1. A note on the endomorphism ring of orthogonal modules

quả này đã tổng quát một kết quả của S.[r]

Vinh University Journal of Science, Vol 48, No 2A (2019), pp 5-8

A NOTE ON THE ENDOMORPHISM RING OF ORTHOGONAL MODULES

Le Van An, Nguyen Thi Hai Anh

Department of Education, Ha Tinh University, Ha Tinh City, Vietnam

Received on 25/4/2019, accepted for publication on 13/6/2019

Abstract: In this paper, we extend MohamedMuăllers results [2, Lemma 3.3] about the endomorphism ring of a module M = ⊕i∈IMi, whereMi and Mj are

orthogonal for all distinct elementsi, j∈I

1 Introduction

All rings are associated with identity, and all modules are unital right modules The

endomorphism ring of M are denoted End(M) A submodule N of M is said to be an

essential(notationally N ⊂eM) if N ∩K 6= 0 for every nonzero submodule K of M Two

modulesM and N are called orthogonal if they have no nonzero isomorphic submodules LetN be a rightR−module A moduleM is said to beN−injectiveif for every submodule

X ofN, any homomorphismϕ:X−→M can be extended to a homomorphismψ:N −→

M Two modules M and N are called relatively injective if M is N−injective and N is

M−injective In [2, Lemma 3.3], S H Mohamed and B J Muăller proved that:

Let M =M1⊕M2 If M1 and M2 are orthogonal, then

S/∆∼=S1/∆1×S2/∆2

The converse holds ifM1 and M2 are relatively injective, where

S=End(M), Si=End(Mi)(i= 1,2)

and

∆ ={s∈S|Ker(s)⊂eM},∆

i ={si ∈Si |Ker(si)⊂eMi}(i= 1,2)

In this paper, we study [2, Lemma 3.3] in generalized case We have:

Theorem A (i) Let M = ⊕i∈IMi be a direct sum of submodules such that Mi and

Mj are orthogonal for anyi, j of I and i6=j, then Qi∈ISi/∆i is embedded into S/∆

In particular, if I is a finite set, Q

i∈ISi/∆i∼=S/∆

(ii) LetM =⊕i∈IMi be a direct sum of submodules such that Mi andMj are relatively

injective for any i, j of I, i 6= j and Q

i∈ISi/∆i ∼=S/∆, then Mi and Mj are orthogonal

with i, j of I and i 6= j, where S = End(M), Si = End(Mi)(i ∈ I) and ∆ = {s ∈ S |

Ker(s)⊂eM}, ∆

i ={si ∈Si |Ker(si)⊂eMi}(i∈I) 1)

Email: an.levan@htu.edu.vn (L V An)

Le Van An, Nguyen Thi Hai Anh / A note on the endomorphism ring of orthogonal modules

2 Proof of Theorem A

(i) Lets be an element of the endomorphism ring S and x an element of the module

M, then x = P

i∈Ixi with xi 6= for every i ∈ I

0 (where I0 is the finite subset of I),

s(x) = P

i∈Is(xi) Because s(xi) is an element of M, thus s(xi) = Pj∈Isij(xi) with

sij(xi) =pj◦s(xi) is an element ofMj (wherepj :M −→Mj is a natural homomorphism,

sij(xi) 6= for every j ∈ I0, I0 is finite and I0 is a subset of I) We consider the matrix

s= [sij]I×I withsij :Mi −→Mj being homomorphism Note that,sij is an endomorphism

ofM because sij(Pj∈Ixj) = + + +sij(xi) + +

Claim Ker(sij) is an essential submodule of M for every i, j belonging to I and

i6=j

LetN be a nonzero submodule of M andKer(sij)∩N = 0, thensij |N is a

monomor-phism, thusN ∼=sij(N) withsij(N)being a submodule of Mj Butsij(⊕k6=iMk) = 0, thus

⊕k6=iMkis a submodule ofKer(sij) Hence⊕k6=iMk∩N = 0,(⊕k6=iMk)⊕N is a submodule

ofM = (⊕k6=iMk)⊕Mi Thus

N ∼= ((⊕k6=iMk)⊕N)/(⊕k6=iMk)⊂M/(⊕k6=iMk)∼=Mi

Let sij(N) = Y be a submodule of Mj, there exists a submodule X of Mi such that

X∼=N ∼=Y This is a contradiction to the fact thatMi andMj are orthogonal Therefore,

Ker(sij) is an essential submodule ofM for everyi, j that are elements ofI and i6=j

Claim

Ker(s)∩Mi =∩j∈IKer(sij),

for everyiofI

Lets:⊕i∈IMi −→ ⊕i∈IMi, and let xbe an element of ⊕i∈IMi, thenx=Pi∈Ixi with

xi∈Mi,xi 6= for everyi∈I0 (whereI0 is finite andI0 is subset ofI) Thus

s(x) =s(X

i∈I

xi) =

X

i∈I

s(xi) =

X

i∈I

X

j∈I

sij(xi) = [sij]TI×I.[xi]I×1,

with[sij]TI×I is the transposet matrix of[xij]I×I Letxbe an element ofKer(s)∩Mi, then

x is an element ofMi and s(x) = Thusx =Pj∈Ixj =xi with xj being an element of

Mj for every j of I, and sij(xi) = for every j of I Hence, xi is an element of Ker(sij)

for everyI, it follows thatx is an element of∩j∈IKer(sij), thusKer(s)∩Mi is a subset of

∩j∈IKer(sij) Ifx is an element of∩j∈IKer(sij) thenxis an element ofMi andsij(x) =

for everyj inI Thuss(x) =s(P

j∈Ixj) =s(xi) =

P

j∈Isij(xi) = 0, hencexis an element

of Kers, i.e., x is an element of Ker(s)∩Mi It follows that ∩j∈IKer(sij) is a subset of

Ker(s)∩Mi Thus,

Ker(s)∩Mi =∩j∈IKer(sij),

for everyiofI

Claim 3.If sis an element of∆thensi is an element of ∆i, for every iofI

Let s be an element of ∆, then Ker(s) is an essential submodule of M By Claim and [1, Proposition 5.16],Ker(s)∩Mi =∩j∈IKer(sij) is an essential submodule ofMi for

Vinh University Journal of Science, Vol 48, No 2A (2019), pp 5-8

everyiof I ThusKer(si) is an essential submodule ofMi It follows thatsi is an element

of∆i, for everyiof I

Claim 4.IfI is a finite set andsi is an element of∆i for everyiof I thensis also an

element of∆

By Claim 1,Keri6=j(sij)is an essential submodule ofM for everyiofI, thusKeri6=j(sij)∩

Mi is also an essential submodule ofMi SinceI is the finite set and by [1, Proposition 5.16],

∩i6=jKer(sij) is an essential submodule ofMi Because, si is an element of ∆i,Ker(si) is

an essential submodule ofMi, thus ∩j∈IKer(sij)is an essential submodule ofMi for every

iofI Hence Ker(s)∩Mi is an essential submodule ofMi (by Claim 2),⊕i∈I(Ker(s)∩Mi)

is an essential submodule ofM =⊕i∈IMi Thus Ker(s) is also an essential submodule of

M It follows that sis an element of∆ By Claim 1, Claim 2, Claim 3,

S/∆ = (Aij)I×I

with Aij = Si/∆i if i = j and Aij = if i 6= j Let ϕ : Qi∈ISi/∆i −→ S/∆ be a

homomorphism such thatϕ((si+ ∆i)) = [sij]I×I with sij is an element of Aij Note that

Ker(ϕ) = {(si + ∆i) | s = [sij]I×I ∈ ∆} = {(si + ∆i) | si ∈ ∆i} = (0), thus ϕ is a

monomorphism Hence,Q

i∈ISi/∆i∼=X withX is a submodule ofS/∆

If I is a finite set, then s is an element of ∆ if and only if si is an element of ∆i for

everyiofI Hence S/∆ = [Aij]I×I ∼=Qi∈ISi/∆i

(ii) Assume that,Q

i∈ISi/∆i =∼S/∆with Mi and Mj are relatively injective for every

i, jare elements ofI and i6=j, we will show thatMi andMj are orthogonal for anyi, j of

I and i6=j

Assume that, there are two elementsαand β ofI and α=6 β such thatMα and Mβ are

not orthogonal There exist two submodules Eα of Mα and Eβ of Mβ withEα ∼=Eβ Let

fαβ :Eα −→Eβ be an isomorphism, thenfαβ :Eα−→Mβ is a monomorphism SinceMβ

isMα−injective, there existgαβ :Mα−→Mβ is an extending offαβ Note thatKer(gαβ)is

an essential submodule ofM thusKer(gαβ)∩Eα 6= There exists element xα of Eα with

xα 6= 0and gαβ(xα) =fαβ(xα) = 0, this is the contradiction Sincef is a monomorphism

Hence,Mi and Mj are orthogonal for anyi, j of I and i6=j

By the Theorem A, we have the Corollary B

Corollary B (i) Let M = ⊕n

i=1Mi be a direct sum of submodules such that Mi and

Mj are orthogonal for anyi, j of {1,2, , n} andi=6 j, then Qni=1Si/∆i ∼=S/∆

(ii) LetM =⊕n

i=1Mi be a direct sum of submodules such thatMi andMj are relatively

injective for any i, j of {1,2, , n}, i 6= j and Qn

i=1Si/∆i ∼= S/∆, then Mi and Mj are

orthogonal with i, j of {1,2, , n} and i 6= j, where S = End(M), Si = End(Mi)(i =

1,2, , n) and∆ ={s∈S |Ker(s)⊂eM},∆

i={si∈Si|Ker(si)⊂eMi}(i= 1,2, , n)

Note that, Regarding Corollary B, in case n= 2, we have [2, Lemma 3.3]

Acknowledgment

This research was supported by Ministry of Education and Training, grant no B2018-HHT-02

Le Van An, Nguyen Thi Hai Anh / A note on the endomorphism ring of orthogonal modules

REFERENCES

[1] F W Anderson and K R Fuller,Ring and Categories of Modules, Springer - Verlag, New York - Heidelberg - Berlin, 1974

[2] S H Mohamed and B J Muăller,Continuous and Discrete Modules,London Math Soc Lecture Note Series147, Cambridge Univ Press, 1990

TÓM TẮT

MỘT CHÚ Ý VỀ VÀNH CÁC TỰ ĐỒNG CẤU CỦA MƠĐUN TRỰC GIAO

Trong báo chúng tơi đưa kết vành tự đồng cấu mơđun

M =⊕i∈IMi Mi vàMj trực giao lẫn với bất kỳi, j I i6=j Kết

quả tổng quát mt kt qu ca S H Mohamed v B J.Muăller [2, Lemma 3.3]