1 Introduction - Vietnam Journals Online

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 Mohamed−M¨uller’s 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. M¨uller 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 ={s_i ∈S_i |Ker(s_i)⊂^eM_i}(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 Q

i∈ISi/∆i is embedded into S/∆.

In particular, if I is a finite set, Q

i∈ISi/∆i∼=S/∆.

(ii). LetM =⊕_i∈IM_i be a direct sum of submodules such that M_i andM_j are relatively injective for any i, j of I, i 6= j and Q

i∈IS_i/∆_i ∼=S/∆, then M_i and M_j 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 ={s_i ∈S_i |Ker(s_i)⊂^eM_i}(i∈I).

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

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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= 0 for every i ∈ I⁰ (where I⁰ is the finite subset of I), s(x) = P

i∈Is(x_i). Because s(x_i) is an element of M, thus s(x_i) = P

j∈Is_ij(x_i) with s_ij(x_i) =p_j◦s(x_i) is an element ofM_j (wherep_j :M −→M_j is a natural homomorphism, sij(xi) 6= 0 for every j ∈ I0, I0 is finite and I0 is a subset of I). We consider the matrix s= [s_ij]I×I withs_ij :M_i −→M_j being homomorphism. Note that,s_ij is an endomorphism ofM because s_ij(P

j∈Ix_j) = 0 +...+ 0 +s_ij(x_i) + 0 +...

Claim 1. Ker(s_ij) is an essential submodule of M for every i, j belonging to I and i6=j.

LetN be a nonzero submodule of M andKer(s_ij)∩N = 0, thens_ij |_N is a monomor- phism, thusN ∼=sij(N) withsij(N)being a submodule of Mj. Butsij(⊕_k6=iMk) = 0, thus

⊕_k6=iM_kis a submodule ofKer(sij). Hence⊕_k6=iM_k∩N = 0,(⊕_k6=iM_k)⊕N is a submodule ofM = (⊕_k6=iM_k)⊕M_i. Thus

N ∼= ((⊕_k6=iM_k)⊕N)/(⊕_k6=iM_k)⊂M/(⊕_k6=iM_k)∼=M_i.

Let s_ij(N) = Y be a submodule of M_j, there exists a submodule X of M_i such that X∼=N ∼=Y. This is a contradiction to the fact thatM_i andM_j are orthogonal. Therefore, Ker(sij) is an essential submodule ofM for everyi, j that are elements ofI and i6=j.

Claim 2.

Ker(s)∩M_i =∩_j∈IKer(s_ij), for everyiofI.

Lets:⊕_i∈IM_i −→ ⊕_i∈IM_i, and let xbe an element of ⊕_i∈IM_i, thenx=P

i∈Ix_i with xi∈Mi,xi 6= 0 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]^T_I×I.[xi]I×1,

with[s_ij]^T_I×I is the transposet matrix of[x_ij]I×I. Letxbe an element ofKer(s)∩M_i, then x is an element ofM_i and s(x) = 0. Thusx =P

j∈Ix_j =x_i with x_j being an element of Mj for every j of I, and sij(xi) = 0 for every j of I. Hence, xi is an element of Ker(sij) for everyI, it follows thatx is an element of∩_j∈IKer(s_ij), thusKer(s)∩M_i is a subset of

∩_j∈IKer(s_ij). Ifx is an element of∩_j∈IKer(s_ij) thenxis an element ofM_i ands_ij(x) = 0 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)∩M_i. It follows that ∩_j∈IKer(s_ij) is a subset of Ker(s)∩M_i. Thus,

Ker(s)∩Mi =∩_j∈IKer(sij), for everyiofI.

Claim 3.If sis an element of∆thens_i 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 2 and [1, Proposition 5.16],Ker(s)∩M_i =∩_j∈IKer(s_ij) is an essential submodule ofM_i for 6

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

everyiof I. ThusKer(s_i) is an essential submodule ofM_i. It follows thats_i 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(s_ij)is an essential submodule ofM for everyiofI, thusKeri6=j(s_ij)∩

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

∩_i6=jKer(s_ij) is an essential submodule ofM_i. Because, s_i is an element of ∆_i,Ker(s_i) is an essential submodule ofM_i, thus ∩_j∈IKer(s_ij)is an essential submodule ofM_i 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∈IM_i. 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 = 0 if i 6= j. Let ϕ : Q

i∈ISi/∆i −→ S/∆ be a homomorphism such thatϕ((s_i+ ∆_i)) = [s_ij]I×I with s_ij is an element of A_ij. Note that Ker(ϕ) = {(s_i + ∆_i) | s = [s_ij]I×I ∈ ∆} = {(s_i + ∆_i) | s_i ∈ ∆_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 s_i is an element of ∆_i for everyiofI. Hence S/∆ = [A_ij]I×I ∼=Q

i∈IS_i/∆_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 α6=β such thatMα and Mβ are not orthogonal. There exist two submodules Eα of Mα and E_β of M_β withEα ∼=E_β. Let f_αβ :E_α −→E_β be an isomorphism, thenf_αβ :E_α−→M_β is a monomorphism. SinceM_β isMα−injective, there existgαβ :Mα−→Mβ is an extending offαβ. Note thatKer(gαβ)is an essential submodule ofM thusKer(g_αβ)∩Eα 6= 0. There exists element xα of Eα with x_α 6= 0and g_αβ(x_α) =f_αβ(x_α) = 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=1M_i be a direct sum of submodules such that M_i and Mj are orthogonal for anyi, j of {1,2, ..., n} andi6=j, then Qn

i=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=1S_i/∆_i ∼= S/∆, then M_i and M_j 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={s_i∈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.

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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. M¨uller,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 MÔĐUN TRỰC GIAO

Trong bài báo này chúng tôi đưa ra một kết quả về vành các tự đồng cấu của môđun M =⊕i∈IMi trong đó Mi vàMj là trực giao lẫn nhau với bất kỳi, j của I và i6=j. Kết quả này đã tổng quát một kết quả của S. H. Mohamed và B. J.M¨uller trong [2, Lemma 3.3].

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