Some Recent Developments in the Theory of Baer and Rickart Modules

(1)

Developments in the Theory of Baer and Rickart Modules

S. Tariq Rizvi

Baer and Rickart Rings Module Theoretic Versions of Baer and Rickart Rings Connections to the Endomorphism Ring The Direct Sum Problem Free Baer Modules and Classes of Rings Free Rickart Modules and Classes of Rings IMPORTANT!

Some Recent Developments in the Theory of Baer and Rickart Modules

(joint work with C. Roman and G. Lee)

S. Tariq Rizvi

The Ohio State University

(ICRA, Taipei, Taiwan- In honor of Prof. P.-H. Lee, July 11)

(2)

an idempotent, i.e. rR(K) ={x|Kx = 0}=eR ∃e²=efor any nonempty subsetK ⊆R.

⇔the left annihilator of every subset inR is generated by an idempotent, i.e. lR(K) ={x|xK = 0}=Re ∃e²=e for any nonemptyK ⊆R (i.e. the notion is left-right symmetric)

Example

- Roots in Functional Analysis- the Algebra of bounded operators on a Hilbert space, in fact any von Neumann Algebra is a Baer (*-)ring;

- any domain;

- any right noetherian hereditary ring is a Baer ring;

(3)

Some Recent Developments in the Theory of Baer and Rickart Modules

S. Tariq Rizvi

Definition

A ringR is called aright Rickart ring(also known as aright PP ring) if the right annihilator of any element ofR is generated by an idempotent as a right ideal, i.e.,∀a∈R,r_R(a) =eR,

∃e²=e∈R.

Example

- von Neumann regular rings;

- Baer rings, (For example, every right nonsingular (hence regular) right self injective ring);

- right (semi-)hereditary rings;

-EndR(R^(I)) withRa right hereditary ring and I an index set, is a right Rickart ring. - A right hereditary ring which is not a Baer ring is right Rickart but not Baer.

* left Rickart (p.p.) rings are defined similarly.

* There is a well-known example of Chase of a right Rickart ring which is not left Rickart.(i.e. the notion is NOT left-right symmetric)

(4)

* Throughout this talk, letM be a rightR-module and

S =End(M).(We will see that this notion heavily depends onS) Definition(T. Rizvi and C. Roman, 2004)

A rightR-module M is calleda Baer module if∀ N≤M,lS(N) =Se withe²=e∈S.

⇔if∀ SI ≤S,r_M(I) =eM withe²=e∈S

Recall that the annihilators are given bylS(N) ={φ∈S|φN= 0}

andrM(I) ={m∈M|Im= 0}.

Examples: (i) IfR is a Baer ring ande²=e∈R theneR is a BaerR-module, (as isRR.) (ii) Any free module of countable rank over a PIDR, is a BaerR-module. In particular,Zⁿis a Baer

(5)

S. Tariq Rizvi

Rickart Modules

* To obtain a module theoretic analogue for right Rickart rings via endomorphism rings we restrict the Baer module definition to single elements. One motivation for our study is the question: IfR is a right Rickart ring then what can be said about the rightR module eR?

Definition

LetM be a rightR-module and letS =EndR(M).

ThenM is called aRickart moduleif the right annihilator inM of any single element ofS is generated by an idempotent inS.

Equivalently,∀ϕ∈S,rM(ϕ) =Kerϕ=eM for some e²=e∈S. Note thatKerϕ=rM(ϕ) =rM(Sϕ) forϕ∈S.

(6)

Examples

-RR is a Rickart module ifRis a Baer ring, a von Neumann regular ring, or a right hereditary ring; right Rickart ring

- any semisimple module; any Baer module (e.g., every n.s. CS);

- The free Z-moduleZ^(I), for any index setφ6=I,(whileZ^(I) is not a BaerZ-module ifI is uncountable). In particular,Z⁽^R⁾ is a RickartZ-module but is not a BaerZ-module.

- every proj. module over a rt. hered. ring, is Rickart but may not be Baer module.

(7)

S. Tariq Rizvi

Known: Every Baer/Rickart ring is nonsingular. Similarly, every Baer/Rickart module satisfies a certain type of nonsingularity which we callK-nonsingularity.

Definition

We say a module M K-nonsingulariffKerϕ=rM(ϕ)≤^e M impliesϕ= 0, for allϕ∈End(M).

Proposition

M is a nonsingular module⇒M is polyform⇒M is K-nonsingular.

(A moduleM is called non-M-singular or polyform if∀N≤M, 06=ϕ:N→M,Kerϕis not esential inN.)

-The reverse implications are not true. The Z-moduleZp (p∈Z any prime), is aK-nonsingular module which is not nonsingular.

-For the case of a ringR, the notions coincide.

(8)

* Some properties of a Baer module:

Theorem

(i) Every direct summand of a Baer module is a Baer module.

(ii) Every Baer module isK-nonsingular.

(iii) Every Baer M module satisfies the SSIP

(the intersection of any family of direct summands of M is a direct summand).

(iv) A finitely generatedZ-module M is Baer if and only if M is semisimple or torsion-free.

(9)

S. Tariq Rizvi

* Some properties of a Rickart module:

Theorem

(i) Every direct summand of a Rickart module is a Rickart module.

(ii) Every Rickart module isK-nonsingular.

(iii) Every Rickart module has the SIP

(the intersection of any two direct summands is a direct summand).

Corollary

If R is a (Baer) right Rickart ring then M =eR, e²=e, is a (Baer) Rickart R-module.

This answers a question posed earlier.

(10)

A connection between Baer and Rickart Modules:

Theorem

A module M is Baer if and only if M has the strong summand intersection property and Ker(ϕ)≤^⊕M, ∀ϕ∈S (i.e. M is Baer iff M satisfies the SSIP and M is Rickart).

(11)

S. Tariq Rizvi

A well-known result of L. Small can be extended to a general module theoretic setting as follows. Recall:

Theorem

(Small) Let R be a ring that has no infinite set of orthogonal nonzero idempotents. Then the following conditions are equivalent:

(a) R is a right Rickart ring;

(b) R is a Baer ring.

Using the module theoretic methods we prove:

Theorem

Let M be a module and let S =EndR(M)have no infinite set of orthogonal nonzero idempotents. Then the following conditions are equivalent:

(a) M is a Rickart module;

(b) M is a Baer module.

(12)

There exist close links between Baer modules and extending modules viaK-nonsingularity.

First recallChatters-Khuri Theorem: A ringRis a right nonsingular right extending ring iffR is Baer and right

co-nonsingular ring. (R is right cononsingular if rt. annihilator of a nonessential rt. ideal is nonzero.)

(13)

S. Tariq Rizvi

Module Theoretic Versions of Baer and Rickart Rings

Proposition

Every Baer module isK-nonsingular.

Proposition

K-nonsingular extending modules are Baer.

(This provides a rich source of examples of Baer modules, e.g. any nonsingular injective or extending module is Baer. In general M/Z2(M) is a Baer module for any extending moduleM, where Z2(M) is the second singular submodule ofM)

Now an important result:

A module isK-nonsingular, extending if and only if it is Baer and K-cononsingular.

(A moduleM is calledK-cononsingular if∀N≤M,ϕN6= 0 for all 06=ϕ∈S impliesN≤^eM.)

(14)

Proposition

K-nonsingular extending modules are Baer.

(This provides a rich source of examples of Baer modules, e.g. any nonsingular injective or extending module is Baer. In general M/Z2(M) is a Baer module for any extending moduleM, where Z2(M) is the second singular submodule ofM)

Now an important result:

Theorem(R-R, 2004)

A module isK-nonsingular, extending if and only if it is Baer and

(15)

S. Tariq Rizvi

Connections of Baer and Rickart modules to their Endomorphism Rings

Theorem

Let M be a Baer (respectively, Rickart) module. Then S =End(M)is a Baer (respectively, right Rickart) ring.

Converse not true:

Example

LetM =Zp^∞, considered as a right Z-module. Then it is well-known thatS =End_Z(M) is the ring ofp-adic integers, a commutative domain. Thus it is a Baer (and hence Rickart) ring.

HoweverM=Zp^∞ is not (even a Rickart) a Baer module.

(16)

*A characterization for the case of Baer Modules Theorem(Rizvi-Roman, J. Algebra 2009)

A module M is a Baer module if and only if its endomorphism ring S is a Baer ring and M is quasi-retractable.

(A moduleM is calledquasi-retractable ifHom(M,rM(I))6= 0,∀ 06=rM(I),I ≤SS (or, equivalently, ifrM(I)6= 0 thenrS(I)6= 0,∀ I ≤SS)).

Corollary

Rⁿ is a Baer module iff Mn(R)is a Baer ring (n∈N).

(17)

S. Tariq Rizvi

*A characterization for the case of Rickart Modules Theorem(Lee-Rizvi-Roman, Comm. Algebra 2011)

The following conditions are equivalent:

(a) M is a Rickart module;

(b) S =End(M)is a right Rickart ring and M is k-local-retractable.

(M is calledk-local-retractableif for anyϕ∈S =End(M)and any nonzero element m∈rM(ϕ)in M, there exists a nonzero homomorphismψm∈S such that m∈ψm(M)⊆rM(ϕ) =Kerϕ) Corollary

The free R-module R⁽ⁿ⁾ is Rickart if and only ifMn(R)is a right Rickart ring for any n∈N.

(18)

We have seen earlier that any direct summand of a Baer or Rickart module inherits the property however a direct sum of Baer

(resp.,Rickart) modules does not:

Example

ZandZ2are BaerZ-modules (Zis a domain;Z2 is simple). By HoweverZ⊕Z2is not a Baer (or even a Rickart) module (For ϕ(n,m) = ˆˆ ntheKer(ϕ) = 2Z⊕Z2, is not a direct summand since 2Z⊕Z²6=Z⊕Z², asZis uniform).

Proposition

If M is an indecomposable Rickart module which has a nonzero

(19)

S. Tariq Rizvi

*When are Direct Sums of Baer (resp., Rickart) Modules, Baer (resp., Rickart)?

Finding necessary and sufficient conditions for direct sums of extending modules to be extending, has been an open question for well over 2 decades. A sufficient condition for a finite direct sum of extending modules to be extending is that each summand be relatively injective to all others (Harmanci-Smith). We show that an analogue holds true also for the case of Baer modules.

Theorem

Let{Mi}_1≤i≤nbe a class of Baer (resp., Rickart) modules, where n∈N. Assume that Mi is Mj-injective for all i<j . Then Ln

i=1Mi is a Baer (resp., Rickart) module iff Mi is Mj-Rickart.

(We call a moduleMi relatively Rickart to N(orN-Rickart) if, for every homomorphism ϕ:M_i→N,Kerϕ≤^⊕M_i.)

(20)

Theorem

Let{M_i}_i∈I be a class of right R-modules where

I ={1,2,· · · ,n}. Assume that M_i is M_j-C₂for all i,j∈ I. Then T.F.A.E:

(a) Ln

i=1 Mi is a Rickart module;

(b) Mi is Mj-Rickart for all i,j ∈ I.

Proposition Let MiEL

i∈IMi,∀i∈ I,I is an arbitrary index set. Then

(21)

S. Tariq Rizvi

As a consequence, we get: (can also be obtained by using results of [ABT] and [RR])

Theorem

Let M be a nonsingular extending module and E be a nonsingular injective module. Then M and E are relatively Rickart to each other and M⊕E is a Baer (hence, also Rickart) module.

Corollary

Let M be a nonsingular extending module. Then M and E(M)are relatively Rickart to each other and E(M)⊕M is a Rickart module. In this case, E(M)⊕M is a Baer module.

We have examples that show thatneither the extending condition nor the nonsingularcondition is superfluous in the previous corollary.

Remark

IfM is a nonsingular extending module thenE(M)⁽ⁿ⁾⊕M is a Baer module for anyn∈N.

(22)

*The case of Free Baer modules: Our next result provides a characterization of ringsR for which every free rightR-module is Baer.

Theorem(Rizvi-Roman, J Algebra 2009)

The following statements are equivalent for a ring R.

1 every free right R-module is a Baer module;

2 every projective right R-module is a Baer module;

3 R is a semiprimary, hereditary (Baer) ring.

Since condition (3) is left-right symmetric, the left-handed versions of (1) and (2) also hold.

(23)

S. Tariq Rizvi

Carl Faith, in “Embedding Torsionless Modules in Projectives”, Publ. Mat. 1990, showed that for a von Neumann regular ringR, every f.g. torsionless rightR-module embeds in a free right R-module (FGTF property) iffM_n(R) is a Baer ring for every n∈N.

Our next characterization of ringsR for which every f.g. free right R-module is Baer, extends the result of Carl Faith by effectively dropping the requirement ofvon Neumann regularityof the ringR.

(Recall that a module M is calledtorsionlessif it can be

embedded in a direct product of copies of the base ring andM is said to be finitely presented if there exists a short exact sequence 0→K →F →M →0 withK andF finitely generated.)

(24)

Theorem

Let R be a ring. The following statements are equivalent.

1 every f.g. free (projective) right module over R is a Baer module;

2 every f.g. torsionless right R-module is projective;

3 R is left semihereditary and rightΠ-coherent (i.e. every finitely generated torsionless right R-module is finitely presented);

4 Mn(R)is Baer ring for every n∈N.

In particular, a ring R satisfying these equivalent conditions is right and left semihereditary.

(25)

S. Tariq Rizvi

The next result illustrates an application to the case when the base ringR is a commutative integral domain.

It is well-known that ann×nmatrix ring over a commutative integral domainR is Baer if and only if every f.g. ideal of R is invertible i.e., ifR is a Pr¨ufer domain (Kaplansky).

We obtain the following for a finite rank free module over a commutative domain.

Theorem

Let R be a commutative integral domain and M a free R-module of finite rank>1. Then M is Baer if and only if R is a Pr¨ufer domain.

(26)

*The case of Free Rickart modules

The class of right hereditary ringsR =every free R-module is Rickart:

Theorem

The following conditions are equivalent for a ring R:

(a) every free (projective) right R-module is Rickart;

(b) every column finite matrix ring, CFM(R), is a right Rickart ring;

(c) the free right R-module R^(R) is Rickart;

(d) R is a right hereditary ring.

(27)

S. Tariq Rizvi

The class of right semihereditary ringsR =every f.g. free R-module is Rickart

Theorem

The following conditions are equivalent for a ring R:

(a) every f.g. free (projective) right R-module is Rickart;

(b) Mn(R)is a right Rickart ring for all n∈N;

(c) Mk(R)is a right semihereditary ring for some k∈N; (d) R is a right semihereditary ring.

(28)

general result as a consequence of the previous result:

Proposition

Let R be a right (semi)hereditary ring which is not a Baer ring.

Then every (finitely generated) free right R-module is Rickart but not Baer.

Example LetA=Q∞

n=1Z2. Then the ringAis commutative, von Neumann regular, and Baer. Consider R={(an)^∞_n=1∈A| an is eventually constant}, a subring ofA. ThenR is a von Neumann regular ring which is not a Baer ring. Thus, every finitely generated free right

(29)

S. Tariq Rizvi

Example

Zis a right hereditary ring but is not a semiprimary ring. Z⁽^R⁾ is a RickartZ-module but is not a BaerZ-module.

The class of semisimple artinian ringsR =every R-moduleis Rickart:

(30)

SHEY SHEY!!

(31)

S. Tariq Rizvi

∗ SOME APPLICATIONS:

An alternate proof of a result of Small can be obtained via the theory of Rickart modules.

Proposition

For any k ∈N, R is a right hereditary ring iffMk(R)is a right hereditary ring.

For a commutative domain we get:

Proposition

Let R be a commutative domain. Then the following conditions are equivalent:

(a) every finitely generated free (projective) right R-module is Rickart;

(b) the free right R-module R⁽²⁾ is a Rickart module;

(c) M2(R)is a right Rickart ring;

(d) R is a Pr¨ufer domain.