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  本文選題：雙半環(huán) + 乘法含幺雙半環(huán)　； 參考：《山東師范大學(xué)》2012年碩士論文

【摘要】：本文主要研究雙半環(huán)的結(jié)構(gòu)和同余,給出了乘法含幺雙半環(huán)同余和強理想之間的關(guān)系,刻畫了加法可逆雙半環(huán)同余對,找到幾個通過加入幺元使得不含有乘法單位元的雙半環(huán)變成乘法含幺雙半環(huán)的充要條件,并討論了乘法含幺雙半環(huán)擬分配格的結(jié)構(gòu).本文共分四章： 第一章給出引言和預(yù)備知識. 第二章主要研究了乘法含幺雙半環(huán)的同余和強理想,并找到他們的一個一對應(yīng),刻畫了加法可逆雙半環(huán)的同余對,主要結(jié)論如下： 引理2.1.2設(shè)R為乘法含幺雙半環(huán)s的強理想,在s上定義二元關(guān)系σ如下：σ={(x,y)∈S×SlxR=yR}則σ為s的同余,且σ的核K=R. 定理2.1.4設(shè)(S,+,·,*)為乘法含幺雙半環(huán),且滿足(S,·)為群,對(?)s∈S有s=ss+s=s+ss.s=ss*s=s*ss則s的同余集和s的強理想之間有一一對應(yīng)關(guān)系. 定理2.2.3設(shè)(S,+,·,*)為乘法含幺雙半環(huán),且(S,·)為逆半群,定義S上二元關(guān)系如下：ρ={(x,y)∈S×S|(?)e,(?)∈E(?),使得xe=yf},其中E(?)={e∈S|ee=e}.則ρ為s上的最小擬雙環(huán)同余. 定理2.2.4設(shè)(s,+,·,*)為乘法含幺雙半環(huán),并滿足(S,·)冪等可換,對(?)s∈S有1=s+1,1=1+s,1=1*s,1=s*1,且(s,*),(s,+)滿足消去率,則集合E[·]={e∈S[ee=e)為s的一個理想.此外若(s,·)為逆半群,并滿足消去律,則集合E[·]為s的強理想. 定理2.3.9設(shè)(S,+,·,*)加法可逆雙半環(huán),ρ為S上的雙半環(huán)同余,則(Kerρ,trρ)為S上的同余對；反之,若(N,τ)為s上的同余對,則關(guān)系ρ(N,τ)={(a,b)∈S×S|(a'+a,b'+b)∈τ,a+b'∈N}為S上的雙半環(huán)同余,且Kerp(N,τ)=N,trp(N,τ)=τ,ρ(Kerp,trp)=p. 第三章主要討論如何由不含乘法幺元的雙半環(huán)變成乘法含幺雙半環(huán),并給出了乘法含幺雙半環(huán)擬分配格的一個機構(gòu)定理,主要結(jié)論如下： 定理3.1.1設(shè)(S,+,·,*)為雙半環(huán),1(?)S,且滿足：(1)對Vs∈S∪{1},1s=s1=s；(2)對Vs∈S∪{1},1+s=s+1=1(3)對Vs∈S∪{1},1*s=s*1=1則(S U{1},+,·,*)為乘法含幺雙半環(huán),當(dāng)且僅當(dāng)S滿足對Vs,x∈Ss=sx+s=xs+s=s+sz=s+xs=sx*s=xs*s=s*sx=s*xs 定理3.1.2設(shè)(S,+,·,*)為雙半環(huán),1(?)S,且滿足：(1)對Vs∈S∪{1},1s=s1=s；(2)對Vs∈S∪{1},1+s=s+1=s(3)對Vs∈S∪{1},1*s=s*1=s則(S U{1},+,·,*)為乘法含幺雙半環(huán),當(dāng)且僅當(dāng)S滿足對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設(shè)(S,+,·,*)為雙半環(huán),1(?)S,且滿足：(1)對Vs∈S∪{1},1s=s1=s；(2)對Vs∈S∪{1},1+s=1,s+1=s(3)對Vs∈S∪{1},1*s=1,s*1=s則(S∪{1},+,·,*)為乘法含幺雙半環(huán),當(dāng)且僅當(dāng)S滿足對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設(shè)S=[D；Sα],則(S,+,·,*)為雙半環(huán). 定理3.2.3設(shè)S=[D;Sα],(?)α∈Sα,b∈Sβ,(α,β∈D),若a·1αβ=b·1αβ,則(?)δ≤αβ,a·1δ=b·1δ(C4)；若(?)δ≤αβ,a·1δ=b·1δ,則a·1αβ=b·1αβ(C5)；在S上定義關(guān)系(?)α∈Sα,b∈Sβαρb(?)α·1αβ=b·(1αβ),( )；則ρ為S上的雙半環(huán)同余,且S為分配格D和雙半環(huán)S/ρ的擬次直積；反之,若S=[D；Sα]上存在形如( )定義的同余ρ,且1αβ=1α·1β,(α,β∈D),則S滿足(C4),(C5). 定理3.2.4設(shè)S=[D；Sα],若(?)α,β∈D,1α·1β=1αβ,則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.
【學(xué)位授予單位】：山東師范大學(xué)
【學(xué)位級別】：碩士
【學(xué)位授予年份】：2012
【分類號】：O153.3

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