand ofM= [13].

THEOREM 2.1. Let M be an automorphism-invariant module and E(M)be its injective envelope. Then

(a) There is a canonical local morphism

W:End(M)!End(E(M))=J(End(E(M)))

with kernel J(End(M)), so that W induces an embedding W, as a rationally closed subring, of the von Neumann regular ring

End(M)=J(End(M)) into the von Neumann regular right self-in- jective ring

End(E(M))=J(End(E(M))):

(b) For every invertible element v of the ringEnd(E(M))=J(End(E(M))), there exists an invertible element u ofEnd(M)=J(End(M))such that W(u)ˆv.

(c) For every idempotent element f of the ringEnd(E(M))=J(End(E(M))) there exists an idempotent element e of End(M)=J(End(M)) such thatW(e)ˆf if and only if the module M is quasi-injective.

(d)If M is quasi-injective, thenWis an isomorphism.

PROOF. (a) For any rightR-moduleM, the morphism W:End(M)!End(E(M))=J(End(E(M)))

is defined as follows. Iff 2End(M), let~fbe an endomorphism ofE(M) that extendsf. ThenW(f)ˆ~f ‡J(End(E(M))) [8, §4, p. 412]. It is easily seen that Wis a well-defined ring morphism. Moreover,W is a local morphism, because if f 2End(M) and W(f) is invertible in the ring End(E(M))=

J(End(E(M))), then ~f is an automorphism of E(M). Since M is auto- morphism-invariant, it follows that ~f(M)ˆM; that is, f(M)ˆM. This proves thatfis onto. Moreover,~fis an automorphism ofE(M) implies that its restrictionfis an injective endomorphism ofM. Thusf is an automorphism, and the ring morphismW is a local morphism. It follows that the injective morphismW:End(M)=ker (W)!End(E(M))=J(End(E(M))) induced byWis a local morphism as well. Moreover, ker (W)ˆD(M;M)ˆJ(End(M)).

(b) If v is an invertible element of End(E(M))=J(End(E(M))), then vˆv⁰‡J(End(E(M))) for some element v⁰2End(E(M)), necessarily invertible. Therefore v⁰ is an automorphism of E(M). Since M is auto- morphism-invariant, the restrictionu⁰ofv⁰toMis an automorphism ofM.

Thusu:ˆu⁰‡J(End(M)) is an invertible element of End(M)=J(End(M)) andW(u)ˆv.

(c) If M is quasi-injective, for everyf 2End(E(M)), the restriction f⁰ of f to M is an endomorphism of M. Thus W(f⁰‡J(End(M)))ˆ f‡J(End(E(M))). Hence Wis onto, and (a) allows us the conclusion.

(d) Assume that for every idempotent element f 2End(E(M))=

J(End(E(M))) there exists an idempotent element e of End(M)=

J(End(M)) withW(e)ˆf. In order to show that M is quasi-injective, we will prove that it satisfies Condition (C₁). LetNbe a submodule ofM. We must show thatNis essential in a direct summand ofM. NowE(M) has a

direct-sum decompositionE(M)ˆE(N)E. Thus there is an idempotent e2End(E(M)) with E(N)ˆeE(M) and Eˆ(1 e)E(M). By hypothesis, there exists an idempotent e2End(M)=J(End(M)) with W(e)ˆ e‡J(End(E(M))). As idempotents lift modulo J(End(M)), there is an idempotent e⁰2End(M) such that eˆe⁰‡J(End(M)). The idem- potente⁰2End(M) corresponds to a direct-sum decomposition Mˆ e⁰M(1 e⁰)M. This direct-sum decomposition of M induces a direct- sum decomposition E(M)ˆE(e⁰M)E((1 e⁰)M). Thus there is an idempotent e⁰⁰2End(E(M)) with E(e⁰M)ˆe⁰⁰E(M) and E((1 e⁰)M)ˆ (1 e⁰⁰)E(M). We claim that endomorphism e⁰⁰ of E(M) extends the endomorphism e⁰ of M. To prove this claim, it suffices to show that e⁰⁰(x)ˆx for every x2e⁰M and e⁰⁰(y)ˆ0 for every y2(1 e⁰)M.

Now e⁰ME(e⁰M)ˆe⁰⁰E(M), so that e⁰⁰(x)ˆx for every x2e⁰M.

Similarly (1 e⁰)ME((1 e⁰)M)ˆ(1 e⁰⁰)E(M), so that for every y2(1 e⁰)M one has that y2(1 e⁰⁰)E(M). Hence e⁰⁰(y)ˆ0. This proves the claim. ThusW(e⁰‡J(End(M))ˆe⁰⁰‡J(End(E(M_R))). ButW(e)ˆ e‡J(End(E(M))) and eˆe⁰‡J(End(M)), so that W(e⁰‡J(End(M)))ˆ e‡J(End(E(M))). It follows thate⁰⁰‡J(End(E(M)))ˆe‡J(End(E(M)));

that is,e⁰⁰ e2J(End(E(M))), so 1 e⁰⁰‡eis an automorphism ofE(M). As M is automorphism-invariant, we have that (1 e⁰⁰‡e)(M)ˆM. Thus e(M)(1 e⁰⁰‡e)(M)‡1(M)‡e⁰⁰(M)ˆM‡M‡e⁰(M)ˆM. It follows thaterestricts to an idempotent endomorphism ofM. In particular,e(M) is a direct summand ofM. Moreover,NE(N)\MˆeE(M)\Mˆe(M), so thatNis a submodule ofe(M). It remains to show thatNis essential ine(M).

This follows immediately from the fact thate(M)eE(M)ˆE(N) andN is essential in E(N). This proves that M satisfies Condition (C₁), and hence is quasi-injective [13].

The converse follows immediately from (d), noting that the inverse image of an idempotent via an injective morphism is necessarily

idempotent. p

Notice that if W:R!S is a local morphism and S is local (resp., semilocal), that is, a division ring (resp., a semisimple artinian ring) modulo the Jacobson radical, then R is local (resp., semilocal) [4, Theo- rem 1]. Nevertheless, for any ringR there exist a von Neumann regular right self-injective ringSand a local morphismsx:R!S. For example, it suffices to take the first part 0!R_R!E₁!E₂of an injective resolution of the R-module RR, the von Neumann regular right self-injective ring S:ˆEnd(E1)=J(End(E1))End(E2)=J(End(E2)) and the local morphism x defined in [8, Theorem 5.3].

PROPOSITION 2.2. Let M be an automorphism-invariant module.

Then

(a)If M is indecomposable, thenEnd(M)is a local ring.

(b) If M has finite Goldie dimension, then every injective endomor- phism of M is an automorphism of M and the endomorphism ring End(M)is a semiperfect ring.

PROOF. (a) Automorphism-invariant modules have the exchange prop- erty [11], and indecomposable modules with the exchange property have a local endomorphism ring [7, Theorem 2.8].

(b) Let M be an automorphism-invariant module of finite Goldie di- mension and let W:M!M be an injective endomorphism ofM. Then W extends to an endomorphism W₀:E(M)!E(M), which is necessarily in- jective. AsMhas finite Goldie dimension,W₀is an automorphism ofE(M).

ButMis automorphism-invariant, soW₀(M)ˆM. ThusW(M)ˆM, that is, the endomorphismWis also surjective.

Finally, every module of finite Goldie dimension is a direct sum of in- decomposable modules. Thus if MˆM₁. . .M_n is automorphism-in- variant and the M_i are indecomposable, then the modules M_i are auto- morphism-invariant. Hence they have a local endomorphism ring by (a).

ThereforeMis a direct sum of modulesMiwith local endomorphism rings,

so that End(M) is semiperfect. p

COROLLARY2.3. If M;N are two automorphism-invariant R-modules of finite Goldie dimensions isomorphic to submodules of each other, then M is isomorphic to N.

PROOF. By the hypothesis, there exists two monomorphisms f :M!N andg:N!M. So fg2End(N) and fg is injective. Hence fg is an automorphism by Proposition 2.2(b). Thus f is onto. Since f is a

monomorphism, f is an isomorphism. p

We now recall a theorem by Guil Asensio and Srivastava that will be repeatedly used in the sequel.

THEOREM 2.4 [10, Theorem 3]. Let M be a right module such that End(M)has no factor isomorphic toF₂. Then M is quasi-injective if and only if M is automorphism-invariant.

PROPOSITION 2.5. If R is a ring of odd characteristic, then every automorphism-invariant R-module is quasi-injective.

PROOF. Suppose thatRis a ring of odd characteristicnwith a module M that is automorphism-invariant but not quasi-injective. By Theo- rem 2.4, the endomorphism ring End(M) has a factor End(M)=M iso- morphic to F₂. Then nRˆ0, so that nMˆ0. Hence nEnd(M)ˆ0, so that n(End(M)=M)ˆ0. Thus nF₂ˆ0, which is a contradiction because

n is odd. p

LEMMA 2.6. Let M be an automorphism-invariant module. If

MˆM1M2. . .Mn, where each M_iis a quasi-injective module, then

M is quasi-injective.

PROOF. It clearly suffices to prove the case nˆ2. Assume that MˆM₁M₂ is automorphism-invariant, where M₁ and M₂ are quasi- injective. By [13, Theorem 5], M1 isM2-injective andM2 isM1-injective.

Since M1 andM2 are quasi-injective, M is quasi-injective by [1, Propo-

sitions 16.10 and 16.13]. p

PROPOSITION 2.7. Let M₁;M₂;. . .;M_n be uniform modules. If M:ˆ

M1M2. . .Mnis automorphism-invariant, then M is quasi-injective.

PROOF. By the previous lemma, it suffices to show that eachMiis quasi- injective. On the one hand, eachM_iis uniform, and eachM_isatisfies (C₁).

On the other hand, each M_i is automorphism-invariant, being a direct summand of an automorphism-invariant module. By [13, Corollary 13], everyMiis quasi-injective. Now apply Lemma 2.6. p PROPOSITION2.8. The following conditions are equivalent for a ring R.

(1) Every automorphism-invariant R-module of finite Goldie dimen- sion is quasi-injective.

(2) Every automorphism-invariant indecomposable R-module of finite Goldie dimension is uniform.

(3) Every automorphism-invariant indecomposable R-module of finite Goldie dimension is quasi-injective.

PROOF. (1))(2) An automorphism-invariant indecomposable module M of finite Goldie dimension is quasi-injective by (1). Hence it satisfies Condition (C₁). ThereforeMis uniform.

(2))(3) LetMbe an automorphism-invariant indecomposable module of finite Goldie dimension. ThenMis uniform by (2), and hence it satisfies Condition (C1). By [13, Corollary 13],Mis quasi-injective.

(3))(1) LetMbe an automorphism-invariant module of finite Goldie dimension. Then End(M) is semilocal by Proposition 2.2(b). So

MˆM1M2. . .Mn, where each M_i is an automorphism-invariant

indecomposable module of finite Goldie dimension. By (3), every M_i is quasi-injective. From Lemma 2.6, it follows thatMis quasi-injective. p By Proposition 2.5, every ring R of odd characteristic satisfies the equivalent conditions of Proposition 2.8.

LetE(M) be the injective envelope of a moduleM. It is easily seen that X

W2Aut(E(M))

W(M)

is the smallest automorphism-invariant submodule ofE(M) containingM.

We call it the automorphism-invariant envelopeof M, and denote it by AI(M). Clearly, a module is automorphism-invariant if and only if MˆAI(M).

LEMMA 2.9. Let M;N be arbitrary R-modules. Then every mono- morphism M!N extends to a monomorphism AI(M)!AI(N).

PROOF. A monomorphism W:M!N extends to a monomorphism W⁰:E(M)!E(N), which is necessarily a split monomorphism. Thus there is a direct-sum decompositionE(N)ˆW⁰(E(M))Cand, with respect to this direct-sum decomposition,W⁰:E(M)!W⁰(E(M))Ccan be written in ma- trix form as W⁰ˆa

0

, where a:E(M)!W⁰(E(M)) is an isomorphism. It suffices to show that W⁰(AI(M))AI(N): Let f be an automorphism of E(M). Then

afa ¹ 0

0 1

is an automorphism ofW⁰(E(M))CˆE(N). Thus W⁰(f(M))ˆaf(M)ˆ(afa ¹)(a(M))

afa ¹ 0

0 1

(a(M)) afa ¹

0 0 1

(N) AI…N†:

ThereforeW⁰(AI(M))AI(N). p

3. Boolean rings

Recall that a non-zero ringRis aboolean ringif every element ofRis idempotent. Every boolean ring is necessarily a commutative ring of characteristic 2. A ring is boolean if and only if it is isomorphic to a subring ofF^X₂, whereXis a non-empty set andF₂is the field with two elements. If a

ringRis boolean, thenE(RR) is the Dedekind-McNeil completion ofR, and the Dedekind-McNeil completion of R is a boolean ring, which is iso- morphic, as an R-module, to the total ring of quotients ofR ([2, Corol- lary 2] and [18, p. 79]). If R is a boolean ring, the identity is the unique automorphism of the R-module E(R_R), so that every R-submodule of E(R_R) is automorphism-invariant, butR_Ris not quasi-injective (except for the caseRRˆE(RR), that is,Ris complete).

The condition ``the endomorphism ring S of a module does not have maximal idealsMwithS=M F₂'' has recently received a lot of attention (Theorem 2.4 and [12]).

LEMMA3.1. Let T be a ring and I the two-sided ideal of T generated by the subsetft t²jt2Tgof T. Then

(a) The ideal I is the smallest ideal of T with T=I a boolean ring or the zero ring.

(b) The ideal I is the intersection of all maximal two-sided idealsMof T with T=M F₂.

(c) The ideal I contains the Jacobson radical J(T)of T.

(d) The kernel of every ring morphism T!F2 contains I.

(e) I is a proper ideal of T if and only if there exists a ring morphism T!F₂, if and only if T has a maximal two-sided ideal Mwith T=M F₂.

PROOF. (a) is trivial.

(b) Let us check that

Iˆ \

T=MF2

M:

() SinceIis generated by the elementst t², it suffices to show that t t²2 M for every t2T and every maximal two-sided ideal M with T=M F₂. Now F₂ is boolean, so that T=M is boolean, hencet‡ M ˆ t²‡ M. It follows thatt t²2 M.

() By (a), the ringT=Iis boolean. Boolean rings are isomorphic to subrings of F^X₂ for some set X. Let e:T=I!F^X₂ be an embedding and p_x:F^X₂ !F₂ (x2X), p:T!T=I be the canonical projections. Then the morphisms W_x:ˆp_xep:T!F₂ have kernels ker W_x, which are maximal two-sided ideals of T, T=ker W_xF2 and T

x2Xker W_xˆI. Thus T

T=MF2M T

x2Xker W_xˆI.

(c) By (b),Iis the intersection of all maximal two-sided idealsMofT withT=M F₂, and all maximal two-sided idealsMofTwithT=M F₂

are maximal right ideals ofT. HenceIis an intersection of maximal right ideals ofT, so thatIJ(T).

(d) The kernel of every ring morphismT!F2is a maximal two-sided ideal ofTwithT=M F₂. Thus (d) follows from (b).

(e) is now trivial. p

Notice that 1_T is an element of T, so that ( 1_T) ( 1_T)²ˆ 21_T2I. ThereforeIcontains the two-sided ideal 2TofT. For example, forTˆZ, the ideal ofZgenerated by alln n², wherenranges inZ, is exactly the ideal 2Z.

Recall that two modules are said to beorthogonalto each other if they do not contain non-zero isomorphic submodules. The following Lemma appears in [15, Lemma 3.3] and will be repeatedly used in the sequel.

LEMMA3.2[15, Lemma 3.3]. Let MˆM₁M₂. If M₁and M₂are ortho- gonal, thenEnd(M)=D(M;M)End(M1)=D(M1;M1)End(M2)=D(M2;M2).

The converse holds if M1and M2are relatively injective.

COROLLARY3.3. Let MˆM1M2be an automorphism-invariant R- module where M₁ and M₂ are orthogonal. Then End(M) has no factor isomorphic to F₂ if and only if each End(M_i) (iˆ1;2) has no factor isormophic toF2.

PROOF. LetIbe the two-sided ideal of End(M) generated by the set fx x²jx2End(M)g. By Lemma 3.2, End(M)=D(M;M)End(M1)=

D(M₁;M₁)End(M₂)=D(M₂;M₂). AsD(M;M)ˆJ(End(M)) for any auto- morphism-invariant R-module M, it follows that End(M)=J(End(M)) End(M₁)=J(End(M₁))End(M₂)=J(End(M₂)) in a canonical way. Thus there is a homomorphism End(M)!F2 if and only if there is a homo- morphism End(M)=J(End(M))!F2, if and only if there is a homomorph- ism End(M_i)= J(End(M_i))!F₂ for ani equal to 1 or 2. The conclusion

follows immediately. p

Recall that a module issquare-freeif it does not contain a direct sum of two non-zero isomorphic submodules. The following important result is due to Er, Singh and Srivastava.

THEOREM3.4 [6, Theorem 3]. Every automorphism-invariant mod- ule M decomposes as a direct sum MˆXY, where X is quasi-injective, Y is a square-free module orthogonal to X, and X and Y are relatively injective modules.

The previous Theorem reduces our study to considering automorph- ism-invariant square-free modules. Notice that ifMis any rightR-module such that End(M) has no factor isomorphic toF2, thenMis quasi-injective if and only ifMis automorphism-invariant (Theorem 2.4).

LEMMA3.5. If M₁;M₂are two right modules over a ring R and M₁;M₂ have isomorphic injective envelopes, which are non-zero modules, then M₁ and M2have non-zero isomorphic submodules.

PROOF. Let f:E(M1)!E(M2) be an isomorphism. Then M1 and f ¹(M₂) are essential submodules of E(M₁). Hence M₁\f ¹(M₂) is an essential submodule ofE(M₁). It follows that M₁\f ¹(M₂) is a non-zero submodule ofM₁. Via the isomorphismf, we find thatf(M₁\f ¹(M₂)) is an essential submodule of E(M2) isomorphic to M1\f ¹(M2). But f(M1\f ¹(M2))ˆf(M1)\M2 is a submodule ofM2. p COROLLARY3.6. A module M is square-free if and only if its injective envelope E(M)is square-free.

PROOF. If M is not square-free, then it contains a submodule iso- morphic toNNfor some non-zero moduleN. Hence the same holds for E(M), that is,E(M) is not square-free.

Conversely, assume thatE(M) is not square-free. ThenE(M) contains a submodule isomorphic toNN for some non-zero moduleN. It follows that E(M)ˆE₁E₂E₃ with E₁E₂6ˆ0. Then M\E_i is a non-zero essential submodule of Ei foriˆ1;2. In particular,M\E1 andM\E2

have isomorphic injective envelopes, which are non-zero modules. By Lemma 3.5, M\E1 and M\E2 have non-zero isomorphic submodules.

ThusMis not square-free. p

COROLLARY 3.7. If M is an automorphism-invariant square-free module, then every injective endomorphism of M is an automorphism of M.

PROOF. Let M be an automorphism-invariant square-free module and let W:M!M be an injective endomorphism ofM. Then Wextends to an endomorphism W₀:E(M)!E(M), which is necessarily injective.

Then E(M)ˆW₀(E(M))C, so that E(M)ˆW²₀(E(M))W₀(C)C with W₀(C)C. By Corollary 3.6,E(M) is square-free, soCˆ0. This proves thatW₀is an automorphism of E(M). ButM is automorphism-invariant, so W₀(M)ˆM. Thus W(M)ˆM, that is, the endomorphism W of M is

also surjective. p

Arguing as in Corollary 2.3, we find that:

COROLLARY3.8. If M;N are two automorphism-invariant square-free R-modules isomorphic to submodules of each other, then M is isomorphic to N.

COROLLARY 3.9. Let M be an automorphism-invariant R-module.

Assume that MˆM1M2, where M1and M2are orthogonal. Let E_ibe an injective envelope of M_i. Then E₁ is orthogonal to E₂.

PROOF. Assume that there exists 06ˆN₁E₁ and 06ˆN₂E₂ such thatN1N2. LetE⁰_ibe an injective envelope ofNi. ThenE1ˆE⁰₁E⁰⁰₁and E2ˆE⁰₂E⁰⁰₂ where E⁰₁E⁰₂. Set E_R:ˆE1E2ˆE⁰₁E⁰₂(E⁰⁰₁E⁰⁰₂).

ThenEis an injective envelope ofM. Since Mis automorphism-invariant and E⁰₁E⁰₂, we get that Mˆ(M\E⁰₁)(M\E⁰₂)(M\(E⁰⁰₁E⁰⁰₂)) [17, Lemma 7]. We will show that M\E⁰₁M₁. Let x2M\E⁰₁, then xˆx1‡x2 where xi2Mi and x2E⁰₁E1. Hence x2 ˆx x12M2\ E1E2\E1ˆ0. Thereforexˆx12M1. By a similar argument, we get thatM\E⁰₂M2. AsM\E⁰_iis essential inE⁰_iandE⁰_iis injective,E⁰_iis an injective envelope ofM\E⁰_i. Moreover,E⁰₁E⁰₂. Hence, by Lemma 3.5, there exist non-zero submodules P_!1M\E⁰₁M₁ and P₂ M\E⁰₂M2 such that P1P2. ThereforeM1 is not orthogonal toM2.

This is a contradiction. p

PROPOSITION3.10. Let M be an automorphism-invariant module and E(M)be its injective envelope. The following conditions are equivalent:

(a) M is square-free.

(b) E(M)is square-free.

(c) The von Neumann regular ringEnd(M)=J(End(M))is abelian.

(d) The von Neumann regular right self-injective ring End(E(M))=

J(End(E(M)))is abelian.

PROOF. (a),(b) has been proved in Corollary 3.6.

(b))(d) follows from the fact that D(E;E)ˆJ(End(M)) for any in- jective moduleEand [15, Lemma 3.4].

(d))(c) follows from the fact that every subring of an abelian ring is an abelian ring and Theorem 2.1.

(c))(a) Assume that (c) holds. Set S:ˆEnd(M). Suppose that M contains a direct sum XY of two isomorphic submodules. Taking the

injective envelopes in E(M), one finds thatE(M)ˆE(X)E(Y)C. If W:X!Yis an isomorphism,Wextends to an isomorphismc:E(X)!E(Y).

Thus there is an isomorphism v:ˆ 0 c ¹ 0

c 0 0

0 0 1_C

0

@

1

A:E(M)ˆE(X)E(Y)C!E(M)ˆE(X)E(Y)C:

The automorphismvof E(M) restricts to an automorphismv⁰ ofM be- causeMis automorphism-invariant. From [17, Lemma 7], we know that Mˆ (M\E(X))(M\E(Y))(M\C). Thus Mˆe1Me2Me3M for orthogonal idempotents e_i2S, where e1MˆM\E(X) and e₂MˆM\E(Y). Nowv⁰(M\E(X))ˆv(M\E(X))ˆv(M)\v(E(X))ˆ M\E(Y). ThusM\E(X))M\E(Y), that is,e₁Me₂M. Applying the functor Hom(M; ):Mod-R!Mod-S, one finds thatS_S ˆe₁Se₂Se₃S and e1SSe2SS [7, Theorem 4.7]. If ei is the image of ei in S=J(S), then S=J(S)ˆe1S=J(S)e2S=J(S)e3S=J(S) and e1S=J(S)e2S=J(S) [1, Exercise 7.2]. But S=J(S) is abelian, so that e₁S=J(S)e₂S=J(S) implies e₁S=J(S)ˆe₂S=J(S) [9, Theorem 3.4]. From this we get that 1 e₁e₂ˆ0, i.e., e₂ˆe₁e₂. Similarly e₁ˆe₂e₁. Thus e₁ e₂ is an idem- potent in J(S), from which e1ˆe2. Thus M\E(X)ˆM\E(Y), and

XˆYˆ0. p

The next Corollary generalizes [14]. Recall that a ring isduoif all its right ideals and all its left ideals are two-sided ideals. A ring isquasi-duoif all its maximal right ideals and all its maximal left ideals are two-sided ideals.

COROLLARY 3.11. The endomorphism ring of an automorphism-in- variant square-free module is quasi-duo.

PROOF. LetMbe an automorphism-invariant square-free module. By Proposition 3.10, End(M)=J(End(M)) is an abelian von Neumann regular ring. One-sided ideals of abelian regular rings are generated by central idempotents, hence all of them are two-sided. Thus End(M)=J(End(M)) is a duo ring. The conclusion now follows from the fact that a ringSis quasi-duo

if and only ifS=J(S) is quasi-duo. p

THEOREM3.12. Let M be an automorphism-invariant module and let E(M)be its injective envelope.

(a) If M is quasi-injective andEnd(M)has a factor isomorphic toF₂, thenEnd(E(M))has a factor isomorphic toF2.

(b) If M has finite Goldie dimension and End(M) has a factor iso- morphic toF2, then the following conditions hold.

(i) End(E(M))has a factor isomorphic toF2.

(ii) E(M)has a direct-sum decomposition E(M)ˆEC with E orthogonal to C, E an indecomposable R-module andEnd(E)=

J(End(E))F2.

(iii) Aut(E)ˆ1‡J(End(E)), so that every automorphism of the R- module E is the identity on an essential R-submodule of E.

(iv) E is the injective envelope of its non-zero R-submoduleannE(2).

PROOF. (a) IfMis quasi-injective, the mapping

W:End(M)=J(End(M))!End(E(M))=J(End(E(M)))

is an isomorphism by Theorem 2.1(d). Thus End(E(M)) has a factor iso- morphic toF₂.

(i) We first consider the case of M indecomposable. If M is auto- morphism-invariant indecomposable, then End(M) is local by Proposi- tion 2.2. If End(M) also has a factor isomorphic toF2, then

End(M)=J(End(M))F₂:

SinceM is automorphism-invariant, we get thatMˆNP, whereN is quasi-injective andPis square-free (Theorem 3.4). ButMis indecomposable, so that eitherMˆNorMˆP. IfMˆNis quasi-injective, End(E(M)) has a factor isomorphic toF₂by (a). In the other case,MˆPis square-free, so that End(E(M))=J(End(E(M)) is abelian. AsMhas finite Goldie dimension, E(M) has finite Goldie dimension. Hence End(E(M)) is semilocal. Therefore End(E(M))=J(End(E(M))D1D2. . .Dn, where each D_i is a divi- sion ring. Consider the mappingW:End(M)!End(E(M))=J(End(E(M)) of Theorem 2.1. From ker WˆJ(End(M)), it follows that imWEnd(M)=

J(End(M))F₂. Moreover, the group of units of End(E(M))=J(End(E(M)) is contained in im(W), becauseMis automorphism-invariant. Hence the group of units of

End(E(M)=J(End(E(M))

has one element. Since it is isomorphic toD1nf g 0 . . .Dnnf g, it fol-0 lows that DiF2 for every iˆ1;. . .;n. So End(E(M)) has a factor iso- morphic toF2. This concludes the proof of (i) forMindecomposable.

Now let M be an arbitrary automorphism-invariant module of finite Goldie dimension and assume that End(M) has a factor isomorphic toF₂.

The proof will be by induction on the Goldie dimensionnofM. Ifnˆ1, then M is indecomposable, and we are done. Suppose n>1. SinceM is automorphism-invariant, we have that MˆNP, where N is quasi-in- jective,Pis square-free andN;Pare orthogonal. IfPˆ0, thenMis quasi- injective, and we conclude by (a). IfNˆ0, thenMis square-free. IfMis indecomposable, we are done, as we have seen in the previous paragraph.

Otherwise MˆM1M2 for suitable non-zero submodules M1;M2. The modules M1;M2 are orthogonal because M is square-free. By Corollary 3.3, either End(M₁) or End(M₂) has a factor isomorphic toF₂. Without loss of generality, we can assume that End(M₁) has a factor isomorphic toF₂. Let E_i be an injective envelope ofM_i, so that E(M)ˆE₁E₂. By the inductive hypothesis, we get that End(E1) has a factor isomorphic toF2. Moreover, E1;E2 are orthogonal by Corollary 3.9. Thus End(E) has a factor isomorphic toF₂by Corollary 3.3, and we are done.

It remains to consider the case MˆNP with bothN andP non- zero. Then E(M)ˆE(N)E(P). Then E(N) and E(P) are orthogonal (Corollary 3.9), and either End(N) or End(P) has a factor isomorphic to F2(Corollary 3.3). By the inductive hypothesis, End(E(N)) or End(E(P)) has a factor isomorphic to F2. The conclusion follows by Corollary 3.3.

(ii) SinceMis of finite Goldie dimension,E(M) decomposes asE(M)ˆ E₁:::E_n, where theE_iare indecomposable injectiveR-modules. Now End(M) is semiperfect (Proposition 2.2(b)), hence semilocal. By the hy- pothesis, there exists a ring morphism End(M)!F2, so that there exists a ring morphism End(M)=J(End(M))!F2. The semisimple artinian ring End(M)=J(End(M)) is a finite direct product of rings of matricesM_nj(D_j) over division rings D_j. The kernel of the ring morphism End(M)=

J(End(M))!F2is a maximal ideal of this finite direct product of rings of matrices Mnj(Dj). It follows that there exists an indexj withnj ˆ1 and D_jF2. Thus, in the direct-sum decomposition E(M)ˆE1:::En, there exists an indexiwithE_i6E_k for everykˆ1;. . .ndifferent fromi and End(E_i)=J(End(E_i))F₂. Set E:ˆE_i and C:ˆE₁. . .E_{i 1}

Ei‡1. . .En. In order to conclude the proof of (ii), it suffices to show

that E is orthogonal to C. Assume the contrary. Then there exist iso- morphic non-zero submodules A of E and B of C. Thus E(B) is an in- decomposable direct summand ofCisomorphic toE(A)E. By the Krull- Schmidt-Azumaya Theorem, the moduleE(B) must be isomorphic to one of the modulesE₁;. . .;E_{i 1};E_i‡1;. . .;E_n. This is a contradiction.

(iii) IfW2Aut(E), we have thatW‡J(End(E)) is an invertible element in the ring End(E)=J(End(E)). But End(E)=J(End(E))F2, so that W‡J(End(E))ˆ1‡J(End(E)). ThusW21‡J(End(E)). This proves that

Aut(E)ˆ1‡J(End(E)). In particular, every automorphism of the R- moduleEis the identity on an essentialR-submodule ofE.

(iv) From End(E)=J(End(E))F2, it follows that 1_E‡1_E2J(End(E));

that is 2 annihilates an essential submodule ofE. Therefore ann_E(2) is a non-zeroR-submodule ofE. B utEis uniform. p Theorem 3.12(b) does not hold whenMis not automorphism-invariant.

To see this, takeRˆZandMˆZZ. ThenQZis an injective envelope of Z_Z. The endomorphism ring ofZ_Z is isomorphic toZ. So it has a factor isomorphic to F₂. But the endomorphism ring of Q_Z has no factor iso- morphic toF₂.

REMARK3.13. LetMbe any rightR-module, letE(M) be its injective envelope andS:ˆEnd(E(M)) be the endomorphism ring ofE(M), so that E(M) turns out to be a S-R-bimodule. Let Ibe the two-sided ideal of S generated by the set fs s²js2Sg. Then the annihilator ann_E(M)I:ˆ

fe2E(M)jIeˆ0gis anS-R-subbimodule ofSE(M)_R, as is easily seen.

Thus there is anR-module direct-sum decompositionE(M)_RˆE1E2, whereE1is an injective envelope E(ann_E(M)I) of ann_E(M)Iin E(M)_R and E₂is a complement ofE₁inE(M)_R, so that no non-zero element ofE₂is annihilated byI, i.e.,e₂2E₂andIe₂ˆ0 implye₂ˆ0. Assume there are two non-zero R-submodules A1;A2 such that A1E1, A2E2 and A1A2. Then their injective envelopesE(A1);E(A2) are isomorphic and eachE(A_i) is a direct summand of E_i. SoE(M) decomposes as a direct sum E(M)ˆe₁E(M)e₂E(M)e₃E(M) for orthogonal idempotents e_i2End(E(M)) where e_iE(M)ˆE(A_i)(iˆ1;2). Since E(A₁)E(A₂), e1E(M)e2E(M). Applying the functor Hom(E(M); ):Mod-R!Mod-S, one finds that SSˆe1Se2Se3S and e1SSe2SS [7, Theorem 4.7], where SˆEnd(E(M)). So there exists a unit element u2S such that e₁ˆu ¹e₂u. As e₂ann_E(M)Iˆ0 and e₁ˆu ¹e₂u, it follows that e₁ann_E(M)Iˆ0. But this contradictse₁ann_E(M)I6ˆ0, becausee₁ann_E(M)Iˆ E(A1)\annE(M)I6ˆ0:Therefore twoR-modulesE1andE2are orthogonal.

By Lemma 3.2,S=D(E(M);E(M))S1=D(E1;E1)S2=D(E2;E2), whereSi

denotes the endomorphism ring of the R-module E_i. As D(E;E)ˆ J(End(E)) for any injective R-module E, it follows that S=J(S) S₁=J(S₁)S₂=J(S₂) in a canonical way. IfI_idenotes the two-sided ideal of S_igenerated by allx x² withx2S_i, thenI=J(S)I₁=J(S₁)I₂=J(S₂).

Now consider the ring morphismr:S!End(annE(M)I) that associates to anyf 2S its restrictionfj_annE(M)I to ann_E(M)I. The ring morphism ris well defined because ann_E(M)I is a left S-submodule of E(M). The

morphism r is clearly an onto mapping, and its kernel is kerr:ˆ

ff 2Sjf(annE(M)I)ˆ0g. In particularIkerr. SinceS=Iis a boolean ring, the ring End(ann_E(M)I) is also boolean. Moreover,J(S)Ikerr, so that r induces a ring morphism r:S=J(S)!End(ann_E(M)I). As S=J(S)S₁=J(S₁)S₂=J(S₂) and the elements of S₂=J(S₂) are clearly mapped to 0 by r, we get that 0S₂=J(S₂)ker (r). Thus there is a surjective ring morphism S1=I1!End(annE(M)I).

From Remark 3.13, we get in particular that:

PROPOSITION 3.14. Let M be an R-module, S:ˆEnd(E(M)) be the endomorphism ring of E(M)and I be the two-sided ideal of S generated by the setfs s²js2Sg.

(a) Ifann_E(M)I6ˆ0, thenEnd(M)has a factor isomorphic toF₂. (b) If M is automorphism-invariant and annE(M)I is an essential

submodule of the R-module E(M), then the ringEnd(M)=J(End(M)) is a boolean ring.

PROOF. Compose the ring morphism W:End(M)!S=J(S) of Theo- rem 2.1 with the morphism r:S=J(S)!End(ann_E(M)I) in Remark 3.13, obtaining a morphismrW:End(M)!End(ann_E(M)I), where End(ann_E(M)I) is a boolean ring. If ann_E(M)I6ˆ0, then End(ann_E(M)I) is a non-zero boolean ring, so that there is a morphism End(ann_E(M)I)!F₂. Thus there is a morphism End(M)!F2, necessarily surjective. Hence End(M) has a factor isomorphic toF2. This concludes the proof of (a).

If M is automorphism-invariant and ann_E(M)I is essential in E(M), then in Remark 3.13 we have that E(M)ˆE₁, E₂ˆ0 and kerr D(E(M);E(M))ˆJ(S). As Ikerr and J(S)I, it follows that Iˆ kerrˆJ(S). Thus S=J(S)End(annE(M)I) is a boolean ring. By Theo- rem(a) 2.1, the ring End(M)=J(End(M)) is isomorphic to a subring of the ring End(E(M))=J(End(E(M)))ˆS=J(S). Thus End(M)=J(End(M)) is

boolean. p

PROPOSITION3.15. Let M be an automorphism-invariant square-free module of finite Goldie dimension. Then M decomposes as a direct sum MˆNP, where N is a module orthogonal to P,End(N)has no factor isomorphic toF₂, andEnd(P)=J(End(P))is isomorphic to a boolean ring Fⁿ₂ for some n.

PROOF. The automorphism-invariant moduleMof finite Goldie dimen- sion, decomposes as a direct sum MˆM₁. . .M_t of indecomposable

modules, necessarily automorphism-invariants. Lete1;. . .;et2End(M) be the orthogonal idempotents corresponding to this direct-sum decomposition of M. Then e1;. . .;et2End(M)=D(M;M) are orthogonal idempotents of End(M)=D(M;M), which is an abelian ring by Proposition 3.10. Thus the idempotentse₁;. . .;e_t of End(M)=D(M;M)ˆEnd(M)=J(End(M)) are cen- tral, so that

End(M)=J(End(M))

e₁End(M)=J(End(M))e₁. . .e_tEnd(M)=J(End(M))e_t 

End(M₁)=J(End(M₁)). . .End(M_t)=J(End(M_t));

is isomorphic to the direct product of the residue division rings End(Mi)=J(End(Mi)). LetNbe the direct sum of theMiwith the residue division rings End(Mi)=J(End(Mi)) not isomorphic toF2andPbe the di- rect sum of theM_iwith the residue division rings End(M_i)=J(End(M_i)) isomorphic toF₂. ThenMˆNP, End(N) has no factor isomorphic toF₂, because End(N)=J(End(N)) is a direct product of finitely many division rings not isomorphic toF2, and End(P)=J(End(P)) isomorphic to a direct product of finitely many copies ofF2.

Finally,N andPare relatively injective by [13, Theorem 5]. As End(M)=D(M;M)End(N)=D(N;N)End(P)=D(P;P);

we conclude thatN andPare orthogonal (Lemma 3.2). p

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Manoscritto pervenuto in redazione il 27 Dicembre 2013.