Báo cáo nghiên cứu khoa học: "Môđun tựa cấu xạ." pps

Then, we consider a general problem: For a quasi-morphic module M, when is endM left quasi-morphic, and conversely?. Using this result, we show that a ringRis regular if and only if it

Quasi-morphic modules

Ngo Sy Tung (a), Tran Giang Nam(b) Ngo Ha Chau Loan(c)

Abstract In this paper, we prove that the matrix ring Mn(R) is left quasi-morphic

if and only if the left R-module Rn is quasi-morphic Then, we consider a general problem: For a quasi-morphic module M, when is end(M ) left quasi-morphic, and conversely? Using this result, we show that a ringRis regular if and only if it is a left quasi{morphic, leftP P ring.

1 Introduction

Nicholson - Campos ( [5], p 2630) call a left moduleM morphic if M/Im(α) ∼= ker(α) for all endomorphism α in end(M ), equivalently if there exists β ∈ end(M )

such that Im(β) = ker(α) and Im(α) = ker(β) In this paper, we only need the existence of β andγ such thatIm(β) = ker(α) andIm(α) = ker(γ),and we call M quasi-morphic if for every element α inend(M ),αsatisfies the above condition We use the notion to characterize the classes of quasi{morphic rings{these rings were introduced and studied by Camillo - Nicholson in [2] More precisely, we answer the question: For each a ring R, when is the ring Mn(R) left quasi{morphic? And we obtain that the matrix ring Mn(R) is left quasi-morphic if and only if Rn is quasi-morphic as a left R-module (Theorem 2.2) More generally, we investigate when

M being quasi-morphic implies that end(M ) is left quasi-morphic, and conversely (Proposition 3.1) Furthermore, we also investigate when M being quasi-morphic implies thatend(M ) is regular (Theorem 3.1) Then, applying this result, we obtain that a ringRis regular if and only if it is a left quasi{morphic, left PP ring (Corollary 3.4)

Throughout this paper, every ringRis associative with unity and all modules are unitary We denote left annihilator of a set X ⊆ R bylR(X)

Finally, all notions used here without any comments, can be found in [4]

( 1) NhËn bµi ngµy 08/4/2009 Söa ch÷a xong ngµy 03/6/2009

2 Quasi morphic modules

Recall ([5], p 2630) that a left moduleM is calledmorphic ifM/Im(α) ∼= ker(α)

for all endomorphismα inend(M ); equivalently if for eachα ∈ end(M )there exists

β ∈ end(M ) such that Im(β) = ker(α) and Im(α) = ker(β) (see [5], Lemma 1) More generally, we call an endomorphism α of a left module M quasi–morphic if there exist β and γ in end(M ) such that Im(β) = ker(α) andIm(α) = ker(γ) The left moduleM is calledquasi–morphicif every endomorphism ofM is quasi{morphic Clearly, every morphic module is quasi{morphic, but the converse is false In order to prove this fact, we first need the following result{a characterization of a quasi-morphic module in terms of the lattice of submodules

Theorem 2.1 LetM be a left module Then the following conditions are equivalent: (1)M is quasi{morphic;

(2) If M/K ∼= N where K and N are submodules of M, then there exist β, γ ∈ end(M ) such thatK = Im(β) and N = ker(γ).

Proof (1) =⇒ (2) If η : M/K → N is an isomorphism, define α : M −→ M by

α(m) = η(m + K) It is easy to see that α is an endomorphism of M Since M

is quasi{morphic, so there exist β and γ in end(M ) such that Im(β) = ker(α) and

Im(α) = ker(γ) Then,Im(β) = K andIm(α) = N

(2) =⇒ (1).It is clear

Using Theorem 2.1, we immediately have that Z2 ⊕ Z4 is a quasi-morphic Z -module, but it is well known thatZ2⊕ Z4isn't a morphicZ-module (see [5], p 2631)

Corollary 2.1 Every semisimple left module is quasi{morphic.

Proof Let M be a semisimple left module Then, for each submodule N of M, since N is a direct summand of M, so that there exist α, β ∈ end(M ) such that

K = Im(α) andN = ker(β) Using this and Theorem 2.1, we immediately have M

is quasi-morphic

Recall ([5], Theorem 23) that every direct summand of a morphic left module is again morphic Is this true for the class of quasi-morphic modules? We can't answer this question yet, however we partially answer by using Theorem 2.1 as follows:

Corollary 2.2 Let M and N be left R-modules such that HomR(M, N ) = 0 = HomR(N, M ) Then, M ⊕ N is quasi-morphic if and only if M and N are quasi-morphic.

It is well known that there exists a natural isomorphism ϕ : R −→ end(RR) is defined by ϕ(r)(x) = xrfor allr, x ∈ R.Using this, we immediately obtain that ifR

is a ring, RR is quasi-morphic if and only if for eacha ∈ R there existb, c ∈ R such that Ra = l(b)andl(a) = Rc These rings were introduced and studied by Camillo -Nicholson in [2], and called left quasi-morphic

Camillo - Nicholson showed that there exists the ringRsuch that the ringMn(R)

is not left (right) quasi{morphic (see, [2], Example 4) In light of this fact, it is natural to bring up the following problem: When is Mn(R) left quasi{morphic? In the following theorem, we give the answer to the question

Theorem 2.2 Let R be a ring Then, the following conditions are equivalent: (1)RRn is quasi-morphic;

(2)Mn(R)is left quasi{morphic.

Proof We prove it for n = 2; the general case is analogous

(1) =⇒ (2) Let K and N be left ideals of S = M2(R) such that S/N ∼= K By ([5], p 2635), so there exist submodules X and Y of RR2 such that K = X

X

 and

N =  Y

Y



Since S/N ∼= K and ([5], Lemma 16), X ∼= R2/Y By RR2 is quasi-morphic and Theorem 2.1, so there exist β, γ ∈ end(RR2) such that Y = Im(β) and X = ker(γ) Then, there are the homomorphisms α, δ : S −→ S defined by α( x y

z t



) = β(x, y)

β(z, t)

 and δ( x y

z t

 ) =  γ(x, y)

γ(z, t)

 It is easy to see that Im(α) = N and ker(δ) = K Using again Theorem 2.1, we have S is left-morphic (2) =⇒ (1) It is similarly proved as the direction (1) =⇒ (2)

3 Endomorphism rings

LetM be a left R-module The set of all endomorphisms of M will be denoted by

end(M ) Further, end(M ) has the structure of a ring, when we define (f + g)(x) =

f (x) + g(x)andf g(x) = g(f (x)) for allx ∈ M andf, g ∈ end(M )

Recall (see, for example, [5], p 2640) that a left module M is called image– projective if, whenever Im(γ) ⊆ Im(α) where γ, α ∈ E = end(M ), then γ ∈ Eα

Of course, every projective left module is image{projective Also, in ([5], p 2641)

a left module M is said to generate a submodule K ⊆ M if K = P{Im(λ)|λ ∈

end(M ), Im(λ) ⊆ K}, and we say thatM generates its kernelsifM generatesker(β)

for all β ∈ end(M ).One has the following remark

Remark 3.1 Let R be a ring and F a free left R-module Then, F generates all its submodules In particular,F generates its kernels

Proof Assume K is a submodule of F Since F is free, so that F ∼= R(I) for some set I For each x ∈ K, there exists the homomorphism fx : F −→ F defined

by fx((ri)i∈I) = P

i∈Irix for all (ri)i∈I ∈ F Clearly, x ∈ Im(fx) Hence, K ⊆ P

x∈KIm(fx) ⊆P{Im(λ)|λ ∈ end(F ), Im(λ) ⊆ K} It follows that M generates K

Proposition 3.1 LetM be a leftR-module and E = end(M ).Then,

(1) IfM is quasi-morphic and image{projective, then E is left quasi{morphic; (2) IfM is quasi-morphic, then it generates its kernels;

(3) If E is left quasi{morphic and M generates its kernels, then M is quasi-morphic.

Proof (1) If M is quasi-morphic and image–projective, and given α ∈ E, choose

β, γ ∈ E such that Im(α) = ker(β) and Im(γ) = ker(α) Then, since αβ = 0,

Eα ⊆ lE(β) Conversely, if ϕ ∈ lE(β) then Im(ϕ) ⊆ ker(β) = Im(α), so ϕ ∈ Eα because M is image–projective Thus Eα = lE(β), and Eγ = lE(α) follows in the same way Hence E is left quasi–morphic

(2) It is clear

(3) Give α ∈ E, choose β, γ ∈ E such that Eα = lE(β) and Eγ = lE(α) Then Im(α) ⊆ ker(β) because αβ = 0 By M generates its kernels, so ker(β) = P{Im(λ)|λ ∈ end(M ), Im(λ) ⊆ ker(β)} But Im(λ) ⊆ ker(β) implies λ ∈ lE(β) =

Eα, say λ = δα, δ ∈ E Hence, Im(λ) = Im(δα) ⊆ Im(α) It follows that ker(β) ⊆ Im(α) Thus ker(β) = Im(α), and ker(α) = Im(γ) is proven in the same way

Remark 3.2 Let R be a ring It is well known thatMn(R) ∼= end(RRn) Further, since Rn is free leftR-module, it is image{projective And, applying Remark 3.1, we haveRngenerates its kernels Therefore, takingM = Rnin Proposition 3.1 provides another proof of Theorem 2.2

Let M be a left R-module and α ∈ end(M ) In [1], Azumaya proved that α is regular if and only if bothIm(α)andker(α)are direct summands ofM" Using this fact, we immediately obtain the following remark

Remark 3.3LetM be a leftR-module andE = end(M ).Then, for anyα ∈ end(M ),

if αis regular then it is quasi-morphic In particular, if E is a regular ring then M

is quasi-morphic

In the next theorem, we can give a partial converse for Remark 3.3 First, recall ([5], p 2643) that a left module M is called kernel-direct ifker(α) is a summand of

M for every α ∈ end(M ), and callM image-direct ifIm(α) is a summand of M for everyα ∈ end(M )

Theorem 3.1 The following conditions are equivalent for a left module M:

(1)end(M ) is regular;

(2)M is quasi{mophic and kernel-direct;

(3)M is quasi{mophic and image-direct.

Proof (1) =⇒ (2) This follows from Remark 3.3

(2) =⇒ (3) This follows from the definitions

(3) =⇒ (1) Given α ∈ end(M ), Im(α) is a summand of M by (3) Since M is quasi-morphic, there exists β ∈ end(M ) such that ker(α) = Im(β) Using again (3), Im(β) is a summand of M ; hence ker(α) is also Thus α is regular

If R is a ring then RR is kernel-direct if and only if lR(a) is a direct summand

of RR for all a ∈ R,that is if and only if every principal left ideal Ra is projective These are called left P P rings Using this fact and Theorem 3.1, we immediately have the following corollary

Corollary 3.2 A ring R is regular if and only if it is a left quasi{morphic, left PP ring.

References

[1] G Azumaya, On generalized semi-primary rings and Krull-Remak-Schmidt's

theorem, Japan J Math., 19, 1960, 525 - 547.

[2] V Camillo and W K Nicholson, Quasi-morphic rings, Journal of Algebra and

its Appl., Vol 6, No 5, 2007, 789 - 799

[3] G Erlich, Units and one-sided units in regular rings, Trans A.M.S 216, 1976

81-90

[4] T Y Lam, A first course in noncommutative rings, 2nd Ed., Springer-Verlag,

New York-Berlin, 2001

[5] W K Nicholson and E S Campos, Morphic modules, Comm in Algebra, 33,

2005, 2629 - 2647

Tóm tắt

Môđun tựa cấu xạ

Trong bài báo này, chúng tôi chứng minh rằng vành ma trận Mn(R) là tựa cấu xạ khi và chỉ khi R-môđun trái Rnlà tựa cấu xạ Tiếp đó chúng tôi xét bài toán tổng quát: với môđun tựa cấu xạ M, khi nào vành end(M ) là tựa cấu xạ trái, và ng−ợc lại? áp dụng kết quả này, chúng tôi chỉ ra rằng một vành R là chính qui khi

và chỉ khi nó là tựa cấu xạ trái, P P trái

(a)Departrment of Mathematics, University of Vinh

(b)Department of Mathematics, University of Dong Thap

(c)46A, Departrment of Mathematics, University of Vinh.