1988
DOI: 10.1016/s0747-7171(88)80054-3
|View full text |Cite
|
Sign up to set email alerts
|

On recognisable properties of associative algebras

Help me understand this report

Search citation statements

Order By: Relevance

Paper Sections

Select...
2
1
1
1

Citation Types

0
10
0

Year Published

1989
1989
2011
2011

Publication Types

Select...
6
1
1

Relationship

0
8

Authors

Journals

citations
Cited by 24 publications
(10 citation statements)
references
References 4 publications
0
10
0
Order By: Relevance
“…This theorem was proved by T. Gateva-Ivanova [1] and [2] in the finitely presented case (see also [4,Proposition 15,Chapter 24] Proof. Let k be a natural number and w any subword of W with |u;| = (k + 2)|w|.…”
Section: Definitionmentioning
confidence: 94%
“…This theorem was proved by T. Gateva-Ivanova [1] and [2] in the finitely presented case (see also [4,Proposition 15,Chapter 24] Proof. Let k be a natural number and w any subword of W with |u;| = (k + 2)|w|.…”
Section: Definitionmentioning
confidence: 94%
“…Then Π has polynomial growth if and only if each nonzero element of Π that is not nilpotent can be represented by a word whose primitive root labels some loop of the Ufnarovsky graph of the presentation. This follows from [6] where the syntactic structure of the Jacobson radical was described (also see this description in [24, Section 7.6, Theorem 3]) and the trivial fact that Π embeds in an associative monomial algebra with the same set of generators and defining relations (see also [13,Chapter 24]). PROOF.…”
Section: Bounded Heightmentioning
confidence: 99%
“…ExampIe: k(x, y 1 xy = y2 = x4 = 0, x3 = yx) is nondegenerate, but k(z 1 z2 = az) and L (z 1 z2 = 1) are not. Remark 1: The dimension finiteness implies that condition (2) in Definition 1 can be replaced by condition (2') which says that there is not a nonzero a E k such that vu = au. Remark 2: A monomial algebra is a non-degenerate binomial algebra.…”
Section: The Word Posetsmentioning
confidence: 99%
“…Let us call such a path a linked path. Here p(A) z p(B) as word posets w there exists a poset isomorphism u: p(A) + p(B) such that (1) the restriction of 0 to X gives a bijection from X into Y (therefore #X = fly), and (2) if uu is in p(A), then a( is in p(B) and a(u)o(v) = u(uv).…”
Section: The Word Posetsmentioning
confidence: 99%
See 1 more Smart Citation