Conference in Research and Practice in Information Technology - Style Guide

 

Power Module

 

Fang Gang[1]

  

 

Abstract:  In this paper, we discuss the upgrade problem of module, and introduce the concepts of the power module, regular power module and uniform power module. We give some results of them

Key words:  power group; power module; regular power module; uniform power module

 


 

1.   Introduction

 

The notion of the hypergroup was first introduced by LI Hong-xing and WANG Pei-zhuang in 1985(LI, DUAN & WANG, 1985). Afterward, in 1988, LI Hong-xing and WANG Pei-zhuang emphasized on the upgrade problem for algebraic group, and first introduced the notion of power group (LI & WANG, 1988). In 1990, ZHONG Yu-bin investiaged further the structures of hypergroups(ZHONG, 1990). In 1988, LI Hong-xing established the HX ring in hypergroups (power groups) (LI, 1988). Yao Bing-xue and LI Hong-xing introduced the concept of power ring and improved some results of HX rings (YAO & LI, 2000). ZHANG Zhen-liang described the normal power rings and uniform power rings in 2001(ZHANG, 2001). Nowdays, it has been seen that the upgrade of all kinds of structures such as algebraic structures, ordered strucutres, topological structures, measurable structures is important on the development of fuzzy mathematics.

In this paper, we shall extend hypergroup and power ring to module theory by introducing the notions of so called power module, regular power module and uniform power module in a R-module.

Through the paper, we always assume that the ring R is a commutative with identity. By a left R-module M, we shall mean an abelian group (M, +) together with a left action R M M, described by (r, x)rx, such that for all r, s in R, and all x, y in M, we have

(1) r(x+y) = rx+ry ;

(2) (r+s) x = rx+sx ;

(3) (rs) x = r (sx) ;

(4) 1x = x, where 1 is multiplicative identity element of R.

Suppose M is a left R-module and N is a subgroup of M. Then N  is a

submodule (or R-submodule) if, for any n in N and any r in R, the product rn is in N.

In the following, we shall introduce the power sets into module. For the more details for Module theory we refer the reader to [8, 9].

Let(M)={A | A Í M} and (M)=(M)-{f}. For every A, BÎ(M) and ÎR, the sum of A and B is defined by

A+B ={ a+b | aÎA, bÎB},                              (*)

and the product of B and number (ÎR) is defined by

B ={b | bÎB }.                                     (**)

Clearly, we have

Proposition1.1  Let A, B, C(M) and , ÎR. Then we have

(1) A+B=B+A

(2) (A+B)+C=A+(B+C)

(3) (˦) C= (C)

(4) (B+C)=B+C

(5) (+) A Í A+A

(6) 1A=A, ( where 1 is multiplicative identity element of R)

(7) If A Í B, then A+C Í B+C, and A Í B.

Definition 1.1  Let M be a non-empty subset of (M). If M forms a left R-module M under the operation (*) and (**), then M is called a power module ( or R-power module ) on M, whose null element is denoted by Q and the nagative element of A is denoted by -A. The set ={x | -xÎAxÎM } is called the inverse of A.

Clearlyfor every ÎR we have Q=Q.

Example 1  Let S be a submodule of M. Then {{x} | xÎS} is an R-power module on M.

Example 2  Let S be a submodule of M. Then the factor (quotient) module M/S={x+S | xÎM} is an R-power module on M.

Example 3  Let R be a real field and A={X | f X Í Z, where Z is the set of integers}. Then R is a left R-module,  and A is a hypergroup relative to the operation (*). Let 0.7ÎR and {1, 2}ÎA we have that 0.7{1, 2}={0.7, 1.4}A, hence A is not a R-power module of R. Thus we can see that not all hypergroups are power modules.

Definition 1.2  Let M be a power module of M.Then M is a regular power module of M , if 0ÎQ (0 is the zero element of M ), and then M is an uniform power module of M if -A= for every AÎM.

Definition 1.3  Let M be an R-power module on M. For every AÎM, the set ={a | aÎA, -aÎ-A} is called the kernel of A.

 

2.  Power Module

 

Theorem 2.1  Let M be a power module of M. Then we have

(1) For every AÎM, we have | A |=| Q |;

(2) For all A, BÎMif ABf then | A |=| AB |;

(3) For all A, BÎMif A Í B then -B Í -A.

Proof  (1) Since Q is zero element of M, so for every aÎM we have a+Q Í A+Q =A, and hence |Q|=| a+Q | | A |.

Conversely, since -A+A=Qthen for every b-A we have that b+A Í -A+A=Q and | A |=| b+A | | Q|. Consequencely, | A |=| B |=| Q |.

2Since ABf then we have | AB | | A |. Moreover, if zÎAB then we have zÎA and zÎBhence we see that z+Q Í A+Q=A and z+Q Í B+Q=B. Thus we obtain z+Q Í AB, so | A | = | Q |=| z+Q | | AB |. Consequently, we have | A |=| AB |.

3By A Í B we have -A-B+A Í B-A-B, so CB Í -A.

Theorem 2.2  Let M be a power module of M. If 0ÎAÎMthen -A Í Q Í A.

Proof  Since 0Athen Q=0+Q Í A+Q=A. On the other hand, we have that -A =0+(-A) Í A+(-A)=Q.  Thus we have CA Í Q Í A.

Corollary 2.1  Let M be a power module of M. If 0ÎAÎM and |M | is finite, then -A=Q=A.

Proof  Since |M | is finite, so are A and Q. By Theorem 2.2 we see CA Í Q Í A, hence |-A | | Q | | A |. Moreover, we have | A |=| -A |, thus |A|=| -A |=| Q |. Consequently, we obtain  -A=Q=A.

Theorem 2.3  Let M be a power module of M and the zero element Q be a submodule of M. Then  Í .

ProofLet bÎ, then -bÎA , and for every cÎ we have -b+ cÎA- A= Q. Since Q is a submodule of M, then there exists a nagative element tÎQ such that c-b+t=0 , namely t=b-c, whence b=c+ tÎ+Q =, and consequently, we have  Í .

Theorem 2.4  Let M be a power module of M, then M is a regular power module Ûf, for every AM.

ProofÜ  If f, then there exist aÎA and -aÎ-A, so we have 0=a-aÎA-A=Q. Consequently, M is a regular power module.

Þ  For every AÎ we have A-A=Q. Since M is a regular power module, then 0ÎQ so there exist aÎA and bÎ-A, such that a+ b= 0. Hence  -a= bÎ-A, and whence  aÎf.

Theorem 2.5  Let f : L1L2  be an R-morphism and L11 be a R-power module on L1, then L22={ f(A) | AÎL11} is an R-power module on L2 and L11~L22.

ProofIt is easy to see from [7] that L22={ f(A) | AÎL11} forms an additive power group with null element f (Q) and for all A, BÎL11, we have f(A)+f(B) = f (A+B , -f(A) = f(-A). Moreover, let ÎR , AÎL11 and tÎf(A) then we have that t1ÎA  and t=f(t1)=f(t1)Îf(A), hence f(A) Í f(A). For the converse inclusion, we let hÎf(A),  then h1ÎA and h=f(h1)=f(h1)Îf(A). Hence f(A) Í f(A). Thus we obtain f(A1)=f(A1). Consequently, L22={ f(A) | AÎ L11 } is an R-power module on L2 and g: L11L22 defined by g(A) = f (A) is an R-epimorphism.

Theorem 2.6  Let f : L1L2  be an R-epimorphism and L22 be a R-power module on L2, then f -1(L22)={ f -1(A) | AÎ L22} is an R-power module on L1 and f -1(L22)L22.

Proof  It is clear from [7] that  f -1(L22) forms an additive power group. For A L22 and ÎR ,we have f -1(A) = f -1 (A). Hence  f -1(L22) is an R-power module on L1 and g: f -1(L22)L22 defined by g (f -1(A))=A is an R-isomorphism.

Theorem 2.7  Let M be a power module of M and Q be a subgroup of M, then M*={N | NÎM } is a left R-module.

Proof  By Theorem 2.1 in [4] and Theorem 2.2, we have that M* is a subgroup of M.

For every NÎM* and ÎR , by Definition 1.1 we have NÎM, and so NÎM*. Thus M* is a left R-module.

 

3.  Regular power module and uniform power module

 

Theorem 3.1  Let M be a regular power module of M. If AÎM and aÎA , then aÎÛ A=a+Q.

ProofÜ Since 0ÎQ, then a=a+0ÎA+Q=A and Q=A-A= (a+Q) -A =a+(Q-A)=a-A. Moreover, by 0ÎQ and aÎA, we obtain 0Î a-A, then there exists bÎ-A such that 0=a+b, namely -a=bÎ-A. Thus aÎ.

Þ We can see clearly that a+Q Í A+Q=A. By the definition 1.3 and aÎ, we have that aA and -aÎ -A. Let bÎA, then we have that b=0+b=(a-a)+b=a+(b-a) Îa+A+(-A)=a+Q. Hence we have A Í a+Q and whence A=a+Q.

Theorem 3.2 Let M be a regular power module of M. If AÎM and aÎA, then aÎÛ =a+.

ProofÜ It follows that 0Î and hence a=a+0Îa+=.

Þ For evey xÎ, then xÎA and -xÎ-A, and by Theorem 3.1 we have xÎA=a+Q. Hence there exists bÎQ such that x=a+b. Since b=x-a and so -b=a-xÎ (-A)+A =Q. We thus have bÎ, and x=a+bÎa+, namelyÍ a+. For the converse inclusion we let yÎa+. Then by Theorem 3.1, there exists bÎÍ Q such that y=a+bÎa+Q=A. Since bÎ and aÎ, then -bÎ-Q=Q and -aÎ-A, hence -y= (-b)+(-a)ÎQ-A=-A. It follows that yÎ, and so a+Í . Consequently =a+.

Theorem 3.3  Let M be a regular power module of M. Then M is an uniform power module of M Û =A, for every AÎM.

ProofÞ Since M is a regular power module of M, for every AÎM and every aÎA we have -aÎ=-A, namely aÎ, hence A Í. Moreover, it is clear to see that Í A. Thus we have =A.

Ü Since AÎM and =A, then for every aÎA we have aÎ, and by theorem 3.1, we have a+Q=A and -(a+Q)=-A, thus -(a+Q)= - a+Q = -A.

Now, we shall verify that =-A. Let bÎ then -bÎA=a+Q. Hence there exists sÎQ such that -b=a+s. Since for every AÎ, we see = A. Then =Q and so there exists -sÎQ such that b= -a-s Î- a+Q = -A. Thus Í -A. On the other hand, if bÎ-A, by -A=-a+Q we have bÎ- a+Q, and hence there exists tÎQ such that b=t-a. Similarly, there exists -tÎQ such that  -b=a-tÎa+Q =A. Hence we have bÎ, and so -A Í. Thus  =-A, and consequently, M is an uniform power module of M.

Theorem 3.4  Let M be an uniform power module of M. Then M*={A | AÎM } is a submodule of M.

Proof  By Definition 1.2 and Theorem 2.1 in [4], we have that M* is an additive subgroup of M. For every ÎR and aÎ*,  there exists AÎM  such that aÎA,  so aÎA Í A. Hence aÎM*. Thus M*={ A | AÎM } is a submodule of M.

Theorem 3.5  is an uniform power module of M Û Q is a submodule of M.

ProofÞ By Definition 1.2 and Theorem 3.1 in [4], we see that Q is an additive subgroup of M. If ÎR then Q=Q Í M, so that Q is a submodule of M.

Ü Since Q is a submodule of M, then we have that Q is an additive subgroup of M. For every AÎM,  we obtain by Theorem 2.2 in [4] that -A=. Thus M is an uniform power module of M.

Theorem 3.6  (Structure theorem 1) Let M be a regular power module of M. Then M ={ a+Q | aÎ Í M** }, where M**={| AÎM }.

Proof  By Definition 1.3, if aÎ then we have that a+QÍ A+Q=A. On the other hand,  for bÎA , since aÎ, we have that b=0+b=(a-a)+b=a+(b-a) Îa-A+A=a+Q, and hence AÍ a+Q. Consequently, A=a+Q, and then M={ a+Q | aÎÍ M** }.

Corollary 3.1 (Structure theorem 2) Let M be an uniform power module of M. Then M ={ a+Q | aÎA Í M* }, where M*={ A | AÎM }.

Proof  By Theorem 3.5, we see that Q is a submodule of M, hence 0ÎQ. Namely, M is a regular power module of M, and by Theorem 3.3, we have that =A. Consequently, we have by Theorem 3.6 that M={ a+Q | aÎA Í M* }.

 

References

[1] LI Hong-xing, DUAN Qin-zhi, WANG Pei-zhuang. (1985). Hypergroups[J]. BUSERAL, 23: 22-29.

[2] LI Hong-xing. (1988). HX Rings[J]. BUSERAL, 34:3-8

[3] LI Hong-xing. (1991). HX Rings[J]. Chinese Quarterly Journal of Mathematics, 6(1): l5-20.

[4] ZHONG Yu-bin. (1990). The Structure and Relationship on Hypergroup[J]. Chinese Quarterly Journal of Mathematics, 5(4): 102-106.

[5] YAO Bing-xue, LI Hong-xing. (2000). Power Ring[J]. Fuzzy Systems and Mathematics, 14(2): 15-20.

[6] ZHANG Zhen-liang. (2001). Normal power ring and uniform power ring[J]. Pure and Applied Mathematics, 17(1): 6-13.

[7] LI Hongxing,WANG Pei-zhuang. (1988). The Power Group[J]. Mathematics Applicate, 1988,1(1): 1-4.

[8] Anderson F W, Fuller K R. (1992). Rings and Categories of Modules[M]. New York Heidelberg Berlin Spinger-Verlag.

[9] T.S. Blyth. Module theory[M]. (1990). Oxford University Press, Oxford.

 



[1] Vice professor, School of Computer Science, Guangdong Normal Polytechnic University, Guangzhou, 510665, China.

* Received 5 February  2009;  accepted 25 April 2009



DOI: http://dx.doi.org/10.3968%2Fg795

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