Management Science and Engineering

 

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 | -xÎA��xÎM } is called the inverse of A.

Clearly��for 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, BÎM��if A��B��f then | A |=| A��B |;

(3) For all A, BÎM��if 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=Q��then for every b��-A we have that b+A Í -A+A=Q and | A |=| b+A | �� | Q|. Consequencely, | A |=| B |=| Q |.

��2��Since A��B��f then we have | A��B |�� | A |. Moreover, if zÎA��B then we have zÎA and zÎB��hence we see that z+Q Í A+Q=A and z+Q Í B+Q=B. Thus we obtain z+Q Í A��B, so | A | = | Q |=| z+Q | �� | A��B |. Consequently, we have | A |=| A��B |.

��3��By 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 0ÎAÎM��then -A Í Q Í A.

Proof  Since 0��A��then 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 .

Proof��Let 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 Û��f, for every A��M.

Proof����Ü ��  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  aÎ��f.

Theorem 2.5  Let f : L₁��L₂  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.

Proof��It 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 tÎ��f(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)Î��f(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: L₁₁��L₂₂ defined by g(A) = f (A) is an R-epimorphism.

Theorem 2.6  Let f : L₁��L₂  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.

Proof����Ü �� 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 a��A 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+.

Proof����Ü �� 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.

Proof����Þ �� 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 ��aÎ��A 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.

Proof����Þ �� 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.