某些双半环的结构和同余

发布时间：2018-06-10 06:54

本文选题：双半环 + 乘法含幺双半环　；参考：《山东师范大学》2012年硕士论文

【摘要】：本文主要研究双半环的结构和同余,给出了乘法含幺双半环同余和强理想之间的关系,刻画了加法可逆双半环同余对,找到几个通过加入幺元使得不含有乘法单位元的双半环变成乘法含幺双半环的充要条件,并讨论了乘法含幺双半环拟分配格的结构.本文共分四章：第一章给出引言和预备知识. 第二章主要研究了乘法含幺双半环的同余和强理想,并找到他们的一个一对应,刻画了加法可逆双半环的同余对,主要结论如下：引理2.1.2设R为乘法含幺双半环s的强理想,在s上定义二元关系σ如下：σ={(x,y)∈S×SlxR=yR}则σ为s的同余,且σ的核K=R. 定理2.1.4设(S,+,·,*)为乘法含幺双半环,且满足(S,·)为群,对(?)s∈S有s=ss+s=s+ss.s=ss*s=s*ss则s的同余集和s的强理想之间有一一对应关系. 定理2.2.3设(S,+,·,*)为乘法含幺双半环,且(S,·)为逆半群,定义S上二元关系如下：ρ={(x,y)∈S×S|(?)e,(?)∈E(?),使得xe=yf},其中E(?)={e∈S|ee=e}.则ρ为s上的最小拟双环同余. 定理2.2.4设(s,+,·,*)为乘法含幺双半环,并满足(S,·)幂等可换,对(?)s∈S有1=s+1,1=1+s,1=1*s,1=s*1,且(s,*),(s,+)满足消去率,则集合E[·]={e∈S[ee=e)为s的一个理想.此外若(s,·)为逆半群,并满足消去律,则集合E[·]为s的强理想. 定理2.3.9设(S,+,·,*)加法可逆双半环,ρ为S上的双半环同余,则(Kerρ,trρ)为S上的同余对；反之,若(N,τ)为s上的同余对,则关系ρ(N,τ)={(a,b)∈S×S|(a'+a,b'+b)∈τ,a+b'∈N}为S上的双半环同余,且Kerp(N,τ)=N,trp(N,τ)=τ,ρ(Kerp,trp)=p. 第三章主要讨论如何由不含乘法幺元的双半环变成乘法含幺双半环,并给出了乘法含幺双半环拟分配格的一个机构定理,主要结论如下：定理3.1.1设(S,+,·,*)为双半环,1(?)S,且满足：(1)对Vs∈S∪{1},1s=s1=s；(2)对Vs∈S∪{1},1+s=s+1=1(3)对Vs∈S∪{1},1*s=s*1=1则(S U{1},+,·,*)为乘法含幺双半环,当且仅当S满足对Vs,x∈Ss=sx+s=xs+s=s+sz=s+xs=sx*s=xs*s=s*sx=s*xs 定理3.1.2设(S,+,·,*)为双半环,1(?)S,且满足：(1)对Vs∈S∪{1},1s=s1=s；(2)对Vs∈S∪{1},1+s=s+1=s(3)对Vs∈S∪{1},1*s=s*1=s则(S U{1},+,·,*)为乘法含幺双半环,当且仅当S满足对Vs,x∈Ssx=x+sx=sx+x=s+sx=sx+s=x*sx=sx*x=s*sx=sx*s; s*x=s+(s*x)=(s*x)+s=x+(s*x)=(s*x)+x；s+x=s*(s+x)=(s+zx)*s=x*(s+x)=(s+x)*x. 定理3.1.3设(S,+,·,*)为双半环,1(?)S,且满足：(1)对Vs∈S∪{1},1s=s1=s；(2)对Vs∈S∪{1},1+s=1,s+1=s(3)对Vs∈S∪{1},1*s=1,s*1=s则(S∪{1},+,·,*)为乘法含幺双半环,当且仅当S满足对Vs,x∈Ss=s+sx=s+xs=s*sx=s*xs; sx=sx*x=sx*s=sx+x=sx+s; s*x=(s*x)+s=(s*x)+x; s+x=(s+x)*s=(s+x)*x. 引理3.2.2设S=[D；Sα],则(S,+,·,*)为双半环. 定理3.2.3设S=[D;Sα],(?)α∈Sα,b∈Sβ,(α,β∈D),若a·1αβ=b·1αβ,则(?)δ≤αβ,a·1δ=b·1δ(C4)；若(?)δ≤αβ,a·1δ=b·1δ,则a·1αβ=b·1αβ(C5)；在S上定义关系(?)α∈Sα,b∈Sβαρb(?)α·1αβ=b·(1αβ),( )；则ρ为S上的双半环同余,且S为分配格D和双半环S/ρ的拟次直积；反之,若S=[D；Sα]上存在形如( )定义的同余ρ,且1αβ=1α·1β,(α,β∈D),则S满足(C4),(C5). 定理3.2.4设S=[D；Sα],若(?)α,β∈D,1α·1β=1αβ,则S=D;Sα,ψα,β.
[Abstract]:In this paper, we mainly study the structure and congruence of a double semiring, give the relation between the congruence and strong ideal of a multiplicative double semi ring, depict a sufficient and necessary condition for the addition of a reversible double semiring congruence pair, and find several necessary and sufficient conditions for the double half ring which does not contain the multiplicative unit element to be a multiplicative double half ring, and discuss the multiplication with a unitary double half ring. The structure of the quasi distributive lattice is divided into four chapters in this paper.
The first chapter gives the introduction and the preparatory knowledge.
In the second chapter, we mainly study the congruence and strong ideals of the multiplicative double semirings, and find one of their one correspondence and depict the congruence pairs of the additive reversible double semirings. The main conclusions are as follows:
Lemma 2.1.2 set R to be a strong ideal of multiplication s containing unitary double semicircular rings, and define the relation of two variables on S, which is as follows: sigma = (x, y) S, SlxR=yR} SlxR=yR} is the congruence of S, and the kernel K=R. of sigma.
Theorem 2.1.4 (S, +, *) is a multiplicative double half loop and satisfies (S,) as a group, and there is a one-to-one correspondence between the congruence set of S and the strong ideal of s for (?) s S s=ss+s=s+ss.s=ss*s=s*ss s.
Theorem 2.2.3 (S, +, *) is a multiplicative double semiring, and (S, /) is an inverse semigroup, and defines the two element relation on S as follows: P = = = (x, y) S x S| (?) e, (?) E (?), which makes xe=yf}, where E (?) is the smallest quasi double ring congruence.
Theorem 2.2.4 (s, +, *, *) is a multiplicative double half loop and satisfies (S,.) idempotent, and (?) s S has 1=s+1,1=1+s, 1=1*s, 1=s*1, and (s, *), (s, +) satisfies the elimination rate, and then the set E[]={e) is an ideal. Besides, if it is the inverse semigroup and satisfies the elimination law, then the set is a strong ideal.
Theorem 2.3.9 (S, +, *, *) add a reversible double half ring, and Rho is the congruence of the double half ring on S, then (Ker rho, TR rho) is the congruence on S; and conversely, if (N, tau) is the congruence on S (N, [Tau] = = [a, b)]
In the third chapter, we mainly discuss how to transform a double half ring without a multiplicative unitary into a multiplicative double half ring, and give a mechanism theorem for the multiplicative quasi distributive lattice with a single double half ring. The main conclusions are as follows:
Theorem 3.1.1 (S, +, *, *) is a double half loop, 1 (?) S, and satisfies: (1) Vs S {1}, 1s=s1=s; (2) Vs S S {1}, 1+s=s+1=1 (3) is a multiplicative double half loop
Theorem 3.1.2 (S, +, *, *) is a double half loop, 1 (?) S, and satisfies: (1) Vs S {1}, 1s=s1=s; (2) Vs S S {1}, 1+s=s+1=s (3) is a multiplicative double half loop. S* (s+x) = (s+zx) *s=x* (s+x) = (s+x) *x.
Theorem 3.1.3 (S, +, *, *) is a double half loop, 1 (?) S, and satisfies: (1) Vs S {1}, 1s=s1=s; (2) Vs S {1}, 1+s=1, s+1=s (+, *) is a multiplicative double half loop. S+x) *x.
Lemma 3.2.2 sets S=[D, S]], then (S, +, *, *) is a double semiring.
Theorem 3.2.3 S=[D; S alpha], (?) alpha S alpha, B S beta, (alpha, beta D), if a 1 alpha beta =b 1 alpha beta, (?) (?) < < alpha beta, a 1 delta =b. 1 delta (C4); if (?) [?] < < alpha beta, 1 delta 1 delta, then 1 alpha beta 1 alpha beta;
(2) if S is a double semiring congruence, and S is a quasi direct product of distributive lattice D and double semiring S/ p; otherwise, if S=[D, S alpha] exists,
The definition of congruence P, and 1 alpha beta =1 alpha, 1 beta, (alpha, beta D), then S satisfies (C4), (C5).
Theorem 3.2.4 set S=[D; S]; if (a), alpha, D, 1 alpha, 1 beta =1 alpha beta, then S=D, S alpha, alpha, beta.
【学位授予单位】：山东师范大学
【学位级别】：硕士
【学位授予年份】：2012
【分类号】：O153.3

【参考文献】