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| Multiplicative Lattice with Maximal Condition |
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Citation： |
Xie Bingzhang.Multiplicative Lattice with Maximal Condition[J].Chinese Annals of Mathematics B,1982,3(6):765~772 |
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Net amount： 809 |
Authors： |
Xie Bingzhang; |
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| Abstract： |
This paper defines multiplicative lattice. The conception of multiplicative lattice is abstracted from several lattices, which may be composed of ideals of the associative ring, ideals of the non-associative ring, or ideals of the both non-assooiatiye and non-distributive ring and which may be composed of normal subgroups of the group as well.
Definition 1. The multiplicative lattice L is a lattice with following conditions:
1. L is complemented and Dedekind's, the least upper bound of L equals 1 and the greatest lower bound of L equals 0.
2. The multiplication is closed and ab\leq a \cap b, where a\cap b denotes the greatest lower bound of a and b.
3.a(b+c)=ab+ac,(b+c)=ba+bc, where b+c denotes the least upper bound of b and c.
Suppose L is multiplicative lattice with maximal condition, the following theorems are proved:
Theorem 1. Every element a of L has normal (right) third decomposition, that means every element is the intersection of finite number of (right) third elements, whose (right) third roots are different from each other and if $a=T_1\cap \cdots \cap T_m=T_1'\cap \cdots\cap T_n'$ are two different normal (right) third decomposition of a, then m=n and their third roots are equal correspondingly by rearranging properly
Theorem 2. If A is an element of L and has property that c^2\leq A \rightarrow c\leq A, then A is the intersection of finite number of prime elements.
Theorem 3. Every element of L has primary decomposition if and only if L
satisfies Artin-Rees condition.
Definition 2. u is called a solvable radical of the multiplicative lattice with maximal condition if u is the least upper bound, of the subset of L whose every element a? satisfies that there exists a group of positive integers n_1,\cdots,n_r such that $X^n_1,\cdots,n_r=0$.
Theorem 4. u is an intersection of finite number of prime elements.
Theorem 5. If L is semi-simple, i. e. $O=u=\cap\limits_{i=1}^n P_i$, and if $P_i+P_j=1(i\ne j)$ and $1\cdot 1=1$ then 1 is the direct sum in R_1,\cdots,R_n, where
$R_i=P_1\cap \cdots \cap P_i-1\cap P_i+1\cap \cdots \cap P_n$ |
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