A rank for right congruences on inverse semigroups
A rank for right congruences on inverse semigroups
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逆半群上右同余的等级
DOI：
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发表时间：
2007
影响因子：
0.7
通讯作者：
V. Gould
中科院分区：
文献类型：
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作者：
V. Gould
半群S上的右同余ρ的S-秩（其中“S”代替“ρ”）是ρ在S的右同余格中关于我们称之为有限型拓扑的Cantor-Bendixson秩。如果每个ρ都有S-秩，则S是秩的。已知每个右Noether半群都是秩的，每个秩逆半群都是弱右Noether的。此外，如果S是秩，则S的每个极大子群也是秩。证明了Brandt半群<$0（G，I）是秩的当且仅当G是秩的且I是有限的.建立了链E上的同余格与任意幂等元半格E（S）<$E的逆半群S上的最小群同余所包含的右同余格之间的对应关系.因此，我们认为（逆）双圈幺半群B是不秩的，而且，一个秩半群不能包含双圈幺半群类。另一方面，B是弱右诺特的，并且具有平凡（因此是秩）子群。我们的秩的概念来自于考虑右凝聚幺半群S上存在闭（右）S-集理论Ts的稳定性。右凝聚的性质保证了存在闭S-集形成一个可公理化的类。我们认为B是右相干的。因此，从已知的结果可以得出，TB是一个超稳定但不完全超越的B-集理论。
The S-rank (where ‘S’ abbreviates ‘sandwich’) of a right congruence ρ on a semigroup S is the Cantor-Bendixson rank of ρ in the lattice of right congruences ℛ of S with respect to a topology we call the finite type topology. If every ρ ϵ ℛ possesses S-rank, then S is ranked. It is known that every right Noetherian semigroup is ranked and every ranked inverse semigroup is weakly right Noetherian. Moreover, if S is ranked, then so is every maximal subgroup of S. We show that a Brandt semigroup 0(G, I) is ranked if and only if G is ranked and I is finite. We establish a correspondence between the lattice of congruences on a chain E, and the lattice of right congruences contained within the least group congruence on any inverse semigroup S with semilattice of idempotents E(S) ≅ E. Consequently we argue that the (inverse) bicyclic monoid B is not ranked; moreover, a ranked semigroup cannot contain a bicyclic -class. On the other hand, B is weakly right Noetherian, and possesses trivial (hence ranked) subgroups. Our notion of rank arose from considering stability properties of the theory Ts of existentially closed (right) S-sets over a right coherent monoid S. The property of right coherence guarantees that the existentially closed S-sets form an axiomatisable class. We argue that B is right coherent. As a consequence, it follows from known results that TB is a theory of B-sets that is superstable but not totally transcendental.